Question

The perimeter of an isosceles triangle is 32cm.The ratio of the equal side to it's base is 3:2.Find the area of the triangle

Answer 1

Answer

The area of triangle is 32√2cm

Given

• The perimeter of an isosceles triangle is 32cm
• The ratio of the equal side to its base is 3:2

Solution

Let us consider the equal sides of the triangle be x cm and the base be y cm

The perimeter is 32cm , i.e.

$\sf \implies2x + y= 32 \: \: .......(1)$

And again by question ,

$\sf \implies x: y = 3 : 2 \\ \\ \sf\implies \frac{x}{y} = \frac{3}{2} \\ \\ \sf \implies x = \frac{3}{2} y \: \: \: .......(2)$

Using the value of (2) from (1)

$\sf \implies2 \times \dfrac{3}{2} y + y = 32 \\ \\ \sf\implies4y = 32 \\ \\ \implies \sf y = 8$

Thus , the base is of 8cm

Putting the value of y in (2)

$\implies \sf x = \dfrac{3}{2} \times 8 \\ \\ \implies \sf x = 12$

The equal sides are of 12cm length

Now let the height of the isosceles triangle be k cm,

from Pythagorean triplet let us calculate the base of the triangle, let

$\sf \implies {k}^{2} + 4 {}^{2} = {12}^{2} \\ \\ \implies \sf {k}^{2} = 144 - 16 \\ \\ \implies \sf {k}^{2} = 128 \\ \\ \implies \sf k = 8 \sqrt{2}$

Thus , the height is 8√2cm

Now area of triangle is :

$\sf Area \: of \: triangle = \frac{1}{2} \times base \times height \\ \\ \sf\implies Area \: of \: triangle = \frac{1}{2} \times 8 \times 8 \sqrt{2} \\ \\ \sf\implies Area \: of \: triangle = 4 \times 8 \sqrt{2} \\ \\ \sf\implies Area \: of \: triangle = 32 \sqrt{2}$

Therefore , area of the triangle is 32√2cm

Answer 2

Answer:

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