Question

In the adjoining figure $AB=12cm$, $CD=8cm$, $BD=20cm$; $\angle ABD=\angle AEC=\angle EDC={90}^o$. If $BE=x$, then:

• A. $x$ has two possible values whose difference is $4$
• B. $x$ has two possible values whose sum is $28$
• C. $x$ has only one value and $x\ge 12$
• D. $x$ cannot be determined with the given information

Answer 1

Answer: (A)

$x$ has two possible values whose difference is $4$

In $\Delta AEC$ ${AC}^2={AE}^2+{EC}^2$

In $\Delta ABE,{AE}^2={AB}^2+{BE}^2$

In $\Delta DEC,{EC}^2={ED}^2+{CD}^2$

$\therefore {AC}^2={AB}^2+{BE}^2+{ED}^2+{CD}^2$

${AC}^2=144+x^2+{(20-x)}^2+\mathrm{64....}\left(i\right)$

Also from the figure

${AC}^2={(20-0)}^2+{(8-12)}^2{AC}^2={\left(20\right)}^2+{\left(4\right)}^2.....\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$

$144+x^2+{(20-x)}^2+64={\left(20\right)}^2+{\left(4\right)}^{2}2x^2-40x+192=0x^2-20x+96=0(x-12)(x-8)=0x=12,8$

The difference of the two values is $4$.

Hence, option A is correct.