Question

In the adjoining figure $AB=12cm,CD=8cm,BD=20cm$; $\angle ABD=\angle AEC=\angle EDC={90}^{\circ }$. 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 $4$

Given

$\angle ABD$ $=$ $\angle AEC$ $=$ $\angle EDC$ $=$ $90°$

$BE$ $=$ $x$

Applying pythagoras theorem,

In $△ABE$,

${AB}^2$ $+$ ${BE}^2$ $=$ ${AE}^2$

${AE}^2$ $=$ ${12}^2$ $+$ $x^2$

In $△CDE$,

${CE}^2$ $=$ ${CD}^2$ $+$ ${DE}^2$

${CE}^2$ $=$ $8^2$ $+$ ${(20-x)}^2$

Now, Draw a line through ${}^{\prime }C"$ parallel to $BD$ such that it cuts $AB$ at ${}^{\prime }{F}^{\prime }$ and $CF$ $=$ $BD$ $=$ $20$ and $AF$ $=$ $4$,

Join the points $A$ and $C$ by a line

In $△AFC$,

${AC}^2$ $=$ ${AF}^2$ $+$ ${CF}^2$

${AC}^2$ $=$ $4^2$ $+$ ${20}^2$

${AC}^2$ $=$ $416$

In $△ACE$,

${AC}^2$ $=$ ${AE}^2$ $+$ ${CE}^2$

$416$ $=$ ${12}^2$ $+$ $x^2$ $+$ $8^2$ $+$ ${(20-x)}^2$

$2x^2$ $-$ $40x$ $+$ $192$ $=$ $0$

$x^2$ $-$ $20x$ $+$ $96$ $=$ $0$

$x^2$ $-$ $12x$ $-$ $8x$ $+$ $12\times 8$ $=$ $0$

$x(x$ $-$ $12)$ $-8(x$ $-$ $12)$ $=$ $0$

$(x$ $-$ $12)$ $(x$ $-$ $8)$ $=$ $0$

$x=12,x=8$

Therefore, ${}^{\prime }{x}^{\prime }$ has two possible values whose difference $(12$ $-$ $8)$ is ${}^{\prime }{4}^{\prime }$.