Question

In the given figure, find $a+b+c+d$.

• A. ${360}^{\circ }$
• B. ${100}^{\circ }$
• C. ${210}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

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

In the quadrilateral PQRS,
$\angle P+\angle Q+\angle R+\angle S={360}^{\circ }$
$\Rightarrow {120}^{\circ }+{80}^{\circ }+{60}^{\circ }+\angle S={360}^{\circ }$
$\Rightarrow \angle S={100}^{\circ }$

$\angle P+ext\angle P={180}^{\circ }$ [Linear Pair]
$\Rightarrow {120}^{\circ }+a={180}^{\circ }$
$\Rightarrow a={60}^{\circ }$
Similarly,

$\angle Q+ext\angle Q={180}^{\circ }$ [Linear Pair]
$\Rightarrow {80}^{\circ }+b={180}^{\circ }$
$\Rightarrow a={100}^{\circ }$

And,
$\angle R+ext\angle R={180}^{\circ }$ [Linear Pair]
$\Rightarrow {60}^{\circ }+c={180}^{\circ }$
$\Rightarrow a={120}^{\circ }$

$\angle S+ext\angle S={180}^{\circ }$ [Linear Pair]
$\Rightarrow {100}^{\circ }+d={180}^{\circ }$
$\Rightarrow d={80}^{\circ }$

$\Rightarrow a+b+c+d={60}^{\circ }+{100}^{\circ }+{120}^{\circ }+{80}^{\circ }$

$\Rightarrow a+b+c+d={360}^{\circ }$