Unit 11: Perimeter and Area

In the Questions $1$ to $37,$ there are four options, out of which one is correct. Choose the correct one.

Observe the shapes $1,2,3$ and $4$ in the figures. Which of the following statements is not correct?

a.    Shapes $1,3$ and $4$ have different areas and different perimeters.

b.   Shapes $1$ and $4$ have the same area as well as the same perimeter.

c.    Shapes $1,2$ and $4$ have the same area.

d.   Shapes $1,3$ and $4$ have the same perimeter.

(a)

Shape $1$:

 

$\begin{array}{l}\text{Perimeter}=1+1+1+1+1+1+1+1+1+1+1\\ +1+1+1+1+1+1+1+1+1+1+1\text{ units}\\ \text{= 22 units}\end{array}$ 

Area $=18\times 1=18$ sq. units

Shape $2$:

$\begin{array}{l}\text{Perimeter}=1+1+1+1+1+1+1+1+1+1\\ +1+1+1+1+1+1+1+1\text{ units}\\ \text{= 18 units}\end{array}$ 

Area $=18\times 1=18$ sq. units

Shape $3$:

$\begin{array}{l} \text{Perimeter}=1+1+1+1+1+1+1+1+1+1+1\\ +1+1+1+1+1+1+1+1+1+1+1\text{ units}\\ \text{= 22 units}\end{array}$ 

Area $=16\times 1=16$ sq. units

Shape $4$:

 

$\begin{array}{l}\text{Perimeter}=1+1+1+1+1+1+1+1+1+1+1\\ +1+1+1+1+1+1+1+1+1+1+1\text{ units}\\ \text{= 22 units}\end{array}$

Area $=18\times 1=18$ sq. units

So option (a) is incorrect.

A rectangular piece of dimensions $3\text{ cm}\times 2\text{ cm}$ was cut from a rectangular sheet of paper of dimensions $6\text{ cm}\times 5\text{ cm}$ (Fig. 9.14).

Area of remaining sheet of paper is

Fig. 9.14

a.  $30{\text{ cm}}^2$ 

b. $36{\text{ cm}}^2$

c.  $24{\text{ cm}}^2$

d. $22{\text{ cm}}^2$

(c)

Given, length and breadth of the bigger rectangle as $6\text{ cm}$ and $5$ cm.

Area of bigger rectangle $=6\text{cm}\times 5\text{cm}=30{\text{cm}}^2$

Also given, length and breadth of the smaller rectangle as $3\text{ cm}$ and $2\text{ cm}$.

Area of smaller rectangle $=3\text{ cm}\times 2\text{ cm}=6{\text{ cm}}^2$

Area of remaining sheet $=$ Area of bigger rectangle $–$ Area of smaller rectangle

$=(30-6){\text{ cm}}^2=24{\text{ cm}}^2$

$36$ unit squares are joined to form a rectangle with the least perimeter. Perimeter of the rectangle is

a.  $12$ units

b. $26$ units

c.  $24$ units

d. $36$ units

(b)

Area of rectangle formed $=36$ square units

We can write,             $36=6\times 6$

   $=2\times 3\times 2\times 3$

   $=2^2\times 3^2$

   $=4\times 9$

The possible pairs of sides can be:

1)              Length = 1, Breadth = 36

2)              Length = 2, Breadth = 18

3)              Length = 4, Breadth = 9

4)              Length = 12, Breadth = 3

5)              Length = 6, Breadth = 6

 

Now, for each pair of sides we will find the perimeter:

1)           Perimeter $=2\left(l+b\right)$

                                    $\begin{array}{l}=2\left(1+36\right)\\ =2\times 37\\ =74\end{array}$

2)           Perimeter $=2\left(l+b\right)$

$\begin{array}{l}=2\left(2+18\right)\\ =2\times 20\\ =40\end{array}$

3)           Perimeter   $=2\left(l+b\right)$

$\begin{array}{l}=2\left(4+9\right)\\ =2\times 13\\ =26\end{array}$

4)           Perimeter $=2\left(l+b\right)$

                $\begin{array}{l}=2\left(12+3\right)\\ =2\times 15\\ =30\end{array}$

5)           Perimeter $=2\left(l+b\right)$

           $\begin{array}{l}=2\left(6+6\right)\\ =2\times 12\\ =24\end{array}$

So, the rectangle with sides 6 units and 6 units will have the least perimeter which is 24 units.

A wire is bent to form a square of side $22$ cm. If the wire is rebent to form a circle, its radius is

a.  $22$ cm

b. $14$ cm

c.  $11$ cm

d. $7$ cm

(b)

Given, side of square $=22\text{ cm}$

$\because$ The wire has same length.

$\therefore$ Perimeter of square $=$ circumference of circle

$4\times \text{side}=2\times \pi \times r$

$4\times 22=2\times \frac{22}{7}\times r$

$r=\frac{4\times 22\times 7}{2\times 22}=14\text{cm}$

Hence, the radius is $14\text{ cm}\text{.}$

Area of the circle obtained in Question 4 is

a.    $196{\text{ cm}}^2$

b.   $212{\text{ cm}}^2$

c.    $616{\text{cm}}^2$ 

d.   $644{\text{ cm}}^2$ 

(c)

Area of the circle $=\pi r^2$

                             $=\frac{22}{7}\times {14}^2=616{\text{ cm}}^2$

So, the correct option is (c)

Area of a rectangle and the area of a circle are equal. If the dimensions of the rectangle are $14\text{ cm}\times 11\text{ cm}$ 14 cm × 11 cm, then radius of the circle is

a.    $21$ cm

b.   $10.5$ cm

c.    $14$ cm

d.   $7$ cm.

(d)

Given, dimensions of rectangle $l=14\text{ cm and }b=11\text{ cm}$

According to the question,

Area of rectangle $=$ Area of circle

$l\times b=\pi r^2$

$14\times 11=\frac{22}{7}\times r^2$

$r^2=\frac{14\times 11\times 7}{22}$

$r^2=49$

$r=\sqrt{49}=7\text{cm}$

Hence, the radius of circle is $7\text{ cm}$.

Area of shaded portion in Fig. 9.15 is

a.    $25{\text{ cm}}^2$ 

b.   $15{\text{ cm}}^2$ 

c.    $14{\text{ cm}}^2$ 

d.   $10{\text{ cm}}^2$ 

Fig. 9.15

(d)

From the given figure,

Length of rectangle $(l)=5\text{cm}$ 

Breadth of rectangle $\left(b\right)=3+1=4\text{cm}$

Area of shaded portion

$\begin{array}{l}=\frac{1}{2}\times \text{Area of rectangle}\\ =\frac{1}{2}\times (l\times b)\end{array}$

$\begin{array}{l}=\frac{1}{2}\times 5\times 4\\ =10{\text{cm}}^2\end{array}$

Area of parallelogram ABCD (Fig. 9.16) is not equal to

a.    DE $\times$ DC

b.   BE $\times$ AD

c.    BF $\times$ DC

d.   BE $\times$ BC

Fig. 9.16

(a)

We know that, Area of parallelogram
$=$ Base $\times$ Corresponding Height

So, Area of parallelogram
$\text{ABCD}=\text{AD}\times \text{BE}=\text{BC}\times \text{BE}$            $[\because \text{AD}=\text{BC}]$

Or, Area of parallelogram $\text{ABCD}=\text{DC}\times \text{BF}$

Area of triangle MNO of Fig. 9.17 is

Fig. 9.17

a.    $\frac{1}{2}\text{MN}\times \text{NO}$ 

b.   $\frac{1}{2}\text{NO}\times \text{MO}$ 

c.    $\frac{1}{2}\text{MN}\times \text{OQ}$ 

d.   $\frac{1}{2}\text{NO}\times \text{OQ}$ 

(d)

We know that,

Area of triangle
$=\frac{1}{2}\times \text{Base}\times \text{Height}$

$=\frac{1}{2}\text{NO}\times \text{OQ}$

Ratio of area of $\Delta \text{MNO}$ to the area of parallelogram MNOP in the same figure 9.17 is

a.    $2:3$ 

b.   $1:1$ 

c.    $1:2$ 

d.   $2:1$ 

(c)

Area of $\Delta \text{MNO}:$ Area of parallelogram MNOP

$\frac{1}{2}\times \text{Base}\times \text{Height : Base}\times \text{Corresponding}\text{Height}$

$\frac{1}{2}\times \text{NO}:\text{MP}$

$\frac{1}{2}\times \text{NO}:\text{NO (}\because \text{ MP = NO)}$

Hence, the required ratio is $1:2$

Ratio of areas of $\Delta \text{MNO}$, $\Delta \text{MOP}$ and $\Delta \text{MPQ}$ in Fig. 9.18 is

a.    $2:1:3$ 

b.   $1:3:2$ 

c.    $2:3:1$ 

d.   $1:2:3$ 

Fig. 9.18

(a)

From the given figure,

Area of $\Delta \text{MNO}=\frac{1}{2}\times \text{NO}\times \text{MO}$

$=\frac{1}{2}\times 4\times 5=10{\text{cm}}^2$

$\begin{array}{l}\text{Area of}\Delta \text{MOP}=\frac{1}{2}\times \text{OP}\times \text{MO}\\ =\frac{1}{2}\times 2\times 5\\ =5{\text{cm}}^2\end{array}$$\begin{array}{l}\text{Area of}\Delta \text{MPQ}=\frac{1}{2}\times \text{PQ}\times \text{MO}\\ =\frac{1}{2}\times 6\times 5\\ =15{\text{cm}}^2\end{array}$  

Hence, required ratio is $\begin{array}{l}=10:5:15\\ =2:1:3\end{array}$

In Fig. 9.19, EFGH is a parallelogram, altitudes FK and FI are 8 cm and 4cm respectively. If $\text{EF}=10$ cm, then area of EFGH is

a.    $20{\text{ cm}}^2$

b.   $32{\text{ cm}}^2$ 

c.    $40{\text{ cm}}^2$ 

d.   $80{\text{ cm}}^2$ 

Fig. 9.19

(c)

In parallelogram EFGH, EF $=$ HG $=\text{1}0\text{cm}$  [Given]

Area of parallelogram EFGH
$=$ Base $\times$ Corresponding height

= HG $\times$ IF

$=10\times 4=40{\text{cm}}^2$

In reference to a circle the value of $\pi$ is equal to

a.    $\frac{\text{Area}}{\text{Circumference}}$ 

b.   $\frac{\text{Area}}{\text{Diameter}}$ 

c.    $\frac{\text{Circumference}}{\text{Diameter}}$ 

d.   $\frac{\text{Circumference}}{\text{Radius}}$ 

(c)

We Know that,

Circumference of a circle $=2\pi r$ (r, the radius of circle)

$\Rightarrow \text{Circumference}=\pi \times \text{Diameter}$

$\Rightarrow \pi =\frac{\text{Circumference}}{\text{Diameter}}$

Circumference of a circle is always

a.    more than three times of its diameter

b.   three times of its diameter

c.    less than three times of its diameter

d.   three times of its radius

(a)

We know that,

Circumference of a circle $=2\pi r$

$\therefore$ Circumference $=2\times 3.14\times r$         $[\because \pi =3.14]$

$\Rightarrow$ Circumference $=3.14\times d$             $[\because d=2r]$

So, circumference of circle is always more than three times of its diameter.

