RD Sharma solution class 8 chapter 17 Understanding Shapes III(Understanding Special type of Quadrilaterals) Exercise 17.1

Exercise 17.1

Page-17.9

Question 1:

Given below is a parallelogram ABCD. Complete each statement along with the definition or property used.
(i) AD =
(ii) ∠DCB =
(iii) OC =
(iv) ∠DAB + ∠CDA =

Answer 1:

The correct figure is

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{AD}=\mathrm{BC} (\mathrm{opposite} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \left(\mathrm{ii}\right) \\ \angle \mathrm{DCB}=\angle \mathrm{BAD} \left(\mathrm{opposite} \mathrm{angles} \mathrm{are} \mathrm{equal}\right) \\ \left(\mathrm{iii}\right) \\ \mathrm{OC}=\mathrm{OA} \left(\mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{prallelogram} \mathrm{bisect} \mathrm{each} \mathrm{other}\right) \\ \left(\mathrm{iv}\right) \\ \angle \mathrm{DAB}+\angle \mathrm{CDA}=180° \left(\mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{two} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{is} {180}^0\right) \end{gathered}$$

Question 2:

The following figures are parallelograms. Find the degree values of the unknowns x, y, z.

Answer 2:

$$\begin{gathered} \left(\mathrm{i}\right) \\ O\mathrm{pposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{same}. \\ \therefore \mathrm{x}=\mathrm{z} \mathrm{and} \mathrm{y}=100° \\ \mathrm{Also}, \mathrm{y}+\mathrm{z}=180° (s\mathrm{um} \mathrm{of} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{quadrilateral} \mathrm{is} 180°) \\ \mathrm{z}+100°=180° \\ \mathrm{x}=180°-100° \\ \mathrm{x}=80° \\ \therefore \mathrm{x}=80°, \mathrm{y}=100° \mathrm{and} \mathrm{z}=80° \\ \left(\mathrm{ii}\right) \\ O\mathrm{pposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{same}. \\ \therefore \mathrm{x}=\mathrm{y} \mathrm{and} \angle \mathrm{RQP}=100° \\ \angle \mathrm{PSR}+\angle \mathrm{SRQ}=180° \\ \mathrm{y}+50°=180° \\ \mathrm{x}=180°-50° \\ \mathrm{x}=130° \\ \therefore \mathrm{x}=130°, \mathrm{y}=130° \\ Since \mathrm{y} \mathrm{and} \mathrm{z} \mathrm{are} \mathrm{alternate} \mathrm{angles}, \mathrm{z}=130°. \\ \left(\mathrm{iii}\right) \\ S\mathrm{um} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{in} \mathrm{a} \mathrm{triangle} \mathrm{is} {180}^{°}. \\ \therefore 30°+90°+\mathrm{z}=180° \\ \mathrm{z}=60° \\ O\mathrm{pposite} \mathrm{angles} \mathrm{are} \mathrm{equal} \mathrm{in} \mathrm{parallelogram}. \\ \therefore \mathrm{y}=\mathrm{z}=60° \\ \mathrm{and} \mathrm{x}=30° (\mathrm{alternate} \mathrm{angles}) \\ \left(\mathrm{iv}\right)x=90° (\mathrm{vertically} \mathrm{opposite} \mathrm{angle}) \\ S\mathrm{um} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{in} \mathrm{a} \mathrm{triangle} \mathrm{is} {180}^{°}. \\ \therefore \mathrm{y}+90°+30°=180° \\ \mathrm{y}=180°-(90°+30°)=60° \\ \mathrm{y}=\mathrm{z}=60° (a\mathrm{lternate} \mathrm{angles}) \\ \left(\mathrm{v}\right) \\ O\mathrm{pposite} \mathrm{angles} \mathrm{are} \mathrm{equal} \mathrm{in} a \mathrm{parallelogram}. \\ \therefore \mathrm{y}=80° \\ \mathrm{y}+\mathrm{x}=180° \\ \mathrm{x}=180°-100°=80° \\ \mathrm{z}=\mathrm{y}=80° (\mathrm{alternate} \mathrm{angles}) \\ \left(\mathrm{vi}\right) \\ \mathrm{y}=112° (\mathrm{opposite} \mathrm{angles} \mathrm{are} \mathrm{equal} \mathrm{in} \mathrm{a} \mathrm{parallelogram}) \\ \mathrm{In}∆ \mathrm{UTW} : \\ \mathrm{x}+\mathrm{y}+40°=180° (a\mathrm{ngle} \mathrm{sum} \mathrm{property} \mathrm{of} a \mathrm{triangle}) \\ \mathrm{x}=180°-(112°-40°)=28° \\ B\mathrm{ottom} \mathrm{left} \mathrm{vertex}=180°-112°=68° \\ \therefore \mathrm{z}=\mathrm{x}=28° (\mathrm{alternate} \mathrm{angles}) \end{gathered}$$

Page-17.10

Question 3:

Can the following figures be parallelograms. Justify your answer.

