DAV Class 8 Maths Chapter 11 Brain Teasers
Understanding Quadrilaterals Brain Teasers
1. A. Tick (✓) the correct option.
(a)
The number of diagonals a pentagon has is— \(
\begin{aligned}
(i)&\ 2 \\
(ii)&\ 4 \\
(iii)&\ 5 \\
(iv)&\ 3
\end{aligned}
\)
Answer \( {\color{orange}(iii)}\ \color{red}{5} \)
(b)
A parallelogram having its adjacent sides equal is a— \(
\begin{aligned}
(i)&\ \text{trapezium}\\
(ii)&\ \text{rhombus}\\
(iii)&\ \text{triangle}\\
(iv)&\ \text{rectangle}
\end{aligned}
\)
Answer \( {\color{orange}(ii)}\ \color{red}{\text{rhombus}} \)
(c)
A seven sided polygon is called a— \(
\begin{aligned}
(i)&\ \text{pentagon}\\
(ii)&\ \text{octagon}\\
(iii)&\ \text{hexagon}\\
(iv)&\ \text{heptagon}
\end{aligned}
\)
Answer \( {\color{orange}(iv)}\ \color{red}{\text{heptagon}} \)
(d)
Regular polygons are— \(
\begin{aligned}
(i)&\ \text{equiangular only}\\
(ii)&\ \text{equilateral only}\\
(iii)&\ \text{equiangular and equilateral}\\
(iv)&\ \text{none of the options}
\end{aligned}
\)
Answer \( {\color{orange}(iii)}\ \color{red}{\text{equiangular and equilateral}} \)
(e)
Which of the given polygons have equal diagonals? \(
\begin{aligned}
(i)&\ \text{square}\\
(ii)&\ \text{rhombus}\\
(iii)&\ \text{parallelogram}\\
(iv)&\ \text{trapezium}
\end{aligned}
\)
Answer \( {\color{orange}(i)}\ \color{red}{\text{square}} \)
B. Answer the following questions.
(a) Name two polygons whose diagonals bisect each other at right angles.
Solution
- In a square, diagonals are equal and bisect each other at right angles.
- In a rhombus, diagonals bisect each other at right angles (though not equal in general).
Answer \( \color{red}{\text{Square and Rhombus}} \)
(b) In a parallelogram \(ABCD\), if \(\angle B = 75^\circ\), find the measure of \(\angle C\).
Solution
In a parallelogram, adjacent angles are supplementary
\[ \begin{aligned} \angle B + \angle C &= 180^\circ \\[4pt] 75^\circ + \angle C &= 180^\circ \\[4pt] \angle C &= 180^\circ - 75^\circ \\[4pt] \angle C &= 105^\circ \end{aligned} \]Answer \( \angle C = \color{red}{105^\circ} \)
(c) What is the measure of each exterior angle of a regular octagon?
Solution
\[ \begin{aligned} (\text{regular octagon}) \ n &= 8 \\[4pt] \text{Each exterior angle} &= \frac{360^\circ}{n} \\[4pt] &= \frac{360^\circ}{8} \\[4pt] &= 45^\circ \end{aligned} \]Answer Each exterior angle \(=\ \color{red}{45^\circ}\)
(d) Find the measure of \(x\) in the diagram.
Solution
Sum of measures of exterior angles of any polygon \( = 360^\circ \) \[ \begin{aligned} 110^\circ + 65^\circ + 80^\circ + x & = 360^\circ \\ 255^\circ + x & = 360^\circ \\ x & = 360^\circ - 255^\circ \\ x & = 105^\circ \\ \end{aligned} \]Answer \( x = \color{red}{105^\circ} \)
(e) Find each angle of a regular pentagon.
Solution
\[ \begin{aligned} n &= 5 \quad (\text{pentagon}) \\[4pt] \text{Each interior angle} &= \frac{(n-2)\times 180^\circ}{n} \\[4pt] &= \frac{(5-2)\times 180^\circ}{5} \\[4pt] &= \frac{3\times 180^\circ}{5} \\[4pt] &= \frac{540^\circ}{5} \\[4pt] &= 108^\circ \end{aligned} \]Answer Each interior angle of a regular pentagon \( = \color{red}{108^\circ}\)
2.
\(ABCD\) is a parallelogram. \(\overline{AP}\) bisects \(\angle A\) and \(\overline{CQ}\) bisects \(\angle C\).
\(P\) lies on \(\overline{CD}\) and \(Q\) lies on \(\overline{AB}\). Show that—
(i) \(\overline{AP} \parallel \overline{CQ}\)
(ii) \(AQCP\) is a parallelogram.
