Question

The perimeter of an isosceles triangle is 72cm and its base is 1.6 times each of the equal sides. Find the area of triangle and its height.

Answer 1

Let the base of the triangle is b, the height is h and the two equal sides are a and c where a=c
then by the given question, 2a+b=72 and
b=1.6a
∴, 2a+1.6a=72
or, 3.6a=72
or, a=72/3.6
or, a=20 cm
∴, c=20 cm and b=1.6×20=32 cm
h divides the isosceles triangle into 2 right angled triangles.
then by Pythagoras's theorem,
h²+(b/2)²=a²
or, h²=a²-(b/2)²
or, h²=20²-(32/2)²
or, h²=400-16²
or, h²=400-256
or, h²=144
or, h=12 cm
∴, the area of the triangle =(base×height)/2
                                         =(32×12)/2
                                         =384/2
                                         =192 cm²
 ∴, The height of the isosceles triangle is 12 cm and the area is 192 cm².

Answer 2

Let the two equal sides be x cm
the base = 1.6x
Perimeter = 72
x + x + 1.6x = 72
3.6x = 72
x = 20
The length of equal sides = 20 cm
Length of the base = 1.6*20 = 32 cm

Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}$
Here s = (20+20+32)/2 = 72/2 = 36

Area of triangle = $\sqrt{36(36-20)(36-20)(36-32)}$
                          = $\sqrt{36(16)(16)(4)}$
                          = $\sqrt{(144)(256)}$
                          = 192 sq cm

But the area of the triangle = 1/2 * base * height

1/2 * 32 * height = 192

16 * height = 192

height = 12 cm