Chapter 11: Perimeter and Area

# Question: 1

The length and the breadth of a rectangular piece of land are  and  respectively. Find

(i)             its area

(ii)          the cost of the land, if  of the land costs $Rs\text{\hspace{0.17em}}10,000.$

## Solution

Given:

Length of the rectangular piece of land

Breadth of the rectangular piece of land  we Know that,

(i)             Area of the rectangular piece of land

$=\text{Length}×\text{Breadth}$

$=500×300$

(ii)          Since, the cost of

Therefore, the cost of  land

$=10,000×1,50,000$

$=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}1,50,00,00,000$

# Question: 2

Find the area of a square park whose perimeter is .

## Solution

It is given in the question that,

We know that,
Perimeter of square park $=4\text{\hspace{0.17em}}×\text{\hspace{0.17em}}$ Length of the side of park

Perimeter of square park

Now, Area of square park

Thus, the area of square park is

# Question: 3

Find the breadth of a rectangular plot of land, if its area is  and the length is . Also find its perimeter.

## Solution

Length of the rectangular park

Area of rectangular park

We know that,

Now, Perimeter of rectangular park

Thus, the perimeter of rectangular park is $84$ m.

# Question: 4

The perimeter of a rectangular sheet is . If the length is , find its breadth. Also find the area.

## Solution

Length of the rectangular sheet $=35$ cm

Perimeter of the rectangular sheet

We know that,

Perimeter of rectangle $=2\left(\text{length}×\text{breadth)}$

Now, Area of rectangular sheet

$=\text{length}×\text{breadth}$

Thus, breadth and area of rectangular sheet are  and  respectively.

# Question: 5

The area of a square park is the same as of a rectangular park. If the side of the square park is  and the length of the rectangular park is  find the breadth of the rectangular park.

## Solution

It is given in the question that, side of the square park

Length of the rectangular park

According to the question,

Area of square park $=$ Area of rectangular park

We know that,

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

Area of the rectangle $=\text{length}×\text{breadth}$

So, according to question

Thus, the breadth of the rectangular park is

# Question: 6

A wire is in the shape of a rectangle. Its length is  and breadth is . If the same wire is rebent in the shape of a square, what will be the measure of each side. Also find which shape encloses more area?

## Solution

Length of the rectangle shape wire $=40$ cm

Breadth of the rectangle shape wire $=22$ cm

According to the question,

Perimeter of square $=$ Perimeter of rectangle

Thus, the side of the square is $31$ cm.

Now, Area of rectangle $=$ length $×$ breadth

and Area of square $=$ side $×$ side

Therefore, on comparing, the area of square is greater than that of rectangle.

That is, area of square $>$ area of rectangle.

# Question: 7

The perimeter of a rectangle is . If the breadth of the rectangle is , find its length. Also find the area of the rectangle.

## Solution

It is given in the question that, Breadth of rectangle $=30$ cm

Perimeter of rectangle

We know that,

Perimeter of rectangle $=2$ (Length $+$ Breadth)

Therefore, length of rectangle $=35$ cm

Now, area of rectangle $=$ length $×$ breadth

Thus, the area of rectangle is

# Question: 8

A door of length  and breadth  is fitted in a wall. The length of the wall is  and the breadth is  (Fig. 11.6). Find the cost of white washing the wall, if the rate of white washing the wall is Fig. 11.6

## Solution

Given:

Length and breadth of the door are   and   respectively.

$\therefore$ Area of rectangular door $=$ length $×$ breadth

It is given in the question that, Length and breadth of the wall are $4.5$ m and $3.6$ m respectively.

$\therefore$ Area of wall including door $=$ length $×$ breadth

Now, Area of wall excluding door

$=$ Area of wall including door $-$ Area of door

Since, the rate of white washing  of the wall $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}20$

Therefore, the rate of white washing  of the wall

$=20×14.2$

$=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}284$

Thus, the cost of white washing the wall excluding the door is $Rs\text{\hspace{0.17em}}284$.