Area of triangle PQR is $100{\text{ cm}}^2$ (Fig. 9.20). If altitude QN is $10\text{ cm}$, then its base PR is

a.    $20\text{ cm}$

b.   $15\text{ cm}$

c.    $10\text{ cm}$

d.   $5\text{ cm}$

Fig. 9.20

(a)

Given, area of $\Delta \text{QPR}=100{\text{ cm}}^2$

We know that,

Area of triangle $=\frac{1}{2}\times \text{Base}\times \text{Height}$

$\therefore$ Area of $\Delta \text{QPR}=\frac{1}{2}\times \text{PR}\times \text{QN}$

                      $100=\frac{1}{2}\times \text{PR}\times 10$

                       $\text{PR}=\frac{100\times 2}{10}=20\text{cm}$

In Fig. 9.21, if $\text{PR}=12\text{ cm},$ $\text{QR}=6\text{ cm}$ and $\text{PL}=8\text{ cm},$ then QM is -

Fig. 9.21

a.    $6\text{ cm}$

b.   $9\text{ cm}$

c.    $4\text{ cm}$

d.   $2\text{ cm}$

(c)

Given that, $\text{PR}=12\text{ cm},$$\text{QR}=6\text{ cm}$ and $\text{PL}=8\text{ cm}$

Now, using Pythagoras theorem in right angled $\Delta$ PLR,

${\left(\text{Hypotenuse}\right)}^2={\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2$

${\text{PR}}^2={\text{PL}}^2+{\text{LR}}^2$

${\text{LR}}^2={\text{PR}}^2-{\text{PL}}^2$

${\text{LR}}^2={(12)}^2-{(8)}^2$

${\text{LR}}^2=144-64=80$

$\text{LR}=\sqrt{80}=4\sqrt{5}\text{cm}$

$\text{LR}=\text{LQ}+\text{QR}$

$\text{LQ}=\text{LR}-\text{QR}=(4\sqrt{5}-6)\text{cm}$

Area of $\Delta \text{PLR,}$ ${\text{A}}_1=\frac{1}{2}\times \text{LR}\times \text{PL}$

     $=\frac{1}{2}\times (4\sqrt{5})\times 8$

     $=16\sqrt{5}{\text{cm}}^2$

$\text{Again},\text{ area of}\text{ΔPLQ,}$

${\text{A}}_2=\frac{1}{2}\times \text{LQ}\times \text{PL}$

      $=\frac{1}{2}\times (4\sqrt{5}-6)\times 8$

     $=(16\sqrt{5}-24){\text{ cm}}^2$

Area of $\Delta \text{PLR}=\text{Area}\text{of}\Delta \text{PLQ}+\text{Area}\text{of}\Delta \text{PQR}$

                $16\sqrt{5}=(16\sqrt{5}-24)+\text{Area of }\Delta \text{PQR}$

$\text{Area of }\Delta \text{PQR}=24{\text{ cm}}^2$

$\frac{1}{2}\times \text{PR}\times \text{QM}=24$

$\frac{1}{2}\times 12\times \text{QM}=24\text{cm}$

            $\text{QM}=4\text{cm}$

In Fig. 9.22 $\Delta \text{MNO}$ is a right-angled triangle. Its legs are $6\text{ cm}$ and $8\text{ cm}$ long. Length of perpendicular NP on the side MO is

Fig. 9.22

a.    $4.8\text{ cm}$

b.   $3.6\text{ cm}$

c.    $2.4\text{ cm}$

d.   $1.2\text{ cm}$

(a)

Given, $\Delta \text{MNO}$ is a right-angled triangle.

According to Pythagoras theorem,

${\text{MO}}^2={\text{MN}}^2+{\text{NO}}^2$

         $=6^2+8^2=36+64$

$\begin{array}{l}{\text{MO}}^2=100\\ \text{MO}=\sqrt{100}\end{array}$

$\text{MO}=10\text{cm}$

$\text{Area of}\Delta \text{MNO}=\frac{1}{2}\times \text{Base}\times \text{Height}$

$\frac{1}{2}\times \text{MN}\times \text{NO}=\frac{1}{2}\times \text{MO}\times \text{NP}$

$\frac{1}{2}\times 6\times 8=\frac{1}{2}\times 10\times \text{NP}$

$\text{NP}=\frac{24}{5}=4.8\text{ cm}$

Area of a right-angled triangle is $30{\text{ cm}}^2.$ If its smallest side is $5\text{cm},$ then its hypotenuse is

a.    $14\text{ cm}$

b.   $13\text{ cm}$

c.    $12\text{ cm}$

d.   $11\text{cm}$

(b)

Given, area of right-angled triangle $=30{\text{cm}}^2$

and smallest side i.e., base $=5\text{ cm}$

Area of right-angled triangle $=\frac{1}{2}\times \text{Base}\times \text{Height}$

$30=\frac{1}{2}\times 5\times \text{Height}$

$\text{Height}=\frac{30\times 2}{5}=12\text{ cm}$

Now, according to Pythagoras theorem,

${(\text{Hypotenuse})}^2={(\text{Perpendicular})}^2+{(\text{Base})}^2$

${(\text{Hypotenuse})}^2={(12)}^2+{(5)}^2$

${(\text{Hypotenuse})}^2=144+25$

${(\text{Hypotenuse})}^2=169$

$\text{Hypotenuse }=\sqrt{169}=13\text{ cm}$

Circumference of a circle of diameter $5\text{ cm}$ is

a.    $3.14\text{ cm}$

b.   $31.4\text{ cm}$

c.    $15.7\text{ cm}$

d.   $1.57\text{ cm}$

(c)

Diameter of the given circle $=5$ cm

$\therefore$ Radius $=\frac{5}{2}\text{cm}$

Now, circumference of the circle

$\begin{array}{l}=2\pi r\\ =2\times \frac{22}{7}\times \frac{5}{2}\end{array}$

$=\frac{110}{7}=15.7\text{cm}$ 

Circumference of a circle disc is 88 cm. Its radius is

a.    $8\text{ cm}$

b.   $11\text{ cm}$

c.    $14\text{ cm}$

d.   $44\text{ cm}$

(c)

Given, circumference of the circular disc $=2\pi r$

$88=2\times \frac{22}{7}\times r$

$r=\frac{88\times 7}{2\times 22}$

$r=14\text{cm}$

Hence, the radius of the disc is $14\text{ cm}$.

Length of tape required to cover the edges of a semicircular disc of radius $10\text{ cm}$ is

a.    $62.8\text{ cm}$

b.   $51.4\text{ cm}$

c.    $31.4\text{ cm}$

d.   $15.7\text{ cm}$

(b)

In order to find the length of tape required to cover the edges of a semi-circular disc, we have to find the perimeter of semi-circle.

Perimeter of semi-circle $=$ circumference of semi-circle $+$ Diameter

Circumference of semi-circle $=\frac{2\pi r}{2}=\pi \times r$

   $=\frac{22}{7}\times 10=\frac{220}{7}=31.4\text{cm}$

Length of tape required $=31.4+(2\times 10)=51.4\text{cm}$ 

Area of circular garden with diameter $8\text{ m}$ is

a.    $12.56{\text{ m}}^2$

b.   $25.12{\text{ m}}^2$

c.    $50.24{\text{ m}}^2$

d.   $200.96{\text{ m}}^2$

(c)

Given, diameter $=8\text{cm}$

So, radius $=\frac{8}{2}=4\text{cm}$

Area of circular garden $=\pi r^2=\frac{22}{7}\times 4\times 4$

                                                $=50.24{\text{m}}^2$ 

Area of a circle with diameter $‘m’$,  radius $‘n’$ and circumference $‘p’$ is -

a.    $2\pi n$

b.   $\pi m^2$

c.    $\pi p^2$

d.   $\pi n^2$

(d)

Given, diameter $=m,$ radius $=n$ and circumference $=p$

We know that, area of circle

 $\begin{array}{l}=\pi {(}^r\\ =\pi n^2\end{array}$

A table top is semicircular in shape with diameter $2.8\text{ m}.$ Area of this table top is -

a.    $3.08{\text{ m}}^2$

b.   $6.16{\text{ m}}^2$

c.    $12.32{\text{ m}}^2$

d.   $24.64{\text{ m}}^2$

(a)

Given, diameter $=2.8\text{ m}$

Now, radius $=\frac{2.8}{2}\text{m}=1.4\text{m}$

Area of table top $=$ Area of semi-circle $=\frac{\pi r^2}{2}$

$=\frac{22}{7}\times \frac{1.4}{2}\times \frac{1.4}{2}=3.08{\text{ m}}^2$

If $1{\text{ m}}^2= x{\text{ mm}}^2,$ then the value of $x$ is

a.    $1000$

b.   $10000$

c.    $100000$

d.   $1000000$

(d)

Given, $1{\text{m}}^2=x{\text{ mm}}^2$

${(1000\text{ mm})}^2=x{\text{ mm}}^2$                [ $1\text{ m}=1000\text{ mm}$ ]

$1000000{\text{ mm}}^2=x{\text{mm}}^2$

$x=1000000$

If $p$ squares of each side $1$ mm makes a square of side $1\text{ cm}$, then $p$ is equal to

a.    $10$

b.   $100$

c.    $1000$

d.   $10000$

(b)

Area of $1$ square of side $1$ mm $=1\times 1{\text{ mm}}^2$

Area of square of side $1$ cm $=1\times 1{\text{ cm}}^2$ $=$ $1{\text{ cm}}^2$

Number of squares with side $1\text{ mm}=p$

According to question,

$p$ $\times$ Area of square of side 1 mm $=\text{Area of square of side }1\text{ cm}$

$p\times 1{\text{mm}}^2=1{\text{ cm}}^2$

$p{\text{mm}}^2=1{\text{cm}}^2$

$p{\text{mm}}^2={(10\text{mm})}^2$

$p{\text{mm}}^2=100{\text{ mm}}^2$

$p=100$

$12{\text{ m}}^2$ is the area of

a.    a square with side $12\text{ m}$

b.   $12$ squares with side $1\text{ m}$ each

c.    $3$ squares with side $4\text{ m}$ each

d.   $4$ squares with side $3\text{ m}$ each

(b)

For option (a),

Area of a square with side $12\text{m}=12\times 12=144{\text{m}}^2$

From option (b),

Area of $12$ squares with side $1\text{ m}$ each $=12\times 1\times 1=12{\text{m}}^2$

For option (c),

Area of $3$ squares with side $4\text{m}$ each $=3\times \text{Area of square with side}4\text{m}$

$=3\times 4\times 4=48{\text{m}}^2$

For option (d),

Area of $4$ squares with side $3\text{ m}=4\times$ Area of square side $3\text{ m}$

$=4\times 3\times 3=36{\text{m}}^2$

Hence, option (b) is correct.

If each side of a rhombus is doubled, how much will its area increase?

a.    $1.5$ times

b.   $2$ times

c.    $3$ times

d.   $4$ times

(b)

Let b be the side and h be the height of a rhombus.

$\therefore$ Area of rhombus $=b\times h$ $[ \text{area of rhombus}=\text{base}\times \text{corresponding height}]$

If side is doubled, then side of new rhombus $=2b$

Now, area of new rhombus $=2b\times h=2(b\times h)=2\times (\text{area of original rhombus)}$  

Hence, the area of new rhombus will be increased by $2$ times.

If the sides of a parallelogram are increased to twice its original lengths, how much will the perimeter of the new parallelogram?

a.    $1.5$ times

b.   $2$ times

c.    $3$ times

d.   $4$ times

(b)

Let the length and breadth of the parallelogram be $l$ and $b,$ respectively.

Then, perimeter

$=2(l+b)$ [ $\because$ perimeter of parallelogram

$=2\times$ (length $+$ breadth)]

If both sides are increased twice, then new length and breadth will be $2l\text{and}2b$ respectively.

Now, new perimeter

$=\text{2}\left(\text{2}l+\text{2}b\right)=\text{2}\times \text{2}\left(l + b\right)=\text{2}$ times of original perimeter.

Hence, the perimeter of original parallelogram will increase by $2$ times.

If radius of a circle is increased to twice its original length, how much will the area of the circle increase?

a.    $1.4$ times

b.   $2$ times

c.    $3$ times

d.   $4$ times

(d)

Let $r$ be the radius of the circle.

$\therefore$ Area of circle $=\pi r^2$

If radius is increased to twice its original length, then new radius will be $2r.$ So, area of new circle $=\pi {(2r)}^2=4\pi r^2=4\times \pi r^2=4$ times of original area

Hence, the area of new circle will be increased by $4$ times.