Answer 3:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{No}. This is because the \mathrm{oppos}ite \mathrm{angles} \mathrm{are} \mathrm{not} \mathrm{equal}. \\ \left(\mathrm{ii}\right) \\ \mathrm{Yes}. This is because the opposite sides are equal. \\ \left(\mathrm{ii}\right) \\ \mathrm{No}, This is because the \mathrm{diagonals} \mathrm{do} \mathrm{not} \mathrm{bisect} \mathrm{each} \mathrm{other}. \end{gathered}$$

Question 4:

In the adjacent figure HOPE is a parallelogram. Find the angle measures x,y and z. State the geometrical truths you use to find them.

Answer 4:

$$\begin{gathered} \angle \mathrm{HOP}+70°=180° (l\mathrm{inear} \mathrm{pair}) \\ \angle \mathrm{HOP}=180°-70°=110° \\ \mathrm{x}=\angle \mathrm{HOP}=110° (o\mathrm{pposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \angle \mathrm{EHP}+\angle \mathrm{HEP}=180° (s\mathrm{um} \mathrm{of} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{is} 180°) \\ 110°+40°+\mathrm{z}=180° \\ \mathrm{z}=180°-150°=30° \\ \mathrm{y}=40° \left(\mathrm{alternate} \mathrm{angle}s \right) \end{gathered}$$

Question 5:

In the following figures GUNS and RUNS are  parallelograms. Find x and y.

Answer 5:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Opposite} \mathrm{sides} \mathrm{are} \mathrm{equal} \mathrm{in} \mathrm{a} \mathrm{parallelogram}. \\ \therefore 3\mathrm{y}-1=26 \\ 3\mathrm{y}=27 \\ \mathrm{y}=9 \\ \mathrm{Similarly}, 3\mathrm{x}=18 \\ \mathrm{x}=6 \\ \left(\mathrm{ii}\right) \\ \mathrm{Diagonals} \mathrm{bisect} \mathrm{each} \mathrm{other} \mathrm{in} \mathrm{a} \mathrm{parallelogram}. \\ \therefore \mathrm{y}-7=20 \\ \mathrm{y}=27 \\ \mathrm{x}-\mathrm{y}=16 \\ \mathrm{x}-27=16 \\ \mathrm{x}=43 \end{gathered}$$

Question 6:

In the following figure RISK and CLUE are parallelograms. Find the measure of x.

Answer 6:

$$\begin{gathered} \mathrm{In} \mathrm{the} \mathrm{parallelogram} \mathrm{RISK}: \\ \angle \mathrm{ISK}+\angle \mathrm{RKS}=180° (\mathrm{sum} \mathrm{of} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{is} 180°) \\ \angle \mathrm{ISK}=180°-120°=60° \\ \mathrm{Similarly}, \mathrm{in} \mathrm{parallelogram} \mathrm{CLUE}: \\ \angle \mathrm{CEU}=\angle \mathrm{CLU}=70° (\mathrm{opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \mathrm{In} \mathrm{the} \mathrm{triangle}: \\ x+\angle \mathrm{ISK}+\angle \mathrm{CEU}=180° \\ x=180°-\left(70°+60°\right) \\ x=180°-\left(70°+60°\right)=50° \end{gathered}$$

Question 7:

Two opposite angles of a parallelogram are (3x − 2)° and (50 − x)°. Find the measure of each angle of the parallelogram.

Answer 7:

$$\begin{gathered} O\mathrm{ppostie} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{congurent}. \\ \therefore \left(3x-2\right)°=\left(50-x\right)° \\ 3x°-2°=50°-x° \\ 3x°+x°=50°+2° \\ 4x°=52° \\ x°=13° \\ Putting \mathrm{the} \mathrm{value} \mathrm{of} x \mathrm{in} \mathrm{one} \mathrm{angle}: \\ 3x°-2°=39°-2° \\ =37° \\ O\mathrm{pposite} \mathrm{angles} \mathrm{are} \mathrm{congurent}: \\ \therefore 50-x° \\ =37° \\ \mathrm{Let} \mathrm{the} \mathrm{remaining} \mathrm{two} \mathrm{angles} \mathrm{be} y and z. \\ \mathrm{Angles} y \mathrm{and} z \mathrm{are} \mathrm{congurent} \mathrm{because} \mathrm{they} \mathrm{are} \mathrm{also} \mathrm{opposite} \mathrm{angles}. \\ \therefore y=z \\ \mathrm{The} \mathrm{sum} \mathrm{of} \mathrm{adjacent} angles \mathrm{of} a paralle\log ram \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \therefore 37°+y=180° \\ y=180°-37° \\ y=143° \\ \mathrm{So}, \mathrm{the} \mathrm{anlges} \mathrm{measure} \mathrm{are}: \\ 37°, 37°, 143° and 143° \end{gathered}$$

Question 8:

If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.

