Class 8: Lines and Angles – Exercise 28A – ICSE / ISC / CBSE Mathematics Portal for K12 Students

281 Question 1: In the adjourning figure, name:

i)six points

ii) five line segments

iii) four rays

iv) four lines

v) four collinear points

Answer:

i) The point are $M, N, P, Q , X, Y$

ii) Five line segments are $\overline{XM},\overline{MP},\overline{YN},\overline{NQ},\overline{MN},\overline{PQ}$

iii) Four Rays are $\overrightarrow{PB}, \overrightarrow{QD}, \overrightarrow{XA}, \overrightarrow{YC}$

iv) Four lines are $\overleftrightarrow{AB}, \overleftrightarrow{CD}, \overleftrightarrow{EF}, \overleftrightarrow{HG}$

v) Four Collinear points are $A, X, M, P \ or\ C, Y, N, Q \ or\ X, M, P, B$

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282 Question 2: In the adjoining figure, name :

i) two pairs of intersecting lines and their corresponding points of intersection

ii) three concurrent lines and their points of concurrence

iii) three rays

iv) two line segments

Answer:

i) Two points of intersecting lines are

$\overleftrightarrow{EF}, \overleftrightarrow{GH}$ intersecting at R

$\overleftrightarrow{CD}, \overleftrightarrow{GH}$ intersecting at Q.

ii) Three concurrent lines and their point of concurrence.

$\overleftrightarrow{AB}, \overleftrightarrow{EF}, \overleftrightarrow{GH}$ and point of concurrence is R

iii) Three rays are, $\overrightarrow{QH}, \overrightarrow{PB},\overrightarrow{PD}$ some other rays are $\overrightarrow{RA}, \overrightarrow{RE},\overrightarrow{RG}$ , etc.

iv) Two line segments are $\overline{QR}, \overline{PQ},\overline{RP}$

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Question 3: State whether the following statements are true or false :

Answer:

i) A ray has no end Point: $False$

ii) A line AB is the same as line BA: $True$

iii) A ray AB is the same as BA: $False$

iv) A line has a definite length: $False$

v) Two planes always meet in a line: $True$

vi) A plane has length and breadth but no thickness: $True$

vii) Two distinct points always determine a unique line: $True$

viii) Two lines may intersect in two points: $False$

ix) Two intersecting lines cannot be both parallel to the same line: $True$

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Question 4: Two adjacent angles on a straight line are $x^{\circ} \ and\ (2x - 21)^{\circ}$ Find i) the value of $x$ ii) the measure of each angle

Answer: 283

i) $\angle AOB+\angle COB=180^{\circ}$

$2x-21+x=180^{\circ}$

$3x=201$

$x=67^{\circ}$

ii) Hence $\angle COB=67^{\circ} \ and\ \angle AOB=113^{\circ}$

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Question 5: Two adjacent angles on a straight line are $(3x - 2)^{\circ} \ and\ 4(x + 7)^{\circ}$ – Find : i) the value of $x$ ii) the measure of each angle

Answer:

i) $3x-2+4(x+7)=180^{\circ}$

$3x+4x+26=180^{\circ}$

$7x=154$

$or x=22$

ii) The measure of angles

$Angle_1 = 3\times 22-2=64^{\circ}$

$Angle_2 = 4\times (22+7)=116^{\circ}$

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Question 6: Two adjacent angles on a straight line are in the ratio $3:2$ . Find the measure of each angle:

Answer:

The ratio of angles $= 3\colon 2$

Therefore the angles are $3x$ and $2x$

$3x+2x=180^{\circ}$

$\Rightarrow 5x=180^{\circ}$

$\Rightarrow x=36^{\circ}$

The two angles are $108^{\circ} \ and\ 72^{\circ}$

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Question 7: In the adjoining figure, $AOB$ is a straight line. Find the value $x$ . Hence ,find, $\angle AOC \ and\ \angle BOD$

Answer: 284

$\angle AOC+\angle COD+\angle DOB=180^{\circ}$

$\Rightarrow 3x-5+55+x+20=180^{\circ}$

$\Rightarrow 4x=110^{\circ}$

$\Rightarrow x=27.5^{\circ}$

Therefore $\angle AOC = 3\times 22-5=77.5^{\circ}$

And $\angle BOD = 22+20=47.5^{\circ}$

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Question 8: In the adjoining figure, AOB is a straight line. If $x \colon y \colon z = 6 \colon 5 \colon 4$ , find the values of $x, \ y$ and $z$ .

