Properties of Triangles (Chapter 6) and Congruence of Triangles (Chapter 7)

In each of the questions 1 to 49, four options are given, out of which only one is correct. Choose the correct one.

The sides of a triangle have lengths (in cm) $10,$ $6.5$ and $a$, where $a$ is a whole number. The minimum value that $a$ can take is

a.    $6$ 

b.   $5$

c.    $3$ 

d.   $4$ 

(d)

As we know, sum of any two sides in a triangle is always greater than the third side. So, only $4$ is the minimum value that satisfies as a side in triangle.

$\left\{\begin{array}{c}10<6.5+4\\ 6.5<10+4\\ 4<10+6.5\end{array}\right\}$

Triangle DEF of Fig. 6.6 is a right triangle with $\angle \text{E}=90°$. What type of angles are $\angle \text{D}$ and $\angle \text{F?}$

Fig. 6.6

a.    They are equal angles

b.   They form a pair of adjacent angles

c.    They are complementary angles

d.   They are supplementary angles

(c)

Since, $\angle \text{D}$ and $\angle \text{F}$ are complementary angles.
In $\Delta \text{DEF,}$

$\angle \text{D}+\angle \text{E}+\angle \text{F}=180°$

$\angle \text{D}+90°+\angle \text{F}=180°$

$\angle \text{D}+\angle \text{F}=180°-90°$

$\angle \text{D}+\angle \text{F}=90°$

In Fig. 6.7, $\text{PQ}=\text{PS}$. The value of $x$ is

Fig. 6.7

a.    $35°$

b.   $45°$

c.    $55°$

d.   $70°$

(b)

In $\Delta \text{PQS},$

$110°+\angle 1=180°$ [Linear pair]

$\angle 1=180°-110°$

$\angle 1=70°$

Also, $\angle 1=\angle 2=70°[\because PQ=PS]$

The measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

$\angle 2=x+25°$

$70°=x+25°$

$x=70°-25°$

$x=45°$

In a right-angled triangle, the angles other than the right angle are

a.    obtuse

b.   right

c.    acute

d.   straight

(c)

In right angled $\Delta \text{ABC},$

$\angle \text{B}=90°$

$\angle \text{A}+\angle \text{B}+\angle \text{C}=180°$

$\angle \text{A}+90°+\angle \text{C}=180°$

$\angle \text{A}+\angle \text{C}=180°-90°=90°$

Hence, in a right-angled triangle, the angles other than the right angle are acute.

In an isosceles triangle, one angle is $70°.$ The other two angles are of

(i)             $55°$ and $55°$ 

(ii)         $70°$ and $40°$ 

(iii)      any measure

In the given option(s) which of the above statement(s) are true?

a.    (i) only

b.   (ii) only

c.    (iii) only

d.   (i) and (ii)

(d)

The sum of the interior angles of a triangle is $180$.

(i)             According to the question,

$70°+55°+55°=180°$

(ii)         According to the question,

$70°+70°+40°=180°$

(iii)      Not possible, because two angles must be equal in an isosceles triangle.

In a triangle, one angle is of $90°.$ Then

(i)             The other two angles are of $45°$ each

(ii)         In remaining two angles, one angle is $90°$ and other is $45°$

(iii)      Remaining two angles are complementary

In the given option(s) which is true?

a.    (i) only

b.   (ii) only

c.    (iii) only

d.   (i) and (ii)

(c)

In a right angled $\Delta \text{ABC}$

$\angle \text{B}=90$

Now, $\angle \text{A}+\angle \text{B}+\angle \text{C}=180°$

$\angle \text{A}+90°+\angle \text{C}=180°$

$\angle \text{A}+\angle \text{C}=180°-90°=90°$

Hence, remaining two angles are complementary.

Lengths of sides of a triangle are $3\text{ cm,}$ $4\text{ cm}$ and
$5\text{ cm}\text{.}$ The triangle is

a.    Obtuse-angled triangle

b.   Acute-angled triangle

c.    Right-angled triangle

d.    An Isosceles right triangle

(c)

Since, these sides satisfy the Pythagoras theorem, therefore it is right-angled triangle. Lengths of the sides of a triangle are $3\text{ cm,}$ $4\text{ cm}$ and $5\text{ cm}\text{.}$

According to Pythagoras theorem,

$3^2+4^2=5^2$

$9+16=25$

$25=25$

Note: The area of the square built upon the hypotenuse of a right angled triangle is equal to the sum of the areas of the squares upon the remaining sides is known as Pythagoras theorem.

In Fig. 6.8, $\text{PB}=\text{PD}$. The value of $x$ is

Fig. 6.8

a.    $85°$

b.   $90°$

c.    $25°$

d.   $35°$

(c)

In a right angled $\Delta \text{PBD},$

$\angle 1+120°=180°$ [Linear pair]

Now, $\angle 1=180°-120°=60°$

$\angle 1=\angle 2=60°[\because PB=PD]$

Also, $\angle 2+\angle 3=180°$ [Linear pair]

$60°+\angle 3=180°$

$\angle 3=180°-60°$

$\angle 3=120°$

 

In $\Delta \text{PDC},$ $35°+\angle 3+x=180°$

$35°+120°+x=180°$

$x=180°-155°=25°$

$x=25°$

In $\Delta \text{PQR}$,

a.    $\text{PQ}-\text{QR}>\text{PR}$

b.   $\text{PQ}+\text{QR}<\text{PR}$

c.    $\text{PQ}-\text{QR}<\text{PR}$

d.   $\text{PQ}+\text{PR}<\text{QR}$

(c)

As we know, sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

In $\Delta \text{PQR,}$

$\text{PR}+\text{QR}>\text{PQ}$

$\text{PR}>\text{PQ}-\text{QR}$

$\text{PQ}-\text{QR}<\text{PR}$

In $\Delta \text{ABC}$,

a.    $\text{AB}+\text{BC}>\text{AC}$ 

b.   $\text{AB}+\text{BC}<\text{AC}$

c.    $\text{AB}+\text{AC}<\text{BC}$ 

d.   $\text{AC}+\text{BC}<\text{AB}$

(a)

As we know, sum of any two sides in a triangle is always greater than the third side. In $\Delta \text{ABC,}$

$\text{AB}+\text{BC}>\text{AC}$

The top of a broken tree touches the ground at a distance of $12\text{m}$ from its base. If the tree is broken at a height of $5\text{m}$ from the ground then the actual height of the tree is

a.    $25\text{m}$

b.   $13\text{m}$

c.    $18\text{m}$

d.   $17\text{m}$

(c)

Let $\text{AB}$ be the given that tree of height $\text{h}$ $\text{m,}$ which is broken at D which is $12\text{m}$ away from its base and the height of remaining part, i.e. CB is $5\text{m}\text{.}$

In the given figure, BC represents the unbroken part of the tree.

Point C represents the point where the tree broke and CA represents the broken part of the tree.

Triangle ABC, thus formed, is right-angled at B.

Applying Pythagoras theorem in $\Delta \text{ABC,}$

${\text{AC}}^2={\text{BC}}^2+{\text{AB}}^2$

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

${\text{AC}}^2=25{\text{m}}^2+144{\text{m}}^2=169{\text{m}}^2$

$\text{AC}=13\text{m}$

Thus, original height of the tree $=\text{AC}+\text{CB}=13\text{m}+5\text{m}=18\text{m}$

The trianlge $\text{ABC}$ formed by $\text{AB}=5\text{cm,}$ $\text{BC}=8\text{cm,}$ $\text{AC}=4\text{cm}$ is

a.    an isosceles triangle only

b.   a scalene triangle only

c.    an isosceles right triangle

d.   scalene as well as a right triangle

(b)

(i)             It’s not isosceles triangle as all the sides are of different measure.

(ii)         It’s not right triangle, since it does not follow Pythagoras theorem.

 

$4^2+5^2=8^2$

$16+25=64$

$41\ne 64$          (not satisfied)

Hence, it is a scalene triangle as all the sides are of different measure.

Two trees $7\text{m}$ and $4\text{m}$ high stand upright on a ground. If their bases (roots) are $4\text{m}$ apart, then the distance between their tops is

a.    $3\text{m}$

b.   $5\text{m}$

c.    $4\text{m}$

d.   $11\text{m}$

(b)

Let $\text{BE}$ be the smaller tree and $\text{AD}$ be the bigger tree. Now, we have to find $\text{AB}$ (i.e. the distance between their tops).

Given:

$\text{ED}=\text{BC}=4\text{m}$ and $\text{BE}=\text{CD}=4\text{m}$

In $\Delta \text{ABC, BC}=4\text{m}$

$\text{AC}=\text{AD}-\text{CD}=\left(7-4\right)\text{m}=3\text{m}$

In right-angled $\Delta \text{ABC,}$

By Pythagoras theorem,

${\text{AB}}^2={\text{AC}}^2+{\text{BC}}^2=4^2+3^2$

$=16+9$

${\text{AB}}^2={25}^2$

$\text{AB}=5\text{m}$

$\therefore$ Distance between the tops of two trees $=5\text{m}$.

If in an isosceles triangle, each of the base angles is $40°,$ then the triangle is

a.    Right-angled triangle

b.   Acute angled triangle

c.    Obtuse angled triangle

d.   Isosceles right-angled triangle

(c)

As we know, the sum of the interior angles of a triangle is $180°.$

In $\Delta \text{ABC,}$

$\angle \text{A}+\angle \text{B}+\angle \text{C}=180°$

$\angle \text{A}+40°+40°=180°$

$\angle \text{A}=180°-80°$

$\angle \text{A}=100°$

Therefore, it is an obtuse angled triangle. Since, it has one angle which is greater than $90°.$

If two angles of a triangle are $60°$ each, then the triangle is

a.    Isosceles but not equilateral

b.   Scalene

c.    Equilateral

d.   Right-angled

(c)

In $\Delta \text{ABC,}$

$\angle \text{A}+\angle \text{B}+\angle \text{C}=180°$

$\angle \text{A}+60°+60°=180°$

$\angle \text{A}=180°-120°$

$\angle \text{A}=60°$

Since, all the angles are of $60.$ So, it is an equilateral triangle.

