Question

8. Find the area of an isosceles triangle whose perimeter is 32cm and its base is 12cm.
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Answer 1

Answer:

Sum of all sides

Explanation:

Do 32+12

Answer 2

Answer:

Let the congruent sides of the isosceles triangle =a and the base =b.

Given, b=12cm

Perimeter of the triangle =Sum of all sides = a + a + b =32

⇒2a+12=32

On solving, a =10cm

Since the height of the triangle, divides it into two right angled triangles,  

$a^{2}$=  $h^{2}$  +($\frac{b}{2}$)2=$10^{2}$ = $h^{2}$+$6^{2}$

 $h^{2}$ =100−36=64

h= $\sqrt{64}$   =8 cm  

Area of triangle = $\frac{1}{2}$   × base × height= $\frac{1}{2}$  × 12 × 8 = 48 sq cm