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

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Question

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

(b) When \(n = 8\)

Then Angle sum of a polygon \(= (n - 2) \times {180^{ \rm{ \circ}}} = (8 - 2) \times {180^{\rm{\circ}}} = 6 \times{180^{\rm{\circ}}} = {1080^{\rm{\circ}}}\)

(c) When \(n = 10\)

Then Angle sum of a polygon \( = (n - 2) \times {180^{\rm{\circ}}} = (10 - 2) \times {180^{\rm{\circ}}} = 8 \times {180^{\rm{\circ}}} = {1440^{\rm{\circ}}}\)

(d) When \(n = n\)

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

  
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