The perimeter of the rectangle whose length is $60\text{cm}$ and a diagonal is $61\text{cm}$ is

a.    $120\text{cm}$ 

b.   $122\text{cm}$ 

c.    $71\text{cm}$

d.   $142\text{cm}$

(d)

${\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2$

${\text{BC}}^2={\text{AC}}^2-{\text{AB}}^2$

        $={(61)}^2-{(60)}^2$

        $=3721-3600$

${\text{BC}}^2=121$

${\text{BC}}^2=\sqrt{121}=11$

 

So, perimeter of rectangle $=2(l+b)$

                                            $=2(60+11)$

                                     $=2(71)$

                                     $=142\text{cm}$

In $\Delta \text{PQR,}$ if $\text{PQ}=\text{QR}$ and $\angle \text{Q}=100°,$ then $\angle \text{R}$ is equal to

a.    $40°$ 

b.   $80°$ 

c.    $120°$ 

d.   $50°$

(a)

In $\Delta \text{PQR,}$               $\text{PQ}=\text{QR}$

Let         $\angle \text{P}=\angle \text{R}=x$

$\angle \text{P}+\angle \text{Q}+\angle \text{R}=180°$

$x+100+x=180°$

$2x+100=180°$

$2x=80°$

$x=40°$

Hence, $\angle \text{P}=\angle \text{R}=40°$

Which of the following statements is not correct?

a.    The sum of any two sides of a triangle is greater than the third side

b.   A triangle can have all its angles acute

c.    A right-angled triangle cannot be equilateral

d.   Difference of any two sides of a triangle is greater than the third side

(d)

The difference of the length of any two sides of a triangle is always smaller than the length of the third side.

In Fig. 6.9, $\text{BC}=\text{CA}$ and $\angle \text{A}=40$. Then, $\angle \text{ACD}$ is equal to

Fig. 6.9

a.    $40°$

b.   $80°$

c.    $120°$ 

d.   $60°$

(b)

Given, $\text{BC}=\text{CA}$

$\angle \text{B}=\angle \text{A}=40°$

The measure of any exterior angle of a triangle is equal to the sum of the measure of its two interior opposite angles.

So,           $\angle \text{ACD}=\angle A+\angle B=40°+40°$

                $\angle \text{ACD}=80°$

The length of two sides of a triangle are $7\text{cm}$ and $9\text{cm}\text{.}$ The length of the third side may lie between

(a) $1\text{cm}$ and $10\text{cm}$

(b) $2\text{cm}$ and $8\text{cm}$

(c) $3\text{cm}$ and $16\text{cm}$

(d) $1\text{cm}$ and $16\text{cm}$ 

(c)

The third side must be greater than the difference between two sides and less than the sum of two sides.

Sum of two sides $=7+9=16\text{cm}$

Difference of two sides $=9-7=2\text{cm}$

So, length of the third side must lie between $2\text{cm}$ and $16\text{cm}$.

From given options, c fits best.

From Fig. 6.10, the value of $x$ is

Fig. 6.10

a.    $75°$

b.   $90°$

c.    $120°$

d.   $60°$

(c)

In $\Delta \text{ABC,}$

$\angle \text{CAB}+\angle \text{ABC}+\angle \text{BCA}=180°$

$25+35+\angle \text{BCA}=180°$

$\angle \text{BCA}=180°-60°=120°$

$\angle \text{BCA}=120°$

Also, $\angle \text{BCA}$ is an exterior angle,

$\angle \text{BCA}=\angle \text{D}+y$

$y=\angle \text{BCA}-\angle \text{D}=120°-60°$

$y=60°$

Now, $\angle x\text{ and }\angle y$ form a linear pair

$x+y=180°$

$x+60°=180°$

$x+180°-60°=120°$

In Fig. 6.11, the value of $\angle \text{A}+\angle \text{B}+\angle \text{C}+\angle \text{D}+\angle \text{E}+\angle \text{F}$ is

Fig. 6.11

a.    $190°$

b.   $540°$

c.    $360°$

d.   $180°$

(c)

As we know, sum of all the interior angles of a triangle is $180°$

In $\Delta \text{ABC,}\angle \text{A}+\angle \text{B}+\angle \text{C}=180°$ …(i)

In $\Delta \text{DEF,}\angle \text{D}+\angle \text{E}+\angle \text{F}=180°$ …(ii)

On adding eq. (i) and (ii), we get

$\angle \text{A}+\angle \text{B}+\angle \text{C}+\angle \text{D}+\angle \text{E}+\angle \text{F}=180°+180°=360°$

In Fig. 6.12, $\text{PQ}=\text{PR, RS}=\text{RQ}$ and $\text{ST}\left|\right|\text{QR}$. If the exterior angle $\text{RPU}$ is $140°\text{,}$ then the measure of angle $\text{TSR}$ is

Fig. 6.12

a.    $55°$

b.   $40°$

c.    $50°$

d.   $45°$

(b)

$\angle 1+\angle \text{UPT}=180°$

$\angle 1+140°=180°$

$\angle 1=180°-140°=40°$

Since, $\text{PQ}=\text{PR}$

$\angle \text{Q}=\angle \text{R}=x$

In $\Delta \text{PQR,}\angle \text{P}+\angle \text{Q}+\angle \text{R}=180°$

$40°+x+x=180°$

$2x=180°-40°$

$x=\frac{140°}{2}=70$

So, $\angle \text{Q =}\angle \text{R}=70°$

Given that, $\text{RS}=\text{RQ}$

$\angle 2=\angle 3=70°$

In $\Delta \text{SQR, }\angle 2+\angle 3+\angle 4=180°$

$70°+70°+\angle 4=180°$

$\angle 4=180°-140°=40°$

Also,        $\text{ST}\left|\right|\text{QR}$

Now,       $\angle 4=\angle 6=40°$ [alternate interior angles]

                $\angle \text{TSR}=40°$

In Fig. 6.13, $\angle \text{BAC}=90°\text{,}$ $\text{AD}\perp \text{BC}$ and $\angle \text{BAD}=50°\text{,}$ then $\angle \text{ACD}$ is

Fig. 6.13

a.    $50°$

b.   $40°$

c.    $70°$

d.   $60°$

(a)

Given, $\angle \text{BAC}=90°\text{, AD}\perp \text{BC and }\angle \text{BAD}=50°$

In $\Delta \text{ABD,}$                        $\angle \text{ABD}+\angle \text{DAB}+\angle \text{ADB}=180$

$\angle ABD+50+90=180$

$\angle \text{ABD}+140=180$

$\angle \text{ABD}=180-140=40$

Now, in $\Delta \text{ABD,}$       $\angle \text{A}+\angle \text{B}+\angle \text{C}=180$

$90+40+\angle \text{C}=180$

$\angle \text{C}=180-130$

$\angle \text{C}=50$

$\angle \text{ACD}=50$

If one angle of a triangle is equal to the sum of the other two angles, the triangle is

a.    obtuse

b.   acute

c.    right

d.   equilateral

(c)

Let $\text{A, B and C}$ be the angles of the triangle. Then, one angle of a triangle is equal to the sum of the other two angles.

i.e.,          $\angle \text{A}=\angle \text{B}+\angle \text{C}$

$\angle \text{A +}\angle \text{B}+\angle \text{C}=180°$

$\angle \text{A +}\angle \text{A}=180°$ 

$2\angle \text{A}=180°$ 

$\angle \text{A}=\frac{180°}{2}$

$\angle \text{A}=90°$ 

Hence, the triangle is right-angled.

If the exterior angle of a triangle is $130°$ and its interior opposite angles are equal, then measure of each interior opposite angle is

a.    $55°$

b.   $65°$

c.    $50°$

d.   $60°$

(b)

As we know, the measure of any exterior angle is equal to the sum of two opposite interior angles.

Let the interior angle be $x\text{.}$

Given that, interior opposite angles are equal.

$130°=x+x$

$130°=2x$

$x=\frac{130°}{2}=65$

Hence, the interior angle is 65°

If one of the angles of a triangle is $110°\text{,}$ then the angle between the bisectors of the other two angles is

a.    $70°$

b.   $110°$

c.    $35°$ 

d.   $145°$ 

(d)

In $\Delta \text{ABC,}$         $\angle \text{A}=110°$

We know that, $\angle \text{A}+\angle \text{B}+\angle \text{C}=180°$

$\angle \text{B}+\angle \text{C}=180°-\angle \text{A}$

$\angle \text{B}+\angle \text{C}=180°-110°$

$\angle \text{B}+\angle \text{C}=70°$

$\frac{1}{2}\angle \text{B}+\frac{1}{2}\angle \text{C}=\frac{70°}{2}=35$

$\frac{1}{2}(\angle \text{B}+\angle \text{C})=35°$

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

$\angle \text{BOC}+\angle \text{OBC}+\angle \text{OCB}=180°$

$\angle \text{BOC}+\frac{1}{2}(\angle \text{B}+\angle \text{C})=180°$

$\angle \text{BOC}+35°=180°$

$\angle \text{BOC}=180°-35°$

$\angle \text{BOC}+145°$

In $\Delta \text{ABC, AD}$ is the bisector of $\angle \text{A}$ meeting $\text{BC}$ at $\text{D}$, $\text{CF}\perp \text{AB}$ and $\text{E}$ is the mid-point of $\text{AC}\text{.}$ Then median of the triangle is

a.    $\text{AD}$

b.   $\text{BE}$

c.    $\text{FC}$

d.   $\text{DE}$

(b)

As we know, median of a triangle bisects the opposite sides.

Hence, the median is $\text{BE}$ as $\text{AE}=\text{EC}$ 

In $\Delta \text{PQR,}$ if $\angle \text{P}=60°\text{, and }\angle \text{Q}=40°\text{,}$ then the exterior angle formed by producing $\text{QR}$ is equal to

1. $60°$ 
2. $120°$ 
3. $100°$
4. $80°$

(c)

As we know, the measure of exterior angle is equal to the sum of opposite two interior angles.

In $\Delta \text{PQR,}$ $\angle x$ is the exterior angle

So, $\angle x=\angle \text{P}+\angle \text{Q}$ 

$=60°+40°=100°$

Which of the following triplets cannot be the angles of a triangle?

a.    $67°\text{, }51°\text{, }62°$

b.   $70°\text{, }83°\text{, }27°$

c.    $90°\text{, }70°\text{, }20°$ 

d.   $40°\text{, }132°\text{, }18°$

(d)

We know that, the sum of interior angles of a triangle is $180.$

Now, we will verify the given triplets:

(a) $67°+51°+62°=180°$

(b) $70°+83°+27°=180°$

(c) $90°+70°+20°=180°$

(d) $40°+132°+18°=190°$

Clearly, triplets in option (d) cannot be the angles of a triangle.

Which of the following can be the length of the third side of a triangle whose two sides measure $18\text{ cm}$ and $14\text{ cm?}$

a.    $4\text{ cm}$ 

b.   $3\text{ cm}$ 

c.    $5\text{ cm}$ 

d.   $32\text{ cm}$

(c)

As we know, sum of any two sides of a triangle is always greater than the third side. Hence, option (c) satisfies the given condition.

Verification

$18+14>5$

$18+5>14$

$5+14>18$

How many altitudes does a triangle have?

a.    $1$ 

b.   $3$ 

c.    $6$ 

d.   $9$

(b)

A triangle has $3$ altitudes.

