RS Aggarwal 2019,2020 solution class 7 chapter 14 Property of Parallel Lines Exercise 14

Exercise 14

Page-177

Question 1:

In the given figure, l || m and t is a transversal.
If ∠5 = 70°, find the measure of each of the angles
∠1, ∠3, ∠4 and ∠8.

Answer 1:

$$\begin{gathered} \mathrm{Given}: \mathrm{l}\parallel \mathrm{m} \\ \mathrm{t} \mathrm{is} \mathrm{a} \mathrm{transversal}. \\ \angle 5 = 70° \\ \angle 5 = \angle 3 =70° (\mathrm{alternate} \mathrm{interior} \mathrm{angles}) \\ \angle 5 +\angle 8 = 180° (\mathrm{linear} \mathrm{pair}) \\ \mathrm{or} 70° + \angle 8 = 180° \\ \angle 8 = 110° \\ \angle 1 = \angle 3 = 70° (\mathrm{vertically} \mathrm{opposite} \mathrm{angles}) \\ \angle 3 +\angle 4 = 180° (\mathrm{linear} \mathrm{pair}) \\ \mathrm{or} \angle 4 = 180-\angle 3 = 180 - 70 = 110° \end{gathered}$$

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Question 2:

In the given figure, l || m and t is a transversal. If ∠1 and ∠2 are in the ratio 5 : 7, find the measure of each of the angles
∠1, ∠2, ∠3 and ∠8.

Answer 2:

$$\begin{gathered} \mathrm{Given}: l\parallel m \\ \mathrm{t} \mathrm{is} \mathrm{a} \mathrm{transversal}. \\ \angle 1:\angle 2 = 5:7 \\ \mathrm{Let} \mathrm{the} \mathrm{angles} \mathrm{measure} 5\mathrm{x} \mathrm{and} 7\mathrm{x}. \\ \angle 1+\angle 2 = 180° (\mathrm{linear} \mathrm{pair}) \\ \therefore 5\mathrm{x} + 7\mathrm{x} = 180 \\ \mathrm{or} 12\mathrm{x} = 180 \\ \mathrm{or} \mathrm{x} = 15 \\ \therefore \angle 1 = 5\mathrm{x} = 5\left(15\right) = 75° \\ \mathrm{and} \angle 2 = 7\mathrm{x} = 7\left(15\right) = 105° \\ \angle 2+\angle 3 = 180° (\mathrm{linear} \mathrm{pair}) \\ \angle 3 = 180-105 = 75° \\ \angle 3+\angle 6 = 180 (\mathrm{interior} \mathrm{angles} \mathrm{on} \mathrm{the} \mathrm{same} \mathrm{side} \mathrm{of} the \mathrm{transversal} \mathrm{are} \mathrm{supplementary}) \\ \angle 6 = 180-\angle 3 = 105° \\ \mathrm{and} \angle 6 =\angle 8 =105° (\mathrm{vertically} \mathrm{opposite} \mathrm{angles}) \\ \therefore \angle 1 = 75° \\ \angle 2 = 105° \\ \angle 3 = 75° \\ \angle 8 = 105° \end{gathered}$$

Question 3:

Two parallel lines l and m cut by a transversal t. If the interior angles of the same side of t be (2x − 8)° and (3x − 7)°, find the measure of each of these angles.

Answer 3:

$$\begin{gathered} \mathrm{Given}: \mathrm{l}\parallel \mathrm{m} \\ \mathrm{t} \mathrm{is} \mathrm{a} \mathrm{transversal}. \\ \mathrm{Let}: \\ \angle 1 =(2\mathrm{x}-8)° \\ \angle 2 = (3\mathrm{x}-7)° \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{consecutive} \mathrm{interior} \mathrm{angles} \mathrm{are} \mathrm{supplementary}. \\ \therefore \angle 1 +\angle 2 =180° \\ \mathrm{or} (2\mathrm{x}-8) + (3\mathrm{x}-7) =180 \\ \mathrm{or} 5\mathrm{x} -15 = 180 \\ \mathrm{or} 5x = 195 \\ \mathrm{or} \mathrm{x} = 39 \\ \angle 1 = (2\mathrm{x}-8) = (2\times 39-8) = 70° \\ \angle 2 =(3\mathrm{x}-7) = (3\times 39-7) = 110° \end{gathered}$$

Question 4:

In the given figure, l || m. If s and t be transversals such that s is not parallel to t. find the values of x and y.

