DAV Class 8 Maths Chapter 11 Practice Worksheet
Understanding Quadrilaterals Practice Worksheet
Section - A
1.
The polygon whose diagonals are equal and bisect each other at right angles is \(
\begin{aligned}
(a)\,& \text{Rhombus} \\
(b)\,& \text{Square} \\
(c)\,& \text{Rectangle} \\
(d)\,& \text{Parallelogram} \\
\end{aligned}
\)
Answer \(\color{orange}{(b)}\ \color{red}{\text{Square}}\)
2.
If \(ABCDEF\) is a regular hexagon, then each exterior angle will be \(
\begin{aligned}
(a)\,& 120^\circ \\
(b)\,& 60^\circ \\
(c)\,& 30^\circ \\
(d)\,& 72^\circ \\
\end{aligned}
\)
Solution
\[ \text{Exterior angle} = \frac{360^\circ}{n} \] \[ \begin{aligned} n &= 6 \ (\text{hexagon}) \\[4pt] \text{Exterior angle} &= \frac{360^\circ}{6} \implies 60^\circ \end{aligned} \]Answer \(\color{orange}{(b)}\ \color{red}{60^\circ}\)
3.
The measure of each exterior angle of a regular octagon is \(
\begin{aligned}
(a)\,& 55^\circ \\
(b)\,& 45^\circ \\
(c)\,& 40^\circ \\
(d)\,& 90^\circ \\
\end{aligned}
\)
Solution
\[ \text{Exterior angle} = \frac{360^\circ}{n} \] \[ \begin{aligned} n &= 8 \ (\text{octagon}) \\[4pt] \text{Exterior angle} &= \frac{360^\circ}{8} \implies 45^\circ \end{aligned} \]Answer \(\color{orange}{(b)}\ \color{red}{45^\circ}\)
4.
If the diagonals of a rectangle are \((2x + 1)\) and \((3x - 1)\), then the value of \(x\) is \(
\begin{aligned}
(a)\,& 1 \\
(b)\,& 2 \\
(c)\,& 3 \\
(d)\,& 4 \\
\end{aligned}
\)
Solution
\[ \begin{aligned} \text{In a rectangle, } & \text{the diagonals are equal}\\ 2x + 1 &= 3x - 1 \\[4pt] 1 + 1 &= 3x - 2x \\[4pt] 2 &= x \end{aligned} \]Answer \(\color{orange}{(b)}\ \color{red}{2}\)
5.
The exterior angle and interior angle of a regular polygon are in the ratio \(2 : 7\).
The number of sides in the polygon is \(
\begin{aligned}
(a)\,& 9 \\
(b)\,& 7 \\
(c)\,& 18 \\
(d)\,& 10 \\
\end{aligned}
\)
Solution
Let exterior angle \(= 2x \) and interior angle \(= 7x \).
\[ \begin{aligned} 2x + 7x &= 180^\circ \\[4pt] 9x &= 180^\circ \\[4pt] x &= 20^\circ \end{aligned} \] \[ \begin{aligned} \text{Exterior angle} &= 2x \implies 40^\circ \\[4pt] \text{Exterior angle} &= \frac{360^\circ}{n} \\[4pt] 40^\circ &= \frac{360^\circ}{n} \\[4pt] n &= \frac{360}{40} \\[4pt] n & = 9 \end{aligned} \]Answer \(\color{orange}{(a)}\ \color{red}{9\ \text{sides}}\)
6.
Two adjacent angles of a parallelogram are \((3x - 4)^\circ\) and \((3x + 10)^\circ\).
The angles of the parallelogram (in degrees) are \(
\begin{aligned}
(a)\,& 60^\circ,\ 40^\circ \\
(b)\,& 60^\circ,\ 120^\circ \\
(c)\,& 97^\circ,\ 83^\circ \\
(d)\,& 93^\circ,\ 87^\circ \\
\end{aligned}
\)
Solution
Adjacent angles of a parallelogram are supplementary.
\[ \begin{aligned} (3x - 4) + (3x + 10) &= 180^\circ \\[4pt] 6x + 6 &= 180^\circ \\[4pt] 6x &= 174^\circ \\[4pt] x &= 29^\circ \\ \\ (3x - 4) &= 3 \times 29 - 4 \\ &= 87 - 4 \\ &= 83^\circ \\[4pt] 3x + 10 &= 3 \times 29 + 10 \\ &= 87 + 10 \\ & = 97^\circ \end{aligned} \]Answer \(\color{orange}{(c)}\ \color{red}{97^\circ,\ 83^\circ}\)
7.