Answer: 285

$x\colon y\colon z=6\colon 5\colon 4$

Therefore

$6a+5a+4a=180^{\circ}$

Or $a=12$ Therefore $x=72^{\circ} , y=60^{\circ} \ \&\ z=48^{\circ}$

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Question 9: In the adjoining figure, what value of $x$ make AOB a straight line?

Answer:

For $\overleftrightarrow{AB}$ to be a straight line

$3x+5+2x-25=180^{\circ}$

$\Rightarrow 5x-20=180^{\circ}$

$\Rightarrow 5x=200^{\circ}$

$\Rightarrow x=40^{\circ}$

$\angle AOC=3\times 40+5=125^{\circ}$

$\angle BOC=2\times 40-25=55^{\circ}$

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Question 10: In the adjoining figure, find the value of $x$ .

Answer: 287

$\angle DOA+\angle AOB+\angle BOC+\angle COD=360^{\circ}$

$x+65+90+120=360$

$\Rightarrow x=360^{\circ} -275^{\circ}$

$\Rightarrow x=85^{\circ}$

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Question 11: In each of the following figures, two lines $AB \ and\ CD$ intersect at a point $O$ . Find the value of $x, \ y \ and \ z$

Answer:

i) $\angle AOD=\angle COB$ (vertically opposite angle)

$\therefore y=75^{\circ}$

$\angle AOD+\angle DOB=180^{\circ}$

$\Rightarrow 75+z=180^{\circ}$

$\Rightarrow z=105^{\circ}$

$\angle AOD+\angle AOC=180^{\circ}$

$\Rightarrow 75+x=180$

$\Rightarrow x=105^{\circ}$

We could have also used

$\angle DOB=\angle AOC$

$\Rightarrow x=105^{\circ}$

ii) $\angle COB=\angle AOB$ (vertically Opposite angles)

$\Rightarrow y=125^{\circ}$

$125+z=180^{\circ}$ (Angles on a straight line are supplementary)

$\Rightarrow z=55^{\circ}$

$\angle BOD=\angle COA$ (Vertically opposite angles)

$\Rightarrow x=55^{\circ}$

iii) $\angle AOC=\angle DOB$ (Vertically opposite angles)

$\Rightarrow y=30^{\circ}$

$\angle COB+\angle BOD=180$ (Angles on a straight line)

$\Rightarrow z=150^{\circ}$

$\angle COB=\angle AOD$ (Vertically opposite angles)

$\Rightarrow x=150^{\circ}$

iv) $\angle AOC+\angle COB+\angle BOD+\angle DOA=360^{\circ}$

$3x-20+x+z+y=360^{\circ} \ldots \ldots i)$

$x=y \$ (Vertically opposite angles)

$3x-20=z \$ (Vertically opposite angles)

$z+y=180^{\circ} \ldots \ldots ii)$

$x+z=180^{\circ} \ldots \ldots iii)$

Substituting (ii) in (i)

$3x-20+x+180=360$

$\Rightarrow 4x=180+20$

$\Rightarrow x=50^{\circ}$

$\angle AOD=50^{\circ} \ \&\ \angle AOC=130^{\circ}$

From iii) $z= 180-50 = 130^{\circ}$

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Question 12: Prove that the bisectors of two adjacent supplementary angles include a right angle.

Answer:

Let $\angle AOC=180-x$

$\Rightarrow \angle BOC=x$

Bisector of $\displaystyle \angle BOC= \frac{x}{2} = \angle DOC$

Bisector of $\displaystyle \angle AOC= \frac{1}{2} (180-x)=90- \frac{x}{2} = \angle COE$

Therefore
$\displaystyle \angle COE+\angle DOC=90- \frac{x}{2} + \frac{x}{2} =90^{\circ}$

Hence
$\angle DOE=90^{\circ} =Right \ angle$

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Question 13: Find the measure of an angle which is (i) equal to its complement (ii) equal to its supplement.