Answer 4:

From the given figure:

$$\begin{gathered} \angle 1 =\angle 3 = 50° (\mathrm{corresponding} \mathrm{angles}) \\ \mathrm{and} \angle 1 + \mathrm{x}° = 180° (\mathrm{linear} \mathrm{pair}) \\ \mathrm{or} \mathrm{x}° = 180° - 50° = 130° \\ \mathrm{or} x = 130 \\ \angle 2 =\angle 4 = 65° (\mathrm{corresponding} \mathrm{angles}) \\ \mathrm{and} \angle 2 + \mathrm{y}° = 180° (\mathrm{linear} \mathrm{pair}) \\ \mathrm{or} \mathrm{y}° = 180° - 65° =115° \\ \mathrm{or} \mathrm{y}= 115 \end{gathered}$$

Question 5:

In the given figure, ∠B = 65° and ∠C = 45° in ∆ABC and DAE || BC. If ∠DAB = x° and ∠EAC = y°, find the values of x and y.

Answer 5:

$$\begin{gathered} \mathrm{Given}: \\ \angle \mathrm{B} = 65° \\ \angle \mathrm{C} = 45° \\ \mathrm{DAE}\parallel \mathrm{BC} \\ \mathrm{The} \mathrm{given} \mathrm{lines} \mathrm{are} \mathrm{parallel}. \\ \therefore \mathrm{x}° = \angle \mathrm{B} = 65° (\mathrm{alternate} \mathrm{angles} \mathrm{when} \mathrm{AB} \mathrm{is} \mathrm{taken} \mathrm{as} \mathrm{the} \mathrm{transversal}) \\ \mathrm{y}° = \angle \mathrm{C} =45° (\mathrm{alternate} \mathrm{angles} \mathrm{when} \mathrm{AC} \mathrm{is} \mathrm{taken} \mathrm{as} \mathrm{the} \mathrm{transversal}) \\ \therefore \mathrm{x} = 65 \\ \mathrm{y} =45 \end{gathered}$$

Question 6:

In the adjoining figure, it is given that CE || BA, ∠BAC = 80° and ∠ECD = 35°.
Find (i) ∠ACE, (ii) ∠ACB,  (iii) ∠ABC.

Answer 6:

$$\begin{gathered} \mathrm{Given}: \mathrm{CE} \parallel \mathrm{BA} \\ \angle \mathrm{BAC}= 80°, \angle \mathrm{ECD} = 35° \\ \left(\mathrm{i}\right) \angle \mathrm{BAC} = \angle \mathrm{ACE} = 80° (\mathrm{alternate} \mathrm{angles} \mathrm{with} \mathrm{AC} \mathrm{as} \mathrm{a} \mathrm{transversal}) \\ \left(\mathrm{ii}\right) \angle \mathrm{ACB} + \angle \mathrm{ACD} = 180° (\mathrm{linear} \mathrm{pair}) \\ \mathrm{or} \angle \mathrm{ACB} + \angle \mathrm{ACE} + \angle \mathrm{ECD} = 180° \\ \mathrm{or}\angle \mathrm{ACB} + 80°+35° =180° \\ \mathrm{or} \angle \mathrm{ACB} = 65° \\ \left(\mathrm{iii}\right) \mathrm{In} ∆\mathrm{ABC}: \\ \angle \mathrm{BAC} + \angle \mathrm{ACB} +\angle \mathrm{ABC} = 180° (\mathrm{angle} \mathrm{sum} \mathrm{property}) \\ 80° +65° + \angle \mathrm{ABC} = 180° \\ \angle \mathrm{ABC} = 35° \end{gathered}$$

Question 7:

In the adjoining figure, it is being given that AO || CD, OB || CE and ∠AOB = 50°
Find the measure of ∠ECD.