A parallelogram having its adjacent sides equal is a \(
\begin{aligned}
(a)\,& \text{Trapezium} \\
(b)\,& \text{Rhombus} \\
(c)\,& \text{Triangle} \\
(d)\,& \text{Rectangle} \\
\end{aligned}
\)
Answer \(\color{orange}{(b)}\ \color{red}{\text{Rhombus}}\)
8.
The sum of the measures of the interior angles of a pentagon is \(
\begin{aligned}
(a)\,& 180^\circ \\
(b)\,& 360^\circ \\
(c)\,& 270^\circ \\
(d)\,& 540^\circ \\
\end{aligned}
\)
Solution
\[ \begin{aligned} \text{Sum of interior angles } & = (n - 2)\times 180^\circ \\ n &= 5 \ (\text{pentagon}) \\[4pt] \text{Sum} &= (5 - 2)\times 180^\circ \\[4pt] &= 3 \times 180^\circ \\[4pt] &= 540^\circ \end{aligned} \]Answer \(\color{orange}{(d)}\ \color{red}{540^\circ}\)
9.
Riya : “A Rhombus is always a Parallelogram.” \(
\begin{aligned}
(A)\,& \text{Only Riya} \\
(B)\,& \text{Only Aryan} \\
(C)\,& \text{Both Riya and Aryan} \\
(D)\,& \text{Both are incorrect}
\end{aligned}
\)
Aryan : “A Parallelogram is always a Rhombus.” Who is correct?
Answer \(\color{orange}{(A)}\ \color{red}{\text{Only Riya}}\)
10.
If two adjacent angles of a parallelogram are in the ratio \(1 : 3\), then the measure of larger angle is \(
\begin{aligned}
(A)\,& 60^\circ \\
(B)\,& 45^\circ \\
(C)\,& 90^\circ \\
(D)\,& 135^\circ
\end{aligned}
\)
Solution
\[ \begin{aligned} \text{Let adjacent angles be }& x \text{ and } 3x\\[4pt] \text{Adjacent angles } & \text{are supplementary} \\ x + 3x &= 180^\circ \\[4pt] 4x &= 180^\circ \\[4pt] x &= \frac{180^\circ}{4} \\[4pt] x & = 45^\circ \\ \\ \text{Larger angle} &= 3x \\ &= 3 \times 45^\circ \\ &= 135^\circ \end{aligned} \]Answer \(\color{orange}{(D)}\ \color{red}{135^\circ}\)
11.
The quadrilateral in which diagonals bisect at right angle is a \(
\begin{aligned}
(A)\,& \text{Rectangle} \\
(B)\,& \text{Rhombus} \\
(C)\,& \text{Trapezium} \\
(D)\,& \text{Parallelogram}
\end{aligned}
\)
Answer \(\color{orange}{(B)}\ \color{red}{\text{Rhombus}}\)
12.
The angles of a quadrilateral are in the ratio \(1 : 2 : 3 : 4\). The smallest angle is \(
\begin{aligned}
(A)\,& 36^\circ \\
(B)\,& 72^\circ \\
(C)\,& 108^\circ \\
(D)\,& 144^\circ
\end{aligned}
\)
Solution
\[ \begin{aligned} x + 2x + 3x + 4x &= 360^\circ \\[4pt] 10x &= 360^\circ \\[8pt] x &= \frac{360^\circ}{10 } \\[8pt] x &= 36^\circ \end{aligned} \]Answer \(\color{orange}{(A)}\ \color{red}{36^\circ}\)
13.
The value of \(x\) in following diagram is \(
\begin{aligned}
(a)\,& 45^\circ \\
(b)\,& 90^\circ \\
(c)\,& 80^\circ \\
(d)\,& 85^\circ \\
\end{aligned}
\)
Solution
\[ \begin{aligned} \text{Sum of exterior angles of a polygon} &= 360^\circ \\[4pt] 90^\circ + 40^\circ + 90^\circ + 60^\circ + x^\circ &= 360^\circ \\[4pt] 280^\circ + x^\circ &= 360^\circ \\[4pt] x^\circ &= 360^\circ - 280^\circ \\[4pt] x^\circ &= 80^\circ \end{aligned} \]Answer \(\color{orange}{(c)}\ \color{red}{80^\circ}\)
14.
