Question

In the given figure, $AC+AB=5AD$ and $AC-AD=8$. Then find the area of the rectangle $ABCD$?

• A. $50sq.units$
• B. $60sq.units$
• C. $70sq.units$
• D. $80sq.units$

Answer 1

The correct option is B $60sq.units$

$AB=x=CD,$

$AD=y=BC,$

$AC=z$

By Pythagoras theorem

$z^2=x^2+y^2$ -------(1)

Given

$AC+AB=5AD$

$2+x=5y$

$z=5y-x$ -------(2)

squaring on both sides

$z^2=25y^2+x^2-10xy$ --------(3)

subtract equation (1) from (3)

we get $z^2=25y^2+x^2-10xy$

$\underset{–}{-z^2=-y^2-x^2-0}$

$24y^2-10xy=0$

$\frac{x}{y}=\frac{12}{5}$

$AC-AD=8$

$z-y=8$

$z=8+5$

$z=13$

Area $=12\times 5=60sq.units$