# Question: 1

Find the area of each of the following parallelograms:

a) b) c) d) e) ## Solution

We know that the area of parallelogram $=$ base $×$ height

a)   Here base $=7$ cm and height

Area of parallelogram

b)  Here base $=5$ cm and height

Area of parallelogram

c)   Here base $=2.5$ cm and height

Area of parallelogram

d)  Here base $=5$ cm and height

Area of parallelogram

e)   Here base $=2$ cm and height

Area of parallelogram

# Question: 2

Find the area of each of the following triangles:    ## Solution

We know that the area of triangle
$=\frac{1}{2}×$ base $×$ height

Here, base  and height

Area of triangle

Here, base  and height

Area of triangle

Here, base  and height

Area of triangle

Here, base  and height

Area of triangle

# Question: 3

Find the missing values:

 S.No. Base Height Area of the Parallelogram a. $20$ cm b. c. d. $15.6$ cm

## Solution

We know that the area of parallelogram
$=$ base $×$ height

a)   Here, base

Area of parallelogram $=$ base $×$ height

b)  Here, height

Area of parallelogram $=$ base $×$ height

c)   Here, height

Area of parallelogram $=$ base $×$ height

d)  Here, base

Area of parallelogram $=$ base $×$ height

Thus, the missing values are:

 S.No. Base Height Area of the Parallelogram a. $20$ cm b. c. d. $15.6$ cm

# Question: 4

Find the missing values:

 Base Height Area of Triangle ______ _____ ______

## Solution

We know that the area of triangle
$=\frac{1}{2}×\text{base}×\text{height}$

In first row, base  and area

In second row, height $=31.4$ mm and area

In third row, base $=22$ cm and area

Thus, the missing values are:

 Base Height Area of Triangle

# Question: 5

PQRS is a parallelogram (Fig 11.23). QM is the height from Q to SR and QN is the height from Q to PS. If SR $=12$ cm and QM $=7.6$ cm. Find: Fig. 11.23

the area of the parallegram PQRS

QN, if PS $=8$ cm

## Solution

Given: SR $=12$ cm, QM $=7.6$ cm, PS $=8$ cm.

Area of parallelogram $=$ base $×$ height

Area of parallelogram $=$ base $×$ height

# Question: 6

DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is , $\text{AB}=35$ cm and AD $=49$ cm, find the length of BM and DL. Fig 11.24

## Solution

Given: Area of parallelogram ABCD

Base (AB) $=35$ cm and base (AD) $=49$ cm

Since Area of parallelogram $=$ base $×$ height

Again, Area of parallelogram $=$ base $×$ height

Thus, the lengths of DL and BM are $42$ cm and $30$ cm respectively.

# Question: 7

$\Delta \text{ABC}$ is right angled at A (Fig 11.25). AD is perpendicular to BC. If AB $=5$ cm, BC $=13$ cm and AC $=12$ cm, Find the area of $\Delta \text{ABC}$. Also find the length of AD. Fig. 11.25

## Solution

In right angled triangle BAC,  and

Area of triangle $=\frac{1}{2}×\text{base}×\text{height}=\frac{1}{2}×\text{AB}×\text{AC}$

Now, in $\Delta \text{ABC}$,

Area of triangle $\text{ABC}=\frac{1}{2}×\text{BC}×\text{AD}$

# Question: 8

$\Delta \text{ABC}$ is isosceles with AB $=$ AC $=7.5$ cm and BC $=9$ cm (Fig 11.26). The height AD from A to BC, is . Find the area of $\Delta \text{ABC}$. What will be the height from C to AB i.e., CE? Fig. 11.26

## Solution

In $\Delta \text{ABC}$, AD $=6$ cm and BC $=9$ cm

Area of triangle $=\frac{1}{2}×\text{base}×\text{height}=\frac{1}{2}×\text{BC}×\text{AD}$

Again, Area of triangle $=\frac{1}{2}×\text{base}×\text{height}=\frac{1}{2}×\text{AB}×\text{CE}$

Thus, height from C to AB i.e., CE is .

# Question: 1

Find the circumference of the circles with the following radius:

a)

b)

c)

## Solution

We know that,

Circumference of the circle $=2\pi r$

We know that,

Circumference of the circle $=2\pi r$

We know that,

Circumference of the circle $=2\pi r$

# Question: 2

Find the area of the following circles, given that:

b)  diameter

## Solution

a)   It is given that,

We know that,

Area of circle $\begin{array}{l}=\pi {r}^{2}\\ =\frac{22}{7}×14×14\end{array}$

b)  It is given that,

Diameter of the circle

We know that,

We know that

Area of circle $\begin{array}{l}=\pi {r}^{2}\\ =\frac{22}{7}×5×5\end{array}$
$=\frac{550}{7}{\text{cm}}^{2}$

$=78.57\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 3

If the circumference of a circular sheet is $154$ m, find its radius. Also find the area of the sheet.