Answer 8:

$$\begin{gathered} \mathrm{Two} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{add} \mathrm{up} \mathrm{to} 180°. \\ \mathrm{Let} x \mathrm{be} \mathrm{the} \mathrm{angle}. \\ \therefore x+\frac{2x}{3}=180° \\ \frac{5x}{3}==180° \\ x=72° \\ \frac{2x}{3}=\frac{2\times 72°}{3}=108° \\ \mathrm{Thus}, \mathrm{two} \mathrm{of} \mathrm{the} \mathrm{angles} \mathrm{in} \mathrm{the} \mathrm{parallelogram} \mathrm{are} 108° \mathrm{and} \mathrm{the} \mathrm{other} \mathrm{two} \mathrm{are} 72°. \end{gathered}$$

Question 9:

The measure of one angle of a parallelogram is 70°. What are the measures of the remaining angles?

Answer 9:

$$\begin{gathered} \mathrm{Given} \mathrm{that} \mathrm{one} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{parallelogram} \mathrm{is} 70°. \\ \mathrm{Since} \mathrm{opposite} \mathrm{angles} \mathrm{have} \mathrm{same} \mathrm{value}, \mathrm{if} \mathrm{one} \mathrm{is} 70°, \mathrm{then} \mathrm{the} \mathrm{one} \mathrm{directly} \mathrm{opposite} \mathrm{will} \mathrm{also} \mathrm{be} 70°. \\ \mathrm{So}, \mathrm{let} \mathrm{one} \mathrm{angle} \mathrm{be} \mathrm{x}°. \\ \mathrm{x}°+70°=180° (\mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{is} 180° ) \\ \mathrm{x}°=180°-70° \\ \mathrm{x}°=110° \\ \mathrm{Thus}, \mathrm{the} \mathrm{remaining} \mathrm{angles} \mathrm{are} 110°, 110° \mathrm{and} 70°. \end{gathered}$$

Question 10:

Two adjacent angles of a parallelogram are as 1 : 2. Find the measures of all the angles of the parallelogram.

Answer 10:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{angle} \mathrm{be} \mathrm{A} \mathrm{and} \mathrm{B}. \\ The a\mathrm{ngles} \mathrm{are} \mathrm{in} \mathrm{the} \mathrm{ratio} \mathrm{of} 1:2. \\ M\mathrm{easures} \mathrm{of} \angle \mathrm{A} \mathrm{and} \angle \mathrm{B} \mathrm{are} \mathrm{x}° \mathrm{and} 2\mathrm{x}°. \\ \mathrm{Then}, \angle \mathrm{C}=\angle \mathrm{A} \mathrm{and} \angle \mathrm{D}=\angle \mathrm{B} (o\mathrm{pposite} \mathrm{angles} \mathrm{of} a \mathrm{parallelogram} \mathrm{are} \mathrm{congruent}) \\ \mathrm{As} \mathrm{we} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{adjacent} \mathrm{angle}s \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{is} 180°. \\ \therefore \angle \mathrm{A}+\angle \mathrm{B}=180° \\ \Rightarrow \mathrm{x}°+2\mathrm{x}°=180° \\ \Rightarrow 3\mathrm{x}°=180° \\ \Rightarrow \mathrm{x}°=\frac{180°}{3}=60° \\ \mathrm{Thus}, \mathrm{measure} \mathrm{of} \angle \mathrm{A}=60°, \angle \mathrm{B}=120°, \angle \mathrm{C}=60° and \angle \mathrm{D}=120°. \end{gathered}$$

Question 11:

In a parallelogram ABCD, ∠D = 135°, determine the measure of ∠A and ∠B.

Answer 11:

$$\begin{gathered} \mathrm{In} a \mathrm{parallelogram}, \mathrm{opposite} \mathrm{angles} \mathrm{have} \mathrm{the} \mathrm{same} \mathrm{value}. \\ \therefore \angle \mathrm{D}=\angle \mathrm{B} \\ =135° \\ \mathrm{Also}, \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360° \\ \angle \mathrm{A}+\angle \mathrm{D}=180° \left(\mathrm{opposite} \mathrm{angles} \mathrm{have} \mathrm{the} \mathrm{same} \mathrm{value}\right) \\ \angle \mathrm{A}=180°-135°=45° \\ \angle \mathrm{A}=45° \end{gathered}$$

Page-17.11

Question 12:

ABCD is a parallelogram in which ∠A = 70°. Compute ∠B, ∠C and ∠D.

