Question

In the diagram $ABCD$ is a rectangle and point $E$ lies on $AB$. Triangle $DEC$ has $\angle DEC={90}^{\circ }$ $DE=3$ and $EC=4$. The length of $AD$ is

• A. $2.4$
• B. $2.8$
• C. $1.8$
• D. $3.2$

Answer 1

The correct option is A $2.4$

In triangle $DEC$, $\angle E={90}^0$

Then $(DC{)}^2=(DE{)}^2+(EC{)}^2$

$\Rightarrow (DC{)}^2=(3{)}^2+(4{)}^2=9+16=25$ .........(1)

$DC=5$

In triangle $EAD$, $\angle A={90}^0$ and let $AE=x$

Then $(AD{)}^2=(DE{)}^2-(Ae{)}^2$

$\Rightarrow (AD{)}^2=(3{)}^2-(x{)}^2=9-x^2$

In triangle $EBC$, $\angle B={90}^0$ and let $EB=5-x$

Then $(BC{)}^2=(AD{)}^2=(EC{)}^2-(EB{)}^2$

$\Rightarrow (AD{)}^2=(4{)}^2-(5-x{)}^2$

$=(AD{)}^2=16-25+10x-x^2=-9+10x-x^2$ ..........(2)

From (1) and (2), we get

$-9+10x-x^2=9-x^2\Rightarrow -9+10x=9\Rightarrow 10x=18\Rightarrow x=1.8$

Put $x=1.8$ in (1), then

$(AD{)}^2=9-(1.8{)}^2=9-3.24=5.76$

$AD=2.4$