The angles of quadrilaterals are in the ratio \(1 : 2 : 3 : 4\). The smallest angle is \(
\begin{aligned}
(a)\,& 36^\circ \\
(b)\,& 72^\circ \\
(c)\,& 108^\circ \\
(d)\,& 144^\circ \\
\end{aligned}
\)
Solution
\[ \begin{aligned} x + 2x + 3x + 4x &= 360^\circ \\[4pt] 10x &= 360^\circ \\[4pt] x &= \frac{360^\circ}{10} \\[4pt] x &= 36^\circ \end{aligned} \]Answer \(\color{orange}{(a)}\ \color{red}{36^\circ}\)
15.
The following statements which is NOT true is \(
\begin{aligned}
(a)\,& \text{Opposite sides of parallelogram are equal} \\
(b)\,& \text{Diagonals of a rectangle are equal} \\
(c)\,& \text{Diagonals of a rhombus are equal} \\
(d)\,& \text{Each square is a rhombus} \\
\end{aligned}
\)
Answer \(\color{orange}{(c)}\ \color{red}{\text{Diagonals of a rhombus are equal}}\)
16.
The measure of \(x\) in the given diagram is \(
\begin{aligned}
(a)\,& 40^\circ \\
(b)\,& 50^\circ \\
(c)\,& 360^\circ \\
(d)\,& 310^\circ \\
\end{aligned}
\)
Solution
\[ \begin{aligned} \text{Sum of exterior angles of any polygon} &= 360^\circ \\[4pt] \Rightarrow 90^\circ + 60^\circ + 90^\circ + 70^\circ + x &= 360^\circ \\[4pt] 310^\circ + x &= 360^\circ \\[4pt] x &= 360^\circ - 310^\circ \\ x &= \color{green}{50^\circ} \end{aligned} \]Answer \(\color{orange}{(b)}\ \color{red}{50^\circ}\)
17.
In the given diagram, \(PQRS\) is a parallelogram with diagonals intersecting at \(O\).
If \(OP = (x + 6)\,\text{cm}\), \(OQ = 12\,\text{cm}\) and \(OS = 2y\,\text{cm}\), then the measure of \((x + y)\) is \(
\begin{aligned}
(a)\,& 8 \\
(b)\,& 14 \\
(c)\,& 12 \\
(d)\,& 10 \\
\end{aligned}
\)
Solution
In a parallelogram, diagonals bisect each other.
\[ \begin{aligned} OP &= OR \\ x + 6 &= 10 \\ x &= 10 - 6 \\ x &= 4 \\[8pt] OS &= OQ \\ 2y & = 12 \\[4pt] y & = \frac{12}{2} \\[4pt] y & = 6 \\[8pt] (x + y) &= 4 + 6 \\ (x + y) &= \color{green}{10} \end{aligned} \]Answer \(\color{orange}{(d)}\ \color{red}{10}\)
18.
\(\angle A\) and \(\angle B\) are two adjacent angles of a parallelogram.
If \(\angle A = 48^\circ\), then the measure of \(\angle B\) is \(
\begin{aligned}
(a)\,& 48^\circ \\
(b)\,& 96^\circ \\
(c)\,& 132^\circ \\
(d)\,& 144^\circ \\
\end{aligned}
\)
Solution
Adjacent angles of a parallelogram are supplementary.
\[ \begin{aligned} \angle A + \angle B &= 180^\circ \\[4pt] 48^\circ + \angle B &= 180^\circ \\[4pt] \angle B &= 180^\circ - 48^\circ \\[4pt] &= \color{green}{132^\circ} \end{aligned} \]Answer \(\color{orange}{(c)}\ \color{red}{132^\circ}\)
19.
Assertion (A): In a parallelogram, the two adjacent angles are \(40^\circ\) and \(130^\circ\). \(
\begin{aligned}
(a)\,& \text{Both A and R are true, and R is the correct explanation of A} \\
(b)\,& \text{Both A and R are true, but R is not the correct explanation of A} \\
(c)\,& \text{A is true, but R is false} \\
(d)\,& \text{A is false, but R is true} \\
\end{aligned}
\)
Reason (R): In a parallelogram, sum of two adjacent angles is \(180^\circ\).
Answer \(\color{orange}{(d)}\ \color{red}{\text{A is false, but R is true}}\)
20.
