Question

The area of a triangle is 80x^5y^3. The height of the triangle is x^4y. What is the length of the base of the triangle?

Answer 1

Area of a triangle = 1/2 * base * height

Area, A = 80x^5 · y^3

Base, b = ?

Height, h = x^4 · y

Therefore, 80x^5 · y^3 = 1/2 * b * x^4 · y

Multiply the equation by 2

2 * 80x^5 · y^3 = 2 * 1/2 * b * x^4 · y

160x^5 · y^3 = b * x^4 · y

b * x^4 · y = 160x^5 · y^3

Divide the equation by x^4 · y

(b * x^4 · y)/ x^4 · y = (160x^5 · y^3) / x^4 · y

b = 160x^(5 – 4) · y^(3 – 1)

b = 160x · y^2

The length of the base of the triangle is 160x · y^2

Answer 2

Area of a triangle is given by 1/2bh where b is the base and h is the perpendicular height of the triangle.
The area is 80x∧5y³ and the height is x∧4y
Thus; 80x∧5y³ = 1/2(x∧4y) b
160x∧5y³ = (x∧4y)b
b = (160x∧5y³)/ x∧4y)
b = 160xy²
Therefore, the base of the triangle is 160xy²