## Solution

Circumference of the circular sheet

We know that,

Circumference of the circular sheet $=$ $2\pi r$

Now, area of circular sheet

Area of the circle $=\pi {r}^{2}$
$=\pi {r}^{2}=\frac{22}{7}×24.5×24.5$

Hence, the radius and area of circular sheet are
$24.5$ m and  respectively.

# Question: 4

A gardener wants to fence a circular garden of diameter $21$ m. Find the length of the rope he needs to purchase, if he makes $2$ rounds of fence. Also find the costs of the rope, if it costs $Rs4$ per meter. ## Solution

Diameter of the circular garden

Now, circumference of circular garden $=2\pi r$

After putting the values we get,

$=2×\frac{22}{7}×\frac{21}{2}$

Now, the length of rope required for fencing the garden will be equal to the two times the circumference of the garden, because the fencing is done in 2 rounds.

$=2×2\pi r$

Since, the cost of $1$ meter rope $=\text{\hspace{0.17em}}Rs4$

Hence, cost of $132$ meter rope $=4×132=\text{\hspace{0.17em}}Rs528$

# Question: 5

From a circular sheet of radius $4$ cm, a circle of radius $3$ cm is removed. Find the area of the remaining sheet. (Take $\pi =3.14$ )

## Solution

It is given in the question that,

Area of remaining sheet

$=$ Area of circular sheet $-$ Area of removed circle

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

$=\pi \left({4}^{2}-{3}^{2}\right)=\pi \left(16-9\right)$

Hence, the area of remaining sheet is

# Question: 6

Saima wants to put a lace on the edge of a circular table cover of diameter . Find the length of the lace required and also find its cost if one meter of the lace costs $Rs15.$ (Take $\pi =3.14$ )

## Solution

Diameter of the circular table cover

Radius of the circular table cover

Circumference of circular table cover

$=2\pi r$

$=2×3.14×\frac{1.5}{2}$

Therefore, the length of required lace is .

Now, the cost of $1$ m lace $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}15$

Then the cost of $4.71$ m lace

$=15×4.71$

$=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}70.65$

Therefore, the cost of $4.71$ m lace is $Rs\text{\hspace{0.17em}}70.65$.

# Question: 7

Find the perimeter of the adjoining figure, which is a semicircle including its diameter. ## Solution

Diameter of the semicircle

We know that perimeter of a complete circle $=2\pi r,$ where r is the radius of the circle. So, the perimeter of a semi-circle $=\pi r$

Perimeter of the figure

$=$ Circumference of semi-circle $+$ diameter

$=\pi r+\text{D}$

$\begin{array}{l}=\frac{22}{7}×5+10\\ =\frac{110}{7}+10\end{array}$

Hence, the perimeter of the given figure is $25.71$ cm.

# Question: 8

Find the cost of polishing a circular table-top of diameter $1.6$ m, if the rate of polishing is $Rs15{\text{/m}}^{2}.$ (Take $\pi =3.14$ )

## Solution

Diameter of the circular table top

Radius of the circular table top

Area of circular table top
$=\pi {r}^{2}$

$=3.14×0.8×0.8$

As per the question,

Cost of polishing  of the table top $=\text{\hspace{0.17em}}Rs15$

Cost of polishing  of table top
$=15×2.0096$
$=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}30.14$ (approx.)

Hence, the cost of polishing a circular table top is $Rs30.14$ (approx.)

# Question: 9

Shazli took a wire of length $44$ cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square?

## Solution

Total length of the wire

This wire has been bent into the shape of a circle.

The circumference of the circle

Now, area of the circle
$=\pi {r}^{2}$

Now the wire is rebent into a square.

Then perimeter of square

Now area of square $={\left(\text{side)}}^{\text{2}}$

Therefore, the area of circle is greater than that of the square, so the circle encloses more area.