Answer 12:

$$\begin{gathered} \mathrm{Opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}. \\ \therefore \angle \mathrm{C}=70°=\angle \mathrm{A}. \\ \angle \mathrm{B}=\angle \mathrm{D} \\ \mathrm{Also}, \mathrm{the} \mathrm{sum} o\mathrm{f} \mathrm{the} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{is} 180°. \\ \therefore \angle \mathrm{A}+\angle \mathrm{B}=180° \\ 70°+\angle \mathrm{B}=180° \\ \angle \mathrm{B}=110° \\ \therefore \angle \mathrm{B}=110°, \angle \mathrm{C}=70° \mathrm{and} \angle \mathrm{D}=110° \end{gathered}$$

Question 13:

The sum of two opposite angles of a parallelogram is 130°. Find all the angles of the parallelogram.

Answer 13:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{angles} \mathrm{be} \mathrm{A}, \mathrm{B}, \mathrm{C} \mathrm{and} \mathrm{D}. \\ It is g\mathrm{iven} \mathrm{that} the \mathrm{sum} \mathrm{of} \mathrm{two} \mathrm{opposite} \mathrm{angles} \mathrm{is} 130°. \\ \therefore \angle \mathrm{A}+\angle \mathrm{C}=130° \\ \angle \mathrm{A}+\angle \mathrm{A}=130° \left(\mathrm{opp}o\mathrm{site} \mathrm{angle}s \mathrm{of} a \mathrm{parallelogram} \mathrm{are} \mathrm{same}\right) \\ \angle \mathrm{A}=65° \\ \mathrm{and} \angle \mathrm{C}=65° \\ The s\mathrm{um} \mathrm{of} adjacent angles of a paralle\log ram is 180°. \\ \angle \mathrm{A}+\angle \mathrm{B}=180° \\ 65°+\angle \mathrm{B}=180° \\ \angle \mathrm{B}=180°-65° \\ \angle \mathrm{B}=115° \\ \angle \mathrm{D}=115° \\ \therefore \angle \mathrm{A}=65°, \mathrm{v}\angle \mathrm{B}=115°, \angle \mathrm{C}=65° \mathrm{and} \angle \mathrm{D}=115°. \end{gathered}$$

Question 14:

All the angles of a quadrilateral are equal to each other. Find the measure of each. Is the quadrilateral a parallelogram? What special type of parallelogram is it?

Answer 14:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{angle} \mathrm{be} \mathrm{x}. \\ A\mathrm{ll} \mathrm{the} \mathrm{angles} \mathrm{are} \mathrm{equal}. \\ \therefore x+x+x+x=360° \\ 4x=360° \\ x=90° \\ \mathrm{So}, \mathrm{each} \mathrm{angle} \mathrm{is} 90° \mathrm{and} \mathrm{quadrilateral} \mathrm{is} \mathrm{a} \mathrm{parallelogram}. \mathrm{It} \mathrm{is} \mathrm{a} \mathrm{rectangle}. \end{gathered}$$

Question 15:

Two adjacent sides of a parallelogram are 4 cm and 3 cm respectively. Find its perimeter.

Answer 15:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{opposite} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}. \\ \mathrm{Two} \mathrm{sides} \mathrm{are} \mathrm{given}, \mathrm{i}.\mathrm{e}. 4 \mathrm{cm} \mathrm{and} 3 \mathrm{cm}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{rest} \mathrm{of} \mathrm{the} \mathrm{sies} \mathrm{will} \mathrm{also} \mathrm{be} 4 \mathrm{cm} \mathrm{and} 3 \mathrm{cm}. \\ \therefore P\mathrm{erimeter} = S\mathrm{um} \mathrm{of} \mathrm{all} the \mathrm{sides} \mathrm{of} a \mathrm{parallelogram} \\ =4+3+4+3 \\ =14 \mathrm{cm} \end{gathered}$$

Question 16:

The perimeter of a parallelogram is 150 cm. One of its sides is greater than the other by 25 cm. Find the length of the sides of the parallelogram.

Answer 16:

$$\begin{gathered} \mathrm{Opposite} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{same}. \\ \mathrm{Le}t \mathrm{two} \mathrm{sides} \mathrm{of} the \mathrm{parallelogram} \mathrm{be} x \mathrm{and} y. \\ \mathrm{Given}: \\ x=y+25 \\ \mathrm{Also}, x+y+x+y=150 (Perimeter= Sum of all the sides of a paralle\log ram) \\ y+25+y+y+25+y=150 \\ 4y=150-50 \\ 4y=100 \\ y=\frac{100}{4}=25 \\ \therefore x=y+25=25+25=50 \\ \mathrm{Thus}, \mathrm{the} \mathrm{length}s \mathrm{of} \mathrm{the} \mathrm{sides} \mathrm{of} \mathrm{the} \mathrm{parallelogram} are 50 \mathrm{cm} \mathrm{and} 25 \mathrm{cm}. \end{gathered}$$

Question 17:

The shorter side of a parallelogram is 4.8 cm and the longer side is half as much again as the shorter side. Find the perimeter of the parallelogram.

