DAV Class 8 Maths Chapter 11 Worksheet 3
Understanding Quadrilaterals Worksheet 3
1.
\(ABCD\) is a quadrilateral in which the diagonals are equal and bisect each other at right angles. What can you say about the figure? Is it—
(i) a rectangle
(ii) a rhombus
(iii) a square?
Solution
(i) Rectangle
In a rectangle, the diagonals are equal and bisect each other, but they do not meet at right angles.(ii) Rhombus
In a rhombus, the diagonals bisect each other at right angles, but they are not equal.(iii) Square
In a square, the diagonals are equal and also bisect each other at right angles.Answer (iii) \(\color{red}{\text{ABCD is a square}}\)
2. The diagonals of a rhombus are in the ratio \(5:12\). If its perimeter is \(104\) cm, find the lengths of the sides and the diagonals.
Solution
\[ \begin{aligned} \text{All the sides} & \text{ are equal in Rhombus} \\ \text{Perimeter} & =104 \ cm \\[6pt] 4 \times \text{Side} &= 104 \\[6pt] \text{Side} &= \frac{104}{4} \\[6pt] \text{Side} & =\color{green}{26\text{ cm}}\\[6pt] AB = BC &= CD = DA \implies \color{green}{26\text{ cm}}\\[6pt] \text{Let the } & \text{diagonals be} \\ BD & = 5x \\ AC & = 12x \\ \text{Diagonals of } & \text{rhombus bisect each other} \\ OB = OD & \implies \frac{5x}{2} \\[6pt] OA = OC & = \frac{12x}{2} \implies 6x \\[6pt] \text{In } & \triangle AOB \\[6pt] \color{magenta} (OB)^2 + (OA)^2 &= \color{magenta} (AB)^2 \\[6pt] \left(\frac{5x}{2} \right)^2 + (6x)^2 &= 26^2 \\[6pt] \frac{25x^2}{4} + 36x^2 &= 26 \times 26 \\[6pt] \frac{25x^2 + 144 x^2}{4} &= 26 \times 26 \\[6pt] \frac{169 x^2}{4} &= 26 \times 26 \\[6pt] x^2 &= \frac{\cancel{26}^2 \times \cancel{26}^2 \times 4}{\cancel{169}_{\cancel{13}_1}} \\[6pt] x^2 &= 16 \\[6pt] x &= 4 \ cm \\[6pt] &Diagonals \\ BD & = 5x \implies 20 \ cm \\ AC & = 12x \implies 48 \ cm \\ \end{aligned} \]Answer Side \(=\color{red}{26\text{ cm }}\) ; Diagonals \(=\color{red}{20\text{ cm and }48\text{ cm}}\)
3. A pair of adjacent sides of a rectangle are in the ratio \(3:4\). If its diagonal is \(20\) cm, find the lengths of the sides and hence, the perimeter of the rectangle.
Solution
\[ \begin{aligned} \text{Let the sides be } & 3x \ and \ 4x \\ AB = DC & \implies 4x \\ AD = BC & \implies 3x \\[6pt] In \ \triangle ABC \\ \color{magenta} (AB)^2+(BC)^2 & = \color{magenta} (AC)^2\\ (4x)^2+(3x)^2 & = 20^2\\ 16x^2 + 9x^2 & = 400 \\ 25x^2 & = 400 \\ x^2 & = \frac{\cancel{400}^{16}}{\cancel{25}_1} \\ x^2 & = 16 \\ x &= \color{green} 4 \ cm \\[6pt] {\color{magenta}Length} = 4x & \implies \color{green} 16 \ cm \\ {\color{magenta}Breadth} = 3x & \implies \color{green} 12 \ cm \\[6pt] \text{Perimeter }&=2(l+b) \\ &=2(16+12)\\ \color{magenta}Perimeter &= \color{green}{56\text{ cm}} \end{aligned} \]Answer Sides \(=\color{red}{12\text{ cm and }16\text{ cm }}\); Perimeter \(=\color{red}{56\text{ cm}}\)
4. \(ABCD\) is a rhombus whose diagonals intersect at \(O\). Show that \(\triangle AOB \cong \triangle COD\).
