SOLUTION: In parallelogram ABCD, line EF is perpendicular to line AB and line GH is perpendicular to BD. Compute the area of the parallelogram, given AD=22, AE=13, EF=12, BG=17, and GH=8.

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Question 1099501: In parallelogram ABCD, line EF is perpendicular to line AB and line GH is perpendicular to BD. Compute the area of the parallelogram, given AD=22, AE=13, EF=12, BG=17, and GH=8.
link for the problem https://gyazo.com/fc882226a5f618fc9931d6a27147f4ac
Found 2 solutions by KMST, ikleyn:
Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!

Right triangle AEF, with a right angle at F,
has hypotenuse AE=13, leg EF=12, and leg AF that we can calculate as
.
Right triangle BGH, with a right angle at H,
has hypotenuse BG=17, leg GH=8, and leg BH that we can calculate as
.

From those two triangles,we could get angles BAD and ADB=DBC,
and with AD=22, we would have angle-side-angle of triangle ABD,
which is half of the parallelogram.
That looks complicated, though.

If we find the height , perpendicular to AD, we could calculate the area as .


The height drawn forms right triangle ABI,
and right triangle BDI.
ABI shares the angle at A angle with right triangle AEF,
so they are similar right triangles.
So,
BDI has angle BDI, congruent with angle GBH,
and that makes right triangles BDI similar to right triangle GBH.
So, .
adding and , we get






As BI-9.6 is the height of the parallelogram relative to base AD=22,
the area of the parallelogram is
Answer by ikleyn(52794)   (Show Source): You can put this solution on YOUR website!
.
1.  From the right-angled triangle AFE  sin( < FAE) = .

    Hence, sin( < BAD) = sin( < FAE) = .


2.  From the right-angled triangle BHG  sin( < GBH) = .

    Hence,  sin( < CBD) = sin( < GBH) = .


3.  Angles CBD and ADB are, obviously, congruent.

    Therefore, sin( < ADB) = sin( < CBD) = .


4.  Thus in triangle ADB we know

    AD = 22,  sin( < BAD) =   and  sin( < ADB) = .


    Having sines of two angles at the base of the triangle ADB, we can find sin( < ABD):

    sin( < ABD) = sin(180 - < BAD - < ADB) = sin( < BAD + < ADB) = sin*cos + cos*sin =  = 

    =  = .


5.  Now apply the sine law theorem to find the side AB:

     =   ====>  |AB| = 22* =  =  =  = 10.4.



6.  Now calculate the area of the parallelogram ABCD by applying the formula

     = |AD|*|AB|*sin( < A) = 22*10.4* = 22*0.8*12 = 211.2.

Answer. The area of the parallelogram ABCD is 211.2 square units.



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