What will be the area of the largest square that can be cut out of a circle of radius $10$ cm?

a.    $100{\text{ cm}}^2$

b.   $200{\text{ cm}}^2$

c.    $300{\text{ cm}}^2$

d.   $400{\text{ cm}}^2$

(b)

Given, radius of circle $=10\text{cm}$

The largest square that can be cut-out of a circle of radius $10\text{cm}$ will have its diagonal equals to the diameter of the circle.

Let the side of square be $x$ cm.

Area of the square is $x\times x=x^2{\text{ cm}}^2$

Now, in right angled $\Delta \text{DAB},$

${(\text{BD})}^2={(\text{AD})}^2+{(\text{AB})}^2$

${(20)}^2=x^2+x^2$

$2x^2=400$

$x^2=200$

${(\text{side})}^2=200$

Hence, the area of the largest square is $200{\text{cm}}^2$.

What is the radius of the largest circle that can be cut out of the rectangle measuring $10\text{ cm}$ in length and $8\text{ cm}$ in breadth?

a.    $4\text{ cm}$

b.   $5\text{ cm}$

c.    $8\text{ cm}$

d.   $10\text{ cm}$

(a)

The largest circle will have its diameter equal to the smaller side of rectangle i.e., $8\text{cm}$.

So, diameter $=8\text{cm}$

Radius $=\frac{\text{Diameter}}{2}=4\text{ cm}$

The perimeter of the figure ABCDEFGHIJ is

a.    $60\text{ cm}$

b.   $30\text{ cm}$

c.    $40\text{ cm}$

d.   $50\text{ cm}$

Fig. 9.23

(a)

We know that,

Perimeter of a closed shape $=$ sum of all its sides

So, $\text{AJ}+\text{JI}+\text{IH}+\text{HG}+\text{GF}+\text{FE}+\text{ED}+\text{CD}+\text{BC}+\text{AB}$

$=(\text{AJ}+\text{IH}+\text{GF}+\text{BC})+3+5+2+20+4+6$

$=\text{DE}+40$

$=20+40=60\text{cm}$

The circumference of a circle whose area is $81\pi r^2,$ is

a.    $9\pi r$

b.   $18\pi r$

c.    $3\pi r$

d.   $81\pi r$

(b)

Let the radius of circle be R

Area of circle $=\pi {\text{R}}^2$

$81\pi r^2=\pi {\text{R}}^2$

$\text{R}=\sqrt{81r^2}$

$\text{R}=9r$

Now, circumference of circle $=2\pi R=2\pi (9r)=18\pi r$

The area of a square is $100{\text{ cm}}^2.$ The circumference (in cm) of the largest circle cut of it is

a.    $5\pi$

b.   $10\pi$

c.    $15\pi$

d.   $20\pi$

(b)

Let the side of square be a cm.

Given, area of square $=100{\text{ cm}}^2$

Area of square $=a^2$

$a^2=100{\text{cm}}^2$

$a=\sqrt{100}$

$a=10\text{cm}$

Now, for the largest circle in the square, diameter of the circle must be equal to the side of square.

Diameter $=$ side of square $= 10\text{ cm}$

$2r=10\text{cm}$

$r=5\text{cm}$

Circumference of circle $=2\pi r=2\times \pi \times r=10\pi \text{ cm}$

If the radius of a circle is tripled, the area becomes

a.    $9$ times

b.   $3$ times

c.    $6$ times

d.   $30$ times

(a)

Let $r$ be the radius of a circle.

$\therefore$ Area of circle $=\pi r^2$

If radius is tripled, then new radius will be $3r.$

$\therefore$ Area of new circle $=\pi {(3r)}^2=9{\pi }^2=9$ times area of original circle

Hence, the area of new circle increases by $9$ times.

The area of a semicircle of radius $4r$ is

a.    $8\pi {\text{r}}^2$

b.   $4\pi r^2$

c.    $12\pi r^2$

d.   $2\pi r^2$

(a)

Given, radius of semi-circle $=4r$

Area of semi-circle $=\frac{1}{2}\times \pi r^2$

$=\frac{1}{2}\times \pi \times {(4r)}^2=\frac{16}{2}\pi r^2=8\pi r^2$

 

In Questions 38 to 56, fill in the blanks to make the statements true.

 

Perimeter of a regular polygon $=$ Length of one side $\times$ ___________.

Perimeter of regular polygon $=$ Length of one side $\times$ Number of sides.

If a wire in the shape of a square is rebent into a rectangle, then the ___________ of both shapes remain same, but ___________ may vary.

 If a wire in the shape of a square is rebent into a rectangle, then the perimeter of both shapes remain same, but area may vary.

Area of the square MNOP of Fig. 9.24 is $144{\text{ cm}}^2.$ Area of each triangle is ___________.

Fig. 9.24

Given, area of square MNOP $=144\text{ cm}$

Since, there are $8$ identical triangles in the given square MNOP.

Hence, area of each triangle $=\frac{1}{8}\times$ Area of square MNOP

$=\frac{1}{8}\times 144=18\text{cm}$

In Fig. 9.25, area of parallelogram BCEF is ___________ ${\text{cm}}^2$ , where ACDF is a rectangle.

Fig. 9.25

Area of a parallelogram
BCEF $=$ Area of ACDF $-2\times$ Area of $\Delta \text{CDE}$

$=10\times 5-2\times \left(\frac{1}{2}\times 3\times 5\right)$

$=50-15=35{\text{cm}}^2$

To find area, any side of a parallelogram can be chosen as ___________ of the parallelogram.

To find area, any side of a parallelogram can be chosen as base of the parallelogram.

 

Perpendicular dropped on the base of a parallelogram from the opposite vertex is known as the corresponding ___________ of the base.

Perpendicular dropped on the base of a parallelogram from the opposite vertex is known as the corresponding height/altitude of the base.

The distance around a circle is its ___________.

The distance around a circle is its circumference.

 

Ratio of the circumference of a circle to its diameter is denoted by symbol ___________.

Circumference of a circle $=2\pi r$   r, the radius of circle

$\text{C}=\pi d$            since, $d=2r$ 

$\frac{C}{d}=\pi$

$\pi =\text{C}:d$

Here, $\pi$ is the answer.

If area of a triangular piece of cardboard is $90{\text{ cm}}^2,$ then the length of altitude corresponding to $20\text{ cm}$ long base is ___________ cm.

We know that,

Area of triangle $=\frac{1}{2}\times \text{Base}\times \text{Height}$

$\therefore 90=\frac{1}{2}\times 20\times h$

Length of altitude $=9\text{ cm}$

Value of $\pi$ is ___________ approximately.

We know that, $\pi =\frac{22}{7}=3.14$

Thus, the approximate value of $\pi =3.14$

Circumference $‘\text{C}’$ of a circle can be found by multiplying diameter $‘\text{d}’$ with ___________.

$\because$ Circumference of a circle $=2\pi r$   $r$, the radius of circle

Since, diameter $(\text{d})=2r$

So, $\text{C}=\pi \times d$

Hence, $\pi$ is the answer.

Circumference $‘\text{C}’$ of a circle is equal to $2\pi \times$ ___________.

Circumference of a circle $=2\pi \times r$   $r$, radius of circle

Hence, $r$ is the answer.

$1{\text{ m}}^2=$ ___________ ${\text{cm}}^2.$

We know that, $1\text{m}=100\text{cm}$

On squaring both sides, we get

$\therefore 1{\text{ m}}^2={(100)}^2{\text{ cm}}^2$

$\Rightarrow 1{\text{ m}}^2=10000{\text{ cm}}^2$

$1{\text{ cm}}^2=$ ___________ ${\text{mm}}^2.$

We know that,

$1\text{ cm}=10\text{ mm}$

On squaring both sides, we get

${\text{1 cm}}^{\text{2}}={\left(\text{1}0\right)}^{\text{2}}{\text{mm}}^{\text{2}}=\text{1}00{\text{ mm}}^{\text{2}}$

$1$ hectare $=$ ___________ ${\text{m}}^2$.

$1\text{ hectare}=\text{1}0000{\text{ m}}^2$

Area of a triangle $=\frac{1}{2}\times \text{base}\times$ ___________.

Area of triangle $=\frac{1}{2}\times \text{Base}\times \text{Height}\text{.}$

$1{\text{ km}}^2=$ ___________ ${\text{m}}^2.$

We know that, $1\text{ km}=1000\text{ m}$

On squaring both sides, we get

$1{\text{ km}}^{\text{2}}={\left(1000\right)}^{\text{2}}{\text{ m}}^{\text{2}}=1000000{\text{ m}}^{\text{2}}$

Area of a square of side $6$ m is equal to the area of ___________ squares of each side $1$ cm.

Let number of squares having side $1\text{ cm}=x$

According to the question,

Area of square with side $6\text{ m}=x$ $\times$ Area of square with side $1\text{ cm}$ 

$\because$ area of square $=$ (side)²

$\therefore {(6\text{m})}^2=x\times {(1\text{cm})}^2$           [ $\because 1\text{m}=100\text{cm}$ ]

$\Rightarrow {(600\text{cm})}^2=x\times {(1\text{ cm})}^2$

$\Rightarrow 360000{\text{cm}}^2=x{\text{cm}}^2$

$\Rightarrow x=360000$

$10{\text{ cm}}^2=$ ___________ ${\text{m}}^2.$

We know that $1\text{ cm}=\frac{1}{100}$ m

$\therefore$$\text{1}0{\text{ cm}}^{\text{2}}=\text{1}0{\left(\frac{1}{100}\right)}^2{\text{m}}^2=\frac{10}{10000}{\text{m}}^2$

$10{\text{cm}}^2=\frac{1}{1000}{\text{m}}^2$

$10{\text{ cm}}^{\text{2}}=0.001{\text{ m}}^2$

 

 

In Questions 57 to 72, state whether the statements are True or False.

 

In Fig. 9.26, perimeter of (ii) is greater than that of (i), but its area is smaller than that of (i).

(i)                                    (ii)

Fig. 9.26

True

We know that, perimeter is the sum of all sides of any polygon.

We also know that, area is the space covered in the polygon.

So, by observing the figures we can say that, perimeter of (ii) is greater than (i) and area is less than that of (i).

In Fig. 9.27,

a.    area of (i) is the same as the area of (ii).

Fig. 9.27

b.   Perimeter of (ii) is the same as (i).

c.    If (ii) is divided into squares of unit length, then its area is $13$ unit squares.

d.   Perimeter of (ii) is $18$ units.

1. True

Since the number of blocks in fig (i) and fig (ii) are same, area of both figures is same.

2. False

Because $2$ new sides are added in (ii).

So, the perimeter of (ii) is greater than (i).

3. False

$\because$ Area of $1$ square $=1\times 1=1$ unit squares

and number of squares $=12$

So, total area $=\text{12}\times 1=12$ unit squares

4. True

Perimeter is the sum of all sides.

When we add the sides of the figure, we get 18 units.

So, the perimeter is $18$ units.

If perimeter of two parallelograms are equal, then their areas are also equal.

False

Even if the perimeter of two parallelograms are equal, their corresponding sides and height may be different. So, their area cannot be equal.

All congruent triangles are equal in area.

True

Congruent triangles have equal shape and size. Hence, their areas are also equal.

All parallelograms having equal areas have same perimeters.

False

It is not necessary that all parallelograms having equal areas have same perimeters as their base and height may be different.

 

 

Observe the figure and answer the statements 62 to 65 as true or false:

 

Observe all the four triangles FAB, EAB, DAB and CAB as shown in Fig. 9.28:

Fig. 9.28

 

All triangles have the same base and the same altitude.

True

It is clear from the figure that all triangles have same base AB and all the vertices lie on the same line, so the distance between vertex and base of triangle (i.e. length of altitude) are equal.

All triangles are congruent.