Assertion (A): A square is a regular polygon. \(
\begin{aligned}
(a)\,& \text{Both A and R are true, and R is the correct explanation of A} \\
(b)\,& \text{Both A and R are true, but R is not the correct explanation of A} \\
(c)\,& \text{A is true, but R is false} \\
(d)\,& \text{A is false, but R is true} \\
\end{aligned}
\)
Reason (R): A regular polygon has all its sides and angles equal.
Answer \(\color{orange}{(a)}\ \color{red}{\text{Both A and R are true, and R explains A}}\)
Section – B
1. The diagonals of a rhombus are in the ratio \(5:12\). If its perimeter is \(104\) cm, then find the lengths of the sides and the diagonals of the rhombus.
Solution
\[ \begin{aligned} \text{All the sides} & \text{ are equal in Rhombus} \\ \text{Perimeter} & =104 \ cm \\[6pt] 4 \times \text{Side} &= 104 \\[6pt] \text{Side} &= \frac{104}{4} \\[6pt] \text{Side} & =\color{green}{26\text{ cm}}\\[6pt] AB = BC &= CD = DA \implies \color{green}{26\text{ cm}}\\[6pt] \text{Let the } & \text{diagonals be} \\ BD & = 5x \\ AC & = 12x \\ \text{Diagonals of } & \text{rhombus bisect each other} \\ OB = OD & \implies \frac{5x}{2} \\[6pt] OA = OC & = \frac{12x}{2} \implies 6x \\[6pt] \text{In right } \triangle AOB & \text{, Using pythagoras theorem} \\[6pt] \color{magenta} (OB)^2 + (OA)^2 &= \color{magenta} (AB)^2 \\[6pt] \left(\frac{5x}{2} \right)^2 + (6x)^2 &= 26^2 \\[6pt] \frac{25x^2}{4} + 36x^2 &= 26 \times 26 \\[6pt] \frac{25x^2 + 144 x^2}{4} &= 26 \times 26 \\[6pt] \frac{169 x^2}{4} &= 26 \times 26 \\[6pt] x^2 &= \frac{\cancel{26}^2 \times \cancel{26}^2 \times 4}{\cancel{169}_{\cancel{13}_1}} \\[6pt] x^2 &= 16 \\[6pt] x &= 4 \ cm \\[6pt] &Diagonals \\ BD & = 5x \implies 20 \ cm \\ AC & = 12x \implies 48 \ cm \\ \end{aligned} \]Answer Side \(=\color{red}{26\ \text{cm}}\); Diagonals \(=\color{red}{20\ \text{cm and }48\ \text{cm}}\)
2. The diagonals of a rectangle \(ABCD\) intersect at \(O\). If \(\angle BOC = 70^\circ\), then find \(\angle ODA\).
Solution
\[ \begin{aligned} \angle BOC&=70^\circ \\ \angle AOD & = \angle BOC \ \color{magenta}\text{(vertically opposite angles)}\\ \angle AOD & = 70^\circ \\ \text{In } &\triangle AOD \\ OA & = OD \begin{cases} \color{magenta} \text{Diagonals of rectangle} \\ \color{magenta} \text{bisect each other}\\ \end{cases}\\[6pt] \therefore\ \angle OAD & = \angle ODA \begin{cases} \color{magenta} \text{Angles opposite to} \\ \color{magenta} \text{equal sides of a}\\ \color{magenta} \text{triangle are also equal}\\ \end{cases}\\[6pt] \triangle AOD & \text{ is an isosceles triangle} \\ \angle OAD + \angle ODA + \angle AOD & = 180^\circ \ \color{magenta} \text{(Angle sum property)} \\ x + x + 70^\circ & = 180^\circ \\ 2x & = 180^\circ - 70^\circ \\ 2x & = 110^\circ \\[6pt] x & = \frac{110^\circ}{2} \\[6pt] x & = 55^\circ \\ \angle ODA & = 55^\circ \end{aligned} \]Answer \(\angle ODA = \color{red}{55^\circ}\)
3.
In the given diagram, \(ABCD\) is a quadrilateral in which \(AB = AD\) and \(BC = DC\). Diagonals \(AC\) and \(BD\) intersect each other at \(O\). Show that:
(i) \(\triangle ABC \cong \triangle ADC\)
(ii) \(\triangle AOB \cong \triangle AOD\)
(iii) \(AC \perp BD\)
(iv) \(AC\) bisects \(BD\).
