Question

Find the area of an isosceles triangle whose base is 16 cm and length of each of the equal sides is 10 cm.

Answer 1

Hi !!!

this is your answer......

Given:-------

Base = 16 cm.

Length of both equal sides = 10cm.

We will use Heron's formula to find the area,

Side 1, a = 16 cm.
Side 2, b = 10 cm.
Side 3, c = 10 cm.

Therefore,
Half perimeter= s = 1/2×(16+10+10)
= 36/2
s = 18 cm.

Using Heron's formula,
$\sqrt{s(s - a)(s - b)(s - c)}$
$= \sqrt{18(18 - 16)(18 - 10)(18- 10)} \\ = \sqrt{18 \times 2 \times 8 \times 8} \\ = \sqrt{2 \times 9 \times 4 \times 2 \times 4 \times 2\times 2} \\ = 3 \times 2 \times 2 \times 2 \times 2 \\ = 48 cm²$
Hence, the area of given triangle is 48 cm².


Hope you got the answer......

Answer 2

Heya,

Equal sides of the isosceles triangle = 10cm
Base = 16cm

Let's use herons formula to find the area.

Area = √s (s - a) (s - b) (s - c)

Where,
a = 10cm
b = 10cm
c = 16cm

To find s:-

s = a+b+c/2
s = 10 + 10 + 16/2
s = 36/2
s = 18

Area = √s (s - a) (s - b) (s - c)

Area = √18 (18 - 10) (18 - 10) (18 - 16)

Area = √18 × 8 × 8 × 2

Area = √3 × 3 × 2 × 8 × 8 × 2

Area = 3 × 2 × 8

Area = 48cm^2

Hope my answer helps you :)

Regards,
Shobana