Answer 7:

$$\begin{gathered} \mathrm{Given}: \mathrm{AO}\parallel \mathrm{CD} \\ \mathrm{OB} \parallel \mathrm{CE} \\ \angle \mathrm{AOB} = 50° \\ \angle \mathrm{AOD} = \angle \mathrm{CDB} = 50° (\mathrm{when} \mathrm{AO}\parallel \mathrm{CD} \mathrm{and} \mathrm{OB} \mathrm{is} \mathrm{the} \mathrm{transversal}) \\ \angle \mathrm{ECD} +\angle \mathrm{CDB} = 180° (\mathrm{consecutive} \mathrm{interior} \mathrm{angles} \mathrm{are} \mathrm{supplementary}, \mathrm{DB} \parallel \mathrm{CE} \mathrm{and} CD \mathrm{is} the \mathrm{transversal}) \\ \angle \mathrm{ECD} = 180°-50° = 130° \end{gathered}$$

Question 8:

In the adjoining figure, it is given that AB || CD, ∠AOB = 50° and ∠CDO = 40°.
Find the measure of ∠BOD.

Answer 8:

$$\begin{gathered} \mathrm{Given}: \mathrm{AB}\parallel \hspace{0.17em}\mathrm{CD} \\ \angle \mathrm{ABO} = 50° \\ \angle \mathrm{CDO} = 40° \\ \mathrm{Construction}: \mathrm{Through} \mathrm{O}, \mathrm{draw} \mathrm{EOF}\parallel \mathrm{AB}. \\ \angle \mathrm{ABO} = \angle \mathrm{BOF} = 50° (\mathrm{alternate} \mathrm{angles}, \mathrm{when} \mathrm{AB}\parallel \mathrm{EF} \mathrm{and} \mathrm{OB} \mathrm{is} \mathrm{a} \mathrm{transversal}) \\ \angle \mathrm{FOD} = \angle \mathrm{ODC} = 40° (\mathrm{alternate} \mathrm{angles}, \mathrm{when} \mathrm{CD}\parallel \mathrm{EF} \mathrm{and} \mathrm{OD} \mathrm{is} \mathrm{a} \mathrm{transversal}) \\ \angle \mathrm{BOD} = \angle \mathrm{BOF} + \angle \mathrm{FOD} \\ \angle \mathrm{BOD} = 50°+40° =90° \end{gathered}$$

Question 9:

In the given figure, AB || CD and a transversal EF cuts them at G and H respectively.
If GL and HM are the bisectors of the alternate angles ∠AGH and ∠GHD respectively, prove that GL || HM.

Answer 9:

$$\begin{gathered} \mathrm{Given}: \mathrm{AB} \parallel \mathrm{CD} \\ \mathrm{GL} \mathrm{and} \mathrm{HM} \mathrm{are} \mathrm{angle} \mathrm{bisectors} \mathrm{of} \angle \mathrm{AGH} \mathrm{and} \angle \mathrm{GHD}, \mathrm{respectively}. \\ \angle \mathrm{AGH}=\angle \mathrm{GHD} (\mathrm{alternate} \mathrm{angles}) \\ \mathrm{or} \frac{1}{2} \angle \mathrm{AGH}=\frac{1}{2}\angle \mathrm{GHD} \\ \mathrm{or} \angle \mathrm{LGH} = \angle \mathrm{GHM} \left(\mathrm{given}\right) \\ \mathrm{Therefore}, \mathrm{GL} \parallel \mathrm{HM} \mathrm{as} \mathrm{we} \mathrm{know} \mathrm{that} \mathrm{if} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{any} \mathrm{pair} \mathrm{of} \mathrm{alternate} \mathrm{interior} \mathrm{angles} \mathrm{are} \mathrm{equal}, \mathrm{then} \mathrm{the} \mathrm{lines} \mathrm{are} \mathrm{parallel}. \end{gathered}$$

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Question 10:

In the given figure, AB || CD,
∠ABE = 120°, ∠ECD = 100° and ∠BEC = x°
Find the value of x.

