Question

find the area of an isosceles triangle, the ratio of whose equal side to its base is 3:2 and the perimeter is 32 cm​

Answer 1

Answer:

32 sq cm

Explanation:

Let the common side be x.So,

6x+2x= 32

x= 32/8=4

a , b and c are sides

area of triangle =

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

s = perimeter/2= 16

s-a = 16- 12= 4

s-b= 16 -12= 4

s-c= 16-8= 8

$\sqrt{16(4)(4)(8)}$

32√ 2 cm square

Answer 2

Solution!!

The ratio of equal side to the base of isosceles triangle is given. The perimeter of the triangle is also given. We have to find the area. To do so, we have to find the sides of the triangle.

Let the sides be 3x, 3x and 2x

Perimeter = 32 cm

Perimeter = Sum of all sides of triangle

32 cm = 3x + 3x + 2x

32 cm = 8x

32 cm ÷ 8 = x

x = 4 cm

3x = 3 × 4 cm = 12 cm

3x = 3 × 4 cm = 12 cm3x = 3 × 4 cm = 12 cm

3x = 3 × 4 cm = 12 cm3x = 3 × 4 cm = 12 cm2x = 2 × 4 cm = 8 cm

As we have got the sides of the triangle, now we can find the area. I'll use Heron's formula.

Semi-perimeter (s) = (a + b + c)/2

s = 32 cm/2

s = 16 cm

Area = √[s(s-a)(s-b)(s-c)]

Area = √[16(16-12)(16-12)(16-8)]

Area = √[16(4)(4)(8)]

Area = 32√2 cm²