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galina1969 [7]
3 years ago
11

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?

Mathematics
2 answers:
svetoff [14.1K]3 years ago
7 0
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
mr_godi [17]3 years ago
7 0

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

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