False

It is clear from the Fig. 9.28,that the lengths of sides other than the base in all the triangles are not equal. Hence, not all triangles are congruent.

All triangles are equal in area.

True

This is true because the triangles on same base and between same parallel lines have equal in area.

All triangles may not have the same perimeter.

True

 

 

It is clear from the Fig. 9.28,that the lengths of sides other than the base in all the triangles are not equal. Hence, all triangles may not have the same perimeter.

 

In Fig. 9.29 ratio of the area of triangle ABC to the area of triangle ACD is the same as the ratio of base BC of triangle ABC to the base CD of triangle ACD.

Fig. 9.29

True

Area of $\Delta \text{ABC}$ : Area of $\Delta \text{ACD}$ $=\frac{1}{2}\times \text{BC}\times \text{AC}:\frac{1}{2}\times \text{CD}\times \text{AC}$

$=\text{BC}:\text{CD}$ 

Triangles having the same base have equal area.

False

$\because$ Area of triangle $=\frac{1}{2}\times \text{Base}\times \text{Height}$

So, area of triangle does not only depend on base, it also depends on height.

Hence, if triangles have equal base and equal height, then only their areas are equal.

Ratio of circumference of a circle to its radius is always $2\pi :1.$

True

We know that, the circumference C of a circle of radius r, is $\text{C}=2\pi r$ 

$\because$ Circumference : Radius $=2\pi r:r=2\pi :1$

$5$ hectare $=500{\text{ m}}^2$

False

As we know that, $1$ hectare $=10000{\text{m}}^2$

So, $5$ hectare $=5\times 10000{\text{m}}^2=50000{\text{m}}^2$

An increase in perimeter of a figure always increases the area of the figure.

False

Perimeter is the sum of sides of any polygon and area is space covered by a polygon.

Two figures can have the same area but different perimeters.

True

Perimeter of (i) $=12\text{ cm}$  

Perimeter of (ii) $=18\text{ cm}$ 

Area of  figure (i) $=12{\text{ cm}}^2$ 

Area of  figure (ii) $=12{\text{ cm}}^2$

So, two figures can have the same area but different perimeters.

See the Q.58

Out of two figures if one has larger area, then its perimeter need not to be larger than the other figure.

True

Perimeter of figure (i) $=4\times 4=16$ cm

Perimeter of figure (ii) $=18$ cm

$\begin{array}{l}\text{Area}=16{\text{ cm}}^2\\ \text{Area}=14{\text{ cm}}^2\end{array}$

Though the first figure has the larger area, its perimeter is less than the second one.

$\Rightarrow$ Out of two figures if one has larger area, then its perimeter need not to be larger than the other figure.

A hedge boundary needs to be planted around a rectangular lawn of size $72\text{ m}\times 18\text{ m}.$ If $3$ shrubs can be planted in a metre of hedge, how many shrubs will be planted in all?

Given, length of rectangular lawn $=72\text{ m}$ and breadth of rectangular lawn $=18\text{ m}$

$\because$ Perimeter of rectangle $=2\times (l\text{+b})$

$\therefore$ Perimeter of rectangular lawn $=2(72+18)=2(90)=180\text{m}$

No. of shrubs to be planted in 1 m of hedge $=3$

No. of shrubs to be planted in $180$ m of hedge

$=3\times 180=540$ 

People of Khejadli village take good care of plants, trees and animals. They say that plants and animals can survive without us, but we cannot survive without them. Inspired by her elders Amrita marked some land for her pets (camel and ox) and plants. Find the ratio of the areas kept for animals and plants to the living area.

 

Fig. 9.30

Area of total rectangular land $=15\text{m}\times 10\text{m}=150{\text{m}}^{\text{2}}$

Area of land covered by plants $=9\text{m}\times 1\text{m}=9{\text{m}}^{\text{2}}$

Area of land covered by camel $=5\text{m}\times 3\text{m}=15{\text{m}}^{\text{2}}$

$\therefore$ Region of land covered by ox is circular area.

So, diameter, $d=2.8\text{m}$

Radius $=\frac{d}{2}=\frac{2.8}{2}=1.4\text{cm}$

Area of land covered by ox

$=\pi r^2=\frac{22}{7}\times 1.4\times 1.4=6.16{\text{m}}^2$

Total area covered by plants, camel, and ox $=9+15+6.16{\text{m}}^2$

Living area $=(150-30.16){\text{m}}^2=119.84{\text{m}}^2$

Ratio of areas kept for animals & plants to the living area

 

$=30.16:119.84=3016:11984=377:1498$

The perimeter of a rectangle is $40\text{ m}.$ Its length is four metres less than five times its breadth. Find the area of the rectangle.

Let breadth of the rectangle $=x$

Then, length of the rectangle $=5x-4$

Perimeter of the rectangle $=2(l+b)$

$40=2(l+b)$

$40=2(5x-4+x)$

$40=2(6x-4)$

$12x-8=40$

$12x=40+8$

$12x=48$

$x=\frac{48}{12}=4$

So, breadth $=x=4\text{m}$ and length $=5x-4=5\times 4-4=16\text{m}$

Area of rectangle $=l\times b=4\times 16=64{\text{m}}^2$

Hence, the area of rectangle is $64{\text{m}}^2$

A wall of a room is of dimensions $5\text{ m}\times 4\text{ m}.$ It has a window of dimensions $1.5\text{ m}\times 1\text{ m}$ and a door of dimensions $2.25\text{ m}\times 1\text{ m}.$ Find the area of the wall which is to be painted.

Length of the room $=5\text{m}$ 

Breadth of the room $=4\text{m}$

$\therefore$ Area of the room $=l\times b=5\times 4=20{\text{m}}^{\text{2}}$

Also, length of the window $=1.5\text{ m}$ and breadth of the window $=1\text{m}$         [given]

$\therefore$ Area of the window $=l\times b=1.5\times 1=1.5{\text{m}}^{\text{2}}$

Now, length of the door $=2.25\text{ m}$ and

breadth of the door $=1\text{m}$

$\therefore$ Area of the door $=l\times b=2.25\times 1=2.25{\text{ m}}^{\text{2}}$

Now, area of the wall to be painted $=$ Area of the room $-$ (Area of the window $+$ Area of the door) $=20-(1.5+2.25)=20-3.75=16.25{\text{m}}^{\text{2}}$

Rectangle MNOP is made up of four congruent rectangles (Fig. 9.31).If the area of one of the rectangles is $8{\text{ m}}^2$ and breadth is $2\text{ m},$ then find the perimeter of MNOP.

Fig. 9.31

Area of $1$ smaller rectangle $=8{\text{ m}}^2$

Breadth of smaller rectangle $=2\text{m}$

We know that, Area of the rectangle $=l\times b$

$l\times b=8$

$l\times 2=8$

$l=4\text{m}$

Perimeter of MNOP $=\text{MN}+\text{NC}+\text{CD}+\text{DO}+\text{PO}+\text{PF}+\text{FA}+\text{MA}$

$=4+2+4+2+4+2+4+2=24\text{m}$

So, the perimeter of MNOP is $24\text{ m}$.

In Fig. 9.32, area of $\Delta \text{AFB}$ is equal to the area of parallelogram ABCD. If altitude EF is $16\text{ cm}$ long, find the altitude of the parallelogram to the base AB of length $10\text{ cm}.$ What is the area of $\Delta \text{DAO},$ where O is the mid point of DC?

Fig. 9.32

Given,

Area of $\Delta \text{AFB}=$ Area of parallelogram ABCD

$\frac{1}{2}\times \text{AB}\times \text{EF}=\text{CD}\times (\text{corresponding}\text{height})$

Let corresponding height be $h$

Then,

$\frac{1}{2}\times 10\times 16=10\times h$

$h=8$

In $\Delta \text{DAO},\text{ DO}=5\text{cm}$

Area of $\Delta \text{DAO}=\frac{1}{2}\times \text{OD}\times h$

$=\frac{1}{2}\times 5\times 8=20{\text{cm}}^2$

Ratio of the area of $\Delta \text{WXY}$ to the area of $\Delta \text{WZY}$ is $3:4$ (Fig. 9.33). If the area of $\Delta \text{WXZ}$ is $56{\text{ cm}}^2$ and WY $=8$ cm, find the lengths of XY and YZ.

Fig. 9.33

We know that, area of a triangle $=\frac{1}{2}\times b\times h$ 

Given, area of $\Delta \text{WXZ}=56{\text{cm}}^2$

$\frac{1}{2}\times \text{WY}\times \text{XZ}=56$

$\frac{1}{2}\times 8\times \text{XZ}=56$

$\text{XZ}=14\text{cm}$

Area of $\Delta \text{WXY}:\text{Area}\text{of}\text{ΔWZY}=3:4$

$\frac{\text{Area}\text{of}\text{ΔWXY}}{\text{Area}\text{of}\text{ΔWZY}}=\frac{3}{4}$

$\frac{\frac{1}{2}\times \text{WY}\times \text{XY}}{\frac{1}{2}\times \text{YZ}\times \text{WY}}=\frac{3}{4}$

$\frac{\text{XY}}{\text{YZ}}=\frac{3}{4}$

$\frac{\text{XY}}{\text{XZ}-\text{XY}}=\frac{3}{4}$

$\frac{\text{XY}}{14-\text{XY}}=\frac{3}{4}$

$4\text{XY}=42-3\text{XY}$

$7\text{XY}=42$

$\text{XY}=6\text{cm}$

So, $\text{YZ}=\text{XZ}-\text{XY}=14-6$

$\text{YZ}=8\text{cm}$

Hence, $\text{XY}=6\text{cm}\text{and}\text{YZ}=8\text{cm}.$

Rani bought a new field that is next to one she already owns (Fig. 9.34). This field is in the shape of a square of side $70\text{ m}.$ She makes a semi circular lawn of maximum area in this field.

(i)             Find the perimeter of the lawn.

(ii)         Find the area of the square field excluding the lawn.

Fig. 9.34

(i)        Given, side of the square field $=70\text{m}$

Diameter of semi-circular lawn $=70\text{m}$

From the figure we understand that, Diameter of semi-circle $=$ side of square

Therefore, Radius $=\frac{70}{2}=35\text{cm}$

Perimeter of lawn $=\pi r+2r=\frac{22}{7}\times 35+2\times 35$

$=110+70=180\text{m}$

(ii)    Area of square $=$ side $\times$ side

$=70\times 70=4900{\text{m}}^2$

Required area $=$ Area of square $-$ Area of semi-circle

$=4900-\frac{1}{2}\times \pi \times {(35)}^2$

$=4900-\frac{1}{2}\times \frac{22}{7}\times 35\times 35=4900-1925$

$=2975{\text{ m}}^2$

In Fig. 9.35, find the area of parallelogram ABCD if the area of shaded triangle is $9{\text{ cm}}^2.$

Fig. 9.35

Given, area of shaded triangle $=9{\text{cm}}^2$

And base of the triangle $=3\text{cm}$

We know that, Area of triangle $=\frac{1}{2}\times \text{base}\times \text{height}$

Therefore, $9=\frac{1}{2}\times 3\times h$

$\frac{18}{3}=h$

$h=6\text{cm}$

Area of parallelogram $=$ height $\times$ base of parallelogram

$=\text{AE}\times \text{BC}$

$=6\times (3+4)=6\times 7=42{\text{ cm}}^2$

Pizza factory has come out with two kinds of pizzas. A square pizza of side $45\text{ cm}$ costs $Rs\text{15}0$ and a circular pizza of diameter $50\text{ cm}$ costs $Rs\text{16}0$ (Fig. 9.36). Which pizza is a better deal?

