Question

9. Find the area of an isosceles triangle each of whose equal sides measures
13 cm and whose base measures 20 cm.​

Answer 1

Solution :-

As we have :-

Two equal sides measures = 13 cm

Base measures = 20 cm

Now let sides be

a = 20 cm

b = 13 cm

c = 13 cm

Now as we have got the equation :-

$\small{ \textsf{ Area of Triangle} = \sqrt{ S (S -a) (S-b) (S-c) }}$

Here

S = Semi Perimeter

Now

S = (20 + 13 + 13) ÷ 2

S = 46 ÷ 2

S = 23 cm

So the area

$= \sqrt{ 23 (23 -20) (23-13) (23-13) }$

$= \sqrt{ 23 .(3).(10).(10)}$

$= \sqrt{23 . (3) . 10^2 }$

$= 10 \sqrt{23.(3)}$

$= 10 \sqrt{69}$

So area of Triangle

$\huge{\boxed{\sf{= 10 \sqrt{69}}}}$

Answer 2

Answer:-

Step-by-step explanation:

Known Terms :-

S = Semi perimeter

Given :-

Two Sides of triangle = 13 cm

Base of Triangle = 20 cm.

To Find :-

Area of triangle.

Formula to be used :-

Area of triangle = $\sqrt{ S (S -a) (S-b) (S-c)$

Solution :-

⇒ $S=\frac{a \: + \: b \: + \: c}{2}$

⇒ $S=\frac{13 \: + \: 13 \: + \: 20}{2}$

⇒ $S=\frac{46}{2}$

⇒ $S=23.$

Putting all the values we get,

$\sqrt{ S (S -a) (S-b) (S-c)$

$= \sqrt{ 23 (23 -20) (23-13) (23-13) }$

$= \sqrt{ 23 .(3).(10).(10)}$

$= \sqrt{23 . (3) . 10^2 }$

$= 10 \sqrt{23.(3)}$

$= 10 \sqrt{69}$

Hence, The Area of Trianle is $\huge{\boxed{\sf{ 10 \sqrt{69}}}}$