The quadrilateral ABCD is a kite. If AB = BC = 20, CD = DA = 15 and BD

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? 6.11.png A. 60 B. 72 C. 84 D. 96 E. 108

chetan2u · Answer

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

nick1816 · Answer

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

PrafulKashyap · Answer

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

MathRevolution · Answer

=> 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

stne · Answer

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.

samarpan.g28 · Answer

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).

The quadrilateral ABCD is a kite. If AB = BC = 20, CD = DA = 15 and BD

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The quadrilateral ABCD is a kite. If AB = BC = 20, CD = DA = 15 and BD

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