Find the area of a kite ABCD if BD= 48cm, AB= 25cm, and BC= 26cm?

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Find the area of a kite ABCD if BD= 48cm, AB= 25cm, and BC= 26cm?

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Answer 1 · 2016-01-04T22:51:27

$S = 408 c m^{2}$ $S=408cm^2$

Explanation:

The formula of a kite's area is
$S = (\frac{1}{2}) d 1 \cdot d 2$ $S=(1/2)d1*d2$
where
$d 1 = kite's long diagonal$ $d1="kite's long diagonal"$
$d 2 = kite's short diagonal$ $d2="kite's short diagonal"$

In the problem, be noticed that once the diagonal BD has an endpoint in B (where 2 segments of different sizes, AB and BD, meet), this diagonal is divided in two equal parts by the other diagonal. Calling E the point where the two diagonals intercept each other, we have:
$B D = 2 \cdot D E = 48$ $BD=2*DE=48$ => $D E = 24$ $DE=24$

In the right triangle ABE we can obtain the segment AE
$A B^{2} = D E^{2} + A E^{2}$ $AB^2=DE^2+AE^2$ => $25^{2} = 24^{2} + A E^{2}$ $25^2=24^2+AE^2$ => $A E = \sqrt{625 - 576} = \sqrt{49} = 7$ $AE=sqrt(625-576)=sqrt(49)=7$

In the right triangle BCE we can obtain the segment CE
$B C^{2} = D E^{2} + C E^{2}$ $BC^2=DE^2+CE^2$ => $26^{2} = 24^{2} + C E^{2}$ $26^2=24^2+CE^2$ => $C E = \sqrt{676 - 576} = \sqrt{100} = 10$ $CE=sqrt(676-576)=sqrt(100)=10$

In this way we discovered the previously unknown diagonal:
$A C = A E + C E = 7 + 10 = 17$ $AC=AE+CE=7+10=17$

Finally,
$S = (\frac{1}{2}) (d 1 + d 2) = \frac{48 + 17}{2} = 408 c m^{2}$ $S=(1/2)(d1+d2)=(48+17)/2=408cm^2$

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Find the area of a kite ABCD if BD= 48cm, AB= 25cm, and BC= 26cm?

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