Ex.3.1 Q4 Understanding Quadrilaterals Solution - NCERT Maths Class 8
Question
Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)
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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 \( = \left( {n - {\rm{ }}2} \right) \times 180^\circ \)