Question

Find the area of an isosceles triangle whose Base is 16 cm and one of its equal side is 10 cm

Answer 1

consider this ∆ABC with AD as perpendicular from A on BC
Since it is an isosceles triangle
BD=DC=8 cm

now applying Pythagoras theorem in ∆ADC,

AD^2+DC^2=AC^2
AD^2=AC^2-DC^2
AD^2

$= {10}^{2} - {8}^{2} \\ = 100 - 64 \\ = 36$
therefore,
AD=6cm

area of ∆ABC=
$\frac{1}{2} \times base \times height$
=1/2×6×16
=48 sq. cm

Answer 2

Answer:

Step-by-step explanation:

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