If we join a vertex to a point on opposite side which divides that side in the ratio $1:1\text{,}$ then what is the special name of that line segment?

a.    Median

b.   Angle bisector

c.    Altitude

d.   Hypotenuse

(a)

Consider $\Delta \text{ABC}$ in which $\text{AD}$ divides $\text{BC}$ in the ratio $1:1.$

Now, $\text{BD}:\text{DC}=1:1$

$\frac{\text{BD}}{\text{DC}}=\frac{1}{1}$

$\text{BD}=\text{DC}$

Since, $\text{AD}$ divides $\text{BC}$ into two equal parts.

Hence, $\text{AD}$ is the median.

The measures of $\angle x$ and $\angle y$ in Fig. 6.14 are respectively

a.    $30°\text{, }60°$ 

b.   $40°\text{, }40°$

c.    $70°\text{, }70°$ 

d.   $70°\text{, }60°$

(d)

As we know,

Measure of exterior angle $=$ sum of the opposite interior angles

$\angle \text{R}=\angle \text{P}+\angle \text{Q}$

$120°=x+50$

$x=120°-50°=70°$

Now, the sum of the interior angles of a triangle is $180°$,

$\begin{array}{l}x+50°+y=180°\\ 70°+50°+y=180°\\ 120°+y=180°\\ y=60°\end{array}$

If length of two sides of a triangle are $6\text{ cm}$ and $10\text{ cm,}$ then the length of the third side can be

a.    $3\text{ cm}$ 

b.   $4\text{ cm}$

c.    $\text{2 cm}$

d.   $6\text{ cm}$

(d)

As we know, sum of any two sides of a triangle is always greater than the third side. So, option (d) satisfy this rule.

Verification

$6+6>10$

$6+10>6$

$10+6>6$

In a right-angled triangle $\text{ABC,}$ if angle $\text{B}=90°\text{,}$ $\text{BC}=3\text{ cm}$ and $\text{AC}=5\text{ cm,}$ then the length of side $\text{AB}$ is

a.    $3\text{ cm}$ 

b.   $4\text{ cm}$ 

c.    $5\text{ cm}$ 

d.   $6\text{ cm}$

(b)

Since, $\Delta \text{ABC}$ is a right angled triangle.

In right angled $\Delta \text{ABC,}$

${\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2$

$5^2={\text{AB}}^2+3^2$

${\text{AB}}^2=25-9$

${\text{AB}}^2=16$

$\text{AB}=4\text{cm}$

In a right-angled triangle $\text{ABC,}$ if angle $\text{B}=90°\text{,}$ then which of the following is true?

a.    ${\text{AB}}^2={\text{BC}}^2+{\text{AC}}^2$

b.   ${\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2$

c.    $\text{AB}=\text{BC}+\text{AC}$

d.   $\text{AC}=\text{AB}+\text{BC}$

(b)

According to Pythagoras theorem,

${\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2$

Which of the following figures will have it’s altitude outside the triangle?

1. 
2. 
3. 
4. 

 

 

Fig. 6.15

 

(d)

As we know, the perpendicular line segment from a vertex of a triangle to its opposite side is called an altitude of the triangle.

In Fig. 6.16, if $\text{AB}\parallel \text{CD,}$ then

Fig. 6.16

1. $\angle 2=\angle 3$
2. $\angle 1=\angle 4$
3. $\angle 4=\angle 1+\angle 2$
4. $\angle 1+\angle 2=\angle 3+\angle 4$

(d)

Given, $\text{AB}\parallel \text{CD}$ and $\text{AC}$ is the transversal.

Measure of exterior angle $=$ sum of the opposite interior angles

In $\Delta \text{ABC, }\angle 3+\angle 4=\angle 1+\angle 2$ 

In $\Delta \text{ABC,}$ $\angle \text{A}=100°\text{,}$ $\text{AD}$ bisects $\angle \text{A}$ and $\text{AD}\perp \text{BC}.$ Then, $\angle \text{B}$ is equal to

1. $80°$ 
2. $20°$ 
3. $40°$ 
4. $30°$

(c)

Given,             $\angle \text{BAD}=\angle \text{DAC}=50°$

$\angle \text{BDA}=\angle \text{ADC}=90°$

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

$\angle \text{ABD}+\angle \text{BAD}+\angle \text{BDA}=180°$

$\angle \text{ABD}+50°+90°=180°$

$\angle \text{ABD}+140°=180°$

$\angle \text{ABD}=180°-140°=40°$

In $\Delta \text{ABC,}$ $\angle \text{A}=50°\text{, }\angle \text{B}=70°$ and bisector of $\angle \text{C}$ meets AB in D (Fig. 6.17). Measure of $\angle \text{ADC}$ is.

Fig. 6.17

(a)    $50°$

(b)   $100°$

(c)    $30°$

(d)   $70°$

(b)

In $\Delta \text{ADC,}$ 

$\angle \text{ADC}+\angle \text{DAC}+\angle \text{ACD}=180°$

$\angle \text{ADC}+50°+\angle \text{ACD}=180°$

$\angle \text{ACD}=130°-\angle \text{ADC}$

In $\Delta \text{DBC,}$                $\angle \text{ADC}=\angle \text{DBC}+\angle \text{BCD}$

$\angle \text{ADC}=70°+\angle \text{ACD}$

$\angle \text{ADC}=70°+130°-\angle \text{ADC}$

$\angle \text{ADC}=200°-\angle \text{ADC}$

$2\angle \text{ADC}=200°$

$\angle \text{ADC}=\frac{200°}{2}=100°$

If for $\Delta \text{ABC}$ and $\Delta \text{DEF,}$ the correspondence $\text{CAB}\leftrightarrow \text{EDF}$ gives a congruence, then which of the following is not true?

a.    $\text{AC}=\text{DE}$

b.   $\text{AB}=\text{EF}$ 

c.    $\angle \text{A}=\angle \text{D}$

d.   $\angle \text{C}=\angle \text{E}$

(b)

Two figures are said to be congruent, if the trace copy of figure 1 fits exactly on that of figure 2

Now, if $\Delta \text{ABC}$ and $\Delta \text{DEF}$ are congruent, then

$\text{AB}=\text{DF, BC}=\text{FE}$

$\text{AC}=\text{DE, }\angle \text{A}=\angle \text{D}$ 

$\angle \text{B}=\angle \text{F, }\angle \text{C}=\angle \text{E}$

Hence, option (b) is not true.

In Fig. 6.18, M is the mid-point of both AC and BD. Then

Fig. 6.18

1. $\angle \text{1}=\angle \text{2}$
2. $\angle \text{1}=\angle \text{4}$
3. $\angle \text{2}=\angle \text{4}$
4. $\angle \text{1}=\angle \text{3}$

(b)

In $\Delta \text{AMB}$ and $\Delta \text{CMD,}$

$\text{AM}=\text{CM}$ [Since M is mid-point]

$\text{BM}=\text{DM}$ [Since M is mid-point]

$\angle \text{AMB}=\angle \text{CMD}$ [Vertically opposite angles]

By $\text{SAS}$ congruence criterion,

$\Delta \text{AMB}\cong \Delta \text{CMD}$

$\angle 1=\angle 4$           [by CPCT]

If D is the mid-point of the side $\text{BC}$ in $\Delta \text{ABC}$ where $\text{AB}=\text{AC,}$ then $\angle \text{ADC}$ is

a.    $60°$ 

b.   $45°$

c.    $120°$ 

d.   $90°$

(d)

In $\Delta \text{ADB}$ and $\Delta \text{ADC}$,

$\text{BD}=\text{DC}$

$\text{AB}=\text{AC}$

$\text{AD}=\text{AD}$

By SSS congruence criterion,

$\Delta \text{ABD}\cong \Delta \text{ACD}$

$\angle \text{ADB}=\angle \text{ADC}$

We know that, $\angle \text{ADB}+\angle \text{ADC}=180°$

$2\angle \text{ADC}=180°$

$\angle \text{ADC}=90°$

Two triangles are congruent, if two angles and the side included between them in one of the triangles are equal to the two angles and the side included between them of the other triangle. This is known as the

1. RHS congruence criterion
2. ASA congruence criterion
3. SAS congruence criterion
4. AAA congruence criterion

(b)

Under ASA congruence criterion, two triangles are congruent, if two angles and the side included between them in one of the triangles are equal to the two angles and the side included between them of the other triangle.

By which congruency criterion, the two triangles in Fig. 6.19 are congruent?

Fig. 6.19

1. $\text{RHS}$ 
2. $\text{ASA}$ 
3. $\text{SSS}$ 
4. $\text{SAS}$ 

(c)

In $\Delta \text{PQR}$ and $\Delta \text{PQS}$,

$\text{PR}=\text{PS}=a\text{ cm}$

$\text{RQ}=\text{SQ}=b\text{ cm}$ 

$\text{PQ}=\text{PQ}=$ common line segment

By SSS Congruence criterion,

$\Delta \text{PQR}\cong \Delta \text{PQS}$

By which of the following criterion two triangles cannot be proved congruent?

1. $\text{AAA}$ 
2. $\text{SSS}$ 
3. $\text{SAS}$ 
4. $\text{ASA}$ 

(a)

AAA is not a congruency criterion, because if all the three angles of two triangles are equal; this does not imply that both the triangles fit exactly on each other.

If $\Delta \text{PQR}$ is congruent to $\Delta \text{STU}$ (Fig. 6.20), then what is the length of $\text{TU}$?

1. $5\text{ cm}$ 
2. $6\text{ cm}$ 
3. $7\text{ cm}$ 
4. cannot be determined

Fig. 6.20

(b)

Given      $\Delta \text{PQR}\cong \Delta \text{STU}$

$\begin{array}{l}\text{PQ}=\text{ST}\\ \text{QR}=\text{TU}\\ \text{PR}=\text{SU}\end{array}$

Hence, $\text{TU}=\text{QR}=6\text{ cm}$

If $\Delta \text{ABC}$ and $\Delta \text{DBC}$ are on the same base $\text{BC,}$ $\text{AB}=\text{DC}$ and $\text{AC}=\text{DB}$ (Fig. 6.21), then which of the following gives a congruence relationship?

Fig. 6.21

1. $\Delta \text{ABC}\cong \Delta \text{DBC}$ 
2. $\Delta \text{ABC}\cong \Delta \text{CBD}$
3. $\Delta \text{ABC}\cong \Delta \text{DCB}$
4. $\Delta \text{ABC}\cong \Delta \text{BCD}$

(c)

Since,              $\text{AB}=\text{DC}$

$\text{AC}=\text{DB}$

$\text{BC}=\text{BC}$

By SSS congruence criterion,

$\Delta \text{ABC}\cong \Delta \text{DCB}$ 

 

In questions 50 to 69, fill in the blanks to make the statements true.

The ________ triangle always has altitude outside itself.

