Question

In the figure(not drawn to scale), ABCD is a rectangle, FA$=$FG, AGE and BGF are straight lines. Find $x+y$.

• A. ${145}^o$
• B. ${124}^o$
• C. ${210}^o$
• D. ${116}^o$

Answer 1

Answer: (B)

${124}^o$

$\Rightarrow$ In given figure ABCD is a rectangle, FA = FG, AGE and BGF are straight lines.

$\Rightarrow$ $\angle DAF={20}^{\circ }$ [Given]

$\Rightarrow$ $\angle ADF={90}^{\circ }$ [All four angle of rectangle is 90${}^{\circ }$]

$\Rightarrow$ So, $\angle DFA={180}^{\circ }-({20}^{\circ }+{90}^{\circ })={70}^{\circ }$

$\Rightarrow$ As DFC and AB are parallel then,

$\Rightarrow$ $\angle DFC=\angle FAB={70}^{\circ }$

$\Rightarrow$ DA and CEB are parallel

$\Rightarrow$ So, $\angle BEA=\angle EAD={74}^{\circ }$

$\Rightarrow$ So, $\angle FAG={70}^{\circ }-{20}^{\circ }={54}^{\circ }$

$\Rightarrow$ $\angle FAG=\angle FGA={54}^{\circ }$ [FAG is isosceles triangle.]

$\Rightarrow$ So, $x={180}^{\circ }-(54+54{)}^{\circ }$

$\Rightarrow$ $x={72}^{\circ }$

$\Rightarrow$ Again $\angle AGF=\angle BGE={54}^{\circ }$

$\Rightarrow$ So, $y={180}^{\circ }-{128}^{\circ }={52}^{\circ }$

$\Rightarrow$ $x+y={72}^{\circ }+{52}^{\circ }={124}^{\circ }$