Question

$\square XYZW$ is the rectangle. If $XY+YZ=17$ and $XZ+YW=26$ then find $XY$ and $YZ$ (where $XY>YZ$)

Answer 1

In a rectangle WXYZ ,

Given that : -

$XY+YZ=17cm$

$XZ+YW=26cm$

To calculate : - Length and breadth of the rectangle.

We know that,

Diagonals of a rectangle are equal.

So, $XZ=YW$

Then, $XZ=YW=26/2=13cm$

Now,

In $\Delta XYZ$,

Let $YZ=P$ , Then, $XY=(17-P)$.

Then, By Pythagoras theorem,

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$\left(P\right)²+(17-P)²=\left(13\right)²$

$=>P²+289-34P+P²=169$

$=>2P²-34P=169-289$

$=>2(P²-17P)=-120$

$=>P²-17P=-120/2$

$=>P²-17P=-60$

$=>P²-17P+60=0$

$=>P²-12P-5P+60=0$

$=>P(P-12)-5(P-12)=0$

$=>(P-12)(P-5)=0$

Now,

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$P-12=0$

$=>P=12cm$

Again,

$P-5=0$

$=>P=5cm$

Now,

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$YZ=P=12cm$ [Because , $YZ$ is the length of the rectangle ,so we will assign it the greatest value of $P$]

Again, $XY=(17-P)=(17-12)cm=5cm$ [Because , $XY$ is the breadth .]