The obtuse angled triangle always has altitude outside itself.

The sum of an exterior angle of a triangle and its adjacent angle is always ________.

The sum of an exterior angle of a triangle and its adjacent angle is always, $180°$, because they form a linear pair.

The longest side of a right-angled triangle is called its ________.

the longest side of a right-angled triangle is called its Hypotenuse.

Median is also called ________ in an equilateral triangle.

Median is also called an altitude in an equilateral triangle.

Measures of each of the angles of an equilateral triangle is ________.

Measures of each of the angles of an equilateral triangle is $60°$ as all the angles in an equilateral triangle are equal.

Let $x$ be the angle of equilateral.

According to the question, the angle sum property of a triangle

$x+x+x=180°$

$3x=180°$

$x=\frac{180°}{3}=60°$

In an isosceles triangle, two angles are always ________.

In an isosceles triangle, two angles are always equal. Since, if two sides are equal, then the angles opposite them are equal.

In an isosceles triangle, angles opposite to equal sides are ________.

In an isosceles triangle, angles opposite to equal sides are equal.

If one angle of a triangle is equal to the sum of other two, then the measure of that angle is ________.

Let the angles of a triangle be $a$, band $c$. It is given that,

$a=b+c$

$a+b+c=180°$

$a+a=180°$

$2a=180°$

$a=\frac{180°}{2}=90°.$

Hence, the measure of that angle is $90°.$

Every triangle has at least ________ acute angle (s).

Every triangle has atleast two acute angles.

Two line segments are congruent, if they are of ________ lengths.

Two line segments are congruent, if they are of equal lengths.

Two angles are said to be ________, if they have equal measures.

Two angles are said to be congruent, if they have equal measures.

Two rectangles are congruent, if they have same ________ and ________.

Two rectangles are congruent, if they have same length and breadth.

Two squares are congruent, if they have same ________.

Two squares are congruent, if they have same side.

If $\Delta \text{PQR}$ and $\Delta \text{XYZ}$ are congruent under the correspondence $\text{QPR}\leftrightarrow \text{XYZ}$, then

(i)             $\angle \text{R}=$ ________

(ii)         $\text{QR}=$ ________

(iii)      $\angle \text{P}=$ ________

(iv)       $\text{QP}=$ ________

(v)          $\angle \text{Q}=$ ________

(vi)       $\text{RP}=$ ________

Given, $\Delta \text{PQR}\cong \Delta \text{XYZ}$

(i)             $\angle \text{R}=\angle Z$ 

(ii)         $\text{QR}=\text{XZ}$ 

(iii)      $\angle \text{P}=\angle \text{Y}$ 

(iv)       $\text{QP}=\text{XY}$ 

(v)          $\angle \text{Q}=\angle \text{X}$ 

(vi)       $\text{RP}=\text{ZY}$ 

In Fig. 6.22, $\Delta \text{PQR}\cong \Delta \text{\_\_\_\_\_\_\_\_}$

Fig. 6.22

In $\Delta \text{PQR and }\Delta \text{XYZ}$

$\text{PQ}=XZ=3.5\text{ cm}$

$\text{QR}=\text{ZY}=5\text{ cm}$

$\angle \text{PQR}=\angle \text{XYZ}=45$

By SAS congruence criterion,

$\Delta \text{PQR}\cong \Delta \text{XZY}$

In Fig. 6.23, $\Delta \text{PQR}\cong \Delta \text{\_\_\_\_\_\_\_\_}$

Fig. 6.23

In $\Delta \text{PQR and }\Delta \text{RSP}$

$\text{QR}=\text{SP}=4.1\text{ cm}$

$\text{PR}=\text{PR}$ 

$\angle \text{SPR}=\angle \text{QRP}=45$

By SAS congruence criterion,

$\Delta \text{PQR}\cong \Delta \text{RSP}$

In Fig. 6.24, $\Delta \text{\_\_\_\_\_\_\_\_ }\cong \Delta \text{PQR}$

Fig. 6.24

In given fig. $\Delta \text{DRQ and }\Delta \text{PQR}$,

$\text{QR}=\text{QR}$

$\angle \text{DRQ}=\angle \text{PQR}=70°$

$\angle \text{DQR}=\angle \text{PRQ}=40°$

By ASA congruence criterion,

$\Delta \text{DRQ}\cong \Delta \text{PQR}$

In Fig. 6.25, $\Delta \text{ARO}\cong \Delta \text{\_\_\_\_\_\_\_\_}$

Fig. 6.25

In $\Delta \text{ARO and }\Delta \text{PQO,}$             $\angle \text{AOR}=\angle \text{POQ}$

                                                $\angle \text{ARO}=\angle \text{PQO}=55$

Now, in $\Delta \text{ARO and }\Delta \text{PQO,}$    $\angle \text{AOR}=\angle \text{POQ}$

                                                $\text{AO}=\text{PO}=2.5\text{ cm}$

                                                $\angle \text{RAO}=\angle \text{QPO}$

By AAS congruence criterion, $\Delta \text{ARO}\cong \Delta \text{PQO,}$

In Fig. 6.26, $\text{AB}=\text{AD and }\angle \text{BAC}=\angle \text{DAC}.$ Then

(i)             $\Delta \text{\_\_\_\_\_\_\_\_}\cong \Delta \text{ABC}\text{.}$

(ii)         $\text{BC}=\text{\_\_\_\_\_\_\_\_}.$

(iii)      $\angle \text{BCA}=\text{\_\_\_\_\_\_\_\_}.$

(iv)       Line segment AC bisects _______ and _______.

Fig. 6.26

(i)             In $\Delta \text{ABC and }\Delta \text{ADC}$

$\text{AB}=\text{AD}$

$\text{AC}=\text{AC}$

$\angle \text{BAC}=\angle \text{DAC}$

By SAS congruence criterion,

$\Delta \text{ADC}\cong \Delta \text{ABC}$

(ii)         $\text{BC}=\text{DC}$ [By CPCT]

(iii)      $\angle \text{BCA}=\angle \text{DCA}$ [By CPCT]

(iv)       Line segment AC bisects $\angle \text{BAD and }\angle \text{BCD}$

Since,      $\angle \text{BAC}=\angle \text{DAC}$

In Fig. 6.27,

(i)             $\angle \text{TPQ}=\angle \_\_\_\_\_\_+\angle \_\_\_\_\_\_$

(ii)         $\angle \text{UQR}=\angle \_\_\_\_\_\_+\angle \_\_\_\_\_\_$

(iii)      $\angle \text{PRS}=\angle \_\_\_\_\_\_+\angle \_\_\_\_\_\_$

Fig. 6.27

Exterior angle property

The measure of an exterior angle is equal to the sum of the two opposite interior angles.

(i)             $\angle \text{TPQ}=\angle \text{PQR}+\angle \text{PRQ}$

(ii)         $\angle \text{UQR}=\angle \text{QRP}+\angle \text{QPR}$

(iii)      $\angle \text{PRS}=\angle \text{RPQ}+\angle \text{RQP}$

 

In questions 70 to 106 state whether the statements are True or False.

In a triangle, sum of squares of two sides is equal to the square of the third side.

False

Only in a right-angled triangle, the sum of two shorter sides is equal to the square of the third side.

Sum of two sides of a triangle is greater than or equal to the third side.

False

Sum of two sides of a triangle is greater than the third side

The difference between the lengths of any two sides of a triangle is smaller than the length of third side.

True

The difference between the lengths of any two sides of a triangle is smaller than the length of third side.

e.g., for any triangle ABC, $\text{AB}-\text{BC}<\text{AC}$

In $\Delta \text{ABC, AB}=3.5\text{ cm,}$ $\text{AC}=5\text{ cm, BC}=6\text{ cm}$ and in $\Delta \text{PQR, PR}=3.5\text{ cm,}$ $\text{PQ}=5\text{cm, RQ}=6\text{ cm}\text{.}$ Then $\Delta \text{ABC}\cong \Delta \text{PQR}\text{.}$

False

In $\Delta \text{ABC and }\Delta \text{PRQ,}$

$\text{AB}=\text{PR}=3.5\text{ cm,}$ $\text{BC}=\text{RQ}=6\text{ cm}$ and $\text{AC}=\text{PQ}=5\text{ cm}$

By SSS congruence criterion, $\Delta \text{ABC}\cong \Delta \text{PRQ}$

Sum of any two angles of a triangle is always greater than the third angle.

False

It is not necessary that sum of any two angles of a triangle is always greater than the third angle, e.g.

Let the angles of a triangle be $20°$, $50°$ and $110°$, respectively.

Hence, $20°+50°=70°$, which is less than $110°$.

The sum of the measures of three angles of a triangle is greater than $180°$.

False

The sum of the measures of three angles of a triangle is always equal to $180°$.

It is possible to have a right-angled equilateral triangle.

False

In a right-angled triangle, one angle is equal to $90°$ and in equilateral triangle, all angles are equal to $60°$.

If M is the mid-point of a line segment AB, then we can say that AM and MB are congruent.

True

Given, M is the mid-point of a line segment AB

$\text{AM}=\text{MB}$

Two line segments are congruent, that’s why they are of same lengths.

It is possible to have a triangle in which two of the angles are right angles.

False

If in a triangle two angles are right angles, then third angle $=180°-(90°+90°)=0°$, which is not possible.

It is possible to have a triangle in which two of the angles are obtuse.

False

Obtuse angles are those angles which are greater than $90°$. So, sum of two obtuse angles will be greater than $180°$, which is not possible as the sum of all the angles of a triangle is $180°$.

It is possible to have a triangle in which two angles are acute.

True

In a triangle, atleast two angles must be acute angle.

It is possible to have a triangle in which each angle is less than $60°$.

False

The sum of all angles in a triangle is equal to $180°$. So, all three angles can never be less than $60°$.

It is possible to have a triangle in which each angle is greater than $60°$.

False

If all the angles are greater than $60°$ in a triangle, then the sum of all the three angles with exceed $180°$, which cannot be possible in case of triangle.

It is possible to have a triangle in which each angle is equal to $60°$.

True

The triangle in which each angle is equal to $60°$ is called an equilateral triangle.

A right-angled triangle may have all sides equal.

False

Hypotenuse is always the greater than the other two sides of the right-angled triangle

If two angles of a triangle are equal, the third angle is also equal to each of the other two angles.

False

In an isosceles triangle, always two angles are equal and not the third one.

In Fig. 6.28, two triangles are congruent by $\text{RHS}$.

