Question

Find the altitude and area of an isosceles triangle whose perimeter is 64 cm and whose base is 24.

Answer 1

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Answer 2

The altitude of an isosceles triangle is 16 cm.

The area of an isosceles triangle is 192 cm².

Step-by-step explanation:

Given : An isosceles triangle whose perimeter is 64 cm and whose base is 24.

To find : The altitude and area of an isosceles triangle ?

Solution :

Assume a be two sides, b is the base and h be the altitude of isosceles triangle.

The perimeter of isosceles triangle is $P=2a+b$

The altitude of the isosceles triangle is $h=\sqrt{a^2-\frac{b^2}{4}}$

Now, the formula of h in terms of P and b is

$h=\frac{1}{2}\sqrt{P(P-2b)}$

Here, P=64 cm and b=24 cm.

$h=\frac{1}{2}\sqrt{64(64-2(24))}$

$h=\frac{1}{2}\sqrt{64(64-48)}$

$h=\frac{1}{2}\sqrt{64(16)}$

$h=\frac{1}{2}\sqrt{1024}$

$h=\frac{1}{2}\times 32$

$h=16$

So, the altitude of an isosceles triangle is 16 cm.

The area of the triangle is given by,

$A=\frac{1}{2}\times b\times h$

$A=\frac{1}{2}\times 24\times 16$

$A=192\ cm^2$

So, the area of an isosceles triangle is 192 cm².

#Learn more

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