Fig. 9.36

Given, side of square pizza $=45\text{cm}$

Area of a square pizza $={(\text{side})}^2={(45)}^2=2025{\text{cm}}^2$

Diameter of circular pizza $=50\text{cm}$

Radius $=\frac{50}{2}=25\text{cm}$

Now, area of the circular pizza $=\frac{22}{7}\times 25\times 25$

$=\frac{22}{7}\times 625$

$=\frac{13750}{7}=1964.28{\text{cm}}^2$

Price of $1$ sq. cm. pizza $=\frac{2025}{150}=Rs13.5$

Price of $1$ cm circular pizza $=\frac{1964.24}{160}=Rs12.27$

Since price of $1{\text{ cm}}^2$ of circular pizza is lesser, hence circular pizza is a better deal.

Three squares are attached to each other as shown in Fig. 9.37. Each square is attached at the mid-point of the side of the square to its right. Find the perimeter of the complete figure.

Fig. 9.37

Perimeter of the complete figure

 

Side of first square $=6\text{ m}$ 

Side of second square $=\frac{6}{2}=3\text{ m}$ 

Side of third square $=\frac{3}{2}=1.5\text{ m}$ 

$=6+6+6+3+1.5+1.5+1.5+3+3+1.5=33\text{m}$

In Fig. 9.38, ABCD is a square with AB $=15$ cm. Find the area of the square BDFE.

Fig. 9.38

Given, ABCD is a square and AB $=15\text{cm}$

Diagonal of square $\text{ABCD}=\sqrt{2a}=\sqrt{2}\times 15=15\sqrt{2}\text{cm}$  (a, the side of square ABCD)

Diagonal of square ABCD is the side of square BDEF

Area of the square $\text{BDFE}={(\text{side})}^2$

$={(15\sqrt{2})}^2=15\times 15\times \sqrt{2}\times \sqrt{2}$

$=225\times 2=450{\text{cm}}^2$

In the given triangles of Fig. 9.39, perimeter of $\Delta \text{ABC}$$=$ perimeter of $\Delta \text{PQR}.$ Find the area of $\Delta \text{ABC}.$

 

Fig. 9.39

Given, Perimeter of $\Delta \text{ABC}=\text{Perimeter}\text{of}\Delta \text{PQR}$

$\text{AB}+\text{BC}+\text{CA}=\text{PQ}+\text{QR}+\text{RP}$

$\text{AB}+5+13=6+10+14$

$\text{AB}+18=30$

$\therefore \text{ AB}=30–18=12\text{ cm}$

Area of $\Delta \text{ABC}=\frac{1}{2}\times \text{base}\times \text{height}$

$=\frac{1}{2}\times 5\times 12=5\times 6=30{\text{cm}}^2$

Altitudes MN and MO of parallelogram MGHK are $8$ cm and $4$ cm long respectively (Fig. 9.40). One side GH is $6$ cm long. Find the perimeter of MGHK.

Fig. 9.40

Given, MGHK is a parallelogram where $\text{MN=}8\text{cm},\text{ MO}=4\text{cm}\text{and}\text{GH}=6\text{cm}$

We know that, area of parallelogram $=$ base $\times$ height

Area of parallelogram MGHK, when base is GH

$=\text{GH}\times \text{MN}$

$=6\times 8{\text{cm}}^2=48{\text{cm}}^2$

Area of parallelogram MGHK, when base is HK

$=\text{HK}\times \text{MO}$

$48=\text{HK}\times 4$

$\text{HK}=12\text{cm}$

In parallelogram, opposite sides are equal

So, $\text{GH}=\text{MK}=6\text{cm and MG}=\text{HK}=12\text{cm}$

Perimeter of parallelogram $\text{MGHK}=(6+6+12+12)\text{cm}=36\text{cm}$

In Fig. 9.41, area of $\Delta \text{PQR}$ is $20{\text{ cm}}^2$ and area of $\Delta \text{PQS}$ is $44$ ${\text{cm}}^2.$ Find the length RS, if PQ is perpendicular to QS and QR is $5\text{cm}.$

Fig. 9.41

Given, area of $\Delta \text{PQR}=20{\text{cm}}^2$ 

area of $\Delta \text{PQS}=44{\text{cm}}^2$

Area of triangle $=\frac{1}{2}\times \text{base}\times \text{height}$

Area of $\Delta \text{PQR}=\frac{1}{2}\times \text{PQ}\times \text{QR}$

$20=\frac{1}{2}\times \text{PQ}\times 5$

$\frac{20\times 2}{5}=\text{PQ}$

$\text{PQ}=8\text{cm}$

Area of $\Delta \text{PQS}=\frac{1}{2}\times \text{PQ}\times \text{QS}$

$44=\frac{1}{2}\times 8\times \text{QS}$

$\text{QS}=\frac{44\times 2}{8}=11\text{cm}$

Now, $\text{RS}=\text{QS}-\text{QR}=11-5=6\text{cm}$

Area of an isosceles triangle is $48{\text{ cm}}^2.$ If the altitudes corresponding to the base of the triangle is $8\text{ cm},$ find the perimeter of the triangle.

Given, area of $\Delta \text{ABC}=48{\text{cm}}^2$ and altitude $=8\text{cm}$

$\Delta \text{ABC}$ is an isosceles triangle, where $\text{AB}=\text{AC}$

area of $\Delta \text{ABC}=\frac{1}{2}\times \text{BC}\times \text{AD}$

$48=\frac{1}{2}\times \text{BC}\times \text{AD}$

$\frac{1}{2}\times \text{BC}\times 8=48$

$\text{BC}=\frac{48\times 2}{8}$

$\text{BC}=12\text{cm}$

In an isosceles triangle, $\text{BD}=\text{DC}=6\text{cm}$

Using Pythagoras theorem in right angled $\Delta \text{ADB},$

${\text{AB}}^2={\text{BD}}^2+{\text{AD}}^2$

${\text{AB}}^2=6^2+8^2$

$=36+64$

${\text{AB}}^2=100$

$\text{AB}=10\text{cm}$

Perimeter of triangle $=\text{AB}+\text{AC}+\text{BC}=\text{AB}+\text{AB}+\text{BC}$

$=10+10+12=32\text{cm}$

Perimeter of a parallelogram shaped land is $96\text{ m}$ and its area is $270$ square meters. If one of the sides of this parallelogram is $18\text{m},$ find the length of the other side. Also, find the lengths of altitudes l and m (Fig. 9.42).

Fig. 9.42

Perimeter of parallelogram ABCD $=96\text{m}$

 Area of parallelogram ABCD $=270{\text{m}}^2$

Perimeter of parallelogram $\text{ABCD}=\text{AB}+\text{BC}+\text{CD}+\text{DA}$

$96=18+\text{AD}+18+\text{AD}$

$96=36+2\text{AD}$

$2\text{AD}=60$

$\text{AD}=30\text{cm}$

So, $\text{AD}=\text{BC}=30\text{cm}$

Now, area of parallelogram $\text{ABCD}=\text{Base}\times \text{Corresponding}\text{Height}$

$270=\text{AB}\times \text{DE}$

$270=18\times \text{DE}$

$\frac{270}{18}=\text{DE}$

$\text{DE}=15\text{cm}$

Also, area of parallelogram $\text{ABCD}=\text{AD}\times \text{BF}$

$270=30\times l$

$l=\frac{270}{30}=9\text{m}$

Hence, altitudes of parallelogram $l=9\text{cm}$ and $\text{m}=15\text{cm}$

Area of a triangle PQR right-angled at Q is $60{\text{ cm}}^2$ (Fig. 9.43). If the smallest side is $8\text{cm}$ long, find the length of the other two sides.

Fig. 9.43

Area of $\Delta \text{PQR}=60{\text{cm}}^2\text{ and side}\text{PQ}=8\text{cm}$

Area of $\Delta \text{PQR}=\frac{1}{2}\times \text{PQ}\times \text{QR}$

$60=\frac{1}{2}\times 8\times \text{QR}$

$\text{QR}=\frac{60\times 2}{8}$

$\text{QR}=15\text{cm}$

In right angled $\Delta \text{PQR},{\text{ PR}}^2={\text{PQ}}^2+{\text{QR}}^2$ (using Pythagoras theorem)

${\text{PR}}^2=8^2+{15}^2$

$=64+225$

${\text{PR}}^2=289=\sqrt{289}=17\text{cm}$

Hence, the length of two sides are $15\text{ cm}$ and $17\text{ cm}$.

In Fig. 9.44 a rectangle with perimeter $264\text{ cm}$ is divided into five congruent rectangles. Find the perimeter of one of the rectangles.

Fig. 9.44

Let $l\text{and}b$ be the length of each smaller rectangle

Given, perimeter of the bigger rectangle $=264\text{cm}$

According to the figure, $4l+5b=264\dots (\text{i})$

& $2l=3b\dots (\text{ii})$

Put the value of $3\text{b}$ from eq. (ii) in eq.(ii),

$2\left(2l\right)+5b=264$

$2\times 3b+5b=264$

$6b+5b=264$

$11b=264$

$b=\frac{264}{11}=24\text{cm}$

$l=\frac{3b}{2}=\frac{3\times 24}{2}=36\text{cm}$

 

Perimeter of a rectangle $=2(l+b)$

Hence, perimeter of smaller rectangle

$=2(36+24)=120\text{cm}$

Find the area of a square inscribed in a circle whose radius is $7\text{ cm}$ (Fig. 9.45).

[Hint: Four right-angled triangles joined at right angles to form a square]

Fig. 9.45

ABCD is a square                      (given)

According to the question,

Area of square $\text{ABCD}=4\times \text{Area}\text{of}\Delta \text{AOB}$

$=4\times \left(\frac{1}{2}\times \text{AO}\times \text{BO}\right)$

$=2\times 7\times 7=98{\text{cm}}^2$

Hence, area of inscribed square is $98{\text{cm}}^2$

Find the area of the shaded portion in question $92.$

Area of shaded option $=$ area of circle $-$ area of square

$=\pi r^2-98$

$=\frac{22}{7}\times 7\times 7-98=154-98=56{\text{cm}}^2$

 

 

In Questions 94 to 97 find the area enclosed by each of the following figures:

 

 

Fig. 9.46

Given shape contains a rectangle and a semi-circle.

Area of rectangle $\begin{array}{l}=l\times b\\ =(10.2\times 1.5){\text{cm}}^2\\ =15.3{\text{cm}}^2\end{array}$

Here, diameter of semi-circle $=(10.2-3.9)\text{cm}=6.3\text{cm}$

So, radius $=\frac{\text{diameter}}{2}=\frac{6.3}{2}=3.15\text{cm}$

Area of semi-circle $=\frac{1}{2}\pi r^2=\frac{22}{7}\times \frac{1}{2}\times 3.15\times 3.15$

$=15.59{\text{cm}}^2$

Total area $=$ area of rectangle $+$ area of semi-circle

$=15.3+15.59=30.89{\text{cm}}^2$

Fig. 9.47

For rectangle, $l=13\text{cm}$ and $b=4\text{cm}$

Area of rectangle $=l\times b=13\times 4=52{\text{cm}}^2$

For triangle, base $(b)=13-8=5\text{cm}$

And height $(h)=(16-4)\text{cm}=12\text{cm}$

Area of triangle $\begin{array}{l}=\frac{1}{2}\times b\times h\\ =\frac{1}{2}\times 5\times 12=30{\text{cm}}^2\end{array}$ 

Total area enclosed by the shape $=(52+30){\text{cm}}^2=82{\text{cm}}^2$

Fig. 9.48

For rectangle, $l=15\text{cm},b=3\text{cm}$

Area of rectangle $=l\times b=15\times 3=45{\text{cm}}^2$

According to questions

 

Now, base of triangle $=15–10=5\text{ cm}$ 

For triangle, base $(b)=5\text{cm}$ and height $(h)=4\text{cm}$

Area of triangle $=\frac{1}{2}\times b\times h=\frac{1}{2}\times 5\times 4=10{\text{cm}}^2$

Total area enclosed by the shape $=(45+10){\text{cm}}^2=55{\text{cm}}^2$

Fig. 9.49

Given shape contains a semi-circle and a triangle

Area of semi-circle $=\frac{1}{2}\pi r^2$

$=\frac{1}{2}\times \frac{22}{7}\times 10\times 10=\frac{1100}{7}{\text{cm}}^2$

Area of triangle $=\frac{1}{2}\times b\times h=\frac{1}{2}\times 20\times 7=70{\text{cm}}^2$

Hence, total area enclosed by the shape = area of semicircle + area of triangle

$=\frac{1100}{7}+70=\frac{1100+490}{7}=\frac{1590}{7}=227.14{\text{cm}}^2$ 

Find the areas of the shaded region:

Fig. 9.50

Let the radius of smaller circle be $r$ and bigger circle be $\text{R}$

From the figure, $r=\frac{7}{2}\text{cm}$ and $\text{R}=\frac{7}{2}+7=\frac{21}{2}\text{cm}$

Area of the shaded region $=$ area of bigger circle – area of smaller circle

$=\pi {\text{R}}^2-\pi r^2$

$=\pi ({\text{R}}^2-r^2)=\pi \left(\frac{21}{2}\times \frac{21}{2}-\frac{7}{2}\times \frac{7}{2}\right)$

$=\pi \left(\frac{441}{4}-\frac{49}{4}\right)=\frac{22}{7}\times \frac{392}{4}=308{\text{cm}}^2$

Hence, the area of shaded region is $308{\text{cm}}^2$.