 

Fig. 6.28

True

In $\Delta \text{ABC}$ $\text{AC}=\sqrt{{\text{AB}}^2+{\text{BC}}^2}=\sqrt{4^2+5^2}=\sqrt{41}\text{ cm}$

In $\Delta \text{PQR}$ $\text{PR}=\sqrt{{\text{PQ}}^2+{\text{QR}}^2}=\sqrt{4^2+5^2}=\sqrt{41}\text{ cm}$

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

$\text{AB}=\text{PQ}=4\text{ cm}$

$\text{AC}=\text{PR}=\sqrt{41}\text{ cm}$

$\angle \text{ABC}=\angle \text{PQR}=90°$

By RHS congruence criterion,

$\Delta \text{ABC}\cong \Delta \text{PQR}$

The congruent figures super impose each other completely.

True

Because congruent figures have same shape and same size.

A one rupee coin is congruent to a five rupee coin.

False

Because they don’t have same shape and same size.

The top and bottom faces of a kaleidoscope are congruent.

True

Because they superimpose each other.

Two acute angles are congruent.

False

Because the measure of two acute angles may be different.

Two right angles are congruent.

True

Since, the measure of right angles is always same which is equal to $90°$.

Two figures are congruent, if they have the same shape.

False

Two figures are congruent, if they have the same shape and same size.

If the areas of two squares is same, they are congruent.

True

Because two squares will have same areas only if their sides are equal and squares with same sides will superimpose each other.

If the areas of two rectangles are same, they are congruent.

False

Because rectangles with the different length and breadth may have equal areas. But they will not superimpose each other.

If the areas of two circles are the same, they are congruent.

True

Because areas of two circles will be equal only if their radii are equal and circle with same radii will superimpose each other.

Two squares having same perimeter are congruent.

True

If two squares have same perimeter, then their sides will be equal. Hence, the squares will superimpose each other.

Two circles having same circumference are congruent.

True

If two circles have same circumference, then their radii will be equal. Hence, the circles will superimpose each other.

If three angles of two triangles are equal, triangles are congruent.

False

Consider two equilateral triangles with different sides.

          

Both $\Delta \text{ABC and }\Delta \text{DEF}$ have same angles but their size is different. So, they are not congruent.

If two legs of a right triangle are equal to two legs of another right triangle, then the right triangles are congruent.

True

If two legs of a right-angled triangle are equal to two legs of another right-angled triangle, then their third leg will also be equal. Hence, they will have same shape and same size.

If two sides and one angle of a triangle are equal to the two sides and angle of another triangle, then the two triangles are congruent.

False

Because if two sides and the angle included between them of the other triangle, only then the two triangles will be congruent.

If two triangles are congruent, then the corresponding angles are equal.

True

Because if two triangles are congruent, then their sides and angles are equal.

If two angles and a side of a triangle are equal to two angles and a side of another triangle, then the triangles are congruent.

False

If two angles and the side included between them of a triangle are equal to two angles and included a side between them of the other triangle, only then the triangles are congruent.

If the hypotenuse of one right triangle is equal to the hypotenuse of another right triangle, then the triangles are congruent.

False

Two right-angled triangles are congruent, if the hypotenuse and a side of one of the triangle are equal to the hypotenuse and one of the side of the other triangle.

If hypotenuse and an acute angle of one right triangle are equal to the hypotenuse and an acute angle of another right triangle, then the triangles are congruent.

True

In $\Delta \text{ABC}$ and $\Delta \text{PQR}$,

$\angle \text{B}=\angle \text{Q}=90°$

$\angle \text{C}=\angle \text{R}$

$\angle \text{A}=\angle \text{P}$

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

$\angle \text{A}=\angle \text{P}$

$\text{AC}=\text{PR}$

$\angle \text{C}=\angle \text{R}$

By ASA congruence criterion, $\Delta \text{ABC}\cong \Delta \text{PQR}$

$\text{AAS}$ congruence criterion is same as $\text{ASA}$ congruence criterion.

False

In ASA congruence criterion, the side ‘S’ included between the two angles of the triangle. In AAS congruence criterion, side ‘S’ is not included between two angles.

In Fig. 6.29, $\text{AD}\perp \text{BC}$ and $\text{AD}$ is the bisector of angle $\text{BAC}$. Then, $\Delta \text{ABD}\cong \Delta \text{ACD by RHS}\text{.}$

Fig. 6.29

False

In $\Delta \text{ABD and }\Delta \text{ACD}$,

$\text{AD}=\text{AD}$ [Common]

$\angle \text{BAD}=\angle \text{CAD}$ [ $\text{AD}$ is the bisector of angle $\text{BAC}$ ]

$\angle \text{ADB}=\angle \text{ADC}=90°$

By AAS congruence criterion,

$\Delta \text{ABD}\cong \Delta \text{ACD}$

The measure of three angles of a triangle are in the ratio $5:3:1.$ Find the measures of these angles.

Let measures of the given angles of a triangle be $5x\text{,}$ $3x$ and $1x\text{.}$

Sum of all the angles in a triangle $=180°$

$5x+3x+1x=180°$

$9x=180°$

$x=\frac{180°}{9}=20°$

So, the angles are $5x=5\times 20=100°\text{,}$ $3x=3\times 20=60°$ and $x=20°$

In Fig. 6.30, find the value of $x$.

Fig. 6.30

We know that, the sum of all three angles in a triangle is equal to $180°$.

So, $x+55°+90°=180°$

$x+145°=180°$

$x=180°-145°$

$x=35°$

In Fig. 6.31(i) and (ii), find the values of $a$, $b$ and $c$.

(i)              

 

 

Fig. 6.31

(ii)          

 

 

Fig. 6.31

In fig. (i), $\angle \text{A}+\angle \text{B}+\angle \text{C}=180°$

$90°+a+70°=180°$

$a+160°=180°$

$a=180°-160°=20°$

Since, $c$ is an exterior angle of $\Delta \text{ABD,}$

$\angle c=a+30°=20°+30°=50°$

Also, $b$ is an exterior angle of $\Delta \text{ADC}$

$\angle b=60°+70°=130°$

In fig. (ii),

In $\Delta \text{PQS,}$ $\angle \text{QPS}+\angle \text{PQS}+\angle \text{PSQ}=180°$

$55°+60°+a=180°$

$115°+a=180°$

$a=180°-115°=65°$

$a+b=180°$ [Linear pair]

$65°+b=180°$

$b=180°-65°=115°$

In $\Delta \text{PSR,}$ $\angle \text{PSR}+\angle \text{SPR}+\angle \text{PRS}=180°$

$115°+c+40°=180°$

$c=180°-155°=25°$

In triangle XYZ, the measure of angle $\angle \text{X}$ is $30°$ greater than the measure of angle Y and angle Z is a right angle. Find the measure of $\angle \text{Y}$.

According to the question,

Measure of $\angle \text{X}=\angle \text{Y}+30$

Measure of $\angle \text{Z}=90$

The sum of all three angles in a triangle is equal to $180$.

$\angle \text{X}+\angle \text{Y}+\angle \text{Z}=180$

$(\angle \text{Y}+30)+\angle \text{Y}+90°=180$

$2\angle Y+120=180°$

$2\angle Y=180-120=60$

$\angle \text{Y}=\frac{60}{2}=30$

In a triangle ABC, the measure of angle A is $40°$ less than the measure of angle B and $50°$ less than that of angle C. Find the measure of $\angle \text{A}$.

According to the question,

Measure of $\angle \text{A}=\angle \text{B}-40$

Measure of $\angle \text{C}=\angle \text{B}-40+50$

The sum of all $3$ angles in a triangle is equal to $180$

$\angle \text{A}+\angle \text{B}+\angle \text{C}=180$

$(\angle \text{B}-40)+\angle \text{B}+(\angle \text{B}-40+50)=180$

$3\angle \text{B}-30=180$

$3\angle \text{B}=210$

$\angle \text{B}=\frac{210}{3}=70$

So, the measure of $\angle \text{A}=70-40=30$

I have three sides. One of my angle measures $15°$. Another has a measure of $60°$. What kind of a polygon am I? If I am a triangle, then what kind of triangle am I?

The polygon with three sides is called triangle.

 

According to the angle sum property of a triangle

$\angle \text{A}+\angle \text{B}+\angle \text{C}=180$

$\angle \text{A}+15+60=180$

$\angle \text{A}=180-75=105$

As one angle in this triangle is greater than $90$, so it is an obtuse angled triangle.

Jiya walks $6\text{ km}$ due east and then $8\text{ km}$ due north. How far is she from her starting place?

As per the given information, we can draw the following figure, which is a right-angled triangle at B.

Distance from starting point to the final position in the hypotenuse of right angled $\Delta \text{ABC,}$

${\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2$

${\left(6\right)}^2+{\left(8\right)}^2={(\text{distance})}^2$

$36+64={(\text{distance})}^2$

$\text{Distance}=\sqrt{100}=10\text{ km}$

Jayanti takes shortest route to her home by walking diagonally across a rectangular park. The park measures $60$ metres $\times 80$ metres. How much shorter is the route across the park than the route around its edges?

As the park is rectangular, all the angles are of $90$.

In right-angled $\Delta \text{ABC,}$

${\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2$

${\text{AC}}^2={\left(60\right)}^2+{\left(80\right)}^2=3600+6400$

${\text{AC}}^2=\sqrt{10000}=100\text{m}$

If she goes through AB and AC, then total distance covered $=\left(60+80\right)\text{m}=140\text{m}$

Difference between two paths $=(140-100)\text{m}=40\text{m}$

In $\Delta \text{PQR}$ of Fig. 6.32, $\text{PQ}=\text{PR}$. Find the measures of $\angle \text{Q}$ and $\angle \text{R}$.

Fig. 6.32

Since, $\text{PQ}=\text{PR}$

$\angle \text{Q}=\angle \text{R}=x$

$\angle \text{P}+\angle \text{Q}+\angle \text{R}=180$

$30+x+x=180$

$2x=150$

$x=\frac{150}{2}=75$

Hence, $\angle \text{Q}=\angle \text{R}=75$

In Fig. 6.33, find the measures of $\angle x$ and $\angle y$.

Fig. 6.33

Since, $\angle y$ and $45$ from a linear pair.

So, $\angle y+45=180$

$\angle y=180-45=135$

The sum of all angles in a triangle is equal to $180$.

So,           $45+60+\angle x=180$

$105+\angle x=180$

$\angle x=180-105=75$

In Fig. 6.34, find the measures of $\angle \text{PON}$ and $\angle \text{NPO}$.

Fig. 6.34

In $\Delta \text{LOM,}$

$\angle \text{LOM}+\angle \text{OML}+\angle \text{LOM}=180$

$70+20+\angle \text{LOM}=180$

$\angle \text{LOM}=180-90=90$

Also, $\angle \text{LOM}=\angle \text{PON}=90$ [Vertically opposite angles]

In $\Delta \text{PON,}$        $\angle \text{PON}+\angle \text{NPO}+\angle \text{ONP}=180$

$90+\angle \text{NPO}+70=180$

$\angle \text{NPO}=180-160=120$

In Fig. 6.35, $\text{QP}\parallel \text{RT}$. Find the values of $x$ and $y$.

