Question

Find the base of an isosceles triangle whose area is 192cm² and the length of one of the equal sides is 20cm.

Answer 1

Area of an isosceles triangle = 192cm²

An isosceles triangle is a triangle with (at least) two equal sides. In the figure above, the two equal sides have length b and the remaining side has length a. This property is equivalent to two angles of the triangle being equal. An isosceles triangle therefore has both two equal sides and two equal angles. Let h be the height of the isosceles triangle as illustrated.

So, h = `sqrt("b"^2 - "a"^2/4`

We know that, Area of a Triangle

= `(1)/(2)("Base" xx "Height")`

= `(1)/(2) xx "a" xx "h"`

= `(1)/(2) xx "a" xx sqrt("b"^2 - "a"^2/4)`

= `(1)/(2) xx "a"^2 xx sqrt("b"^2/"a"^2 - 1/4)`

Here, The area is therefore given by

= `(1)/(2) xx x^2 xx sqrt(20^2/x^2 - 1/4)`

= `x^2/(2)sqrt((1600 - x^2)/(4x^2)`

⇒ 192 = `x^2/(2)sqrt((1600 - x^2)/(4x^2)`

⇒ 192² = `x^4/(4) xx (1600 - x^2)/(4x^2)`

⇒ 192² x 16 = x²(1600 - x²)
⇒ x⁴ - 1600x² + 589824 = 0
⇒ t² - 1600t + 589824 = 0; wherex² = t
⇒ t² - 1024t - 576t + 589824 = 0
⇒ t(t - 1024) - 576(t - 1024) = 0
⇒ t = 1024 or 576
⇒ x² = 1024 pr 576
⇒ x = 32cm or 24cm.