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.

Algebra ->  Parallelograms -> 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.       Log On


   

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) About Me  (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
AF=sqrt%2813%5E2-12%5E2%29=sqrt%2825%29=5 .
Right triangle BGH, with a right angle at H,
has hypotenuse BG=17, leg GH=8, and leg BH that we can calculate as
BH=sqrt%2817%5E2-8%5E2%29=sqrt%28225%29=15 .

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 h, perpendicular to AD, we could calculate the area as h%2AAD .


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, AI%2FBI=AF%2FEF=5%2F12
BDI has angle BDI, congruent with angle GBH,
and that makes right triangles BDI similar to right triangle GBH.
So, ID%2FBI=BH%2FGH=15%2F8 .
adding AI%2FBI=5%2F12 and ID%2FBI=15%2F8 , we get
AI%2FBI%2BID%2FBI=5%2F12%2B15%2F8
%28AI%2BID%29%2FBI=55%2F24
AD%2FBI=55%2F24
22%2FBI=55%2F24
BI%2A55=22%2A24
BI=22%2A24%2F55=9.6
As BI-9.6 is the height of the parallelogram relative to base AD=22,
the area of the parallelogram is
BI%2AAD=9.6%2A22=highlight%28211.2%29

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
1.  From the right-angled triangle AFE  sin( < FAE) = 12%2F13.

    Hence, sin( < BAD) = sin( < FAE) = 12%2F13.


2.  From the right-angled triangle BHG  sin( < GBH) = 8%2F17.

    Hence,  sin( < CBD) = sin( < GBH) = 8%2F17.


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

    Therefore, sin( < ADB) = sin( < CBD) = 8%2F17.


4.  Thus in triangle ADB we know

    AD = 22,  sin( < BAD) = 12%2F13  and  sin( < ADB) = 8%2F17.


    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 =  = 

    = %2812%2F13%29%2A%2815%2F17%29+%2B+%285%2F13%29%2A%288%2F17%29 = 220%2F221.


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

    abs%28AB%29%2Fsin%28ADB%29 = abs%28AD%29%2Fsin%28ABD%29  ====>  |AB| = 22*%28%288%2F17%29%29%2F%28%28220%2F221%29%29 = %288%2A221%29%2F%2810%2A17%29 = %288%2A13%29%2F10 = 104%2F10 = 10.4.



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

    S%5BABCD%5D = |AD|*|AB|*sin( < A) = 22*10.4*%2812%2F13%29 = 22*0.8*12 = 211.2.

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