Fig. 6.35

In the given figure, $\text{QP}\parallel \text{RT,}$ where PR is a transversal line.

So, $\angle x$ and $\angle \text{TRP}$ are alternate interior angles

$\angle x=70°$

The sum of all angles in a triangle is equal to $180°$

So,           $70+30+\angle y=180$

$100+\angle y=180$

$\angle x=180-100=80$

Find the measure of $\angle \text{A}$ in Fig. 6.36.

Fig. 6.36

As we know, the measure of exterior angle is equal to the sum of opposite interior angles.

$115=65+\angle \text{A}$

$\angle A=115-65=50$

In a right-angled triangle if an angle measures $35°$, then find the measure of the third angle.

In a right-angled $\Delta \text{ABC,}$

$\angle \text{A}+\angle \text{B}+\angle \text{C}=180$

$\angle \text{A}+90+35=180$

$\angle \text{A}+125=180$

$\angle \text{A}=180-125=55$

Each of the two equal angles of an isosceles triangle is four times the third angle. Find the angles of the triangle.

Let the third angle be $x$. Then, the other two angles are $4x$ and $4x$, respectively.

We know that, the sum of all three angles in triangle is $180$.

$\angle \text{A}+\angle \text{B}+\angle \text{C}=180$

$x+4x+4x=180$

$9x=180$

$x=\frac{180}{9}=20$

Hence, three angles are $4x=4\times 20=80$, $4x=4\times 20=80$ and $x=20$

The angles of a triangle are in the ratio $2:3:5$. Find the angles.

Let measures of the given angles of a triangle be $2x$, $3x$ and $5x$

Sum of all the angles in a triangle $=180$

$2x+3x+5x=180$

$10x=180$

$x=\frac{180}{10}=18$

So, the angles are $2x=2\times 18=36$, $3x=3\times 18=54$ and $5x=5\times 18=90$

If the sides of a triangle are produced in an order, show that the sum of the exterior angles so formed is $360°$.

In $\Delta \text{ABC}$ by exterior angle property,

Exterior $\angle 1=$ interior $\angle \text{A}+$ interior $\angle \text{B}$

Exterior $\angle 2=$ interior $\angle \text{B}+$ interior $\angle \text{C}$

Exterior $\angle 3=$ interior $\angle \text{C}+$ interior $\angle \text{A}$

On adding eqs. (i), (ii) and (iii), we get

$\angle 1+\angle 2+\angle 3=2(\angle \text{A}+\angle \text{B}+\angle \text{C})$         

$\angle 1+\angle 2+\angle 3=180\times 2$   [by angle sum property]

$\angle 1+\angle 2+\angle 3=360$

Hence, sum of exterior angles is $360$.

In $\Delta \text{ABC}$, if $\angle \text{A}=\angle \text{C}$, and exterior angle $\text{ABX}=140°$, then find the angles of the triangle.

Given, $\angle \text{A}=\angle \text{C}$ and exterior $\angle \text{ABX}=140$

Let $\angle \text{A}=\angle \text{C}=x$

According to exterior angle property,

Exterior $\angle \text{B}=$ interior $\angle \text{A}+$ interior $\angle \text{C}$

$140=x+x$

$140=2x$

$x=\frac{140}{2}=70$

$\angle \text{A}=\angle \text{C}=70$

Now,       $\angle \text{A}+\angle \text{B}+\angle \text{C}=180$

$70+\angle \text{B}+70=180$

$\angle \text{B}=180-140=40$

$\angle B=40$

Hence, all the angles of the triangle are $70\text{, }40$ and $70$.

Find the values of $x$ and $y$ in Fig. 6.37.

Fig. 6.37

In $\Delta \text{TQS,}$

$\angle \text{T}+\angle \text{Q}+\angle \text{S}=180$

$50+30+\angle 1=180$

$80+\angle 1=180$

$\angle 1=180-80=100$

$\angle 1+x=180$ [Linear pair]

$\begin{array}{l}100+x=180\\ x=180-100\end{array}$

$x=80$

In $\Delta \text{TSR,}$

$x+45+\angle 2=180$

$\angle 2=180-80-45=55[\because x=80]$

Now, $50+\angle 2+y=180$ [Linear pair]

$y=180-50-55=75$.

Find the value of $x$ in Fig. 6.38.

Fig. 6.38

In the given figure, $\angle \text{BAC}=80\text{, }\angle \text{ABC}=30\text{,}$ $\angle \text{ACE}=x$ and $\angle \text{ECD}=90$

In $\Delta \text{ABC}$, we know that, exterior angle is equal to the sum of exterior opposite angles.

$\angle \text{ACD}=\angle \text{CAB}+\angle \text{ABC}$

$\angle \text{ACE}+\angle \text{ECD}=80+30$

$x+90=110$

$x=110-90=20$

The angles of a triangle are arranged in descending order of their magnitudes. If the difference between two consecutive angles is $10°$, find the three angles.

Let one of the angles of a triangle be $x$. If angles are arranged in descending order.

Then, angles will be $x\text{, }(x-10°)$ and $(x–20°)$.

We know that, the sum of all angles in a triangle is equal to $180$

So, $x+(x-10)+(x-20)=180$

$x+x+x-30=180$

$3x=180+30$

$3x=210$

$x=\frac{210}{3}=70$

Hence, angles will be $70,70-10$ and $70-20$ i.e., $70,60\text{ and }50$.

In $\Delta \text{ABC}$, $\text{DE}\parallel \text{BC}$ (Fig. 6.39). Find the values of $x$, $y$ and $z$.

Fig. 6.39

In $\Delta \text{ABC,}$ 

$\angle \text{A}+\angle \text{B}+\angle \text{C}=180$

$z+30+40=180$

$z+70=180$

$z=180-70=110$

$\text{DE}\parallel \text{BC}$

Now, $\angle \text{ADE}=\angle \text{ABC}$ [Corresponding angles]

$\angle x=30$ 

$\angle \text{AED}=\angle \text{ACB}$ [Corresponding angles]

$\angle y=40$

In Fig. 6.40, find the values of $x$, $y$ and $z$.

Fig. 6.40

In the given figure, $\angle \text{BAD}=60$, $\angle \text{ABD}=60$, $\angle \text{ADB}=x$, $\angle \text{DAC}=30$, $\angle \text{ADC}=y$ and $\angle \text{ACD}=z$

We know that, the sum of all angles in a triangle is equal to 180

In $\Delta \text{ABD,}$

$\angle \text{BAD}+\angle \text{ABD}+\angle \text{ADB}=180$

$60+60+x=180$

$120+x=180$

$x=180-120=60$

Exterior angle is equal to the sum of exterior opposite angles.

$y=\angle \text{BAD}+\angle \text{ABD}$

$y=60+60$

$y=120$

In $\Delta \text{ADC,}$

$\angle \text{DAC}+\angle \text{ADC}+\angle \text{ACD}=180$

$30+120+z=180$

$150+z=180$

$z=180-150=30$

Hence, $x=60$, $y=120$ and $z=30$

If one angle of a triangle is $60°$ and the other two angles are in the ratio $1:2$, find the angles.

Given, one angle of a triangle is $60$

Let the other two angles be $x$ and $2x$

We know, the sum of all angles in a triangle is equal to $180$

So,

$x+2x+60=180$

$3x=180-60$

$3x=120$

$x=\frac{120}{3}=40$

So, the other two angles will be $x=40$ and $2x=2\times 40=80$

In $\Delta \text{PQR}$, if $3\angle \text{P}=4\angle \text{Q}=6\angle \text{R}$, calculate the angles of the triangle.

Given, $3\angle \text{P}=4\angle \text{Q}=6\angle \text{R}$

Then, $\angle \text{P}=\frac{6}{3}\angle \text{R}=2\angle \text{R}$

$\angle \text{Q}=\frac{6}{4}\angle \text{R}=\frac{3}{2}\angle \text{R}$

In $\Delta \text{PQR,}$

$\angle \text{P}+\angle \text{Q}+\angle \text{R}=180$

$2\angle \text{R}+\frac{3}{2}\angle \text{R}+\angle \text{R}=180$

$3\angle \text{R}+\frac{3}{2}\angle \text{R}=180$

$6\angle \text{R}+3\angle \text{R}=180\times 2$

$9\angle \text{R}=360$

$\angle \text{R}=\frac{360}{9}=40$

$\angle \text{P}=2\angle \text{R}=2\times 40=80$

$\angle \text{Q}=\frac{3}{2}\angle \text{R}=\frac{3}{2}\times 40=60$

Hence, all the angles of the triangle are $80$, $60$ and $40$

In $\Delta \text{DEF}$, $\angle \text{D}=60°$, $\angle \text{E}=70°$ and the bisectors of $\angle \text{E}$ and $\angle \text{F}$ meet at $\text{O}$.

Find (i) $\angle \text{F}$ (ii) $\angle \text{EOF}$.

(i)             As we know,

$\angle \text{D}+\angle \text{E}+\angle \text{F}=180$

$60+70+\angle \text{F}=180$

$\angle \text{F}=180-130=50$

(ii)         Now, as FO is the bisector of $\angle \text{F}$

so,    $\angle \text{EFO}=\frac{\angle \text{F}}{2}=\frac{50}{2}=25$

and          $\angle \text{OEF}=\frac{\angle \text{E}}{2}=\frac{70}{2}=35$

In $\Delta \text{EOF,}$

$\angle \text{EOF}+\angle \text{OEF}+\angle \text{OFE}=180$

$\angle \text{EOF}+35+25=180$

$\angle \text{EOF}=180-60=120$

In Fig. 6.41, $\Delta \text{PQR}$ is right-angled at P. U and T are the points on line QRF. If $\text{QP}\parallel \text{ST}$ and $\text{US}\parallel \text{RP}$, find $\angle \text{S}$.

Fig. 6.41

If $\text{QP}\parallel \text{ST}$ and $\text{QT}$ is a transversal, then $\angle \text{PQR}=\angle \text{STU}$ [alternate interior angles] and if $\text{DS}\parallel \text{RP}$ and $\text{QT}$ is a transversal, then $\angle \text{PRQ}=\angle \text{SUT}$ [alternate interior angles]

Hence, $\angle \text{S}$ must be equal to $\angle \text{P}$ i.e. $90°$.

In each of the given pairs of triangles of Fig. 6.42 (a.-f.), applying only ASA congruence criterion, determine which triangles are congruent. Also, write the congruent triangles in symbolic form.

a.     

b.    

c.     

d.          

e.     

 

f.