Answer 17:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Shorter} \mathrm{side}=4.8 \mathrm{cm} \\ \mathrm{Longer} \mathrm{side}=\frac{4.8}{2}+4.8=7.2 \mathrm{cm} \\ \mathrm{Perimeter}=\mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{sides} \\ =4.8+4.8+7.2+7.2 \\ =24 \mathrm{cm} \end{gathered}$$

Question 18:

Two adjacent angles of a parallelogram are (3x − 4)° and (3x + 10)°. Find the angles of the parallelogram.

Answer 18:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{adjacent} \mathrm{angle}s \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{supplementry}. \\ \mathrm{Hence}, \left(3x+10\right)° and \left(3x-4\right)° \mathrm{are} \mathrm{supplementry}. \\ \left(3x+10\right)°+\left(3x-4\right)°=180° \\ 6x°+6°=180° \\ 6x°=174° \\ x=29° \\ \mathrm{First} \mathrm{angle} = \left(3\mathrm{x}+10\right)°=\left(3\times 29°+10°\right)=97° \\ \mathrm{Second} \mathrm{angle} = \left(3\mathrm{x}-4\right)°=83° \\ \mathrm{Thus}, \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{parallelogram} \mathrm{are} 97°, 83°, 97° \mathrm{and} 83°. \end{gathered}$$

Question 19:

In a parallelogram ABCD, the diagonals bisect each other at O. If ∠ABC = 30°, ∠BDC = 10° and ∠CAB = 70°. Find:
∠DAB, ∠ADC, ∠BCD, ∠AOD, ∠DOC, ∠BOC, ∠AOB, ∠ACD, ∠CAB, ∠ADB, ∠ACB, ∠DBC and ∠DBA.

Answer 19:

$$\begin{gathered} \angle ABC=30° \\ \therefore \angle ADC=30° \left(\mathrm{opposite} \mathrm{angle} \mathrm{of} the \mathrm{parallelogram}\right) \\ \mathrm{and} \angle BDA=\angle \mathrm{ADC}-\angle \mathrm{BDC}=30°-10°=20° \\ \angle BAC=\angle ACD=70° (alternate angle) \\ \mathrm{In} △ \mathrm{ABC}: \\ \angle \mathrm{CAB}+\angle \mathrm{ABC}+\angle \mathrm{BCA}=180° \\ 70°+30°+\angle \mathrm{BCA}=180° \\ \therefore \angle \mathrm{BCA}=80° \\ \angle \mathrm{DAB}=\angle \mathrm{DAC}+\angle \mathrm{CAB}=70°+80°=150° \\ \angle \mathrm{BCD}=150° \left(\mathrm{opposite} \mathrm{angle} \mathrm{of} the \mathrm{parallelogram}\right) \\ \angle DCA=\angle CAB=70° \\ \mathrm{In} △\mathrm{DOC}: \\ \angle \mathrm{ODC}+\angle \mathrm{DOC}+\angle \mathrm{OCD}=180 \\ 10°+70°+\angle \mathrm{DOC}=180° \\ \therefore \angle \mathrm{DOC}=100° \\ \angle \mathrm{DOC}+\angle \mathrm{BOC}=180° \\ \angle \mathrm{BOC}=180°-100° \\ \angle \mathrm{BOC}=80° \\ \angle AOD=\angle BOC=80° \left(\mathrm{ve}\mathrm{rtically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle \mathrm{AOB}=\angle \mathrm{DOC}=100° \left(\mathrm{ve}\mathrm{rtically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle \mathrm{CAB}=70° \left(\mathrm{given}\right) \\ \angle \mathrm{ADB}=20° \\ \angle DBA=\angle BDC=10° (alternate angle) \\ \angle ADB=\angle DBC=20° (alternate angle) \end{gathered}$$

Question 20:

Find the angles marked with a question mark shown in Fig. 17.27

Answer 20:

$$\begin{gathered} \mathrm{In} △\mathrm{CEB}: \\ \angle \mathrm{ECB}+\angle \mathrm{CBE}+\angle \mathrm{BEC}=180° (angle sum property of a triangle) \\ 40°+90°+\angle \mathrm{EBC}=180° \\ \therefore \angle \mathrm{EBC}=50° \\ \mathrm{Also}, \angle \mathrm{EBC}=\angle \mathrm{ADC}=50° \left(\mathrm{opposite} \mathrm{angle} \mathrm{of} \mathrm{a} \mathrm{parallelogram}\right) \\ \mathrm{In} △\mathrm{FDC}: \\ \angle \mathrm{FDC}+\angle \mathrm{DCF}+\angle \mathrm{CFD}=180° \\ 50°+90°+\angle \mathrm{DCF}=180° \\ \therefore \angle \mathrm{DCF}=40° \\ \mathrm{Now}, \angle \mathrm{BCE}+\angle \mathrm{ECF}+\angle \mathrm{FCD}+\angle FDC=180° (\mathrm{in} \mathrm{a} \mathrm{parallelogram}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{alternate} \mathrm{angles} \mathrm{is} 180° ) \\ 50°+40°+\angle \mathrm{ECF}+40°=180° \\ \angle \mathrm{ECF}=180°-50°+40°-40°=50° \end{gathered}$$

Question 21:

The angle between the altitudes of a parallelogram, through the same vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.