Find the area of the shaded region:

Fig. 9.51

Diameter of bigger circle $=14\text{cm}$

Radius of bigger circle $=\frac{14}{2}=7\text{cm}$

So, area of bigger circle $=\pi r^2=\frac{22}{7}\times 7\times 7=154{\text{cm}}^2$

Diameter of smaller circle $=\frac{7}{4}\text{cm}$

Radius of the smaller circle $=\frac{7}{4\times 2}=\frac{7}{8}\text{cm}$

Area of two smaller circles $=2\times \pi r^2=2\times \frac{22}{7}\times \frac{7}{8}\times \frac{7}{8}$

$=\frac{77}{16}={\text{cm}}^2$

Area of shaded region $=$ area of bigger circle $-$ area of two smaller circles

$=154-\frac{77}{16}=\frac{154\times 16-77}{16}$

$=\frac{2464-77}{16}=\frac{2387}{16}=149\frac{3}{16}{\text{cm}}^2$

Hence, area of shaded region is $149\frac{3}{16}{\text{cm}}^2$

A circle with radius $16\text{ cm}$ is cut into four equal parts and rearranged to form another shape as shown in Fig. 9.52: Does the perimeter change? If it does change, by how much does it increase or decrease?

Fig. 9.52

Yes, the perimeter changes.

Perimeter of the circle in first fig includes 4 arcs of equal length.

While in the second fig, perimeter includes 4 arcs as well as radius twice.

So, The perimeter is increased by $2r=2\times 16=32\text{cm}.$

A large square is made by arranging a small square surrounded by four congruent rectangles as shown in Fig. 9.53. If the perimeter of each of the rectangle is $16\text{ cm},$ find the area of the large square.

Fig. 9.53

Let the length & breadth of rectangle be $l\text{ and }b,$ respectively

Given, perimeter of $1$ rectangle $=16\text{cm}$

$2(l+b)=16$

$l+b=8\text{cm}$

Since, the side of large square is $(l+b)$

Hence, area $={(\text{side})}^2={(l+b)}^2={(8)}^2=64{\text{cm}}^2$

ABCD is a parallelogram in which AE is perpendicular to CD (Fig. 9.54). Also $\text{AC}=5\text{ cm},$ $\text{DE}=4\text{ cm},$ and the area of $\Delta \text{AED}=6{\text{ cm}}^2.$ Find the perimeter and area of ABCD.

Fig. 9.54

Area of $\Delta \text{AED}=6{\text{cm}}^2\&\text{AC}=5\text{cm}\text{and}\text{DE}=4\text{cm}$

Area of $\Delta \text{AED}=\frac{1}{2}\times \text{DE}\times \text{AE}$

$=\frac{1}{2}\times 4\times \text{AE}=6$

$\text{AE}=\frac{6\times 2}{4}$

$\text{AE}=3\text{cm}$

In right angled $\Delta \text{AEC},\text{ AE}=3\text{cm and AC}=5\text{cm}$

So,        ${(\text{EC})}^2={(\text{AC})}^2-{(\text{AE})}^2$      using Pythagoras theorem

      ${(\text{EC})}^2=5^2-3^2=25-9$

      $\text{EC}=\sqrt{16}$

      $\text{EC}=4\text{cm}$

$\text{DE}+\text{EC}=\text{DC}$

$\text{DC}=4+4=8\text{cm}$

Now, in right angled $\Delta \text{AED},{\text{ AD}}^2={\text{AE}}^2+{\text{ED}}^2$

${\text{AD}}^2=3^2+4^2=9+16$

$\text{AD}=\sqrt{25}$

$\text{AD}=5\text{cm}$

Perimeter of parallelogram $\text{ABCD}=2(l+b)=2(\text{DC}+\text{AD})$

$=2(8+5)=2\times 13=26\text{cm}$

Area of parallelogram $\text{ABCD}=b\times h=\text{DC}\times \text{AE}$

$=8\times 3=24{\text{cm}}^2$ 

Ishika has designed a small oval race track for her remote control car. Her design is shown in the figure 9.55. What is the total distance around the track? Round your answer to the nearest whole cm.

Fig. 9.55

Total distance around the track $=$ length of $2$ parallel strips $+$ length of two semi-circles

$=2\times 52+2\times \pi \times 16$

$=104+2\times 3.14\times 16$

$=104+100.5009$

$=205\text{cm}\text{(approx}\text{.})$ 

A table cover of dimensions $3\text{ m }25\text{ cm}\times 2\text{ m }30\text{ cm}$ is spread on a table. If $30\text{ cm}$ of the table cover is hanging all around the table, find the area of the table cover which is hanging outside the top of the table. Also find the cost of polishing the table top at $Rs\text{16}$ per square metre.

Length of table cloth $=3\text{ m }25\text{ cm}$ 

Breadth of table cloth $=2\text{ m }30\text{ cm}$ 

Area of the table cloth $=3.25\times 2.30$

Area of the table cloth on the top of table $=(3.25-0.60)\times (2.30-0.60)$

Area of hanging table cloth $=$ Area of the table cloth $-$ Area of the table cloth on the top of table $=\{3.25\times 2.30\}-\{(3.25-0.60)\times (2.30-0.60)\}$

$=7.4750-4.5050$

$=2.97\text{ sq}\text{. metre}$

Since the area of the table top = 4.5050 sq. metre

Cost of polishing table top $=4.5050\times 16$

$=Rs72.08$

The dimensions of a plot are $200\text{ m}\times 150\text{ m}.$ A builder builds $3$ roads which are $3\text{ m}$ wide along the length on either side and one in the middle. On either side of the middle road he builds houses to sell. How much area did he get for building the houses?

Dimension of plot $=200\text{m}\times 150\text{m}$

& width of road $=3\text{m}$

Total area available for houses $=$ area of plot $-$ area of $3$ roads

$=200\times 150-3\times (3\times 200)$

$=30000-1800=28200{\text{m}}^2$

A room is $4.5\text{ m}$ long and $4\text{ m}$ wide. The floor of the room is to be covered with tiles of size $15\text{ cm}$ by $10\text{ cm}.$ Find the cost of covering the floor with tiles at the rate of $Rs\text{4}.\text{5}0$ per tile.

Length of room $=4.5\text{m},$ width of room $=4\text{m}$ and size of tiles $15\text{cm}\times 10\text{cm}$

Area of room $=l\times b=4.5\times 4=18{\text{ m}}^2$

$=18\times {(100)}^2{\text{cm}}^2=180000{\text{cm}}^2$

Area of $1$ tile $=15\times 10=150{\text{cm}}^2$

So, number of tiles $=\frac{\text{area}\text{of}\text{room}}{\text{area}\text{of}1\text{tile}}=\frac{180000}{150}=1200$

Cost of covering the floor with tiles at the rate of $Rs4.50$ per tile $=Rs4.50\times 1200=Rs5400$

Find the total cost of wooden fencing around a circular garden of diameter $28\text{ m},$ if $1\text{ m}$ of fencing costs $Rs\text{3}00.$

Given, diameter of the circular garden $=28\text{m}$

Length of the fencing $=$ circumference of circle

$=\pi d=\frac{22}{7}\times 28=88\text{m}$

Total cost of fencing $=88\times 300=`26400$

Priyanka took a wire and bent it to form a circle of radius $14\text{ cm}.$ Then she bent it into a rectangle with one side $24\text{ cm}$ long. What is the length of the wire? Which figure encloses more area, the circle or the rectangle?

Given, radius of circle $(r)=14\text{cm}$ and length of rectangle $(l)=24\text{cm}$

Length of the wire $=$ circumference of the circle $=2\pi r=2\times \frac{22}{7}\times 14=88\text{m}$

Let $b$ be the width of rectangle

Since, the wire is re-bent in the form of rectangle,

Perimeter of rectangle $=$ circumference of the circle

$2\left(l+b\right)=2\pi r$

$2(24+b)=88$

$24+b=44$

        $b=44-24=20\text{cm}$

Area of circle $=\pi r^2=\frac{22}{7}\times {(14)}^2=616{\text{cm}}^2$

Area of rectangle $=l\times b=24\times 20=480{\text{cm}}^2$

Hence, the circle enclosed more area than rectangle.

How much distance, in metres, a wheel of $25\text{ cm}$ radius will cover if it rotates $350$ times?

Radius of wheel $(r)=25\text{cm}=\frac{25}{100}\text{m}=\frac{1}{4}\text{m}$       $\left[1\text{ cm}=\frac{1}{100}\text{m}\right]$

Distance travelled in $1$ rotation = circumference of circle

$=2\pi r=2\times \frac{22}{7}\times \frac{1}{4}=\frac{1}{4}=\frac{11}{7}\text{m}$ 

Distance travelled in $350$ rotation $=\frac{11}{7}\times \text{35}0=550\text{m}$

Hence, the wheel covers a distance of $550$ m.

A circular pond is surrounded by a $2\text{ m}$ wide circular path. If outer circumference of circular path is $44\text{ m},$ find the inner circumference of the circular path. Also find area of the path.

Let $\text{R}$ and $r$ be the radius of outer circle and inner circle respectively.

Circumference of outer circle is $44\text{ m}$.

$2\pi \text{ R}=44$

$2\times \frac{22}{7}\times \text{R}=44$

$\begin{array}{l}\text{R}=\frac{44}{2\times \frac{22}{7}}\\ =\frac{7\times 44}{2\times 22}=7\text{ m}\end{array}$

Since, $\begin{array}{l}r=(\text{R}-2)\text{m}\\ =(7-2)\text{m}\\ =5\text{m}\end{array}$

Inner circumference of the circular path $=2\pi r=2\times \frac{22}{7}\times 5=31.43\text{m}(\text{approx}\text{.})$

Area of the path $=$ area of outer circle $-$ area of inner circle $=\pi ({\text{R}}^2-r^2)$

                $\begin{array}{l}=\frac{22}{7}(7^2-5^2)=\frac{22}{7}\times 14\\ =44{\text{m}}^2\end{array}$

A carpet of size $5\text{ m}\times 2\text{ m}$ has $25\text{ cm}$ wide red border. The inner part of the carpet is blue in colour (Fig. 9.56). Find the area of blue portion. What is the ratio of areas of red portion to blue portion?

