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?

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

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