Answer 21:

$$\begin{gathered} \mathrm{Draw} \mathrm{a} \mathrm{parallelogram} \mathrm{ABCD}. \\ \mathrm{Drop} \mathrm{a} \mathrm{perpendicular} \mathrm{from} \mathrm{B} \mathrm{to} \mathrm{the} \mathrm{side} \mathrm{AD}, \mathrm{at} \mathrm{the} \mathrm{point} \mathrm{E}. \\ \mathrm{Drop} \mathrm{perpendicular} \mathrm{from} \mathrm{B} \mathrm{to} \mathrm{the} \mathrm{side} \mathrm{CD}, \mathrm{at} \mathrm{the} \mathrm{point} \mathrm{F}. \\ \mathrm{In} \mathrm{the} \mathrm{quadrilateral} \mathrm{BEDF}: \\ \angle \mathrm{EBF}=60°,\angle \mathrm{BED}=90° \\ \angle \mathrm{BFD}=90° \\ \angle \mathrm{EDF}=360°-(60°+90°+90°)=120° \\ \mathrm{In} \mathrm{a} \mathrm{parallelogram}, \mathrm{opposite} \mathrm{angles} \mathrm{are} \mathrm{congruent} \mathrm{and} \mathrm{adjacent} \mathrm{angles} \mathrm{are} \mathrm{supplementary}. \\ \mathrm{In} \mathrm{the} \mathrm{parallelogram} \mathrm{ABCD}: \\ \angle \mathrm{B}=\angle \mathrm{D}=120° \\ \angle \mathrm{A}=\angle \mathrm{C}=180°-120°=60° \end{gathered}$$

Question 22:

In Fig. 17.28, ABCD and AEFG are parallelograms. If ∠C = 55°, what is the measure of ∠F?

Answer 22:

$$\begin{gathered} \mathrm{Both} \mathrm{the} \mathrm{parallelograms} \mathrm{ABCD} \mathrm{and} \mathrm{AEFG} \mathrm{are} \mathrm{similar}. \\ \therefore \angle \mathrm{C}=\angle \mathrm{A}=55° (\mathrm{opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \therefore \angle \mathrm{A}=\angle F=55° (\mathrm{opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \end{gathered}$$

Question 23:

In Fig. 17.29, BDEF and DCEF are each a parallelogram. Is it true that BD = DC? Why or why not?

Answer 23:

$$\begin{gathered} \mathrm{In} \mathrm{parallelogram} \mathrm{BDEF} \\ \therefore \mathrm{BD}=\mathrm{EF} ...\left(\mathrm{i}\right) (\mathrm{opposite} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \mathrm{In} \mathrm{parallelogram} \mathrm{DCEF} \\ \mathrm{CD}=\mathrm{EF} ...\left(\mathrm{ii}\right) (\mathrm{opposite} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \mathrm{From} \mathrm{equations} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \\ \mathrm{BD}=\mathrm{CD} \end{gathered}$$

Page-17.12

Question 24:

In Fig. 17.29, suppose it is known that DE = DF. Then, is ΔABC isosceles? Why or why not?
Fig. 17.29

Answer 24:

$$\begin{gathered} \mathrm{In} ∆\mathrm{FDE}: \\ \mathrm{DE}=\mathrm{DF} \\ \therefore \angle \mathrm{FED}=\angle \mathrm{DFE}.............\left(\mathrm{i}\right) (\mathrm{angles} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{sides}) \\ \mathrm{In} \mathrm{the} {\mathrm{II}}^{\mathrm{gm} }\mathrm{BDEF}: \\ \angle \mathrm{FBD}= \angle \mathrm{FED}.......\left(\mathrm{ii}\right) (\mathrm{opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \mathrm{In} \mathrm{the} {\mathrm{II}}^{\mathrm{gm} } \mathrm{DCEF}: \\ \angle \mathrm{DCE}=\angle \mathrm{DFE}......\left(\mathrm{iii}\right) (\mathrm{opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \mathrm{From} \mathrm{equations} \left(\mathrm{i}\right), \left(\mathrm{ii}\right) \mathrm{and} \left(\mathrm{iii}\right): \\ \angle \mathrm{FBD}=\angle \mathrm{DCE} \\ \mathrm{In} △\mathrm{ABC}: \\ \mathrm{If} \angle \mathrm{FBD}=\angle \mathrm{DCE}, \mathrm{then} \mathrm{AB}=\mathrm{AC} (\mathrm{sides} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{angles}). \\ \mathrm{Hence}, △\mathrm{ABC} \mathrm{is} \mathrm{isosceles}. \end{gathered}$$