Fig. 9.56

Length of the carpet $=5$ m

Breadth of the carpet $=2$ m

Given, size of carpet $=5\text{m}\times 2\text{m}$

& width of border $=25\text{cm}=\frac{25}{100}\times \text{m}=0.25\text{m}$

Area of carpet $\text{ABCD}=\text{AB}\times \text{BC}=5\times 2=10{\text{m}}^2$

So, length of inner blue portion, $\text{EF}=\text{AB}-(2\times 0.25)\text{cm}=5-0.50=4.5\text{m}$

& breadth of inner blue portion $\text{FG}=\text{BC}-(2\times 0.25)=2-0.50=1.5\text{m}$ 

Area of blue portion $=$ area of rectangle $\text{EFGH}=\text{EF}\times \text{FG}=4.5\times 1.5=6.75{\text{m}}^2$

Now, area of red portion $=$ area of ABCD $-$ area of EFGH $=10-6.75=3.25{\text{m}}^2$

Ratio of areas of red portion to blue portion $=3.25:6.75=13:27$

Use the Fig. 9.57 showing the layout of a farm house:

Fig. 9.57

a.    What is the area of land used to grow hay?

b.   It costs $Rs\text{91}$ per ${\text{m}}^2$ to fertilise the vegetable garden. What is the total cost?

c.    A fence is to be enclosed around the house. The dimensions of the house are $18.7\text{ m}\times 12.6$ m. At least how many metres of fencing are needed?

d.   Each banana tree required $1.25{\text{ m}}^2$ of ground space. How many banana trees can there be in the orchard?

a.    Area of land used to grow hay $=17.8\text{m}\times 10.6\text{m}=188.68{\text{m}}^2$

b.   Area of vegetable garden $=49\text{ m} \times 15.2\text{ m} =744.80 {\text{m}}^{\text{2}}$

cost to fertilise $1{\text{ m}}^2$ vegetable garden $=Rs91$

cost to fertilise $744.80{\text{ m}}^2$ vegetable garden $=Rs91\times 744.80= Rs67776.80$

c.    Since, fence is to be enclosed around the house of dimensions $18.7\text{ m}\times 12.6\text{ m}$

Perimeter of the house $=2\times \left(l+b\right)$

Total length of fence $=2\times \left(18.7+12.6\right)\text{ m}=2\times 31.3\text{ m}=62.6\text{ m}$

4. Area covered by banana orchard $=20\text{m}\times 15.7\text{ m}=314{\text{ m}}^2$

Since, $1.25{\text{ m}}^2$ area is required by $1$ banana tree

Therefore, number of banana trees required to be covered in $314{\text{ m}}^2$ area $=\frac{314}{1.25}=251.25\approx 251$ trees

Study the layout given below in Fig. 9.58 and answer the questions:

Fig. 9.58

a.    Write an expression for the total area covered by both the bedrooms and the kitchen.

b.   Write an expression to calculate the perimeter of the living room.

c.    If the cost of carpeting is $Rs\text{5}0/{\text{m}}^2$ write an expression for calculating the total cost of carpeting both the bedrooms and the living room.

d.   If the cost of tiling is $Rs30/{\text{m}}^2$ write an expression for calculating the total cost of floor tiles used for the bathroom and kitchen floors.

e.    If the floor area of each bedroom is $35{\text{ m}}^2,$ then find $x$.

1. Area of both bedrooms & the kitchen $=2\left(\text{area of bedroom}\right)x+\text{area of kitchen}$ $=(5\times x)\times 2+(15-x-2)\times 5$
   $=10x+\left(75-5 x-10\right)=10x+65-5x$
   $=\left(65+5x\right){\text{ m}}^2$
2. Perimeter of the living room $=15+2+5+\left(15-x\right)+5+x+2=44\text{ m}$
3. Total area of both the bedrooms & the living room

$=$ area of bedroom 1 $+$ (area of bedroom 2 $+$ area of living room)

$=5\times x+7\times 15$ 
$=\left(5x+105\right){\text{ m}}^2$ 
Total cost of carpeting $=\left(5 x+105\right)\times 50= Rs250\left(x+21\right)$

4. Total area of bathroom & kitchen $=\left(15-x\right)\times 5{\text{ m}}^2$
   Total cost of tilling $=\left(15-x\right)\times 5\times 30= Rs150\left(15-x\right)$
5. Given, area of floor of each bedroom $=35{\text{ m}}^2$ 
   Area of one bedroom $=5x{\text{ m}}^2$ 
   $5 x=35$ 
   $x=7\text{ m}$ 

A $10\text{ m}$ long and $4\text{ m}$ wide rectangular lawn is in front of a house. Along its three sides a $50\text{ cm}$ wide flower bed is there as shown in Fig. 9.59. Find the area of the remaining portion.

Fig. 9.59

Dimensions of the rectangular lawn $=10\text{ m}\times 4\text{ m}$ and width of flower bed $=50\text{ cm}$ Length of remaining portion,

$=10\text{ m}-100\text{ cm}=10\text{ m}-1\text{ m} =9\text{ m}$

Breadth of remaining portion, $=4 \text{m}-0.5 =3.5\text{ m}$

Area of the remaining portion of the lawn $=$ Area of portion $=9\times 3.5=31.5{\text{ m}}^2$

A school playground is divided by a $2\text{ m}$ wide path which is parallel to the width of the playground, and a $3\text{ m}$ wide path which is parallel to the length of the ground (Fig. 9.60). If the length and width of the playground are  $120\text{ m}$ and $80\text{ m}$ respectively, find the area of the remaining playground.

Fig. 9.60

Total area of the playground $=120\times 80=9,600{\text{ m}}^2$

Area of the vertical road lying above the horizontal road $=77\times 2=154{\text{ m}}^2$

Area of the horizontal road $=120\times 3=360{\text{ m}}^2$

Area of the remaining playground $=9,600–\left(154+360\right)$

= 9086 m²

In a park of dimensions $20\text{ m}\times 15\text{ m},$ there is a L shaped $1\text{ m}$ wide flower bed as shown in Fig. 9.61. Find the total cost of manuring for the flower bed at the rate of $Rs45{\text{ per m}}^2.$

Fig. 9.61

Given, dimensions of the park $=20\text{ m}\times 15\text{ m}$ and width of the flower bed $=1\text{ m}$

From the figure, $\text{FG} =\text{BC}-1\text{ m } =\left(15-1\right)=14\text{ m}$

$\text{EF} =\text{DC}-1\text{ m}=\left(20-1\right)=19\text{ m}$

Area of flower bed $=$ Area of ABCD $-$ Area of EFGD $=(20\times 15)-(19\times 14)=300-266=34{\text{ m}}^2$

Cost of manuring $1{\text{ m}}^2$ of flower bed $=Rs45$

Cost of manuring $34{\text{ m}}^2$ of the flower bed $=Rs34\times 45=Rs1530$

Dimensions of a painting are $60\text{ cm}\times 38\text{ cm}.$ Find the area of the wooden frame of width $6\text{ cm}$ around the painting as shown in Fig. 9.62.

Fig. 9.62

Length of inner rectangle = 60 cm

Breadth of inner rectangle = 38 cm

Area of inner rectangle $=60\times 38=2280{\text{cm}}^2$

 

Length of outer rectangle $=60+6+6=72\text{ cm}$

Breadth of outer rectangle $=38+6+6=50\text{ cm}$

 

Area of outer rectangle $=50\times 72=3600 {\text{cm}}^2$

Now, area of wooden frame

= area of outer rectangle – area of inner rectangle

$=3600-2280$ $=1320{\text{cm}}^2$

A design is made up of four congruent right triangles as shown in Fig. 9.63. Find the area of the shaded portion.

Fig. 9.63

Area of one right triangle

$=\frac{1}{2}\times 10\times 30=150{\text{ cm}}^2$

So, area of $4$ right angled triangles $=4\times 150=600{\text{ cm}}^2$

Area of square $={\left(\text{side}\right)}^2$

Area of square $={\left(30+10\right)}^2={40}^2=1600{\text{ cm}}^2$

Area of shaded portion $=1600-600=1000{\text{ cm}}^2$

A square tile of length $20\text{ cm}$ has four quarter circles at each corner as shown in Fig. 9.64 (i). Find the area of shaded portion. Another tile with same dimensions has a circle in the centre of the tile [Fig. 9.64 (ii)]. If the circle touches all the four sides of the square tile, find the area of the shaded portion. In which tile, area of shaded portion will be more? (Take $\pi =3.14$ )

   (i)                                               (ii)

Fig. 9.64

(i)        Area of shaded portion $=$ area of square  $-4\times$ area of quarter circle $=(20\times 20)-(4\times \frac{\pi r^2}{4})$

$=400-4\times \frac{22}{7}\times \frac{1}{4}\times 10\times 10$

$=400-\frac{2200}{7}=\frac{600}{7}=85.71{\text{ cm}}^{\text{2}}=86{\text{ cm}}^2$

(ii)    Area of shaded portion $=$ Area of square $-$ area of circle $=20\times 20-\frac{22}{7}\times 10\times 10$

$=400-\frac{2200}{7}=\frac{600}{7}=85.71{\text{ cm}}^{\text{2}}=86{\text{ cm}}^2$

Hence, area in both cases is equal is $86{\text{ cm}}^{\text{2}}$.

A rectangular field is $48\text{ m}$ long and $12\text{ m}$ wide. How many right triangular flower beds can be laid in this field, if sides including the right angle measure $2\text{ m}$ and $4\text{ m},$ respectively?

Given, dimensions of the rectangular field $=48\text{ m}\times 12\text{ m}$ .

Height of right-angled triangle = 2 m

Base of the triangle = 4m

 

Number of right triangular flower beds

$=\frac{\text{area}\text{of}\text{field}}{\text{area of right angle}}=\frac{48\times 12 }{\frac{1}{2}\times 2\times 4}$

$=144$    [area of rectangle $=l\times b$ & area of a right-angled triangle $=\frac{1}{2}\times h\times b$ ]

Ramesh grew wheat in a rectangular field that measured $32$ meters long and $26$ meters wide. This year he increased the area for wheat by increasing the length but not the width. He increased the area of the wheat field by $650$ square meters. What is the length of the expanded wheat field?

Let the increase in length be $x$ m.

Give, increase in area of the field $=650{\text{ m}}^2$

Area of expanded wheat field $-$ Area of original wheat field $=650{\text{ m}}^2$

$\left(32+x\right)\times 26–32\times 26=650$

$26\left(32+x–32\right)=650$

$x=\frac{650}{26}=25\text{ m}$

Length of expanded wheat field $=32+x=32+25=57\text{ m}$ 

In Fig. 9.65, triangle AEC is right-angled at E, B is a point on EC, BD is the altitude of triangle ABC, $\text{AC}=25\text{ cm},$ $\text{BC}=7\text{ cm}$ and $\text{AE}=15\text{ cm}.$ Find the area of triangle ABC and the length of DB.

Fig. 9.65

Given, $\text{AC}=25\text{ cm, BC}=7\text{ cm}$ and $\text{AE}=15\text{ cm}$

In $\Delta \text{AEC,}$ using Pythagoras theorem

${\text{AC}}^2={\text{AE}}^2+{\text{EC}}^2$

${\text{EC}}^2={\text{AC}}^2-{\text{AE}}^2$

${\text{EC}}^2={\left(25\right)}^2-{\left(15\right)}^2=625-225$

$=400$

$\text{EC}=\sqrt{400}=20\text{ cm}$

& $\text{EB}=\text{EC}-\text{BC}=20-7$

$=13\text{ cm}$

Area of $\Delta \text{AEC}=\frac{1}{2}\times \text{AE}\times \text{EC}$

$=\frac{1}{2}\times 15\times 20=150{\text{ cm}}^2$

& Area of $\Delta \text{AEB}=\frac{1}{2}\times \text{AE}\times \text{EB}$

$=\frac{1}{2}\times 15\times 13=97.5{\text{ cm}}^2$

Area of $\Delta \text{ABC}=\text{Area of }\Delta \text{AEC}-\text{Area of }\Delta \text{AEB}$ $=150-97.5=52.5{\text{ cm}}^2$

Again, Area of $\Delta \text{ABC}=\frac{1}{2}\times \text{BD}\times \text{AC}$

$52.5=\frac{1}{2}\times \text{BD}\times 25$

$\text{BD}=\frac{52.5\times 2}{25}=4.2\text{ cm}$

Hence, the area of $\Delta \text{ABC}$ is $52.5{\text{ cm}}^2$ and the length of DB is $4.2\text{ cm}$

 

Number of pieces of chocolate $=\frac{\text{area of sheet of chocolate}}{\text{area of} 1\text{ chocolate}}$ 

$\begin{array}{l}=\frac{18\times 18}{1.5\times 2}\\ =\frac{324}{3}\\ =108\end{array}$ 

Calculate the area of shaded region in Fig. 9.66, where all of the short line segments are at right angles to each other and $1\text{ cm}$ long.