Answer:

i) If the $\angle 1=x$ , it’s complement $\angle 2=90-x$

If $x=90-x$

$\Rightarrow x=45^{\circ}$

ii) If the $\angle 1=x$ , its supplement $= 180-x$

$x=180-x$

$\Rightarrow x=90^{\circ}$

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Question 14: Find the angle which is $84^{\circ}$ more than its complement.

Answer:

Let the angle $=x$

Complement $= 90 - x$

Given $x = (90-x) + 34$

$\Rightarrow 2x=124$

$\Rightarrow x=62^{\circ}$

$\\$

Question 15: Find the angle which is $16^{\circ}$ less than its complement.

Answer:

Let the angle $= x$

Complement $=90-x$

Given

$x+16=90-x$

$\Rightarrow 2x=74$

$\Rightarrow x=37^{\circ}$

$\\$

Question 16: Find the angle which is $26^{\circ}$ more than its supplement.

Answer:

Let the angle $= x$

Supplement $= 180-x$

Given

$x=(180-x)+26$

$\Rightarrow 2x=206$

$\Rightarrow x=103^{\circ}$

$\\$

Question 17: Find the angle which is $32^{\circ}$ less than its supplement.

Answer:

Let the angle $= x$

Supplement $= 180-x$

Given,

$x+32=180-x$

$\Rightarrow 2x=148^{\circ}$

$\Rightarrow x=74^{\circ}$

$\\$

Question 18: Find the angle which is four times its complement.

Answer:

Let the angle $=x$

Complement $= 90-x$

Given

$x=4(90-x)$

$\Rightarrow 5x=360$

$\Rightarrow x=72^{\circ}$

$\\$

Question 19: Find the angle which is five times its supplement.

Answer:

Let the angle $=x$

Complement $= 180-x$

Given

$x=5(180-x)$

$\Rightarrow 6x=5\times 180$

$\Rightarrow x=150^{\circ}$

$\\$

Question 20: Find the angle whose supplement is four times its complement.

Answer:

Let the angle $=x$

Complement $= 90-x$

Supplement $= 180-x$

Given,

$180-x=4(90-x)$

$\Rightarrow 3x=180$

$\Rightarrow x=60^{\circ}$

$\\$

Question 21: Find the angle whose complement is one third of its supplement.

Answer:

Let the angle $=x$

Therefore Complement $= 90-x$

Therefore Supplement $=180-x$

Given

$\displaystyle 90-x= \frac{1}{3} (180-x)$

$\Rightarrow 270-3x=180-x$

$\Rightarrow 2x=90$

$\Rightarrow x=45^{\circ}$

$\\$

Question 22: Two complementary angles are in the ratio $7\colon 11$ Find the angles.

Answer:

Let the angle $=x$

Complement $=90-x$

Given

$\displaystyle \frac{x}{(90-x)} = \frac{7}{11}$

$\Rightarrow 18x=630 \ or\ x=35^{\circ}$

The complement $= 35^{\circ}$

Hence the angles are $35^{\circ} \ and\ 55^{\circ}$

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Question 23: Two supplementary angles are in the ratio 7. Find the angles.

Answer:

Let the angle $= x$

Supplement $= 180-x$

Given

$\displaystyle \frac{x}{(180-x)} = \frac{7}{8}$

$\Rightarrow 8x=1260-7x$

$\Rightarrow x=84^{\circ} \ supplement=96^{\circ}$

Hence the angles are $84^{\circ} \ and\ 96^{\circ}$

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Question 24: Find the measure of an angle, if seven times its complement is $10^{\circ}$ less than three times its supplement.

Answer:

Let the angle $=x$

Complement $=90-x$ Supplement $= 180-x$

Given

$7(90-x)+10=3(180-x)$

$\Rightarrow 630-7x+10=540-3x$

$\Rightarrow 4x=100$

$\Rightarrow x=25^{\circ}$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 8: Lines and Angles – Exercise 28A

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