 

 

 

1. Not possible, because the side is not included between two angles.
2. $\Delta \text{ABD}\cong \Delta \text{CDB}$
3. $\Delta \text{XYZ}\cong \Delta \text{LMN}$ 
4. Not possible, because there is not any included side equal.
5. Not possible, because there is no proof one angle to be equal.
6. $\Delta \text{AOD}\cong \Delta \text{BOC}$

In each of the given pairs of triangles of Fig. 6.43, using only RHS congruence criterion, determine which pairs of triangles are congruent. In case of congruence, write the result in symbolic form:

a.   

b.  

 

c.   

 

d.  

 

 

e.   

 

f.

 

 

 

a.    In $\Delta \text{ABD}$ and $\Delta \text{ACD}$,

$\text{AB}=\text{AC}$ [Given]

$\text{AD}=\text{AD}$ [Common]

$\angle \text{ADB}=\angle \text{ADC}=\text{90}$

By RHS congruence criterion, $\Delta \text{ABD}\cong \Delta \text{ACD}$

b.   In $\Delta \text{XYZ}$ and $\Delta \text{UZY}$,

$\angle \text{XYZ}=\angle \text{UZY}=90$

$\text{XZ}=\text{YU}$ [Given]

$\text{ZY}=\text{ZY}$ [Common]

By RHS congruence criterion, $\Delta \text{XYZ}\cong \Delta \text{UZY}$

c.    In $\Delta \text{AEC}$ and $\Delta \text{BED}$,

$\text{CE}=\text{DE}$

$\text{AE}=\text{BE}$

It is not known that $\angle \text{ACE}=\angle \text{BDE}=90$

So, given triangles are not congruent   by RHS congruence criterion.

d.   Here, $\text{CD}=\text{BD}-\text{BC}=14-8=6\text{ cm}$ 

In right angled $\Delta \text{ABC}$ 

$\text{AC}=\sqrt{{\text{AB}}^2+{\text{BC}}^2}=\sqrt{6^2+8^2}=\sqrt{36+64}$

 $=\sqrt{100}=10\text{ cm}$

In right angled $\Delta \text{CDE}$,

$\text{DE}=\sqrt{{\text{CE}}^2-{\text{CD}}^2}=\sqrt{{10}^2-6^2}$

 $=\sqrt{100-36}=\sqrt{64}=8\text{ cm}$

In $\Delta \text{ABC}$ and $\Delta \text{CDE}$,

$\text{AC}=\text{CE}=10\text{ cm}$

$\text{BC}=\text{DE}=8\text{ cm}$ 

$\angle \text{ABC}=\angle \text{CDE}=90$

By RHS congruence criterion, $\Delta \text{ABC}\cong \Delta \text{CDE}$

e.    Not possible, because there is no right angle.

f.     In $\Delta \text{LOM}$ and $\Delta \text{LON}$,

$\text{LM}=\text{LN}=8\text{ cm}$

$\text{LO}=\text{LO}$ [Common]

$\angle \text{LOM}=\angle \text{LON}=90$

By RHS congruence criterion, $\Delta \text{LOM}\cong \Delta \text{LON}$

In Fig. 6.44, if $\text{RP}=\text{RQ}$, find the value of $x$.

Fig. 6.44

Given,     $\text{RP}=\text{RQ}$

Since,      $\angle 1=50$ [Vertically opposite angles]

Also,        $\angle 1=x$ [Since $\text{RP}=\text{RQ}$ ]

So,           $x=50$

In Fig. 6.45, if $\text{ST}=\text{SU}$, then find the values of $x$ and $y$.

 

Fig. 6.45

Given, $\text{ST}=\text{SU}$

$\angle \text{SUT}=\angle \text{STU}=y$

Also, $\angle 1=78$

In $\Delta \text{SUT,}$

$78+y+y=180$

$78+2y=180$

$2y=180-78=102$

$y=\frac{102}{2}=51$

Also, $x+y=180$

$x+51=180$

$x=180-51=129$

Check whether the following measures (in cm) can be the sides of a right-angled triangle or not.

$1.5,3.6,3.9$

For a right-angled triangle, the sum of square of two shorter sides must be equal to the square of the third side.

Now, ${1.5}^2+{3.6}^2=2.25+12.96$

$=15.21$

${\left(3.9\right)}^2=15.21$

${\left(1.5\right)}^2+{\left(3.6\right)}^2={\left(3.9\right)}^2$

Hence, the given sides form right-angled triangle.

Height of a pole is $8\text{ m}$. Find the length of rope tied with its top from a point on the ground at a distance of $6\text{ m}$ from its bottom.

Given, height of a pole is $8\text{ m}$.

Distance between the bottom of the pole and a point on the ground is $6\text{ m}$. Based on given information, we can draw the following figure:

Let the length of the rope be $x\text{ m}$

$\text{AB}=$ height of the pole

$\text{BC}=$ Distance between the bottom of the pole and a point on ground, where rope was tied.

To find length of the rope, we will use Pythagoras theorem, in right angled $\Delta \text{ABC}$

${(\text{AC})}^2={(\text{AB})}^2+{(\text{BC})}^2$

${(x)}^2={\left(8\right)}^2+{\left(6\right)}^2$

$x^2=100$

$x=\sqrt{100}=10\text{m}$

Hence, the length of the rope is $10\text{ m}$.

In Fig. 6.46, if $y$ is five times $x$, find the value of $z$.

Fig. 6.46

Given,     $y=5x$

According to the angle sum property of a triangle

$60+x+y=180$

$60+x+5x=180$

$60+6x=180$

$6x=180-60=120$

$x=\frac{120}{6}=20$

$y=5x=5\times 20=100$

According to the exterior angle property,

$z=60+y$

$=60+100$

$=160$

The lengths of two sides of an isosceles triangle are $9\text{ cm}$ and $20\text{ cm}$. What is the perimeter of the triangle? Give reason.

Third side must be $20\text{ cm}$, because sum of two sides should be greater than the third side.

$\therefore$ Perimeter of the triangle $=$ Sum of all sides

$=\left(9+20+20\right)\text{cm}=49\text{ cm}$

Without drawing the triangles write all six pairs of equal measures in each of the following pairs of congruent triangles.

1. $\Delta \text{STU}\cong \Delta \text{DEF}$
2. $\Delta \text{ABC}\cong \Delta \text{LMN}$
3. $\Delta \text{YZX}\cong \Delta \text{PQR}$ 
4. $\Delta \text{XYZ}\cong \Delta \text{MLN}$

We know that, corresponding parts of congruent triangles are equal.

a.    $\Delta \text{STU}\cong \Delta \text{DEF}$

$\angle \text{S}=\angle \text{D},\angle \text{T}=\angle \text{E}$ and $\angle \text{U}=\angle \text{F; ST}=\text{DE},\text{TU}=\text{EF}$ and $\text{SU}=\text{DF}$

b.   $\Delta \text{ABC}\cong \Delta \text{LMN}$ 

$\angle \text{A}=\angle \text{L},\angle \text{B}=\angle \text{M}$ and $\angle \text{C}=\angle \text{N; AB}=\text{LM},\text{BC}=\text{MN}$ and $\text{AC}=\text{LN}$

c.    $\Delta \text{YZX}\cong \Delta \text{PQR}$ 

$\angle \text{Y}=\angle \text{P},\angle \text{Z}=\angle \text{Q}$ and $\angle \text{X}=\angle \text{R; YZ}=\text{PQ},\text{ZX}=\text{QR}$ and $\text{YX}=\text{PR}$

d.   $\Delta \text{XYZ}\cong \Delta \text{MLN}$

$\angle \text{X}=\angle \text{M},\angle \text{Y}=\angle \text{L}$ and $\angle \text{Z}=\angle \text{N; XY}=\text{ML},\text{YZ}=\text{LN}$ and $\text{XZ}=\text{MN}$

In the following pairs of triangles of Fig. 6.47, the lengths of the sides are indicated along the sides. By applying SSS congruence criterion, determine which triangles are congruent. If congruent, write the results in symbolic form.

a.     

 

b.    

c.     

 

 

 

 

d.    

 

e.     

 

 

f.      

 

g.    

 

 

h.    

 

 

 

Fig. 6.47

 

 

 

 

a.    $\Delta \text{ABC}\cong \Delta \text{NLM}$

b.   $\Delta \text{LMN}\cong \Delta \text{GHI}$

c.    $\Delta \text{LMN}\cong \Delta \text{LON}$

d.   $\Delta \text{ZYX}\cong \Delta \text{WXY}$

e.    $\Delta \text{OAB}\cong \Delta \text{DOE}$

f.     $\Delta \text{STU}\cong \Delta \text{SVU}$

g.   $\Delta \text{PSR}\cong \Delta \text{RQP}$

h.   $\Delta \text{STU}\cong \Delta \text{PQR}$

$\text{ABC}$ is an isosceles triangle with $\text{AB}=\text{AC}$ and D is the mid-point of base BC (Fig. 6.48).

a.    State three pairs of equal parts in the triangles ABD and ACD.

b.   Is $\Delta \text{ABD}\cong \Delta \text{ACD}\text{.}$ If so why?

Fig. 6.48

Given,     $\text{AB}=\text{AC}$

And         $\text{BD}=\text{CD}$

a.    In $\Delta \text{ABD and }\Delta \text{ACD}$

$\text{AB}=\text{AC}$

$\text{BD}=\text{CD}$

$\text{AD}=\text{AD}$

2. Yes, by SSS congruence criterion,

$\Delta \text{ABD}\cong \Delta \text{ACD}$

In Fig. 6.49, it is given that $\text{LM}=\text{ON}$ and $\text{NL}=\text{MO}$

1. State the three pairs of equal parts in the triangles NOM and MLN.
2. Is $\Delta \text{NOM}\cong \Delta \text{MLN}$. Give reason?

Fig. 6.49

a.    In $\Delta \text{NOM and }\Delta \text{MLN}$,

$\text{LM}=\text{ON}$

$\text{MN}=\text{MN}$

$\text{LN}=\text{OM}$

2. Yes, by SSS congruence criterion,

$\Delta \text{NOM}\cong \Delta \text{MLN}$

Triangles DEF and LMN are both isosceles with $\text{DE}=\text{DF}$ and $\text{LM}=\text{LN}$, respectively. If $\text{DE}=\text{LM}$ and $\text{EF}=\text{MN}$, then, are the two triangles congruent? Which condition do you use?

If $\angle \text{E}=40°$, what is the measure of $\angle \text{N}$?

Here,

$\text{DE}=\text{DF}$   …(i)

$\text{LM}=\text{LN}$   …(ii)

$\text{DE}=\text{LM}$  …(iii)

From eqs. (i), (ii) and (iii), we get $\text{DE}=\text{DF}=\text{LM}=\text{LN}$

In $\Delta \text{DEF and }\Delta \text{LMN}$,

$\text{DE}=\text{LM}$

$\text{EF}=\text{MN}$

$\text{DF}=\text{LN}$

     

By SSS congruence criterion, $\Delta \text{DEF}\cong \Delta \text{LMN}$

$\angle \text{E}=\angle \text{M}$

$\angle \text{M}=40$

Also, $\angle \text{M}=\angle \text{N}$

$\angle \text{N}=40$

If $\Delta \text{PQR}$ and $\Delta \text{SQR}$ are both isosceles triangle on a common base QR such that P and S lie on the same side of QR. Are triangles PSQ and PSR congruent? Which condition do you use?

