DAV Class 8 Maths Chapter 11 HOTS
Understanding Quadrilaterals HOTS
1. Prove that the interior angle of a regular pentagon is three times the exterior angle of a regular decagon.
Solution
\[ \begin{aligned} \text{Regular pentagon } & (n=5) \\[6pt] \text{Each interior } & \text{angles of a Regular pentagon} \\[6pt] &= \frac{(n-2)\times 180^\circ}{n} \\[4pt] &= \frac{(5-2)\times 180^\circ}{5} \\[4pt] &= \frac{ 3 \times 180^\circ}{5} \\[4pt] &= \frac{540^\circ}{5} \\[4pt] &= 108^\circ \end{aligned} \]Each Interior angle of a regular pentagon \( \color{green} = 108^\circ \)
\[ \begin{aligned} \text{In a regular decagon } & (n=10) \\[6pt] \text{Exterior angle of any regular polygon} &= \frac{360^\circ}{n} \\[6pt] \text{Each exterior angle of decagon} &= \frac{360^\circ}{10} \\[6pt] &= 36^\circ \end{aligned} \]Each exterior angle of regular decagon \( \color{green} = 36^\circ \)
\[ \begin{aligned} 36^\circ \times 3 & = 108^\circ \\ Hence \ & \ Proved \end{aligned} \]Three times the exterior angle of a regular decagon is equal to interior angle of a regular pentagon.
2. In the diagram, \(ABCDE\) is a regular pentagon. If \(\angle EBD = 36^\circ\), find \(\angle 1\), \(\angle 2\) and \(\angle 3\).
Solution
Since \(ABCDE\) is a regular pentagon,\[ \begin{aligned} \text{Regular pentagon } & (n = 5) \\[6pt] \text{Each interior angle} &= \frac{(n-2)\times 180^\circ}{n} \\[6pt] &= \frac{(5-2)\times 180^\circ}{5} \\[6pt] & = \frac{540^\circ}{5} \\[6pt] & = 108^\circ \\[6pt] \therefore \ \color{green} \angle 1 & = \color{green} 108^\circ \\ \\ \triangle ABE & \ \text{ is an isosceles triangle}\\ AB & = AE \\ \therefore \ \angle 4 & = \angle 2 \\ \angle 1 + \angle 2 + \angle 4 & = 180^\circ \ \color{magenta} \text{ (Angle sum property)}\\ 108^\circ + \angle 2 + \angle 2 & = 180^\circ \\ 2 \angle 2 & = 180^\circ - 108^\circ \\ 2 \angle 2 & = 72^\circ \\[6pt] \angle 2 & = \frac{72^\circ}{2} \\[6pt] \color{green} \angle 2 & = \color{green} 36^\circ \\ \\ \text{Similarly in } & \triangle BCD \\ BC &= CD \\ \angle CBD &= \angle CDB \implies \angle 3 \\\angle C + \angle CBD + \angle CDB &= 180^\circ \ \color{magenta} \text{ (Angle sum property)} \\ 108^\circ + \angle 3 + \angle 3 &= 180^\circ \\ 108^\circ + 2\angle 3 &= 180^\circ \\ 2\angle 3 &= 72^\circ \\[6pt] \angle 3 & = \frac{72^\circ}{2} \\[6pt] \color{green} \angle 3 &= \color{green} 36^\circ \end{aligned} \]Answer \(\angle 1 = {\color{red}{108^\circ}}, \ \angle 2 = {\color{red}{36^\circ}} , \ \angle 3 = {\color{red}{36^\circ}} \)