Ex.3.1 Q6 Understanding Quadrilaterals Solution - NCERT Maths Class 8

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Question

Find the angle measure \(x\) in the following figures:

Text Solution

What is Known?

Sum of the measures of all interior angles of a quadrilateral is \( 360^\circ\) and that of a pentagon is \( 540^\circ\)

What is Unknown?

Angle \(x\) in the above figures  \(a, b, c\) and \(d\)

Steps:

a) The above figure has \(4\) sides and hence it is a quadrilateral.

Using the angle sum property of a quadrilateral,

\[\begin{align}{{50}^{\rm{o}}} + {{130}^{\rm{o}}} + {{120}^{\rm{o}}} + x &= {{360}^{\rm{o}}}\\
{{300}^{\rm{o}}} + x &= {{360}^{\rm{o}}}\\x & = {{360}^{\rm{o}}} - {{300}^{\rm{o}}}\\x &= {{60}^{\rm{o}}}
\end{align}\]

b) Using the angle sum property of a quadrilateral.

\[\begin{align}{{90}^\circ } + {{60}^\circ } + {{70}^\circ } + x& = {{360}^\circ }\\
{{220}^\circ } + x &= {{360}^\circ }\\{220^\circ } + x &= {360^\circ }\\
x& = {360^\circ } - {220^\circ }\\x &= {140^\circ }\end{align}\]

c) The given figure is a pentagon \((n=5)\)

\[\begin{align}{\text{Angle sum of a polygon }}& = \,\,({\rm{n} - 2) \times {{180}^{\rm{o}}}}\\{} &= \,\,(5 - 2) \times {{180}^{\rm{o}}}\\{} & = \,\,3 \times {{180}^{\rm{o}}}\\{} & = \,\,{540}^{\rm{o}}\end{align}\,\]

Sum of the interior angle of pentagon is \(540^\circ\)

Angles at the bottom are linear pair.

\(\begin{align}{\therefore {\text{ First base interior angle i}}{\rm{.e}}{\rm{. a}}}\;&= {{{180}^{\rm{o}}} - {{70}^{\rm{o}}}{\text{ (angle of straight line is }}{180}^{\rm{o}})}\\{} & = \,\,{110}^{\rm{o}}\end{align}\)

\(\begin{align}{\therefore {\text{ Second base interior angle i}}{\rm{.e}}{\rm{. b}}} & = {{180}^{\rm{o}} - {{60}^{\rm{o}}}}\\& = {{120}^{\rm{o}}}\end{align}\)

\[\begin{align}{\text{Angle sum of a polygon}} &= ({\rm{n}} - 2) \times {180^o}\\{\rm{i}}{\rm{.e}}{\rm{.}}{30^o} + x + {110^o} + {120^o} + x &= {540^o}\\{\rm{i}}{\rm{.e}}{\rm{.}}2x + {260^{\rm{o}}}\, &= {540^{\rm{o}}}\\{\rm{i}}{\rm{.e}}{\rm{.}}2x &= {540^{\rm{o}}} - {260^{\rm{o}}}\\{\rm{i}}{\rm{.e}}{\rm{.}}2x &= {280^{\rm{o}}}\\{\rm{i}}{\rm{.e}}{\rm{.}}x &= \frac{{{{280}^{\rm{o}}}}}{2}\\{\rm{i}}{\rm{.e}}{\rm{.}}x &= {140^{\rm{o}}}\end{align}\]

D) The given figure is pentagon \((n-5)\)

\(\begin{align}{{\text{ Angle sum of a polygon }}}& = ({\rm{n}} - 2) \times {{180}^{\rm{o}}}\\& = (5 - 2) \times {{180}^{\rm{o}}}\\& = 3 \times {{180}^{\rm{o}}}\\& = {{540}^{\rm{o}}}\end{align}\)

Sum of the interior angle of pentagon is \(540^\circ\)

\(\begin{align}{{\text{ Angle sum of a polygon }}} &= {x + x + x + x + x = {{540}^{\rm{o}}}}\\{\rm{ i}}{\rm{.e}}{\rm{. }}\ \qquad 5x &= {{540}^{\rm{o}}}\\x &= {\frac{{{{540}^{\rm{o}}}}}{5}}\\x & = {{108}^{\rm{o}}}\end{align}\)

A pentagon is a regular polygon i.e. it is both ‘equilateral’ and ‘equiangular’. Thus, the measure of each interior angle of the pentagon are equal.

Hence each interior angle is \(108^\circ .\)

  
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