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: \( \color{red} AC \perp BD\)
\[ \begin{aligned} \angle 3 &= \angle 4 \ \color{magenta}{\text{(CPCT from (ii))}} \\ \angle 3 + \angle 4 &= 180^\circ \ \color{magenta}{\text{(linear pair)}}\\[4pt] \therefore\ 2\angle 4 &= 180^\circ \\ \angle 4 &= 90^\circ \\ \angle 3 &= 90^\circ \\[6pt] \therefore\ \color{red}{AC} & \color{red} \perp {BD} \end{aligned} \](iv) To prove: \( \color{red} AC \) bisects \( \color{red} BD \)
\[ \begin{aligned} OB &= OD \ \color{magenta}{\text{(CPCT from (ii))}} \\ \color{red}{AC} & \color{red} \text{ bisects } {BD} \end{aligned} \]4. Two adjacent sides of a parallelogram are in the ratio \(3:5\) and its perimeter is \(160\) cm. Find the sides of the parallelogram.
Solution
\[ \begin{aligned} \text{Let the adjacent sides be } 3x \text{ and } 5x \\ \color{magenta}\text{Opposite sides of a parallelogram are equal} \\ \end{aligned} \]\[ \begin{aligned} AD = BC &= 3x \\[6pt] AB = CD &= 5x \\[6pt] Perimeter &= 160 \\[6pt] 2(3x + 5x) & = 160 \\[6pt] 2(8x) & = 160 \\[6pt] 16x & = 160 \\[6pt] x &= \frac{160}{16} \\[6pt] x &= 10 \ cm \end{aligned} \]\[ \begin{aligned} 3x &= 3 \times 10 \implies \color{green}{30\ \text{cm}} \\ 5x &= 5 \times 10 \implies \color{green}{50\ \text{cm}} \end{aligned} \]Answer The sides of the parallelogram are \( \color{red}{30\ \text{cm and }50\ \text{cm}}\)
5.
\(ABCD\) is a parallelogram. \(AP\) bisects \(\angle A\) and \(CQ\) bisects \(\angle C\). \(P\) lies on \(CD\) and \(Q\) lies on \(AB\). Show that:
(i) \(AP \parallel CQ\) (ii) \(AQCP\) is a parallelogram.
Solution
(i) To prove \(AP \parallel CQ\)
\[ \begin{aligned} &ABCD \text{ is a parallelogram} \\[6pt] AP &\text{ bisects } \angle A \\ &\angle 1 = \angle 2 \implies \frac{1}{2}\angle A \\[4pt] CQ &\text{ bisects } \angle C \\ &\angle 3 = \angle 4 \implies \frac{1}{2}\angle C \\[4pt] \Rightarrow\ &\angle A = \angle C \ \color{magenta}{\text{(opposite angles of a parallelogram)}} \\[4pt] &\because \frac{1}{2}\angle A = \frac{1}{2}\angle C \\[6pt] & AB \parallel CD \ , \ CQ \text{ is transversal} \\[6pt] &\angle 4 = \angle 5 \ \color{magenta}{\text{(Alternate interior angles)}} \\[4pt] \Rightarrow \ &\angle 1 = \angle 5 \ \color{magenta}{\text{(Corresponding angles)}} \\[4pt] & \therefore\ \color{red} AP \parallel CQ \end{aligned} \](ii) To prove \(AQCP\) is a parallelogram
\[ \begin{aligned} ABCD & \text{ is a parallelogram} \\ \overline{AB} & \parallel \overline{CD} \\[6pt] Q \text{ lies on } \overline{AB} & \text{ and } P \text{ lies on } \overline{CD} \\ \therefore\ \overline{AQ} &\parallel \overline{CP} \\[6pt] \overline{AP} & \parallel \overline{CQ} \ \textbf{ from part (i)} \\[6pt] \text{Opposite } & \text{sides are parallel} \\[6pt] \therefore\ AQCP & \text{ is a parallelogram} \end{aligned} \]6. Two adjacent sides \(AB\) and \(BC\) of a parallelogram \(ABCD\) are in the ratio \(5:3\). If its perimeter is \(200\) cm, find the lengths of the sides \(AB\) and \(BC\).
