Question

find the base of an isosceles triangle whose area is 12cm2 and one of the equal side is 5cm

Answer 1

Both sides are 5.
The base is x
The area is (1/2)x h=12 
xh=24
The height is one leg of a right triangle, half the base another leg, and the hypotenuse is 5.
This has to be a 3-4-5 right triangle.
The base is 3, but that is the base of half of the isosceles triangle. The whole base is 6.
The height is 4 in any case. 
The area is (1/2)*24=12 cm^2
The height is 4 cm, the base 6cm 

Answer 2

Answer:

Step-by-step explanation:

1. Given, Its an Isosceles triangle and not an right angled isosceles triangle. (So, we cant use Area = 1/2 * Base * Height)

2. Since, neither the perimeter nor the third side is given, so Herons formula is not feasible here.

So, we go by

Area of isosceles triangle = $\frac{a}{4}$  $\sqrt{4b^{2} -a^{2} }$  ( General Formula)

(Given, equal sides = b = 5 cm)

=>  $\frac{a}{4}$  $\sqrt{4b^{2} -a^{2} }$  = 12

=>  $\frac{a}{4}$  $\sqrt{4*5^{2} -a^{2} }$  = 12

=>  $\frac{a}{4}$  $\sqrt{100-a^{2} }$  = 12

Squaring on both sides

=> $a^{2}$ (100 -

=> 100 $a^{2}$ - $a^{4}$ = 2304

=> $a^{4}$ - 100 $a^{2}$ + 2304 = 0

=>  ($a^{2}$ - 64) (

Either $a^{2}$ = 64  => a = 8cm

or, $a^{2}$ = 36 => a = 6 cm

Hence, The base of the isosceles triangle is either 8 cm or 6 cm.

Verification :

When you calculate the area by putting both the values in the formula, you find the area as 12$cm^{2}$.