# Parallelogram Help?

#### Storminnorman

A parallelogram exists within a rectangle which measures 15 inches tall by 20.5 inches wide.

A=15 inches
B=20.5 inches
C=1.5 inches
(C makes a 90° angle with X)

Solve for lengths X and Y and angle Z of the parallelogram and please tell me how you did it!

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#### skeeter

MHF Helper
Area of rectangle = 2(area of triangle) + area of parallelogram

$AB=A(B-Y)+CX$

$AB=AB-AY+CX \implies AY=CX \implies 15Y=1.5X \implies X=10Y$

Using Pythagoras ...

$A^2+(B-Y)^2=X^2$

$15^2+(20.5-Y)^2=(10Y)^2$

Solve the above quadratic equation for $Y$, then you can determine $X$. Finally, you can determine $\angle Z$ using the value of $Y$ and $C$ with an appropriate trig ratio.

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#### DenisB

No need for 3 equations, Mr Dutt.

You can also solve it using larger triangle:
x^2 = a^2 + (b - y)^2 [1]
then similar triangles (as Skeeter did):
c/y = a/x : x = ay / c [2]

[1][2]: (ay / c)^2 = a^2 + (b - y)^2
Solve for y

Hate to admit it, but Skeeter's way is a bit shorter

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