Question

In the figure $\angle P+\angle Q+\angle R+\angle S+\angle T=2$ right angles

• A. True
• B. False

Answer 1

The correct option is A True

Sum of angles on a straight line=${180}^{\circ }$

$⟹\angle PAE={180}^{\circ }-\angle 1$

$Similarly,\angle PEA={180}^{\circ }-\angle 5$

$In△PAE$

$\angle PAE+\angle PEA+\angle APE={180}^{\circ }$

$⟹180-\angle 1+180-\angle 5+\angle APE={180}^{\circ }$

$⟹\angle APE=\angle 1+\angle 5-{180}^{\circ }=\angle P$

$Similarly,\angle BSC=\angle 2+\angle 3-{180}^{\circ }=\angle S$

$\angle DRC=\angle 3+\angle 4-{180}^{\circ }=\angle R$

$\angle DQE=\angle 4+\angle \angle 5-{180}^{\circ }=\angle Q$

$\angle ATS=\angle 1+\angle 2-{180}^{\circ }=\angle T$

$⟹\angle P+\angle Q+\angle R+\angle S+\angle T$

$=\angle 1+\angle 2+\angle 3+\angle 4+\angle 2+\angle 3+\angle 4+\angle 5+\angle 1+\angle 5-(180\times 5)$

$=2(\angle 1+\angle 2+\angle 3+\angle 4+\angle 5)-900-\left(1\right)$

From pentagon ABCDE,

Sum of angles of pentagon=$\left(\right(2\times 5)-4)\times {90}^{\circ }$

($\because$ Sum of angles=$(2n-4)\times {90}^{\circ }$)

$=6\times 90={540}^{\circ }$

$⟹\angle P+\angle Q+\angle R+\angle S+\angle T=2\left({540}^{\circ }\right)-{900}^{\circ }$

$={1080}^{\circ }-{900}^{\circ }$

$={180}^{\circ }$

$=2\times 90$

$=2Rightangles$

$A)True$