Lines And Angles Class 9th Mathematics AP Board Solution

Class 9th Mathematics AP Board Solution

Exercise 4.1
Question 1.

In the given figure, name:

i. any six points

ii. any five line segments

iii. any four rays

iv. any four lines

v. any four collinear points



Answer:

i) A point specifies the exact location and looks like a small dot.


∴ The points in the given figure are A,B,C,D,E,F,G,H,M,N,P,Q,X,Y.


ii) A part of a line with two end points is a line segment.


∴ The line segment in the given figure are AX,XM,MP,AM,AP,AB,XB,XY…etc


iii) A ray has a starting point but no end point. it may go to infinity.


∴ The ray in the given figure are A,B,C,D,E,F,G,H.


iv) A line goes without end in both direction.


∴ The line segment in the given figure are AB,CD,EF,GH.


v) if three or more points lie on the same line, they are called collinear points


∴ The collinear points in the given figure are AXM,EMN,GPQ,CYN…



Question 2.

Observe the following figures and identify the type of angles in them.



Answer:

A) The given figure shows the larger angle area, so it is reflex angle. The angle is more than 180° and less than 360°.


B) The given figure shows the right angle that is B = 90°


C) The given figure shows the smaller angle area, so it is acute angle. The angle is more than 0° and less than 90° .



Question 3.

State whether the following statements are true or false :

i. A ray has no end point.

ii. Line  is the same as line .

iii. A ray  is same as the ray 

iv. A line has a define length.

v. A plane has length and breadth but no thickness.

vi. Two distinct points always determine a unique line.

vii. Two lines may intersect in two points.

viii. Two intersecting lines cannot both be parallel to the same line.


Answer:

(i) False


A ray has a starting point but no end point but it goes to infinity.


(ii) True


A line goes without end in both direction. so,  is same as 


(iii) False


A ray has a starting point but no end point but it goes to infinity. so


a ray  is not same as 


(iv) False


A line goes without end in both direction. so, it does not have a define length.


(v) True


A line goes without end in both direction. so,  is same as 


(vi) True


If two points are joined together then forms a line.


(vi) False


Two lines intersected at one point.


(vii) True


A parallel line does not have any intersecting point.



Question 4.

What is the angle between two hands of a clock when the time in the clock is

(a) 9’O clock

(b) 6’O clock

(c) 7:00 PM


Answer:

(a) Let draw the 9’O clock and find the angle between the lines



∴ The angle is 90°


(b) Let draw the 6’O clock and find the angle between the lines



∴ The angle is 180°


(c) Let draw the 7.00 PM and find the angle between the lines



∴ The angle is 210°




Exercise 4.2
Question 1.

In the given figure three lines  and  intersecting at O. Find the values of x, y and z it is being given that x : y : z = 2 : 3 : 5



Answer:

From the given, the three angles are x,y,z


If the two lines intersect at a point then its vertically opposite angles are equal.


∴ A = x, B = z and C = y


We know that,The sum of all the angles around at a point is equal to 360°


∴ A + B + C + x + y + z = 360°


⇒ x + y + z + x + y + z = 360°


⇒ 2x + 2y + 2z = 360°


⇒ 2(x + y + z) = 360°


⇒ (x + y + z) = 


⇒ x + y + z = 180° ------(1)


Given that, x : y : z = 2 : 3 : 5


Let x = 2m,y = 3m,z = 5m (∵ m = constant)


Substitute these values in equation (1) we get


2m + 3m + 5m = 180


10m = 180


m = 


∴ m = 18


Substituting m = 18 in x,y,z


x = 2m,x = 2(18) = 36°


y = 3m,y = 3(18) = 54°


z = 5m,z = 5(18) = 90°


∴ x = 36°,y = 54°,z = 90°



Question 2.

Find the value of x in the following figures.

i.  ii. 

iii.  iv. 


Answer:

(i) From the given figure,


3x + 18° + 93° = 180°


⇒ 3x + 111° = 180°


⇒ 3x = 180°-111°


⇒ 3x = 69°


⇒ x = 


∴ x = 24°


(ii) From the given figure,


(x-24)° + 29° + 296° = 360°


⇒ (x-24)° = 360°-325°


⇒ (x-24)° = 35°


⇒ x = 35° + 24°


∴ x = 59°


(iii) From the given figure,


(2 + 3x)° = 62°


⇒ 3x = 62°-2° = 60°


⇒ x = 


∴ x = 20°


(iv) From the given figure,


40° + (6x + 2)° = 90°


⇒ (6x + 2)° = 90°-40°


⇒ (6x + 2)° = 50°


⇒ 6x = 50°-2° = 48°


⇒ x = 


∴ x = 8°



Question 3.

In the given figure lines  and  intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.



Answer:

Given that,


The lines  and  intersect at O.


