pemdas wrote:Find area of the shaded region in terms of x and y
A. 4x^2 - y^2
B. sqrt(21)*(4x^2 - y^2)
C. sqrt(21)*4x^2 - y^2
D. sqrt(21)*(2x^2 - y^2)/4
E. sqrt(21)*(4x^2 -y^2)/4
source: made up
note- figure not drawn to scale
Larger triangle:
Let x=1.
Base = 2x = 2(1) = 2.
Hypotenuse = 5x = 5(1) = 5.
Thus, height = √(5² - 2²) = √21.
Area = 1/2(2)(√21) = √21.
If y=0, then the shaded region is the entire larger triangle, with an area of √21. This is our target.
Now we plug x=1 and y=0 into the answers to see which yields our target of √21.
Only
E works:
√21 * (4x²-y²)/4 = √21 * (4*1² - 0²)/4 = √21.
The correct answer is
E.
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