$ \mathrm{XYZW} $ is a rectangle. If $ \mathrm{XY}+\mathrm{YZ}=17 $ and $ \mathrm{XZ}+\mathrm{YW}=26 $, find $ \mathrm{XY} $ and $ \mathrm{YZ} .(\mathrm{XY}>\mathrm{YZ}) $.

Given:

\( \mathrm{XYZW} \) is a rectangle.

\( \mathrm{XY}+\mathrm{YZ}=17 \) and \( \mathrm{XZ}+\mathrm{YW}=26 \).

To do:

We have to find \( \mathrm{XY} \) and \( \mathrm{YZ}.

Solution:

We know that,

Diagonals of a rectangle are equal.

This implies,

$XZ = YW$

$XZ = YW = \frac{26}{2} = 13\ cm$

In $∆XYZ$,

Let $YZ = k$, this implies,

$XY = (17 - k)$

Using Pythagoras theorem,

$13^2 = (17 - k)^2 + k^2$

$169=289 - 34k + k^2+k^2$

$2k^2 - 34k + 289 - 169 = 0$

$2k^2 - 34k + 120 = 0$

$k^2 - 17k + 60 = 0$

$k^2 - 12k - 5k + 60 = 0$

$k(k-12)-5(k-12)=0$

$(k-12)(k-5)=0$

$k = 12$ or $k = 5$

Given that $XY>YZ$

$\Rightarrow YZ=k=5\ cm$

$\Rightarrow XY=17-YZ=17-5=12\ cm$