Ex.3.1 Q4 Understanding Quadrilaterals Solution - NCERT Maths Class 8

Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)

FIGURES
SIDE	3	4	5	6
ANGLE SUM	\(180\)	\(\begin{align} 2 \times {180^{\rm{o}}} & = (4 - 2) \times {180^{\rm{o}}}\\  &= 360^\circ \end{align}\)	\(\begin{align} 3 \times {180^{\rm{o}}}\\  &  = (5 - 2) \times {180^{\rm{o}}}\\  &  = 540 \end{align}\)	\(\begin{align} 4 \times {180^{\rm{o}}}\\  &  = (6 - 2) \times {180^{\rm{o}}}\\  &  = {720^{\rm{o}}} \end{align}\)

What can you say about the angle sum of a convex polygon with number of sides?

(a) \(7\)

(b) \(8\)

(c) \(10\)

(d) \(n\)

Text Solution

From the table, it can be observed that the angle sum of a convex polygon of \(n\) sides is \(\left( {n - 2} \right){\rm{ }} \times {\rm{ }}180^\circ\)

(a) When \(n = 7\)

Then Angle sum of a polygon 

\(\begin{align} &= (n - 2) \times {180^{\rm{\circ}}} \\&= (7 - 2) \times {180^{\rm{\circ}}} \\&= 5 \times {180^{\rm{\circ}}} \\&= {900^{\rm{\circ}}}\end{align}\)

(b) When \(n = 8\)

Then Angle sum of a polygon 

\(\begin{align} &= (n - 2) \times {180^{ \rm{ \circ}}} \\&= (8 - 2) \times {180^{\rm{\circ}}} \\&= 6 \times{180^{\rm{\circ}}} \\&= {1080^{\rm{\circ}}}\end{align}\)

(c) When \(n = 10\)

Then Angle sum of a polygon

\(\begin{align} &= (n - 2) \times {180^{\rm{\circ}}} \\&= (10 - 2) \times {180^{\rm{\circ}}} \\&= 8 \times {180^{\rm{\circ}}} \\&= {1440^{\rm{\circ}}}\end{align}\)

(d) When \(n = n\)

Then Angle sum of a polygon \( = \left( {n - {\rm{ }}2} \right) \times 180^\circ \)