Solution
(i) To prove \(\overline{AP} \parallel \overline{CQ}\)
\[ \begin{aligned} &ABCD \text{ is a parallelogram} \\[6pt] AP &\text{ bisects } \angle A \\ &\angle 1 = \angle 2 \implies \frac{1}{2}\angle A \\[4pt] CQ &\text{ bisects } \angle C \\ &\angle 3 = \angle 4 \implies \frac{1}{2}\angle C \\[4pt] \Rightarrow\ &\angle A = \angle C \ \color{magenta}{\text{(opposite angles of a parallelogram)}} \\[4pt] &\because \frac{1}{2}\angle A = \frac{1}{2}\angle C \\[6pt] & AB \parallel CD \ , \ CQ \text{ is transversal} \\[6pt] &\angle 4 = \angle 5 \ \color{magenta}{\text{(Alternate interior angles)}} \\[4pt] \Rightarrow \ &\angle 1 = \angle 5 \ \color{magenta}{\text{(Corresponding angles)}} \\[4pt] & \therefore\ \color{red} \overline{AP} \parallel \overline{CQ} \end{aligned} \](ii) To prove \(AQCP\) is a parallelogram
\[ \begin{aligned} ABCD & \text{ is a parallelogram} \\ \overline{AB} & \parallel \overline{CD} \\[6pt] Q \text{ lies on } \overline{AB} & \text{ and } P \text{ lies on } \overline{CD} \\ \therefore\ \overline{AQ} &\parallel \overline{CP} \\[6pt] \overline{AP} & \parallel \overline{CQ} \ \textbf{ from part (i)} \\[6pt] \text{Opposite } & \text{sides are parallel} \\[6pt] \therefore\ AQCP & \text{ is a parallelogram} \end{aligned} \]3.
In the figure 11.34, \(ABCD\) is a quadrilateral in which \(\overline{AB} = \overline{AD}\) and \(\overline{BC} = \overline{DC}\).
Diagonals \(\overline{AC}\) and \(\overline{BD}\) intersect each other at \(O\). Show that—
(i) \(\triangle ABC \cong \triangle ADC\)
(ii) \(\triangle AOB \cong \triangle AOD\)
(iii) \(\overline{AC} \perp \overline{BD}\)
(iv) \(\overline{AC}\) bisects \(\overline{BD}\).
Solution
(i) To prove: \( \color{red} \triangle ABC \cong \triangle ADC\)
\[ \begin{aligned} In \ \triangle ABC & \text{ and } \triangle ADC \\[4pt] AB &= AD \ \color{magenta}{\text{(given)}}\\[4pt] BC &= DC \ \color{magenta}{\text{(given)}}\\[4pt] AC &= AC \ \color{magenta}{\text{(common side)}}\\[4pt] \text{By } SSS & \text{ congruence} \\ \therefore\ \color{green} \triangle ABC \ & \ \color{green} \cong \triangle ADC \end{aligned} \](ii) To prove: \( \color{red} \triangle AOB \cong \triangle AOD\)
\[ \begin{aligned} In \ \triangle AOB & \text{ and } \triangle AOD \\[4pt] AB &= AD \ \color{magenta}{\text{(given)}}\\[4pt] \angle 1 &= \angle 2 \ \color{magenta}{\text{(CPCT from (i))}} \\[4pt] AO &= AO \ \color{magenta}{\text{(common side)}}\\[4pt] \text{By } SAS & \text{ congruence} \\[4pt] \therefore\ \triangle AOB &\cong \triangle AOD \end{aligned} \](iii) To prove: \( \color{red} \overline{AC} \perp \overline{BD}\)
\[ \begin{aligned} \angle 3 &= \angle 4 \ \color{magenta}{\text{(CPCT from (ii))}} \\ \angle 3 + \angle 4 &= 180^\circ \ \color{magenta}{\text{(linear pair)}}\\[4pt] \therefore\ 2\angle 4 &= 180^\circ \\ \angle 4 &= 90^\circ \\ \angle 3 &= 90^\circ \\[6pt] \therefore\ \color{red}\overline{AC} & \color{red} \perp \overline{BD} \end{aligned} \](iv) To prove: \( \color{red} \overline{AC}\) bisects \( \color{red} \overline{BD}\)
\[ \begin{aligned} OB &= OD \ \color{magenta}{\text{(CPCT from (ii))}} \\ \color{red}\overline{AC} & \color{red} \text{ bisects } \overline{BD} \end{aligned} \]4. \(ABCD\) is a quadrilateral in which all four sides are equal. Show that both pairs of opposite sides are parallel.
Solution
\[ \begin{aligned} \color{magenta}{\textbf{Given:}} &\ \text{All four sides are equal} \\[4pt] \therefore\ &ABCD \text{ is a rhombus} \\[4pt] \color{magenta}{\textbf{Since}} &\ \text{every rhombus is a parallelogram} \\[4pt] \therefore\ &ABCD \text{ is a parallelogram} \\[4pt] &AB \parallel DC \text{ and } AD \parallel BC \\[4pt] \therefore\ &\text{Both pairs of opposite sides are parallel}\end{aligned} \]5. In a quadrilateral \(ABCD\), \(\angle A + \angle D = 180^\circ\). Does this mean \(\overline{AB} \parallel \overline{DC}\)? Why? What special name does this quadrilateral have?
Solution
\[ \begin{aligned} \angle A + \angle D &= 180^\circ \ \color{magenta}{\text{(given)}}\\[4pt] \angle A \text{ and } \angle D &\text{ are co-interior angles which are supplementary} \\[4pt] \text{Yes, } & \color{red}{\overline{AB} \parallel \overline{DC}} \\[4pt] \therefore\ ABCD & \text{ is a } \color{red} trapezium \end{aligned} \]