DAV Class 8 Maths Chapter 11 Worksheet 2
Understanding Quadrilaterals Worksheet 2
1. PQRS is a trapezium with \(PQ \parallel SR\). If \(\angle P=30^\circ,\ \angle Q=50^\circ\), find \(\angle R\) and \(\angle S\).
Solution
\[ \begin{aligned} \color{magenta}{\textbf{Given: }} & PQRS \text{ is a trapezium} \\ PQ & \parallel SR \\ \angle P & =30^\circ \\ \angle Q & = 50^\circ\\[4pt] \color{magenta}{\textbf{To find: }} & \angle R \ , \ \angle S \\ \color{magenta}{\textbf{Proof: }} & \\[8pt] \angle P + \angle S & = 180^\circ \color{magenta}\text{ (co-interior angles are supplementary)} \\ 30^\circ + \angle S & = 180^\circ \\ \angle S & = 180^\circ - 30^\circ \\ \angle S & = 150^\circ \\[8pt]\angle Q + \angle R & = 180^\circ \color{magenta}\text{ (co-interior angles are supplementary)} \\ 50^\circ + \angle R & = 180^\circ \\ \angle R & = 180^\circ - 50^\circ \\ \angle R & = 130^\circ \end{aligned} \]Answer \(\angle R={\color{red}{130^\circ}},\quad \angle S=\color{red}{150^\circ}\)
2.
\(ABCD\) is a quadrilateral with \(\angle A=80^\circ,\ \angle B=40^\circ,\ \angle C=140^\circ,\ \angle D=100^\circ.\)
(i) Is \(ABCD\) a trapezium?
(ii) Is \(ABCD\) a parallelogram? Justify.
Solution
(i) Trapezium?
\[ \begin{aligned} \angle A+\angle D & = 80^\circ+100^\circ \\ \angle A+\angle D & = \color{green} {180^\circ} \\[6pt] \angle B+\angle C & = 40^\circ+140^\circ\\ \angle B+\angle C & = \color{green} {180^\circ} \\[4pt] \color{magenta}\text{ Co-interior } &\color{magenta}\text{angles are supplementary} \\[4pt] \Rightarrow\ & \color{green}{AB \parallel CD} \\[4pt] \therefore \ \color{red} ABCD & \color{red} \text{ is a Trapezium} \end{aligned} \](ii) Parallelogram?
\[ \begin{aligned} \angle A = 80^\circ \ & ; \ \angle C = 140^\circ \\ \angle A & \ne \angle C \\[6pt] \angle B = 40^\circ \ & ; \ \angle D = 100^\circ \\ \angle B & \ne \angle D \\[6pt] \color{magenta}\text{Opposite } &\color{magenta}\text{angles are not equal.} \\[4pt] \therefore \ \color{red}ABCD & \color{red} \text{ is not a parallelogram.} \end{aligned} \]3. One of the angles of a parallelogram is \(75^\circ\). Find the measures of the remaining angles of the parallelogram.
Solution
\[ \begin{aligned} \color{magenta}{\textbf{Given:}}& \ ABCD \text{ is a Parallelogram} \\ \angle A & =75^\circ\\[4pt]\color{magenta}{\textbf{To find:}}&\ \angle B,\ \angle C,\ \angle D\\[6pt]\color{magenta}{\textbf{Proof:}}&\\ \angle A &= \angle C \ \color{magenta}\text{(opposite angles of a parallelogram are equal)}\\ \implies \angle C & = \color{green}{75^\circ} \\[6pt] \angle A + \angle D &= 180^\circ \ \color{magenta}\text{(co-interior angles are supplementary)}\\ 75^\circ+\angle D &= 180^\circ \\ \angle D &= 180^\circ - 75^\circ\\ \angle D &= \color{green}{105^\circ}\\[6pt] \angle D &= \angle B \ \color{magenta} \text{(opposite angles of a parallelogram are equal)} \\ \implies \angle B & = \color{green}{105^\circ} \end{aligned} \]Answer \(\angle B= {\color{red}{105^\circ}},\ \angle C={\color{red}{75^\circ}},\ \angle D={\color{red}{105^\circ}}\)
4. 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 \( = \color{red}{30^\circ,\,150^\circ,\,30^\circ,\,150^\circ}\)