# Question: 10

From a circular card sheet of radius $14$ cm, two circles of radius $3.5$ cm and a rectangle of length  and breadth 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. ## Solution

Radius of circular sheet (R) $=14$ cm

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

Area of bigger circle $=\frac{22}{7}×14×14$

$=22×2×14$

$=44×14$

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

Area of 2 small circles $=2×\frac{22}{7}×3.5×3.5$

$=44×0.5×3.5$

$=22×3.5$

Length of rectangle $\text{(}l\text{)}=3$ cm and breadth of rectangle

Area of rectangle $=$ Length $×$ Breadth

Area of rectangle $=3×1$

According to question,

Area of remaining sheet $=$ Area of circular sheet– (Area of two smaller circles $+$ Area of rectangle)

$=\pi {\text{R}}^{2}-\left[2\left(\pi {r}^{2}\right)+\left(l×b\right)\right]$

$=616–77–3$

$=616-80$

Hence, the area of remaining sheet is .

# Question: 11

A circle of radius $2$ cm is cut out from a square piece of an aluminium sheet of side $6$ cm. What is the area of the left over aluminium sheet? (Take $\pi =3.14$ )

## Solution

Radius of circle $=2$ cm

As we know,

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

Area $=3.14×2×2$

$=3.14×4$

Side of square $=6$ cm

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

Area of square shaped sheet $={\left(6\right)}^{2}$

and

According to question,

Area of aluminium sheet left $=$ Total area of aluminium sheet $-$ Area of circle

$=36-12.56$

Hence, the area of aluminium sheet left is

# Question: 12

The circumference of a circle is $31.4$ cm. Find the radius and the area of the circle? (Take $\pi =3.14$ )

## Solution

Given,

The circumference of the circle $=31.4$ cm

According to question,

Area of the circle

Hence, the radius and the area of the circle are $5$ cm and  respectively.

# Question: 13

A circular flower bed is surrounded by a path $4$ m wide. The diameter of the flower bed is $66$ m. What is the area of this path? ( $\pi =3.14$ ) ## Solution

Diameter of the circular flower bed

Radius of circular flower bed with  wide path According to the question,

Area of the circular path $=$ Area of bigger circle $-$ Area of smaller circle

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

$=\pi \left[{\left(37\right)}^{2}-{\left(33\right)}^{2}\right]$

$=3.14\left[\left(37+33\right)\left(37-33\right)\right]$

$=3.14×70×4$

Hence, the area of the circular path is

# Question: 14

A circular flower garden has an area of . A sprinkler at the centre of the garden can cover an area that has a radius of $12$ m. Will the sprinkler water the entire garden? (Take $\pi =3.14$ )

## Solution

Radius of the circular sprinkler $=12$ m

Area of the circular sprinkler $=\pi {r}^{2}$

$=3.14×12×12$

$=3.14×144$

Area of the circular flower garden

Since, the area of sprinkler is greater than the area of the circular flower garden, hence the sprinkler will water the entire garden.

# Question: 15

Find the circumference of the inner and the outer circles, shown in the adjoining figure?
(Take $\pi =3.14$ ) ## Solution

We know that,

Circumference of the circle $=2\pi r$

$=2×3.14×19$

$=38×3.14$

As we know that,

Circumference of the circle $=2\pi r$

$=2×3.14×9$

$=18×3.14$

$=56.52$

Hence,

Circumference of outer circle

Circumference of inner circle

# Question: 16

How many times a wheel of radius $28$ cm must rotate to go $352$ m?

## Solution

Let us consider that the wheel rotates $n$ times of its circumference.

Radius of wheel $=28$ cm

Total distance covered

Circumference of the wheel $=2\pi r$

Distance covered by wheel $=n×$ circumference of wheel

Therefore, wheel must rotate $200$ times to cover a distance of .

# Question: 17

The minute hand of a circular clock is $15$ cm long. How far does the tip of the minute hand move in $1$ hour? (Take $\pi =3.14$ )

## Solution

In $1$ hour, minute hand completes one round of the clock which means it makes a circle.

We have to find out that how far will the tip of minute hand move in 1 hour.

For this we have to find out the distance travelled by the tip of minute hand.

Distance travelled by the minute hand in 1 hour $=$ Circumference of the clock

We know that,

Circumference of the circle $=2\pi r$

Circumference of circular clock $=2\pi r$

$=2×3.14×15$

Therefore, the tip of the minute hand moves  in $1$ hour.

# Question: 1

A garden is $90$ m long and $75$ m broad. A path  wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.

