Question

A boy is designing a diamond shaped kite, as shown in the figure where $AE=16cm,EC=81cm$. He wants to use a straight cross bar $BD$. How long should it be?

• A. $27cm$
• B. $72cm$
• C. $48cm$
• D. $56cm$

Answer 1

The correct option is B $72cm$

Here $\angle B$ and $\angle D$ both are ${90}^o$

Now, in right angle $△AEB,$

$(AB{)}^2=(AE{)}^2+(BE{)}^2$ [ By Pythagoras theorem ]

$\Rightarrow$ $(AE{)}^2=(AB{)}^2-(BE{)}^2$

$\Rightarrow$ $(16{)}^2=(AB{)}^2-(BE{)}^2$ ---- ( 1 )

And in right angled $△BEC,$

$\Rightarrow$ $(BC{)}^2=(EC{)}^2+(BE{)}^2$ [ By Pythagoras theorem ]

$\Rightarrow$ $(EC{)}^2=(BC{)}^2-(BE{)}^2$

$\Rightarrow$ $(81{)}^2=(BC{)}^2-(BE{)}^2$ ----- ( 2 )

Adding equation ( 1 ) and ( 2 ),

$\Rightarrow$ $(16{)}^2+(81{)}^2=(AB{)}^2+(BC{)}^2-(BE{)}^2-(BE{)}^2$ ---- ( 3 )

Now in right angle $△ABC,$

$\Rightarrow$ $(AB{)}^2+(BC{)}^2=(AC{)}^2=(16+81{)}^2=(97{)}^2$

Putting this in equation $\left(3\right)$ we get,

$\Rightarrow$ $2(BE{)}^2=(97{)}^2-\left[\right(16{)}^2+\left(81{)}^2\right]$

$\Rightarrow$ $2(BE{)}^2=9409-(256+6561)$

$\Rightarrow$ $2(BE{)}^2=2592$

$\Rightarrow$ $(BE{)}^2=1296$

$\therefore$ $BE=36cm$

$\therefore$ $BD=2BE=2\times 36=72cm$