5. An exterior angle of a parallelogram is \(110^\circ\). Find the angles of the parallelogram.
Solution
\[ \begin{aligned} \color{magenta}{\textbf{Proof:}}& \\ \angle ABC + \angle CBE &=180^\circ \ \color{magenta}\text{(Linear Pair)}\\ 110^\circ + \angle ABC &=180^\circ\\ \angle ABC &=180^\circ-110^\circ\\ \angle ABC &= \color{green}{70^\circ} \\[6pt] \angle C &= \angle CBE \ \color{magenta}\text{(Alternate angles)} \\ \angle C &= \color{green}{110^\circ} \\[6pt] \end{aligned} \] \[ \begin{aligned} \angle A &= \angle C \Rightarrow 110^\circ \\ \angle ABC &= \angle D \Rightarrow 70^\circ \end{aligned} \begin{cases} {\color{magenta}\text{Opposite angles}}\\ {\color{magenta}\text{of a parallelogram}}\\ {\color{magenta}\text{are equal}} \end{cases} \]Answer \(\color{red}{70^\circ,\,110^\circ,\,70^\circ,\,110^\circ}\)
6. Two adjacent sides of a parallelogram are in the ratio \(3:8\) and its perimeter is \(110\) cm. Find the sides of the parallelogram.
Solution
\[ \begin{aligned} \text{Let the adjacent sides be } & 3x\text{ and }8x \\ \color{magenta} \text{Opposite sides of a } & \color{magenta} \text{parallelogram are equal} \\ AD = BC & \implies 3x \\[6pt] AB = DC & \implies 8x \\ \text{Perimeter of parallelogram}&=110 \ cm \\ 2(3x+8x)&=110\\ 2 \times 11x &=110 \\ 22x&=110 \\[6pt] x&= \frac{110}{22} \\[6pt] x&= 5 \ cm \\[6pt] 3x&= 15 \ cm \\ 8x&= 40 \ cm \\[6pt] AD = BC & \implies \color{green} 15 \ cm \\ AB = DC & \implies \color{green} 40 \ cm \end{aligned} \]
Answer \(\color{red}{15\text{ cm and }40\text{ cm}}\)
7. One side of a parallelogram is \(\dfrac{3}{4}\) times its adjacent side. If the perimeter is \(70\) cm, find the sides of the parallelogram.
Solution
Let one side of parallelogram be \( x \ cm \) and the adjacent side be \(\dfrac{3x}{4} \ cm \)
\[ \begin{aligned} \color{magenta}\text{Opposite sides of a } & \color{magenta}\text{parallelogram are equal} \\ AD = BC & \implies x \\[6pt] AB = DC & \implies \frac{3x}{4} \\[6pt] \text{Perimeter of parallelogram} & = 70 \ cm \\[6pt] 2 \left ( x + \frac{3x}{4} \right) & = 70 \\[6pt] 2x + \frac{3x}{2} & = 70 \\[6pt] \frac{4x + 3x}{2} & = 70 \\[6pt] \frac{7x}{2} & = 70 \\[6pt] x & = \cancel{70}^{10} \times \frac{2}{\cancel7_1} \\[6pt] x & = 20 \ cm \\[6pt] \frac{3x}{4} &= \frac{3 \times \cancel{20}^{5}}{\cancel4_1} \\[6pt] & \implies 15 \ cm \\[6pt] AD = BC & \implies \color{green} 20 \ cm \\ AB = DC & \implies \color{green} 15 \ cm \end{aligned} \]Answer \(\color{red}{15\text{ cm and }20\text{ cm}}\)
8. \(ABCD\) is a parallelogram whose diagonals intersect each other at right angles. If the lengths of the diagonals are \(6\) cm and \(8\) cm, find the lengths of all the sides of the parallelogram.
