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655-705 (Hard)|   Geometry|      
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MathRevolution
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The quadrilateral ABCD is a kite. If AB = BC = 20, CD = DA = 15 and BD 11 Jun 2019, 00:31
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C
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65% (hard)

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62% (02:56) correct 38% (02:50) wrong based on 76 sessions

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[GMAT math practice question]

The quadrilateral \(ABCD\) is a kite. If \(AB = BC = 20, CD = DA = 15\) and \(BD = 7\), what is the area of the kite \(ABCD\) shown in the figure below?

Attachment:
6.11.png
6.11.png [ 4.17 KiB | Viewed 5182 times ]

\(A. 60\)

\(B. 72\)

\(C. 84\)

\(D. 96\)

\(E. 108\)
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C
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The quadrilateral ABCD is a kite. If AB = BC = 20, CD = DA = 15 and BD 11 Jun 2019, 01:14
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MathRevolution
[GMAT math practice question]

The quadrilateral \(ABCD\) is a kite. If \(AB = BC = 20, CD = DA = 15\) and \(BD = 7\), what is the area of the kite \(ABCD\) shown in the figure below?

Attachment:
The attachment 6.11.png is no longer available

\(A. 60\)

\(B. 72\)

\(C. 84\)

\(D. 96\)

\(E. 108\)
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Join AC and let BD join AC at E.

POints B and D are equidistant from A and C, so BDE is a perpendicular bisector of AC.

Triangle BEC..
\(CE^2=20^2-(7+x)^2\)
Triangle DEC..
\(CE^2=15^2-(x)^2\)

Therefore, \(15^2-(x)^2=20^2-(7+x)^2........225-x^2=400-x^2-49-14x.........14x=400-274=126...x=9\)

so CE= \(\sqrt{15^2-9^2}=12\). OR sides of BDC is 9-CE-15, which is 3-4-5, so CE = 4*3=12

Area of kite = 2*Area of BDC=2*area(BEC-DEC)=2*\(\frac{1}{2}*12*(7+9-9)=12*7=84\)

C
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The quadrilateral ABCD is a kite. If AB = BC = 20, CD = DA = 15 and BD 11 Jun 2019, 01:26
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Area of kite= BD*AC/2

Let OD=y and AC=x

Median in isosceles triangle perpendicularly is also the altitude.

\(15^2-y^2=(x/2)^2\)......(1)
Also,\(20^2-(7+y)^2=(x/2)^2\).......(2)

From (1) and (2)
we get y=9 and x=24

Area of kite= 7*24/2=84



MathRevolution
[GMAT math practice question]

The quadrilateral \(ABCD\) is a kite. If \(AB = BC = 20, CD = DA = 15\) and \(BD = 7\), what is the area of the kite \(ABCD\) shown in the figure below?

Attachment:
The attachment 6.11.png is no longer available

\(A. 60\)

\(B. 72\)

\(C. 84\)

\(D. 96\)

\(E. 108\)
Show more

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Untitled.png [ 7.92 KiB | Viewed 4971 times ]

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The quadrilateral ABCD is a kite. If AB = BC = 20, CD = DA = 15 and BD 11 Jun 2019, 21:37
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Hero's formula

Area =\(\sqrt{p(p−a)(p−b)(p−c)}\)

where p= \(\frac{a+b+c}{2}\)

Area of ABCD = area of triangle ABD *2 (two triangles are similar)

p=\(\frac{20+15+7}{2}\)=21

area= 2*\(\sqrt{21(21-20)(21-15)(21-7)}\)

area of ABCD= 2*7*3*2= 84


OA=C
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The quadrilateral ABCD is a kite. If AB = BC = 20, CD = DA = 15 and BD 12 Jun 2019, 19:20
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=>

Attachment:
6.13.png
6.13.png [ 5.54 KiB | Viewed 4694 times ]

Let \(x\) and \(y\) be the lengths of \(DE\) and \(CE\), repectively.

From the right triangle \(CDE\), we have \(CD^2 = x^2 + y^2\) or \(x^2+y^2=225.\)

From the right triangle \(BCD\), we have \(BC^2 = (x+7)^2 + y^2\) or \((x+7)^2 + y^2 = 400.\)

Subtracting the first equation from the second yields \((x+7)^2 – x^2 = (x+7+x)(x+7-x) = 7(2x+7) = 175\), and \(x = 9.\)

\(y = 12\) since \(y^2 = 225 – x^2 = 225 – 81 = 144.\)

The area of the kite \(ABCD\) is the area of triangle \(ABC\) - the area of triangle \(ADC = (1/2)(2y)(x + 7) – (1/2)(2y)x = (1/2)(24)(16) – (1/2)(24)(9) = (1/2)(24)(16-9) = 12*7 = 84.\)

Therefore, the answer is C.
Answer: C
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The quadrilateral ABCD is a kite. If AB = BC = 20, CD = DA = 15 and BD 17 Jun 2019, 05:33
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MathRevolution
=>

Attachment:
6.13.png

Let \(x\) and \(y\) be the lengths of \(DE\) and \(CE\), repectively.

From the right triangle \(CDE\), we have \(CD^2 = x^2 + y^2\) or \(x^2+y^2=225.\)

From the right triangle \(BCD\), we have \(BC^2 = (x+7)^2 + y^2\) or \((x+7)^2 + y^2 = 400.\)

Subtracting the first equation from the second yields \((x+7)^2 – x^2 = (x+7+x)(x+7-x) = 7(2x+7) = 175\), and \(x = 9.\)

\(y = 12\) since \(y^2 = 225 – x^2 = 225 – 81 = 144.\)

The area of the kite \(ABCD\) is the area of triangle \(ABC\) - the area of triangle \(ADC = (1/2)(2y)(x + 7) – (1/2)(2y)x = (1/2)(24)(16) – (1/2)(24)(9) = (1/2)(24)(16-9) = 12*7 = 84.\)

Therefore, the answer is C.
Answer: C
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Hi MathRevolution,
Thank you for the awesome question and the solution.
While going through your soln. found a small typo which you may wish to correct to facilitate better understanding of your soln.

Line 3, from the right triangle BCD, except BCD is not a right \(\Delta\), I suppose you meant to write right\(\Delta\) BCE.
Hope you will look into this.
Thank you.
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The quadrilateral ABCD is a kite. If AB = BC = 20, CD = DA = 15 and BD 22 Dec 2023, 06:34
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I used Heron's formula to find the area with semiperimeter, Area = Square root of√s(s - a)(s - b)(s - c) where s is half the perimeter, or (a + b + c)/2. The result will be 42. Simply multiply it by 2. The answer is 84 (C).
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