## Solution

Length of garden

We know that,

Area of rectangle $=$ Length $×$ Breadth

Area of garden $=90×75$

Outer length of rectangular garden with path

Outer breadth of rectangular garden with path Outer area of the rectangular garden $=$ outer length $×$ outer breadth

Now, Area of path $=$ Outer area of the rectangular garden $-$ Inner area of the rectangular garden

$=8,500-6,750$

Since,

Therefore,

# Question: 2

A  wide path runs outside and around a rectangular park of length  and breadth . Find the area of the path.

## Solution

Length of rectangular park

Area of rectangle $=$ Length $×$ Breadth

Area of park $=125×65$

Width of the path

Length of rectangular park with path

Breadth of rectangular park with path

Area of rectangle $=$ Length $×$ Breadth

Area of garden including path $=131×71$ Area of path
$=$ Area of park with path $-$ Area of park without path

$=\left(\text{AB}×\text{AD}\right)-\left(\text{EF}×\text{EH}\right)$

$=9301-8125$

Thus, area of path around the park is .

# Question: 3

A picture is painted on a cardboard $8$ cm long and $5$ cm wide such that there is a margin of $1.5$ cm along each of its sides. Find the total area of the margin.

## Solution

Length of painted cardboard $=8$ cm

Breadth of painted cardboard $=5$ cm

We know that,

Area of rectangle $=$ Length $×$ Breadth

$\therefore$ Area of cardboard including margin $=8×5$

Since, there is a margin of $1.5$ cm long from each of its side.

Therefore, reduced length $\therefore$ Area of cardboard not including margin $=5×2$

Area of margin
$=$ Area of cardboard (ABCD) $-$ Area of cardboard (EFGH)

$=\left(\text{AB}×\text{AD}\right)-\left(\text{EF}×\text{EH}\right)$

$=40-10$

Thus, the total area of margin is .

# Question: 4

A verandah of width $2.25$ m is constructed all along outside a room which is $5.5$ m long and $4$ m wide. Find:

(i)             the area of the verandah.

(ii)          the cost of cementing the floor of the verandah at the rate of

## Solution

(i)             The length of room $=5.5$ m and width of the room $=4$ m

We know that,

Area of rectangle $=$ Length $×$ Breadth

Area of room $=5.5×4$

The length of room with verandah

The width of room with verandah

Area of room including verandah $=10×8.5$ Area of verandah

$=$ Area of room with verandah $-$ Area of room without verandah

$=\left(\text{AB}×\text{AD}\right)-\left(\text{EF}×\text{EH}\right)$

$=85-22$

(ii)          The cost of cementing  of the floor of verandah $=\text{\hspace{0.17em}}Rs200$

The cost of cementing  of the floor of verandah $=200×63=\text{\hspace{0.17em}}Rs12,600$

# Question: 5

A path $1$ m wide is built along the border and inside a square garden of side $30$ m. Find:

(i)             the area of the path

(ii)          the cost of planting grass in the remaining portion of the garden at the rate of

## Solution

Side of the square garden $=30$ m and
We know that,

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

Area of square garden $={\left(30\right)}^{2}$

Width of the path along the border

Side of square garden without path

Area of garden not including path $={\left(28\right)}^{2}$

Now Area of path
$=$ Area of ABCD $-$ Area of EFGH

$=900-784$ (ii)   Area of remaining portion

The cost of planting grass in  of the garden

The cost of planting grass in  of the garden $=\text{\hspace{0.17em}}Rs40×784=\text{\hspace{0.17em}}Rs31,360$

# Question: 6

Two cross roads, each of width $10$ m, cut at right angles through the centre of a rectangular park of length $700$ m and breadth $300$ m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.

## Solution

Length of park

We know that,

Area of rectangle $=$ Length $×$ Breadth

Area of park $=700×300$

Here, PQ $=10$ m and PS $=300$ m, EH $=10$ m and EF $=700$ m

And KL $=10$ m and KN $=10$ m $=$ Area of PQRS $+$ Area of EFGH $-$ Area of KLMN

[ $\because$ KLMN is taken twice, which is to be subtracted]

$=\text{PS}×\text{PQ}+\text{EF}×\text{EH}-\text{KL}×\text{KN}$

$=\left(300×10\right)+\left(700×10\right)-\left(10×10\right)$

$=3000+7000-100$

Now, Area of park excluding cross roads

$=2,10,000-9,900$

# Question: 7

Through a rectangular field of length $90$ m and breadth $60$ m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is $3$ m, find

(i)             the area covered by the roads.

