281Question 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

\\

282Question 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}

\\

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

\\

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}

\\

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}

\\

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}

\\

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}

\\

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}

\\

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

Answer:

For \overleftrightarrow{AB}   to be a straight line286

 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}

\\

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}

\\

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

288

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}

\\

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

\\

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}

\\

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}

\\

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}

\\

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}