Question 25:

Diagonals of parallelogram ABCD intersect at O as shown in Fig. 17.30. XY contains O, and X, Y are points on opposite sides of the parallelogram. Give reasons for each of the following:
(i) OB = OD
(ii) ∠OBY = ∠ODX
(iii) ∠BOY = ∠DOX
(iv) ∆BOY ≅ ∆DOX
Now, state if XY is bisected at O.

Answer 25:

(i) Diagonals of a parallelogram bisect each other.
(ii) Alternate angles
(iii) Vertically opposite angles
(iv)
 $$\begin{gathered} \mathrm{In} ∆\mathrm{BOY} \mathrm{and} ∆\mathrm{DOX}: \\ \mathrm{OB}=\mathrm{OD} (d\mathrm{iagonals} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{bisect} \mathrm{each} \mathrm{o}\mathrm{ther}) \\ \angle \mathrm{OBY}=\angle \mathrm{ODX} (a\mathrm{lternate} \mathrm{a}\mathrm{ngles}) \\ \angle \mathrm{BOY}=\angle \mathrm{DOX} (v\mathrm{ertically} \mathrm{o}\mathrm{pposite} \mathrm{a}\mathrm{ngles}) \end{gathered}$$


ASA congruence:
XO = YO (c.p.c.t)
So, XY is bisected at O.

Question 26:

In Fig. 17.31, ABCD is a parallelogram, CE bisects ∠C and AF bisects ∠A. In each of the following, if the statement is true, give a reason for the same:


(i) ∠A = ∠C
(ii) $\angle FAB=\frac{1}{2}\angle A$
(iii) $\angle DCE=\frac{1}{2}\angle C$
(iv) $\angle CEB=\angle FAB$
(v) CE || AF

Answer 26:

(i) True, since opposite angles of a parallelogram are equal.
(ii) True, as AF is the bisector of $\angle$A.
(iii) True, as CE is the bisector of $\angle$C.
(iv) True
               $\angle$CEB = $\angle$DCE........(i)  (alternate angles)
               $\angle$DCE= $\angle$ FAB.........(ii)    (opposite angles of a parallelogram are equal)
            
  From equations (i) and (ii):
              $\angle$CEB = $\angle$FAB
                
(v) True, as corresponding angles are equal ($\angle$CEB = $\angle$FAB).

Question 27:

Diagonals of a parallelogram ABCD intersect at O. AL and CM are drawn perpendiculars to BD such that L and M lie on BD. Is AL = CM? Why or why not?

Answer 27:

$$\begin{gathered} \mathrm{In} \mathrm{\Delta AOL} \mathrm{and} \mathrm{\Delta CMO}: \\ \angle \mathrm{AOL}=\angle \mathrm{COM}( v\mathrm{ertically} \mathrm{opposite} \mathrm{angle})....\left(i\right) \\ \angle \mathrm{ALO}=\angle \mathrm{CMO}=90° (e\mathrm{ach} \mathrm{right} \mathrm{angle}).....\left(ii\right) \\ \mathrm{Usin}g angle sum property: \\ \angle \mathrm{AOL}+\angle \mathrm{ALO}+\angle LAO=180°..........\left(iii\right) \\ \angle COM+\angle CMO+\angle OCM=180°......\left(iv\right) \\ From equations \left(iii\right) and \left(iv\right): \\ \angle \mathrm{AOL}+\angle \mathrm{ALO}+\angle LAO=\angle COM+\angle CMO+\angle OCM \\ \angle LAO=\angle OCM (from equations (i) and (ii) ) \\ \mathrm{In} \mathrm{\Delta AOL} \mathrm{and} \mathrm{\Delta CMO}: \\ \angle \mathrm{ALO}=\angle \mathrm{CMO} (e\mathrm{ach} \mathrm{right} \mathrm{angle}) \\ \mathrm{AO}=\mathrm{OC} (\mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{bisect} \mathrm{each} \mathrm{other}) \\ \angle LAO=\angle OCM (proved above) \\ \mathrm{So}, \mathrm{\Delta AOL} \mathrm{is} \mathrm{congruent} \mathrm{to} \mathrm{\Delta CMO} \left(S\mathrm{AS}\right). \\ \Rightarrow \mathrm{AL}=\mathrm{CM} \left[\mathrm{cpct}\right] \end{gathered}$$

Question 28:

Points E and F lie on diagonal AC of a parallelogram ABCD such that AE = CF. What type of quadrilateral is BFDE?