∠AOC + ∠BOE = 70° ----(1)


∠BOD = 40° ----(2)


If the two lines intersect at a point then its vertically opposite angles are equal.


∴ ∠AOC = ∠BOD


Substitute (2) in (1)


⇒ 40° + ∠BOE = 70°


⇒∠BOE = 70°-40°


∴∠BOE = 30°


From the figure,AOB is a straight line and its angle is 180°


So, ∠AOC + ∠BOE + ∠COE = 180°


From equation (1)


⇒ 70° + ∠COE = 180°


⇒∠COE = 180°-70°


∴∠COE = 110°


Reflex ∠COE = 360° - 110° = 250°


∴∠BOE = 30° and Reflex ∠COE = 250°



Question 4.

In the given figure lines  and  intersect at O. If ∠POY = 90° and a: b = 2 : 3, find c.



Answer:

Given that,


The lines  and  intersect at O.


From the figure, XOY is a straight line and its angle is 180°


So, ∠XOM + ∠MOP + ∠POY = 180°


From the given, Let ∠a = 2x and ∠b = 3x


⇒∠b + ∠a + ∠POY = 180°----(1)


Given that ∠POY = 90°


Substitute the values in equation (1),


⇒2x + 3x + 90° = 180°


⇒5x + 90° = 180°


⇒5x = 180°-90° = 90°


⇒x = 


∴ x = 18°


⇒ ∠a = 2x = 2×18° = 36°


⇒ ∠b = 3x = 3×18° = 54°


From the figure, MON is a straight line and its angle is 180°


⇒∠b + ∠c = 180°


⇒54° + ∠c = 180°


⇒∠c = 180°-54°


∴ ∠c = 126°



Question 5.

In the given figure ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.



Answer:

In the figure, ST is a straight line and its angle is 180°


So, ∠PQS + ∠PQR = 180° ----(1)


And ∠PRT + ∠PRQ = 180°-----(2)


From the two equations, we get


∠PQS + ∠PQR = ∠PRT + ∠PRQ


Given that,


∠PQR = ∠PRQ


⇒ ∠PQS + ∠PRQ = ∠PRT + ∠PRQ


⇒ ∠PQS = ∠PRT + ∠PRQ -∠PRQ


⇒ ∠PQS = ∠PRT


So, ∠PQS = ∠PRT is proved.



Question 6.

In the given figure, if x + y = w + z, then prove that AOB is a line.



Answer:

In a circle, the sum of all angles is 360°


∴ ∠AOC + ∠BOC + ∠DOB + ∠AOD = 360°


⇒ x + y + w + z = 360°


Given that, x + y = w + z


⇒ w + z + w + z = 360°


⇒ 2w + 2z = 360°


⇒ 2(w + z) = 360°


⇒ w + z = 180° or ∠DOB + ∠AOD = 180°


If the sum of two adjacent angles is 180° then it forms a line.


So AOB is a line.



Question 7.

In the given figure  is a line. Ray  is perpendicular to line is another ray lying between rays  and 



Prove that 


Answer:

Given that, is a line and  is perpendicular to line .


⇒ ∠POR = 90°


The sum of linear pair is always equal to 180°


∴ ∠POS + ∠ROS + ∠POR = 180°


Substitute ∠POR = 90°


⇒90° + ∠POS + ∠ROS = 180°


⇒∠POS + ∠ROS = 90°


∴ ∠ROS = 90°-∠POS----(1)


⇒ ∠QOR = 90°


Given that OS is another ray lying between OP and OR


⇒∠QOS-∠ROS = 90°


∴∠ROS = ∠QOS-90°----(2)


On adding two equations (1) and (2) we get


2∠ROS = ∠QOS-∠POS


⇒ ∠ROS = (∠QOS-∠POS)


So, ∠ROS = (∠QOS-∠POS) is proved.



Question 8.

It is given that ∠XYZ = 64° and XY is produced to point P. A ray YQ bisects ∠ZYP. Draw a figure from the given information. Find ∠XYQ and reflex ∠QYP.


Answer:

Let us draw a figure from the given,



Given that, a ray YQ bisects ∠ZYP


So, ∠QYP = ∠ZYQ


Here, PX is a straight line, so the sum of the angles is equal to 180°


∠XYZ + ∠ZYQ + ∠QYP = 180°


Given that, ∠XYZ = 64° and ∠QYP = ∠ZYQ


⇒ 64° + 2∠QYP = 180°


⇒ 2∠QYP = 180°-64° = 116°


∴ ∠QYP =  = 58°


Also, ∠QYP = ∠ZYQ = 58°


Using the angle of reflection,


∠QYP = 360°-58° = 302°


∠XYQ = ∠XYZ + ∠ZYQ


⇒∠XYQ = 64° + 58° = 122°


∴ Reflex ∠QYP = 302° and ∠XYQ = 122°


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