Question

Find the altitude and area of an isosceles triangle whose perimeter is 64 cm and whose base is24 cm

Answer 1

Answer:

The altitude of isosceles triangle is 16 cm

The Area of isosceles triangle is 192 sq cm .

Step-by-step explanation:

Given as :

The perimeter of isosceles triangle = 64 cm

The base of the triangle = 24 cm

Let The altitude of isosceles triangle = h cm

Let The Area of isosceles triangle = A cm²

According to question

As The isosceles triangle has two sides common

Let The common side = x cm

So, Perimeter of isosceles triangle = x + x + 24

or, x + x + 24 = 64

Or, 2 x = 64 - 24

Or, 2 x = 40

∴       x = $\dfrac{40}{2}$

i.e  x = 20 cm

So, Three sides of isosceles triangle = 20 cm , 20 cm , 24 cm

Height of isosceles triangle = h = $\sqrt{a^{2} - \dfrac{b^{2} }{2} }$

Or,  h = $\sqrt{20^{2} - \dfrac{24^{2} }{2} }$

∴     h = √400-144  = √256

i.e height = 16 cm

Now, Area of isosceles triangle

Area = $\dfrac{1}{2}$ × base × height

or, Area = $\dfrac{1}{2}$ × 24 cm × 16 cm

∴ Area = 192 cm²

Hence, The altitude of isosceles triangle is 16 cm

And The Area of isosceles triangle is 192 sq cm . Answer

Answer 2

Answer:

Step-by-step explanation:

1)draw AP perpendicular to BC

2) now BP = CP = 12 CM ( BY CONSTRUCTION)

3) BY Pythagoras theorem find Altitude 'AP'

4) by using 1/2 x base(bc) x height(AD) find its area

5) answer is solved

(done dana done)

you can also first find area using herons formula and then find its height also