Answer 28:

$$\begin{gathered} \mathrm{In} \mathrm{the} {\mathrm{II}}^{\mathrm{gm}} \mathrm{ABCD}: \\ \mathrm{AO}=\mathrm{OC}......\left(\mathrm{i}\right) (\mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{bisect} \mathrm{each} \mathrm{other}) \\ \mathrm{AE}=\mathrm{CF}.......\left(\mathrm{ii}\right) \left(\mathrm{given}\right) \\ \mathrm{Subtracting} \left(\mathrm{ii}\right) \mathrm{from} \left(\mathrm{i}\right): \\ \mathrm{AO}-\mathrm{AE}=\mathrm{OC}-\mathrm{CF} \\ \mathrm{EO}=\mathrm{OF}......... \left(\mathrm{iii}\right) \\ \mathrm{In} ∆\mathrm{DOE} \mathrm{and} ∆\mathrm{BOF}: \\ \mathrm{EO}=\mathrm{OF} (\mathrm{proved} \mathrm{above}) \\ \mathrm{DO}=\mathrm{OB} (\mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{bisect} \mathrm{each} \mathrm{other}) \\ \angle \mathrm{DOE} =\angle \mathrm{BOF} (\mathrm{vertically} \mathrm{opposite} \mathrm{angles}) \\ \mathrm{By} \mathrm{SAS} \mathrm{congruence}: \\ ∆\mathrm{DOE}\cong ∆\mathrm{BOF} \\ \therefore \mathrm{DE}=\mathrm{BF} (\mathrm{c}.\mathrm{p}.\mathrm{c}.\mathrm{t}) \\ \mathrm{In} ∆\mathrm{BOE} \mathrm{and} ∆\mathrm{DOF}: \\ \mathrm{EO}=\mathrm{OF} (\mathrm{proved} \mathrm{above}) \\ \mathrm{DO}=\mathrm{OB} (\mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{bisect} \mathrm{each} \mathrm{other}) \\ \angle \mathrm{DOF} =\angle \mathrm{BOE} (\mathrm{vertically} \mathrm{opposite} \mathrm{angles}) \\ \mathrm{By} \mathrm{SAS} \mathrm{congruence}: \\ ∆\mathrm{DOE}\cong ∆\mathrm{BOF} \\ \therefore \mathrm{DF}=\mathrm{BE} (\mathrm{c}.\mathrm{p}.\mathrm{c}.\mathrm{t}) \\ \mathrm{Hence}, \mathrm{the} \mathrm{pair} \mathrm{of} \mathrm{opposite} \mathrm{sides} \mathrm{are} \mathrm{equal}. \mathrm{Thus}, \mathrm{DEBF} \mathrm{is} \mathrm{a} \mathrm{parallelogram}. \end{gathered}$$

Question 29:

In a parallelogram ABCD, AB = 10 cm, AD = 6 cm. The bisector of ∠A meets DC in E, AE and BC produced meet at F. Find te length CF.

Answer 29:

$$\begin{gathered} \mathrm{AE} \mathrm{is} \mathrm{the} \mathrm{bisector} \mathrm{of} \angle \mathrm{A}. \\ \therefore \angle \mathrm{DAE}=\angle \mathrm{BAE} =x \\ \angle \mathrm{BAE}=\angle \mathrm{AED}=x (\mathrm{alternate} \mathrm{angles}) \\ \mathrm{Since} \mathrm{opposite} \mathrm{angles} \mathrm{in} ∆\mathrm{ADE} \mathrm{are} \mathrm{equal}, ∆\mathrm{ADE} \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{triangle}. \\ \therefore \mathrm{AD}=\mathrm{DE}= 6 \mathrm{cm} (\mathrm{sides} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{angles}) \\ \mathrm{AB}=\mathrm{CD}=10 \mathrm{cm} \\ \mathrm{CD}= \mathrm{DE}+ \mathrm{EC} \\ \Rightarrow \mathrm{EC}=\mathrm{CD}-\mathrm{DE} \\ \Rightarrow \mathrm{EC}=10-6=4 \mathrm{cm} \\ \angle \mathrm{DEA}=\angle \mathrm{CEF}=x (\mathrm{vertically} \mathrm{opposite} \mathrm{angle}) \\ \angle \mathrm{EAD}=\angle \mathrm{EFC}=x (\mathrm{alternate} \mathrm{angles}) \\ \mathrm{Since} \mathrm{opposite} \mathrm{angles} \mathrm{in} ∆\mathrm{EFC} \mathrm{are} \mathrm{equal}, ∆\mathrm{EFC} \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{triangle}. \\ \therefore \mathrm{CF}=\mathrm{CE}= 4 \mathrm{cm} (\mathrm{sides} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{angles}) \\ \therefore \mathrm{CF}= 4\mathrm{cm} \end{gathered}$$