Solution
\[ \begin{aligned} In \ \triangle AOB & \text{ and } \triangle COD \\ AO & = CO \ \color{magenta}\text{(diagonals bisect each other)} \\ OB & = OD \ \color{magenta}\text{(diagonals bisect each other)} \\ AB & = CD \ \color{magenta}\text{(all sides of a rhombus are equal)} \\[6pt] \text{By } SSS & \text{ congruence condition} \\ \triangle AOB &\cong \triangle COD \end{aligned} \]Answer \(\color{red}{\triangle AOB \cong \triangle COD\ \text{(SSS)}}\)
5. The diagonals of a rectangle \(ABCD\) intersect at \(O\). If \(\angle BOC=70^\circ\), find \(\angle ODA\).
Solution
\[ \begin{aligned} \angle BOC&=70^\circ \\ \angle AOD & = \angle BOC \ \color{magenta}\text{(vertically opposite angles)}\\ \angle AOD & = 70^\circ \\ \text{In } &\triangle AOD \\ OA & = OD \begin{cases} \color{magenta} \text{Diagonals of rectangle} \\ \color{magenta} \text{bisect each other}\\ \end{cases}\\[6pt] \therefore\ \angle OAD & = \angle ODA \begin{cases} \color{magenta} \text{Angles opposite to} \\ \color{magenta} \text{equal sides of a}\\ \color{magenta} \text{triangle are also equal}\\ \end{cases}\\[6pt] \angle OAD + \angle ODA + \angle AOD & = 180^\circ \ \color{magenta} \text{(Angle sum property)} \\ x + x + 70^\circ & = 180^\circ \\ 2x & = 180^\circ - 70^\circ \\ 2x & = 110^\circ \\[6pt] x & = \frac{110^\circ}{2} \\[6pt] x & = 55^\circ \\ \angle ODA & = 55^\circ \end{aligned} \]Answer \(\angle ODA=\color{red}{55^\circ}\)
6. \(ABC\) and \(ADC\) are two equilateral triangles on a common base \(AC\). Find the angles of the resulting quadrilateral. Show that it is a rhombus.
Solution
\[ \begin{aligned} \triangle ABC \ \text{ and } \ \triangle ADC & \text{ are equilateral triangles} \\ \angle B = \angle BAC & = \angle BCA \implies 60^\circ \\ \angle D = \angle DAC & = \angle DCA \implies 60^\circ \\[6pt]\therefore \ \color{green} \angle B = \color{green} \angle D \implies 60^\circ & \ \color{magenta} \text{(Opposite angles are equal)} \\[6pt] \angle A &= \angle BAC + \angle DAC \\ \angle A &= 60^\circ + 60^\circ \\ \color{green}\angle A &= \color{green}120^\circ \\[6pt] \angle C &= \angle BCA + \angle DCA \\ \angle C &= 60^\circ + 60^\circ \\ \color{green} \angle C &= \color{green}120^\circ \\[6pt] \therefore \ \color{green} \angle A = \color{green} \angle C \implies 120^\circ \ & \color{magenta} \text{(Opposite angles are equal)} \\[6pt]AB=BC & =AC \\ AD=DC & =AC \\ \therefore \ \color{green} AB=BC = \color{green} CD & = \color{green} DA \ \color{magenta} \text{(All sides are equal)} \\[6pt] \implies ABCD & \color{magenta} \text{ is a Rhombus} \end{aligned} \]Answer All the sides are equal and opposite angles are equal, hence \(ABCD\) is a \(\color{red}{\text{rhombus}}\).
7.
\(ABCD\) is a rectangle as shown in the diagram in which \(DP\) and \(BQ\) are perpendiculars from \(D\) and \(B\) respectively on diagonal \(AC\). Show that—
(i) \(\triangle ADP \cong \triangle CBQ\)
(ii) \(\angle ADP = \angle CBQ\)
(iii) \(DP = BQ\)
Solution
\[ \begin{aligned} \color{magenta}{\textbf{Given:}}&\ ABCD \ \text{is a rectangle} \\ & DP \perp AC \\ & BQ \perp AC \\[6pt] \color{magenta}{\textbf{To prove:}} & \ \text{(i) }\triangle ADP \cong \triangle CBQ \\ &\text{(ii) }\angle ADP=\angle CBQ\\ &\ \text{(iii) }DP=BQ \\[6pt] \color{magenta}{\textbf{Proof:}}&\\In \ \triangle ADP & \text{ and } \triangle CBQ \\ \angle APD &=\angle CQB \implies 90^\circ\ \color{magenta}\text{(Given)} \\ AD&=CB \ \color{magenta}\text{(opposite sides of a rectangle are equal)}\\ \angle PAD & =\angle QCB \ \color{magenta}\text{(alternate interior angles)}\\[6pt] \implies\ \text{By } & ASA \text{ congruence condition} \\[6pt] \textbf{(i) } \Rightarrow\ \triangle ADP &\cong \triangle CBQ \\ \textbf{(ii) } \Rightarrow\ \angle ADP &=\angle CBQ \ \color{magenta}\text{ (by CPCT)}\\ \textbf{(iii) } \Rightarrow\ DP & = BQ \ \color{magenta}\text{ (by CPCT)} \end{aligned} \]