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
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Let the perpendicular of the smaller traingle is z.
now, perpendicular of the bigger trianlge is sqrt(25^2-4x^2)=sqrt(21)*x
Now since both the smaller triangle and biger trianlge are similar(one angle is 90,one angle is common)
so,2x/y=(sqrt)21*x/z
so, z=(sqrt)21*y/2
Hence, area of the shaded region
=area of the bigger triangle - area of the smaller triangle
=[1/2*2x*(sqrt)21*x]-[1/2*z*y]
=(sqrt)21*x^2-[1/2*(sqrt)21*y/2*y]
=(sqrt)21*x^2-(sqrt)21*y^2/4
=(sqrt)21*(4x^2-y^2)/4
Ans is E[/spoiler]
now, perpendicular of the bigger trianlge is sqrt(25^2-4x^2)=sqrt(21)*x
Now since both the smaller triangle and biger trianlge are similar(one angle is 90,one angle is common)
so,2x/y=(sqrt)21*x/z
so, z=(sqrt)21*y/2
Hence, area of the shaded region
=area of the bigger triangle - area of the smaller triangle
=[1/2*2x*(sqrt)21*x]-[1/2*z*y]
=(sqrt)21*x^2-[1/2*(sqrt)21*y/2*y]
=(sqrt)21*x^2-(sqrt)21*y^2/4
=(sqrt)21*(4x^2-y^2)/4
Ans is E[/spoiler]
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LalaB
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y=(1/2)*2x=x
sqrt(25x^2-4x^2)=xsqrt21
(1/2)*2x*xsqrt21-(1/2)*xsqrt21/2*y=sqrt21*(x^2-(1/4)*y^2)=sqrt21*(4x^2-y^2)/4= sqrt21*3x^2/4
E is the answ
sqrt(25x^2-4x^2)=xsqrt21
(1/2)*2x*xsqrt21-(1/2)*xsqrt21/2*y=sqrt21*(x^2-(1/4)*y^2)=sqrt21*(4x^2-y^2)/4= sqrt21*3x^2/4
E is the answ
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Larger triangle: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
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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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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