Question

Find a + b + c + d + e + f in the given figure if AOD, BOE and FOC are straight lines.

• A. ${360}^{\circ }$
• B. ${540}^{\circ }$
• C. ${420}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is A ${360}^{\circ }$

Here, $AOB,COD,EOF$ are triangles.

Then by angle sum property,

$\angle a+\angle b+\angle AOB={180}^o...\left(i\right)$,
$\angle c+\angle d+\angle COD={180}^o...\left(ii\right)$
and $\angle e+\angle f+\angle EOF={180}^o...\left(iii\right)$.

Also, $AOB,COD$ and $EOF$ are straight lines.

Then, $\angle EOF=\angle BOC$ ...[Vertically opposite angles].

So, $\angle AOB+\angle COD+\angle BOC(=\angle EOF)={180}^o...\left(iv\right)$ ...[Straight line].

Then, $\left(i\right),\left(ii\right)$ and $\left(iii\right)$,
$\angle a+\angle b+\angle AOB+\angle c+\angle d+\angle COD+\angle e+\angle f+\angle EOF={180}^o\times 3={540}^o$
$⟹$ $\angle a+\angle b+\angle c+\angle d+\angle e+\angle f+\angle AOB+\angle COD+\angle EOF={540}^o$ ....[Substitute $\left(iv\right)$]
$⟹$ $\angle a+\angle b+\angle c+\angle d+\angle e+\angle f+{180}^o={540}^o$
$⟹$ $\angle a+\angle b+\angle c+\angle d+\angle e+\angle f={360}^o$.

Therefore, option $A$ is correct.