Solution
\[ \begin{aligned} \text{Let the adjacent sides be } 3x \text{ and } 5x \\ \color{magenta}\text{Opposite sides of a parallelogram are equal} \\ \end{aligned} \]\[ \begin{aligned} AD = BC &= 3x \\[6pt] AB = CD &= 5x \\[6pt] Perimeter &= 200 \\[6pt] 2(3x + 5x) & = 200 \\[6pt] 2(8x) & = 200 \\[6pt] 16x & = 200 \\[6pt] x &= \frac{200}{16} \\[6pt] x &= 12.5 \ cm \\[8pt] AB &= 5x \\ &= 5 \times 12.5 \\ AB &= \color{green}{62.5 \ \text{cm}} \\ \\ BC & = 3x \\ &= 3 \times 12.5 \\ BC & = \color{green}{37.5\ \text{cm}} \end{aligned} \]Answer \(\color{red}{AB = 62.5\ \text{cm}, \ BC = 37.5\ \text{cm}}\)
7. In the given diagram, \(ABCD\) is a rhombus. If \(\angle ADB = 50^\circ\), find the measure of \(\angle C\).
Solution
\[ \begin{aligned} {\color{magenta} \textbf{Given : }} & ABCD \text{ is a rhombus} \\ \angle ADB & = 50^\circ \\[4pt] {\color{magenta} \textbf{To find : }} & \angle C \\[4pt] {\color{magenta} \textbf{Proof : }} & \\[4pt] \angle ADB & = \angle 2 \color{magenta} \text{ (Alternate interior angles)} \\[6pt] \therefore \ \angle 2 & = 50^\circ \\[6pt] In \ & \triangle CBD \\[4pt] CD &= CB \color{magenta} \text{ (In a rhombus, all sides are equal.)}\\[4pt]\therefore\ \angle 1 & = \angle 2 \begin{cases} \color{magenta} \text{Angles opposite to} \\ \color{magenta} \text{equal sides of a}\\ \color{magenta} \text{triangle are equal}\\ \end{cases}\\[6pt]\angle 1 &= 50^\circ \\[6pt]\angle 1 + \angle 2 + \angle C &= 180^\circ \color{magenta} \text{ (Angle sum property)}\\[6pt] 50^\circ + 50^\circ + \angle C &= 180^\circ \\ 100^\circ + \angle C &= 180^\circ \\ \angle C &= 180^\circ - 100^\circ \\ \angle C &= 80^\circ \end{aligned} \]
Answer \(\angle C = \color{red}{80^\circ}\)
8. The diagonals of a rhombus are \(12\) cm and \(16\) cm respectively. Find the length of the sides of the rhombus. Also, find its perimeter.
Solution
\[ \begin{aligned} AC = 12 \ cm \ & , \ BD = 16 \ cm \\ \color{magenta}\text{Diagonals of } & \color{magenta}\text{rhombus bisect each other at right angles} \\ OA = \frac{AC}{2} & \implies 6 \text{ cm} \\[6pt] OB=\frac{BD}{2}& \implies 8 \text{ cm} \\[6pt] \triangle AOB & \text{ is right angled triangle} \\[6pt] (AB)^2 & = (OA)^2 + (OB)^2 \quad \color{magenta}\text{(Pythagoras theorem)}\\[6pt] (AB)^2 & = 6^2 + 8^2 \\[6pt] (AB)^2 & = 36 + 64 \\[6pt] (AB)^2 & = 100 \\[6pt] AB & = 10 \ cm \\[6pt] \Rightarrow AB=BC&=CD=DA=\color{green}{10 \text{ cm}}\\[6pt] Perimeter & = 4 \times Side \\ & = 4 \times 10 \\ Perimeter & = 40 \ cm \end{aligned} \]Answer Side \(=\color{red}{10\text{ cm }} \), Perimeter \(=\color{red}{40\text{ cm}}\).
9. Two adjacent angles of a parallelogram are in the ratio \(1:5\). Find all the angles of the parallelogram.
Solution
\[ \begin{aligned} \text{Let the adjacent } & \text{angles be } x\text{ and }5x\\ \angle A &= x \\ \angle B &= 5x \\[6pt] \angle A + \angle B &= 180^\circ \ \color{magenta}\text{(co-interior angles are supplementary)}\\ x+5x&=180^\circ \\ 6x&=180^\circ \\[6pt] x&= \frac{180^\circ}{6} \\[6pt] x &=\color{green}{30^\circ} \\ 5x & =\color{green}{150^\circ}\\[8pt] \angle A &=\color{green}{30^\circ} \\ \angle B &=\color{green}{150^\circ} \\[8pt] \end{aligned} \]\[ \begin{aligned} \angle A &= \angle C \Rightarrow 30^\circ \\ \angle B &= \angle D \Rightarrow 150^\circ \end{aligned} \begin{cases} {\color{magenta}\text{Opposite angles}}\\ {\color{magenta}\text{of a parallelogram}}\\ {\color{magenta}\text{are equal}} \end{cases} \]Answer All the angles of the parallelogram are \(\color{red}{30^\circ,\ 150^\circ,\ 30^\circ,\ 150^\circ}\)