Answer 10:

$$\begin{gathered} \mathrm{Given}: \mathrm{AB} \parallel \mathrm{CD} \\ \angle \mathrm{ABE} = 120° \\ \angle \mathrm{ECD} = 100° \\ \angle \mathrm{BEC} = \mathrm{x}° \\ \mathrm{Construction}: \mathrm{FEG} \parallel \mathrm{AB} \\ \mathrm{Now}, \sin ce AB\parallel FEG \mathrm{and} AB\parallel CD, FEG\parallel CD \\ \therefore EFG\parallel AB\parallel CD \\ \angle \mathrm{ABE} = \angle \mathrm{BEG} = 120° (\mathrm{alternate} \mathrm{angles}) \\ \mathrm{or} \mathrm{x}°+\mathrm{y}° = 120° ....\left(\mathrm{i}\right) \\ \angle \mathrm{DCE} = \angle \mathrm{CEF} = 100° (\mathrm{alternate} \mathrm{angles}) \\ \mathrm{or} x°+\mathrm{z}° = 100° .....\left(\mathrm{ii}\right) \\ \mathrm{Also}, \mathrm{x}°+\mathrm{y}°+\mathrm{z}° = 180° (\mathrm{FEG} \mathrm{is} \mathrm{a} s\mathrm{traight} \mathrm{line}) ...\left(\mathrm{iii}\right) \\ \mathrm{Adding} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right): \\ 2\mathrm{x}°+\mathrm{y}°+\mathrm{z}° = 220° \\ \mathrm{or}, \mathrm{x}° +180° = 220° (\mathrm{substi}\mathrm{tuting} (\mathrm{iii}\left)\right) \\ \mathrm{x}° = 40° \\ \therefore x = 40 \end{gathered}$$

Question 11:

In the given figure, ABCD is a quadrilateral in which AB || DC and AD || BC.
Prove that ∠ADC = ∠ABC.

Answer 11:

$$\begin{gathered} \mathrm{Given}: \mathrm{AB}\parallel \mathrm{CD} \\ \mathrm{AD}\parallel \mathrm{BC} \\ \angle 1 + \angle 2 = 180° (\mathrm{AB}\parallel \mathrm{CD} \mathrm{and} \mathrm{AD} \mathrm{is} \mathrm{the} \mathrm{transversal}) ...\left(\mathrm{i}\right) \\ \angle 2+ \angle 3 =180° (\mathrm{AD}\parallel \mathrm{BC} \mathrm{and} \mathrm{AB} \mathrm{is} \mathrm{the} \mathrm{transversal}) ...\left(\mathrm{ii}\right) \\ \mathrm{From} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right): \\ \angle 1 + \angle 2 = 180° = \angle 2+ \angle 3 \\ \angle 1 = \angle 3 \\ \angle \mathrm{ADC} = \angle \mathrm{ABC} \end{gathered}$$

Question 12:

In the given figure, l || m and p || q.
Find the measure of each of the angles ∠a, ∠b, ∠c and ∠d.

Answer 12:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{l}\parallel \mathrm{m} \\ \mathrm{p}\parallel \mathrm{q} \\ \angle 1 = 65° \\ \therefore \angle 1 = \angle a = 65° (\mathrm{vertically} \mathrm{opposite} \mathrm{angles}) \\ \angle a + \angle \mathrm{d} = 180° (\mathrm{consecutive} \mathrm{interior} \mathrm{angles} \mathrm{on} \mathrm{the} \mathrm{same} \mathrm{side} \mathrm{of} \mathrm{a} \mathrm{transversal} \mathrm{are} \mathrm{supplementary}) \\ \mathrm{or} \angle \mathrm{d} = 180°-65°= 115° \\ \angle c+ \angle \mathrm{d} = 180° (\mathrm{consecutive} \mathrm{interior} \mathrm{angles} \mathrm{on} \mathrm{the} \mathrm{same} \mathrm{side} \mathrm{of} \mathrm{a} \mathrm{transversal} \mathrm{are} \mathrm{supplementary}) \\ \mathrm{or} \angle \mathrm{c} = 180°-115°= 65° \\ \angle \mathrm{c}+ \angle \mathrm{b} = 180° (\mathrm{consecutive} \mathrm{interior} \mathrm{angles} \mathrm{on} \mathrm{the} \mathrm{same} \mathrm{side} \mathrm{of} \mathrm{a} \mathrm{transversal} \mathrm{are} \mathrm{supplementary}) \\ \mathrm{or} \angle \mathrm{b} = 180°-65°= 115° \\ \therefore \angle a = 65° \\ \angle \mathrm{b} = 115° \\ \angle \mathrm{c}= 65° \\ \angle \mathrm{d}= 115° \end{gathered}$$