Solution
\[ \begin{aligned} AC = 6 \ cm \ & , \ BD = 8 \ cm \\ \color{magenta}\text{Diagonals of } & \color{magenta}\text{parallelogram bisect each other} \\ OA = \frac{AC}{2} & \implies 3\text{ cm} \\[6pt] OB=\frac{BD}{2}& \implies 4\text{ cm} \\[6pt] \triangle AOB & \text{ is right angled triangle} \\[6pt] (AB)^2 & = (OA)^2 + (OB)^2 \\[6pt] (AB)^2 & = 3^2 + 4^2 \\[6pt] (AB)^2 & = 9 + 16 \\[6pt] (AB)^2 & = 25 \\[6pt] AB & = 5 \ cm \\[6pt] \Rightarrow AB=BC&=CD=DA=\color{green}{5\text{ cm}} \end{aligned} \]Answer Length of each side \( =\color{red}{5\text{ cm}}\)
9.
In the diagram, one pair of adjacent sides of a parallelogram is in the ratio \(3:4\). If one of its angles, \(\angle A\), is a right angle and diagonal \(BD=10\) cm, find the—
(i) Lengths of the sides of the parallelogram,
(ii) Perimeter of the parallelogram.
Solution
(i) Lengths of the sides of the parallelogram?
\[ \begin{aligned} \text{Let the adjacent } & \text{sides be } 3x\text{ and }4x \\[6pt] \triangle DAB & \text{ is right angled triangle} \\[6pt] AB^2+AD^2 &= BD^2 \\ (4x)^2+(3x)^2 & = 10^2 \\ 16x^2+9x^2 & = 100\\ 25x^2 &= 100\\[6pt] x^2 & = \frac{100}{25}\\[6pt] x^2 & = 4 \\ x & = 2 \ cm \\ 4x & = \color{green}{8\text{ cm}} \\ 3x & = \color{green}{6\text{ cm}} \\[8pt] \color{magenta} \text{Opposite sides of a } & \color{magenta} \text{parallelogram are equal} \\ AB = DC & \implies \color{green} 8 \ cm \\ AD = BC & \implies \color{green} 6 \ cm \end{aligned} \](ii) Perimeter of the parallelogram.
\[ \begin{aligned} Perimeter & = AB + BC + CD + DA \\[4pt] & = 8+6+8+6 \\[4pt] Perimeter & = \color{red} 28 \ cm \end{aligned} \]10. \(ABCD\) is a quadrilateral in which \(AB=CD\) and \(AD=BC\). Show that it is a parallelogram. [Hint: Draw one of the diagonals.]
Solution
\[ \begin{aligned} \color{magenta}{\textbf{Given:}}&\ AB=CD,\ AD=BC\\ \color{magenta}{\textbf{To prove:}}&\ ABCD \text{ is a parallelogram}.\\[4pt] \color{magenta}{\textbf{Proof:}}&\\ & \text{Draw diagonal } AC\\ \text{Consider } & \triangle ABC \text{ and }\triangle CDA \\ AB&=CD\ (\text{given})\\ BC &= DA\ (\text{given}) \\ CA &= AC \ (\text{common side})\\[6pt] \text{By } SSS & \text{ congruence condition} \\ \therefore\ \triangle ABC &\cong \triangle CDA \\ \\ \angle BAC &= \angle DCA \ \color{magenta}\text{(by CPCT)} \\ \therefore\ AB & \parallel CD \\[6pt] \angle BCA &=\angle DAC \ \color{magenta}\text{(by CPCT)}\\ \therefore\ AD &\parallel BC \\[6pt] & \color{green} \text{Opposite sides are parallel}\\ \Rightarrow \ & \text{\(ABCD\) is a parallelogram.} \end{aligned} \]Answer \(\color{red}{ABCD\ \text{is a parallelogram}}\)