10. In the given diagram, \(PQRS\) is a rhombus. If \(\angle PSQ = 40^\circ\), then find the value of \(\angle SPQ\).
Solution
\[ \begin{aligned} {\color{magenta} \textbf{Given : }} & PQRS \text{ is a rhombus} \\ \angle PSQ & = 40^\circ \\[4pt] {\color{magenta} \textbf{To find : }} & \angle SPQ \\[4pt] {\color{magenta} \textbf{Proof : }} & \\[4pt] In \ & \triangle SPQ \\[4pt] PS &= PQ \color{magenta} \text{ (In a rhombus, all sides are equal.)}\\[4pt]\therefore\ \angle PSQ & = \angle PQS \begin{cases} \color{magenta} \text{Angles opposite to} \\ \color{magenta} \text{equal sides of a}\\ \color{magenta} \text{triangle are equal}\\ \end{cases}\\[6pt]\angle PQS &= 40^\circ \\[6pt]\angle PSQ + \angle PQS + \angle SPQ &= 180^\circ \color{magenta} \text{ (Angle sum property)}\\[6pt] 40^\circ + 40^\circ + \angle SPQ &= 180^\circ \\ 80^\circ + \angle SPQ &= 180^\circ \\ \angle SPQ &= 180^\circ - 80^\circ \\ \angle SPQ &= 100^\circ \end{aligned} \]Answer \(\angle SPQ = \color{red}{100^\circ}\)
11. The adjacent sides of a rectangle are in the ratio \(3:4\). If the diagonal is \(20\) cm long, then find the lengths of the sides. Also, find the perimeter of the rectangle.
Solution
\[ \begin{aligned} \text{Let the sides be } & 3x \ and \ 4x \\ AB = DC & \implies 4x \\ AD = BC & \implies 3x \\[6pt] \text{In right } \ \triangle ABC \\[6pt] \text{Using pythagoras} & \text{ theorem} \\[6pt] \color{magenta} (AB)^2+(BC)^2 & = \color{magenta} (AC)^2\\ (4x)^2+(3x)^2 & = 20^2\\ 16x^2 + 9x^2 & = 400 \\ 25x^2 & = 400 \\ x^2 & = \frac{\cancel{400}^{16}}{\cancel{25}_1} \\ x^2 & = 16 \\ x &= \color{green} 4 \ cm \\[6pt] {\color{magenta}Length} = 4x & \implies \color{green} 16 \ cm \\ {\color{magenta}Breadth} = 3x & \implies \color{green} 12 \ cm \\[6pt] \text{Perimeter }&=2(l+b) \\ &=2(16+12)\\ \color{magenta}Perimeter &= \color{green}{56\text{ cm}} \end{aligned} \]Answer Sides \(=\color{red}{16\text{ cm and }12\text{ cm}}\); Perimeter \(=\color{red}{56\text{ cm}}\)
12. The diagonals of a rhombus are in the ratio \(5:12\). If its perimeter is \(208\) cm, then find the lengths of the sides and diagonals.
Solution
\[ \begin{aligned} \text{All the sides} & \text{ are equal in Rhombus} \\ \text{Perimeter} & =208 \ cm \\[6pt] 4 \times \text{Side} &= 208 \\[6pt] \text{Side} &= \frac{208}{4} \\[6pt] \text{Side} & =\color{green}{52\text{ cm}}\\[6pt] AB = BC &= CD = DA \implies \color{green}{52\text{ cm}}\\[6pt] \text{Let the } & \text{diagonals be} \\ BD & = 5x \\ AC & = 12x \\ \text{Diagonals of } & \text{rhombus bisect each other} \\ OB = OD & \implies \frac{5x}{2} \\[6pt] OA = OC & = \frac{12x}{2} \implies 6x \\[6pt] \text{In right } & \triangle AOB \\[6pt] \text{Using pythagoras} & \text{ theorem} \\[6pt]\color{magenta} (OB)^2 + (OA)^2 &= \color{magenta} (AB)^2 \\[6pt] \left(\frac{5x}{2} \right)^2 + (6x)^2 &= 52^2 \\[6pt] \frac{25x^2}{4} + 36x^2 &= 52 \times 52 \\[6pt] \frac{25x^2 + 144 x^2}{4} &= 52 \times 52 \\[6pt] \frac{169 x^2}{4} &= 52 \times 52 \\[6pt] x^2 &= \frac{\cancel{52}^4 \times \cancel{52}^4 \times 4}{\cancel{169}_{\cancel{13}_1}} \\[6pt] x^2 &= 64 \\[6pt] x &= 8 \ cm \\[6pt] &Diagonals \\ BD & = 5x \implies \color{green}40 \ cm \\ AC & = 12x \implies \color{green} 96 \ cm \\ \end{aligned} \]Answer Side \(=\color{red}{52\text{ cm}}\); Diagonals \(=\color{red}{40\text{ cm and }96\text{ cm}}\)
13. Two adjacent angles of a parallelogram \(ABCD\) are \((6x - 9)^\circ\) and \((13x + 18)^\circ\). Find the ratio of these angles. Also, find the value of \((\angle A - \angle C)\) in a parallelogram.