Fig. 9.66

Length of the larger rectangle $=1\times 9\text{ cm}=9\text{ cm}$

Breadth of the larger rectangle $=1\times 9\text{ cm}=9\text{ cm}$

Area of shaded region $=$ area of larger square $-$ area of $41$ small identical squares $=9\times 9-41\times 1\times 1=81-41=40{\text{cm}}^2$

The plan and measurement for a house are given in Fig. 9.67. The house is surrounded by a path $1\text{m}$ wide.

Fig. 9.67

Find the following:

(i)             Cost of paving the path with bricks at the rate of $Rs120$ per ${\text{m}}^2.$

(ii)         Cost of wooden flooring inside the house except the bathroom at the cost of $Rs1200$ per ${\text{m}}^2.$

(iii)      Area of Living Room.

 

(i)             Area of path $=$ Area of rectangle PQRS $-$ Area of rectangle ADFH

$=(\text{PQ}\times \text{QR)}-(\text{AD}\times \text{DF)}$

$=\left(4+2.5+4+1+1\right)\times \left(3+3+1+1\right)-\left(4+2.5+4\right)\times \left(3+3\right)$ $=12.5\times 8-10.5\times 6$

$=37{\text{ m}}^{\text{2}}$

Cost of paving the path with bricks

$=$ cost per unit ${\text{m}}^{\text{2}}\times$ Total area of path $=120\times 37=Rs4440$

(ii)         Area of the house with wooden flooring $=$ Area of house $-$ Area of bathroom

$=$ Area of rectangle ADFH $-$ Area of rectangle BCLK

$=\left(4+2.5+4\right)\times \left(3+3\right)-2.5\times 2$

$=63-5=5{\text{8 m}}^2$

Cost of wooden flooring $=$ cost per unit ${\text{m}}^{\text{2}}\times$ Total area

$=1200\times 58=Rs69600$

(iii)      Area of living room $=$ Area of rectangle ACGH $-$ Area of rectangle ABJI $-$ Area of rectangle BCLK

$=\left(4+2.5\right)\times \left(3+3\right)-4\times 3-2.5\times 2$

$=39-12-5=22{\text{ m}}^2$

Architects design many types of buildings. They draw plans for houses, such as the plan shown in Fig. 9.68:

Fig. 9.68

An architect wants to install a decorative moulding around the ceilings in all the rooms. The decorative moulding costs $Rs500/\text{metre}$

a.    Find how much moulding will be needed for each room.

(i)          family room

(ii)      living room

(iii)   dining room

(iv)    bedroom $1$

(v)       bedroom $2$

b.   The carpet costs $Rs200/{\text{m}}^2$ Find the cost of carpeting each room.

c.    What is the total cost of moulding for all the five rooms.

a.    To install the decorative moulding around the ceilings in all rooms, we need to find the perimeter of ceiling of each room.

(i)  Given, breadth of the family room $=5.48\text{ m}$

& length of the family room $=4.57\text{ m}$

Perimeter of the family room $=2\left(l+b\right)=2\left(5.48+4.57\right)$

$=2\times 10.05=20.10\text{ m}$

(ii)    Given, length of the living room $=3.81\text{ m}$ 

& breadth of the living room $=7.53\text{ m}$ 

Perimeter of the living room $=2\left(l+b\right)=2\left(3.81+7.53\right)\text{m}$

$=2\times 11.34=22.68\text{ m}$

(iii)Given, breadth of the dining room $=5.48\text{ m}$

& length of the dining room $=5.41\text{ m}$

Perimeter of dining room $=2\left(l+b\right)=2\left(5.41+5.48\right)$

$=2\times 10.89=21.78\text{ m}$

(iv) Given, length of bedroom $1=3.04\text{ m}$

& breadth of bedroom $1=3.04\text{ m}$

Perimeter of the bedroom $\text{1}=\text{2}\left(l+b\right)=\text{2}\left(\text{3}.0\text{4}+\text{3}.0\text{4}\right)$

$=2\times 6.08=12.16\text{ m}$

(v)  Given, breadth of bedroom $2=2.43\text{ m}$

& length of bedroom $2=3.04\text{ m}$

Perimeter of the bedroom $2=2\left(l+b\right)=2\left(3.04+2.43\right)$

$=2\times 5.47=10.94\text{ m}$

2. To find the cost of carpeting the rooms, we need to first find the area of the floor of each room.

for bedroom $1,$

Given, length of bedroom $1=3.04\text{ m}$

& breadth of bedroom $1=3.04\text{ m}$

Area of bedroom $1=l\times b$

Area of bedroom $1=3.04\times 3.04=9.2416$

Cost of carpeting $1{\text{m}}^2=Rs200$ 

Therefore, cost of carpeting 9.2416 sq m = 9.2416 $\times$ 200 = $$$Rs$ 1848.32

For bedroom $2$,

Given, length of bedroom $2=3.04\text{ m}$

& breadth of bedroom $2=2.43\text{ m}$

Area of bedroom $2=l\times b$

$=3.04\times 2.43=7.3872{\text{ m}}^2$

Cost of carpeting $1{\text{ m}}^2=Rs200$ 

Therefore, cost of carpeting $7.3872{\text{ m}}^2=7.3872\times 200=Rs1477$

For living room,

Given, length of living room $=3.81$ and breadth of living room $=7.53\text{ m}$

Area of living room $=3.81\times 7.53=28.6893{\text{ m}}^2$

Cost of carpeting of living room $1{\text{ m}}^2=Rs200$

Therefore, cost of carpeting $28.6893{\text{ m}}^2=Rs200\times 28.6893=Rs5737.86$

For dining room,

Given, length of living room $=3.81$ and breadth of living room $=7.53\text{ m}$

Area of living room $=3.81\times 7.53=28.6893{\text{ m}}^2$

Cost of carpeting of living room $1{\text{ m}}^2=Rs200$

Therefore, cost of carpeting

$28.6893{\text{ m}}^2=Rs200\times 28.6893=Rs5737.86$

For dining room,

Length of dining room $=5.41\text{ m}$

$=5.41\text{ m}$ and breadth of dining room $=5.48\text{ m}$

Area of living room $=5.41\times 5.48=29.6468{\text{ m}}^2$

Therefore, cost of carpeting $29.6468{\text{ m}}^2=29.6468\times 200=Rs5929.36$

For family room,

Given, length of family room $=4.57\text{ m}$

& breadth of family room $=5.48\text{ m}$

Area of family room $=5.48\times 4.57=25.0436{\text{ m}}^2$

So, cost of carpeting family room $=25.0436\times 200=Rs5008.72$

c.    Total perimeter of all the $5$ rooms $\begin{array}{l}=20.10\text{ m}+22.68\text{ m}+21.78\text{ m}+12.16\text{ m}+10.94\text{ m}\\ =87.86\text{ m}\end{array}$ 
Cost of moulding each room $=Rs500$ per m

Total cost of moulding all $5$ rooms $=87.66\times 500=Rs43830$

ABCD is a given rectangle with length as $80\text{ cm}$ and breadth as $60\text{ cm}.$ P, Q, R, S are the mid points of sides AB, BC, CD, DA respectively. A circular rangoli of radius $10\text{ cm}$ is drawn at the centre as shown in Fig. 9.69. Find the area of shaded portion.

Fig. 9.69

Here, $\text{AP}=\frac{1}{2}\text{AB}=\frac{1}{2}\times 80=40\text{ cm}$

Also, $\text{AS}=\frac{1}{2}\text{AD}=\frac{1}{2}\times 60=30\text{ cm}$ 

Area of $\Delta \text{APS}=\frac{1}{2}\times \text{AP}\times \text{AS}=\frac{1}{2}\times 40\times 30=600{\text{ cm}}^2$

Area of portion PQRS
$=$ Area of rectangle ABCD $-4\times$$\text{Area of }\Delta \text{APS}$

$=(80\times 60)-(4\times 600)$

$=4800-2400=2400{\text{ cm}}^2$

Area of circular rangoli $=\pi \times {\left(10\right)}^2=\frac{22}{7}\times 100=314.28{\text{ cm}}^2$

Area of shaded region $=2400-314.28=2085.72{\text{ cm}}^2$

$4$ squares each of side $10\text{ cm}$ have been cut from each corner of a rectangular sheet of paper of size $100\text{ cm}\times 80\text{ cm}.$ From the remaining piece of paper, an isosceles right triangle is removed whose equal sides are each of $10\text{ cm}$ length. Find the area of the remaining part of the paper.

Area of each square $=\left(10\right)\text{ cm}=100{\text{ cm}}^{\text{2}}$
[ $\because$ area of square $=$ (side)²]

Area of rectangular sheet $=100\times 80\text{ cm}$

[ $\because$ area of rectangle $=$ length $\times$ breadth]

$=8000{\text{ cm}}^{\text{2}}$

Area of an isosceles right triangle $=\frac{1}{2}\times 10\times 10=50{\text{cm}}^2$

Area of remaining part of paper $=8000-(4\times 100)-50$

$=7550{\text{ cm}}^2$

A dinner plate is in the form of a circle. A circular region encloses a beautiful design as shown in Fig. 9.70. The inner circumference is $352$ mm and outer is $396$ mm. Find the width of circular design.

Fig. 9.70

Let the radius of inner & outer circle be $r$ and $\text{R}$

Inner circumference $=352\text{ mm}$

$2\pi r=352$

$2\times \frac{22}{7}\times r=352$

$r=\frac{352\times 7}{2\times 22}=\frac{2464}{44}=56\text{ mm}$

& outer circumference $=396\text{ mm}$

$2\pi \text{R}=396$

$2\times \frac{22}{7}\times \text{R}=396$

$\text{R}=\frac{396\times 7}{2\times 22}=63\text{ mm}$

Width of circular design = $\text{R}-r=63-56=7\text{ mm}$

The moon is about $384000\text{ km}$ from earth and its path around the earth is nearly circular. Find the length of path described by moon in one complete revolution. (Take $\pi =3.14$ )

Radius of the circular path = 384000 km

Length of path described by moon in one complete revolution = Circumference of the circular path $=2\pi r$

$=2\times 3.14\times 384000$ 
[ $\because$ radius $=$ distance of moon from the earth]

$=2411520\text{ km}$

A photograph of Billiard/Snooker table has dimensions as  $\frac{1}{10}th$ of its actual size as shown in Fig. 9.71:

Fig. 9.71

The portion excluding six holes each of diameter $0.5\text{ cm}$ needs to be polished at the rate of $`200{\text{ per m}}^2.$ Find the cost of polishing.

Actual length of the table $=25\times 10=250\text{ cm}$

Actual breadth of the table $=10\times 10=100\text{ cm}$

Area of table $=250\times 100=25000{\text{ cm}}^2$

Radius of $1$ hole $=\frac{0.5}{2}=0.25\text{ cm}$

Area of $6$ holes $=6\times \pi r^2$

$=6\times \frac{22}{7}\times 0.25\times 0.25=1.18{\text{ cm}}^2$

Area of the portion of table excluding holes $=25000-1.18=24998.8{\text{ cm}}^2$

Since, 1 m = 100 cm

$1{\text{ m}}^{\text{2}}{\text{ = 10000 cm}}^{\text{2}}$

Hence, Cost of polishing $=Rs\frac{24999}{10000}\times 200=Rs500\left(\text{approx}.\right)$