Question

The Perimeter of an isosceles triangle is 32m. The ratio of the equal sides to its base is 3:2
find the area of triangle​

Answer 1

Answer:

• The Area of the isosceles triangle = 45.25 m².

Given :

• Perimeter of the isosceles triangle = 32 m.

• The ratio of the equal sides to its base is 3:2.

To find :

• The Area of the isosceles triangle = ?

Step-by-step explanation:

The perimeter of isosceles triangle = 32cm [Given]

Let, the length of equal side be 3x and base be 2x.

We know that,

Perimeter of isosceles triangle = 2 × equal side + base

Substituting the values in the above formula, we get,

➮ 2 × 3x + 2x = 32

➮ 6x + 2x = 32

➮ 8x = 32

➮ x = 32/8

➮ x= 4

Hence,

Length of equal side, 3x ➮ 3 × 4= 12 m

Base, 2x ➮ 2 × 4=8 m

Now,

We know that,

Area of isosceles triangle=B × 1/4 × √4a² - b²

Substituting the values in the above formula, we get,

= 8 × 1/4√4 × 12² - 8²

= 8 × 1/4√4 × 144 - 64

= 2√512

= 2 × 22.62

= 45.25.

Thus, The Area of the isosceles triangle = 45.25 m².

Answer 2

Answer : 32√2 m

$\rule{100}2$

Solution:

It is given that the ratio of the equal sides to it's base is 3:2.

Let common factor be ' x ' So, equal sides will be of 3x each and base will be 2x.

$\sf Perimeter\:of\triangle=sum\:of\:three\:sides$

$\sf 32=3x+3x+2x$

$\sf\implies 32= 6x+2x$

$\sf\implies 32=8x$

$\sf\implies x=\dfrac{32}{8}$

$\sf\therefore x = 4$

substitute the value of ' x '. We get,

Equal sides ⇒ 3×4 = 12 m

Base ⇒ 2×4 = 8 m

Now,

$\sf Semi-perimeter\:of\:triangle= \dfrac{perimeter}{2}$

$\sf Semi-perimeter=\dfrac{32}{2}$

$\sf Semi-perimeter = 16$

Also, we know that,

$\sf Area\:of\:triangle=\sqrt{S(S-side_1)(S-side_2)(S-side_3)}$

Substitute the given values. We get,

$\sf = \sqrt{16(16-12)(16-12)(16-8)}$

$\sf =\sqrt{16\times 4\times 4\times 8}$

$\sf =\sqrt{4\times 4\times 4\times 4\times 2\times 2\times 2}$

$\sf = 4\times 4\times 2\sqrt{2}$

$\sf = 32\sqrt{2}\:m$

Hence,

Area of the given triangle is 32√2 m.

$\rule{200}2$