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

Answer:

$\large{\underline{\boxed{\sf 10 \sqrt{69}}}}$

Step-by-step explanation:

Given that, the sides of isosceles triangle are 13 cm, 13 cm, and 20 cm.

• a = 13
• b = 13
• c = 20

Using heron's formula -

S = $\rm \dfrac{a + b + c}{2}$

S = $\rm \dfrac{13 + 13 + 20}{2}$

S = $\rm \dfrac{46}{2}$

S = 23

Now,

$\rm Area = \sqrt{s(s-a)(s-b)(s-c)}$

$\rm Area = \sqrt{23(23 - 13)(23 - 13)(23 - 20)} \\ \\ \rm Area = \sqrt{23 \times 10 \times 10 \times 3} \\ \\ \rm Area = 10 \sqrt{69} \: cm^{2}$

Hence, the area of isosceles triangle is 10√69 cm².

Answer 2

let ABC be isisceles triangle

Draw perpendicular AD from A to BC

As perpendicular drawn will bisect BC

So BD = DC = 10

In ABD

AB = 13

BD = 10

AD = √((13)^2 - (10)^2) = √(169-100) = √69

NOW area of isosceles triangle ABC

= 1/2 ( AD) ( BC)

= 1/2 ( √69) ( 20)

= 10√69 cm^2