Question

The perimeter of isosceles triangle is 32cm. The ratio of equal side to its base is 3:2. Find its area

Answer 1

Let the ratio of equal side to its base be 3x and 2x respectively.
Now, Perimeter of an isosceles = 2a + b
=> P = 2*3x + 2x
=> 32 = 6x + 2x
=> 8x = 32
=> x = 4 cm
Then, equal sides = 3*4 = 12 cm
And, base = 2*4 = 8 cm
Therefore, Height =
$\sqrt{12 {}^{2} - 4 { }^{2} }$
$\sqrt{144 - 16}$
$\sqrt{128}$
$\sqrt{8 \times 8 \times 2}$
$8 \sqrt{2}$
Hence,
Area of an isosceles = 1/2 * b * h sq. ut.
=> Area = 1/2 * 8 * 8 root 2 cm*cm
=> Area = 32 root 2 cm*cm

Answer 2

Let the equal equal side be 3a

Base = 2a

Perimeter = 32

=> 3a + 3a + 2a = 32

=> 8a = 32

=> a = 4

Equal sides = 12 cm

Base = 8 cm

S = 16 cm

S-a = 16 - 12 = 4

S-b = 16 - 12 = 4

S-c = 16 - 8 = 8

Area = sqrt ( s(s-a) (s-b) (s-c)

= sqrt ( 16 * 4 *4 *8)

= sqrt ( 16 * 16 * 4 * 2)

= 16 * 2 root2

= 32 root2 cm^2