Question

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

Answer 1

Answer:

The altitude is of length 16 cm and the area is 192 cm².

Step-by-step explanation:

The perimeter of an isosceles triangle is:

Perimeter  = 2a + b

Here, a = sides and b = base.

b = 24 cm

Perimeter = 64 cm.

The sides are:

$Perimeter=2a+b\\64=2a+24\\2a=40\\a =20$

Thus, the two similar sides of the triangle are 20 cm each.

Consider the triangle below.

Altitude AO is perpendicular to the base CB and divided it in equal part as BO and OC.

Consider the right-angled triangle AOB.

The length of AO can be computed using the Pythagoras theorem as follows:

$AB^{2}=AO^{2}+OB^{2}\\20^{2}=AO^{2}+12^{2}\\AO^{2}=400-144\\AO=\sqrt{256}\\ =16$

Thus, the altitude is of length 16 cm.

The area of the triangle is:

$Area=\frac{1}{2}\times base\ height\\ =\frac{1}{2} \times24\times16\\=192$

Thus, the area is 192 cm².

Answer 2

Answer:

Altitude  isosceles triangle is 16 cm

Area of the triangle is  196 cm^2

Step-by-step explanation:

Let other two  side be x

Perimeter of the isosceles triangle is

$x+x+base= 64\\2x+24=64\\2x=64-24\\2x=40\\x=20 cm$

We can consider right angles triangle with base as 12 cm and hypotenuses as 20 cm , opposite side is taken as altitude= y of the isosceles triangle.

$20^2 =y^2+(\frac{24}{2}) ^2\\y^2=400-144\\y^2=256\\y=\sqrt{256} \\y=16 cm$

Altitude  is 16 cm

Area of the triangle is

$A=\frac{1}{2} base \times altitude\\A=\frac{1}{2} \times 24 \times 20\\A=196 cm^2$