(ii)          the cost of constructing the roads at the rate of

## Solution

(i)             Here,   and $=$ Area of PQRS $+$ Area of EFGH $-$ Area of KLMN

[ $\because$ KLMN is taken twice, which is to be subtracted]

$=\text{PS}×\text{PQ}+\text{EF}×\text{EH}-\text{KL}×\text{KN}$

$=\left(60×3\right)+\left(90×3\right)-\left(3×3\right)$

$=180+270-9$

(ii)          The cost of  constructing  of the roads $=\text{\hspace{0.17em}}Rs110$

The cost of  constructing  of the roads $=\text{\hspace{0.17em}}Rs110×441=\text{\hspace{0.17em}}Rs48,510$

Therefore, the cost of constructing the roads $=\text{\hspace{0.17em}}Rs48,510$

# Question: 8

Pragya wrapped a cord around a circular pipe of radius $4$ cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side $4$ cm (also shown). Did she have any cord left? ( $\pi =3.14$ ) ## Solution

Here, cord wrapped around the circular pipe is equal to the circumference of that pipe.

We know that circumference of the pipe or circle $=2\pi r$

Radius of pipe $=4$ cm

Putting the values, we get:

Again, wrapping cord around a square is equal to the perimeter of the square $=4×\text{side}$

Remaining cord
$=$ Cord wrapped on pipe $-$ Cord wrapped on square

$=25.12-16$

Thus, she has left $9.12$ cm cord.

# Question: 9

The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find: (i)             the area of the whole land

(ii)          the area of the flower bed

(iii)      the area of the lawn excluding the area of the flower bed

(iv)       the circumference of the flower bed.

## Solution

Length of rectangular lawn $=10$ m,

breadth of the rectangular lawn $=5$ m

And radius of the circular flower bed $=2$ m

(i)             Area of the whole land $=$ length $×$ breadth

(ii)          Area of flower bed $=\pi {r}^{2}$

(iii)      Area of lawn excluding the area of the flower bed
$=$ area of lawn $-$ area of flower bed
$=50-12.56$

(iv)       The circumference of the flower bed $=2\pi r$

# Question: 10

In the following figures, find the area of the shaded portions:

(i) (ii) ## Solution

(i)             Here,

$=$ Area of rectangle ABCD $-$ (Area of $\Delta \text{FAE}+$ area of $\Delta \text{EBC}$ )

$=\left(\text{AB}×\text{BC}\right)-\left(\frac{1}{2}×\text{AE}×\text{AF}+\frac{1}{2}×\text{BE}×\text{BC}\right)$

$=\left(18×10\right)-\left(\frac{1}{2}×10×6+\frac{1}{2}×8×10\right)$

$=180–\left(30+40\right)$

$=180-70$

(ii)          Here, SR $=$ SU $+$ UR $=$  cm,
QR $=20$ cm

PQ $=$ SR $=20$ cm, PT $=$ PS $-$ TS $=$  cm

TS $=10$ cm, SU $=10$ cm, QR $=20$ cm and
UR $=10$ cm

$=$ Area of square PQRS $-$ Area of $\Delta \text{QPT}$
$-$ Area of $\Delta \text{TSU}$ $-$ Area of $\Delta \text{UQR}$

$=\left(\text{SR}×\text{QR}\right)-\frac{1}{2}×\text{PQ}×\text{PT}-\frac{1}{2}×\text{ST}×\text{SU}-\frac{1}{2}×\text{QR}×\text{UR}$

$=20×20-\frac{1}{2}×20×10-\frac{1}{2}×10×10-\frac{1}{2}×20×10$

$=400-100-50-100$

# Question: 11

Find the area of the quadrilateral ABCD. Here,    and $\text{BM}\perp \text{AC,}$ $\text{DN}\perp \text{AC}$ ## Solution

Here,

$=\frac{1}{2}×\text{AC}×\text{BM}+\frac{1}{2}×\text{AC}×\text{DN}$
$=\frac{1}{2}×22×3+\frac{1}{2}×22×3$
$=3×11+3×11$
$=33+33$