In $\Delta \text{PSQ}$ and $\Delta \text{PSR}$

$\text{PQ}=\text{PR}$ [Since PQR is an isosceles triangle]

$\text{SQ}=\text{SR}$ [Since SQR is an isosceles triangle]

$\text{PS}=\text{PS}$ [Common]

 

By SSS congruence criterion,

$\Delta \text{PSQ}\cong \Delta \text{PSR}$

In Fig. 6.50, which pairs of triangles are congruent by SAS congruence criterion (condition)? If congruent, write the congruence of the two triangles in symbolic form.

(i)              

(ii)          

(iii)       

 

(iv)        

(v)           

 

(vi)        

(vii)    

(viii)                         

 

 

 

Fig. 6.50

a.    In $\Delta \text{PQR and }\Delta \text{TUS}$,

$\text{PQ}=\text{TU}=3\text{ cm}$

$\text{QR}=\text{US}=5.5\text{ cm}$

$\angle \text{PQR}=\angle \text{TUS}=40$

By SAS congruence criterion,

$\Delta \text{PQR}\cong \Delta \text{TUS}$

b.   Not possible, because angle is not included between two sides.

c.    In $\Delta \text{BCD and }\Delta \text{BAE}$,

$\text{AB}=\text{CB}=5.2\text{ cm}$

$\text{DC}=\text{EA}=5\text{ cm}$

$\angle \text{EAB}=\angle \text{DCB}=50$

By SAS congruence criterion, $\Delta \text{BCD}\cong \Delta \text{BAE}$

d.   In $\Delta \text{STU and }\Delta \text{XYZ}$,

$\text{TU}=\text{ZY}=4\text{ cm}$

$\text{TS}=\text{ZX}=3\text{cm}$

$\angle \text{STU}=\angle \text{XZY}=30$

By SAS congruence criterion,

$\Delta \text{STU}\cong \Delta \text{XZY}$

e.    In $\Delta \text{DOF and }\Delta \text{HOC}$,

$\text{DO}=\text{HO}$

$\text{CO}=\text{FO}$

$\angle \text{DOF}=\angle \text{HOC}$

By SAS congruence criterion,

$\Delta \text{DOF}\cong \Delta \text{HOC}$

f.     Not congruent, because angle is not included between two sides

g.   In $\Delta \text{PSQ and }\Delta \text{RQS}$,

$\text{PS}=\text{QS}=4\text{ cm}$

$\text{SQ}=\text{SQ}$

$\angle \text{PSQ}=\angle \text{RQS}=40$

By SAS congruence criterion,

$\Delta \text{PSQ}\cong \Delta \text{RQS}$

h.   In $\Delta \text{LMN and }\Delta \text{OMN}$,

$\text{LM}=\text{OM}$

$\text{MN}=\text{MN}$

$\angle \text{LMN}=\angle \text{OMN}=40$

By SAS congruence criterion,

$\Delta \text{LMN}\cong \Delta \text{OMN}$

State which of the following pairs of triangles are congruent. If yes, write them in symbolic form (you may draw a rough figure).

1. $\Delta \text{PQR}:\text{PQ}=3.5\text{ cm, QR}=4.0\text{ cm, }\angle \text{Q}=60°$

$\Delta \text{STU}:\text{ST}=3.5\text{ cm, TU}=4\text{ cm, }\angle \text{T}=60°$

2. $\Delta \text{ABC}:\text{AB}=4.8\text{ cm, }\angle \text{A}=90°,\text{ AC}=6.8\text{ cm}$

$\Delta \text{XYZ}:\text{YZ}=6.8\text{ cm, }\angle \text{X}=90°\text{, ZX}=4.8\text{ cm}$

a.    Both the triangles are congruent

$\Delta \text{PQR}\cong \Delta \text{STU}$

 

b.   Both the triangles are not congruent.

In Fig. 6.51, $\text{PQ}=\text{PS}$ and $\angle 1=\angle 2$.

Fig. 6.51

(i)             Is $\Delta \text{PQR}\cong \Delta \text{PSR}$? Give reasons.

(ii)         Is $\text{QR}=\text{SR}$? Give reasons.

Yes

(i)             In $\Delta \text{PQR and }\Delta \text{PSR}$,

$\text{PQ}=\text{PS}$

$\angle 1=\angle 2$ 

$\text{PR}=\text{PR [}\because \text{Common]}$

By SAS congruence criterion,

$\Delta \text{PQR}\cong \Delta \text{PSR}$

(ii)         Yes, by CPCT,

$\text{QR}=\text{SR}$

In Fig. 6.52, $\text{DE}=\text{IH, EG}=\text{FI}$ and $\angle \text{E}=\angle \text{I}$. Is $\Delta \text{DEF}\cong \Delta \text{HIG}$? If yes, by which congruence criterion?

Fig. 6.52

Given, $\text{EG}=\text{FI}$

$\begin{array}{l}\text{EG}+\text{GF}=\text{FI}+\text{GF}\\ \text{EF}=\text{IG}\end{array}$

In $\Delta \text{DEF and }\Delta \text{HIG}$,

$\text{DE}=\text{IH}$

$\text{EF}=\text{IG}$

$\angle \text{E}=\angle \text{I}$

By SAS congruence criterion,

$\Delta \text{DEF}\cong \Delta \text{HIG}$

In Fig. 6.53, $\angle \text{1}=\angle \text{2}$ and $\angle \text{3}=\angle \text{4}$.

a.    Is $\Delta \text{ADC}\cong \Delta \text{ABC}$? Why?

b.   Show that $\text{AD}=\text{AB and CD}=\text{CB}$.

Fig. 6.53

a.    In $\Delta \text{ADC and }\Delta \text{ABC}$,

$\angle \text{1}=\angle \text{2}$

$\text{AC}=\text{AC}$

$\angle \text{3}=\angle \text{4}$

By ASA congruence criterion,

$\Delta \text{ADC}\cong \Delta \text{ABC}$

b.   By CPCT,

$\text{AD}=\text{AB}$

$\text{CD}=\text{CB}$

Observe Fig. 6.54 and state the three pairs of equal parts in triangles ABC and DBC.

a.    Is $\Delta \text{ABC}\cong \Delta \text{DCB}$? Why?

b.   Is $\text{AB}=\text{DC}$? Why?

c.    Is $\text{AC}=\text{DB}$? Why?

Fig. 6.54

a.    In $\Delta \text{ABC and }\Delta \text{DCB}$,

$\text{BC}=\text{BC}$

$\angle \text{ABC}=\angle \text{DCB}=70$

$\angle \text{ACB}=\angle \text{DBC}=30$

By ASA congruence criterion,

$\Delta \text{ABC}\cong \Delta \text{DCB}$.

b.   $\text{AB}=\text{DC}$          [by CPCT]

c.    $\text{AC}=\text{DB}$          [by CPCT]

In Fig. 6.55, $\text{QS}\perp \text{PR, RT}\perp \text{PQ}$ and $\text{QS}=\text{RT}$.

1. Is $\Delta \text{QSR}\cong \Delta \text{RTQ}$? Give reasons.
2. Is $\angle \text{PQR}=\angle \text{PRQ}$? Give reasons.

Fig. 6.55

1. In $\Delta \text{QSR and }\Delta \text{RTQ}$,

$\text{QS}=\text{RT}$

$\angle \text{QSR}=\angle \text{QTR}=90$

$\text{QR}=\text{QR}$

By RHS congruence criterion,

$\Delta \text{QSR}\cong \Delta \text{RTQ}$

2. Yes, by CPCT, $\angle \text{PQR}=\angle \text{PRQ}$

Points A and B are on the opposite edges of a pond as shown in Fig. 6.56. To find the distance between the two points, the surveyor makes a right-angled triangle as shown. Find the distance AB.

Fig. 6.56

Since, $\Delta \text{ACD}$ is a right-angled triangle.

Right angled $\Delta \text{ADC}$, by Pythagoras theorem,

${(\text{AC})}^2={(\text{AD})}^2+{(\text{CD})}^2$

${(\text{AC})}^2={(30)}^2+{(40)}^2$

${(\text{AC})}^2=900+1600$

${(\text{AC})}^2=2500$

$\text{AC}=\sqrt{2500}=50\text{ m}$

Now, $\text{AB}=\text{AC}-\text{BC}=50-12=38\text{ m}$

Hence, distance $\text{AB}$ is $38\text{ m}$

Two poles of $10\text{ m}$ and $15\text{ m}$ stand upright on a plane ground. If the distance between the tops is $13\text{ m}$, find the distance between their feet.

Let           $\text{BC}=x\text{m}$

In right-angled $\Delta \text{ACB}$,

${\text{AB}}^2={\text{AC}}^2+{\text{BC}}^2$

${\left(13\right)}^2={\left(5\right)}^2+{(x)}^2$

$169-25=x^2$

$144=x^2$

$x=\sqrt{144}$

$x=12\text{ m}$

Hence, distance between the feet of two poles is $12\text{ m}$

The foot of a ladder is $6\text{ m}$ away from its wall and its top reaches a window $8\text{ m}$ above the ground,

1. Find the length of the ladder.
2. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?

a.    Let the length of the ladder be $x\text{ m}$

In the right angled $\Delta \text{ABC}$,

${\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2$

${(x)}^2={\left(8\right)}^2+{\left(6\right)}^2$

$=64+36=\sqrt{100}$

$x=10\text{ m}$

Hence, the length of the ladder is $10\text{ m}$

b.   Let the height of the top be $x\text{ m}$

In right angled $\Delta \text{ACB}$,

${\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2$

${\text{AB}}^2={\text{AC}}^2-{\text{BC}}^2$

${(x)}^2={\left(10\right)}^2-{\left(8\right)}^2$

$=100-64=36$

$x=6\text{ m}$

Hence, the height of the top is $6\text{ m}$

In Fig. 6.57, state the three pairs of equal parts in $\Delta \text{ABC}$ and $\Delta \text{EOD}$. Is $\Delta \text{ABC}\cong \Delta \text{EOD}$? Why?

Fig. 6.57

In $\Delta \text{ABC and }\Delta \text{EOD}$,

$\text{AB}=\text{EO}$

$\text{AC}=\text{ED}$

$\angle \text{ABC}=\angle \text{EOD}=90$

By RHS congruence criterion,

$\Delta \text{ABC}\cong \Delta \text{EOD}$