How do you find the area of a triangle if height: ${(4 x^{3} y^{4})}^{2}$ $(4x^3y^4)^2$ and base: ${(x y)}^{2}$ $(xy)^2$ ?

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Answer 1 · 2018-04-01T05:03:15

$A = 8 x^{8} y^{10}$ $A=8x^8y^10$

For a triangle,

$A = \frac{1}{2} b h$ $A=1/2bh$ where $b$ $b$ is the base, $h$ $h$ is the height.

We're given $h = {(4 x^{3} y^{4})}^{2}$ $h=(4x^3y^4)^2$

$b = {(x y)}^{2}$ $b=(xy)^2$

So,

$A = \frac{1}{2} {(4 x^{3} y^{4})}^{2} (x y^{2})$ $A=1/2(4x^3y^4)^2(xy^2)$

Simplify:

$A = \frac{1}{2} (16 x^{6} y^{8}) (x^{2} y^{2})$ $A=1/2(16x^6y^8)(x^2y^2)$

$A = 8 x^{8} y^{10}$ $A=8x^8y^10$

How do you find the area of a triangle if height: (4x3y4)2(4x^3y^4)^2 and base: (xy)2(xy)^2?