Solution
Adjacent angles of a parallelogram are supplementary. \[ \begin{aligned} \angle A + \angle B &= 180^\circ \\ (6x - 9)^\circ + (13x + 18)^\circ &= 180^\circ \\ 6x - 9 + 13x + 18 &= 180 \\ 19x + 9 &= 180 \\ 19x &= 171 \\ x &= \color{green}{9}\\[6pt] \angle A & = (6x - 9)^\circ \\ &= 6 \times 9 - 9 \\ \angle A &= 45^\circ \\[6pt] \angle B & = (13x + 18)^\circ \\ &= 13 \times 9 + 18 \\ \angle B & = 135^\circ \end{aligned} \]\[ \begin{aligned} \textbf{Ratio} & : \\[6pt] \angle A : \angle B & = 45 : 135 \\[6pt] & = \frac{45}{135} \\[6pt] & = \frac{1}{3} \\[6pt] & \implies 1 : 3 \end{aligned} \] \[ \begin{aligned} \angle A &= \angle C \ \textbf{(opposite angles are equal)} \\[6pt] \angle A - \angle C & = 45^\circ - 45^\circ \\[6pt] \angle A - \angle C & = 0^\circ \\[6pt] \end{aligned} \]Answer Ratio \(=\color{red}{1:3} \) , \((\angle A - \angle C) = \color{red}{0^\circ}\)
14. \(ABCD\) is a rectangle. If the length of the diagonal \(AC = 2(3z - 23)\) cm and diagonal \(BD = (64 - 4z)\) cm, then find the value of \(z\) and hence find \(AC\).
Solution
\[ \begin{aligned} \text{In a rectangle, diagonals are equal.} \end{aligned} \] \[ \begin{aligned} AC &= BD \\ 2(3z - 23) &= 64 - 4z \\ 6z - 46 &= 64 - 4z \\ 6z + 4z &= 64 + 46 \\ 10z &= 110 \\ z &= \color{green}{11} \\ \\ AC &= 2(3z - 23) \\ &= 2(3 \times 11 - 23) \\ &= 2(33 - 23) \\ &= 2 \times 10 \\ AC &= \color{green}{20 \text{ cm}} \end{aligned} \]Answer \(z = \color{red}{11} \) and \(AC = \color{red}{20\text{ cm}}\)
15. In a square \(ABCD\), \(AB = (3x - 7)\) cm and \(BC = (x + 3)\) cm. Find the length of diagonal \(BD\).
Solution
\[ \begin{aligned} \text{All sides of a square are equal.} \end{aligned} \] \[ \begin{aligned} AB &= BC \\ 3x - 7 &= x + 3 \\ 3x - x &= 3 + 7 \\ 2x &= 10 \\ x &= \color{green}{5} \\ \\ & \text{Side of the square} \\ BC &= x + 3 \\ &= 5 + 3 \\ BC &= 8 \text{ cm} \\ \\ \text{In right} & \ \triangle ABD \\ \text{Using } & \text{pythagoras theorem} \\ (BD)^2 &= (AB)^2 + (AD)^2 \\ (BD)^2 &= 8^2 + 8^2 \\ (BD)^2 &= 64 + 64 \\ (BD)^2 &= 128 \\ BD &= \sqrt{128} \\ BD&= \sqrt{64 \times 2} \\ BD&= 8\sqrt{2} \text{ cm} \end{aligned} \]
Answer \(BD = \color{red}{8\sqrt{2}\text{ cm}}\)