Question

the base of an isosceles triangle is 10cm and one of it side is 8 cm find the area of triangle​

Answer 1

$question$

the base of an isosceles triangle is 10cm and one of it side is 8 cm find the area of triangle

$to \: find$

area of the triangle

$given$

since it is a isosceles triangle than it's 2 side will equal.

1st side - 8 cm

2nd - 10 cm

3rd - 8cm

$solution$

By herons formula

$area \: = \sqrt{s(s - a) (s - b)(s - c)}$

since ,

$s = \frac{a + b + c}{2}$

$s = \frac{8 + 10 + 8}{2} = \frac{36}{2} = 18$

$= \sqrt{18(18 - 8)(18 - 10)(18 - 8)}$

$= \sqrt{18 \times 10 \times 8 \times 10}$

$= \sqrt{14,400}$

= 120

Therefore area= 120cm

Answer 2

Step-by-step explanation:

A=bhb

2

hb=a2﹣b2

4

Solving forA

A=1

4b4a2﹣b2=1

4·10·4·82﹣102≈31.22499cm²