Question 13:

In the given figure, AB || DC and AD || BC, and AC is a diagonal. If ∠BAC = 35°, ∠CAD = 40°, ∠ACB = x° and ∠ACD = y°, find the value of x and y

Answer 13:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{AB}\parallel \mathrm{DC} \\ \mathrm{AD}\parallel \mathrm{BC} \\ \angle \mathrm{BAC} = 35° \\ \angle \mathrm{CAD} =40° \\ \therefore \angle \mathrm{BAC} = \mathrm{y}= 35° (\mathrm{alternate} \mathrm{angles} \mathrm{when} \mathrm{AB}\parallel \mathrm{DC}) \\ \angle \mathrm{CAD} = \mathrm{x} = 40° (\mathrm{alternate} \mathrm{angles} \mathrm{when} \mathrm{AD}\parallel \mathrm{BC}) \\ \therefore \mathrm{x} = 40 \\ \mathrm{y}= 35 \end{gathered}$$

Question 14:

In the given figure, AB || CD and CA has been produced to E so that ∠BAE = 125°.
If ∠BAC = x°, ∠ABD = x°, ∠BDC = y° and ∠ACD = z°, find the values of x, y, z.

Answer 14:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{AB} \parallel \mathrm{CD} \\ \angle \mathrm{BAE} = 125° \\ \angle \mathrm{CAB} + \angle \mathrm{BAE} =180° \\ \mathrm{or} 125° + \mathrm{x}°= 180° \\ \mathrm{or} \mathrm{x} = 55 \\ \mathrm{x} + \mathrm{z} =180° (\mathrm{c}\mathrm{onsecutive} \mathrm{interior} \mathrm{angles} \mathrm{on} \mathrm{the} \mathrm{same} \mathrm{side} \mathrm{of} \mathrm{transversal} \mathrm{are} \mathrm{s}\mathrm{upplementary}) \\ \mathrm{z} =180-\mathrm{x} = 180 -55 = 125 \\ \mathrm{y} + \mathrm{x} =180° (\mathrm{c}\mathrm{onsecutive} \mathrm{interior} \mathrm{angles} \mathrm{on} \mathrm{the} \mathrm{same} \mathrm{side} \mathrm{of} \mathrm{transversal} \mathrm{are} \mathrm{s}\mathrm{upplementary}) \\ \mathrm{y} =180- \mathrm{x} = 180- 55 = 125 \end{gathered}$$

Question 15:

In each of the given figures, two lines l and m are cut by a transevrsal t.
Find whether l || m.

Answer 15:

$$\begin{gathered} \left(\mathrm{i}\right) \angle 1+\angle 2 = 180 (\mathrm{linear} \mathrm{pair}) \\ \mathrm{or} {130}^{°} + \angle 2 = 180° \\ \mathrm{or} \angle 2 = 50° \ne 40° =\angle 3 \\ \therefore \mathrm{l}\nparallel \mathrm{m} \\ \left(\mathrm{ii}\right) \angle 2+\angle 3 = 180° (\mathrm{linear} \mathrm{pair}) \\ 35°+\angle 3 = 180° \\ \angle 3 = 145°= 145° = \angle 1 \\ \therefore \mathrm{l}\parallel \mathrm{m} \\ \left(\mathrm{iii}\right)\angle 2+\angle 3 = 180 (\mathrm{linear} \mathrm{pair}) \\ \angle 3 =180°- 125° = 55° \\ \angle 3 =55°\ne 60° = \angle 1 \\ \therefore \mathrm{l}\nparallel \mathrm{m} \end{gathered}$$