Selina Solutions Concise Mathematics Class 6 Chapter 25 Properties of Angles and Lines Exercise 25(A)

Selina Solutions Concise Mathematics Class 6 Chapter 25 Properties of Angles and Lines Exercise 25(A) are designed by experts after conducting research on each concept. The sum of two adjacent angles in a straight line is 180⁰, which is the main concept discussed under this exercise. Students can improve problem solving and time management skills by using these solutions. This also helps them to appear for the examination fearlessly. To get their doubts cleared about the concepts, students can refer to Selina Solutions Concise Mathematics Class 6 Chapter 25 Properties of Angles and Lines Exercise 25(A) PDF, from the links provided here for free download.

Selina Solutions Concise Mathematics Class 6 Chapter 25: Properties of Angles and Lines Exercise 25(A) Download PDF

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Exercise 25(B) Solutions

Exercise 25(C) Solutions

Exercise 25(D) Solutions

Access Selina Solutions Concise Mathematics Class 6 Chapter 25: Properties of Angles and Lines Exercise 25(A)

Exercise 25(A)

1. Two straight lines AB and CD intersect each other at a point O and angle AOC = 50⁰; find:

(i) angle BOD

(ii) ∠AOD

(iii) ∠BOC

Solution:

(i) ∠BOD

Given ∠AOC = 50⁰

We know that,

Vertically opposite angles are equal

So, ∠BOD = ∠AOC

Therefore, ∠BOD = 50⁰

(ii) ∠AOD

∠AOD + ∠BOD = 180⁰

∠AOD + 50⁰ = 180⁰ {From (i)}

∠AOD = 180⁰ – 50⁰

We get,

∠AOD = 130⁰

(iii) ∠BOC

We know that,

Vertically opposite angles are equal

So, ∠BOC = ∠AOD

Therefore, ∠BOC = 130⁰

2. The adjoining figure, shows two straight lines AB and CD intersecting at P. If ∠BPC = 4x – 5⁰ and ∠APD = 3x + 15⁰; find:

(i) the value of x

(ii) ∠APD

(iii) ∠BPD

(iv) ∠BPC

Solution:

(i) The value of x is calculated as,

3x + 15⁰ = 4x – 5⁰

3x – 4x = – 5⁰ – 15⁰

– x =- 20⁰

x = 20⁰

(ii) the value of ∠APD is calculated as,

∠APD = 3x + 15⁰

= 3 × 20⁰ + 15⁰

We get,

= 60⁰ + 15⁰

= 75⁰

(iii) The value of ∠BPD is calculated as,

∠BPD = 180 – ∠BPC

= 180⁰ – (4x – 5⁰)

= 180⁰ – (4 × 20⁰ – 5⁰)

= 180⁰ – 80⁰ + 5⁰

We get,

= 105⁰

(iv) The value of ∠BPC is calculated as,

∠BPC = (4x – 5⁰)

= (4 × 20⁰ – 5⁰)

We get,

= 80⁰ – 5⁰

= 75⁰

3. The given diagram, shows two adjacent angles AOB and AOC, whose exterior sides are along the same straight line. Find the value of x.

Solution:

Here, the exterior arms of the adjacent angles are in a straight line,

Hence, the adjacent angles are supplementary

∠AOB + ∠AOC =180⁰

68⁰ + (3x – 20⁰) = 180⁰

68⁰ + 3x – 20⁰ = 180⁰

3x = 180⁰ + 20⁰ – 68⁰

3x = 200⁰ – 68⁰

We get,

3x = 132⁰

x = 132⁰ / 3

x = 44⁰

4. Each figure given below shows a pair of adjacent angles AOB and BOC. Find whether or not the exterior arms OA and OC are in the same straight line.

(i)

(ii)

(iii)

Solution:

(i) We know that,

The sum of adjacent angles AOB and COB = 180⁰

Hence,

∠AOB + ∠COB = 180⁰

(90⁰ – x) + (90⁰ + x) = 180⁰

90⁰ – x + 90⁰ + x = 180⁰

We get,

180⁰ = 180⁰

The exterior arms OA and OC are in the same straight line

(ii) ∠AOB + ∠BOC = 97⁰ + 83⁰

= 180⁰

The sum of adjacent angles AOB and BOC is 180⁰

Hence, the exterior arms OA and OC are in the same straight line

(iii) ∠COB + ∠AOB = 88⁰ + 112⁰

We get,

= 200⁰ [which is not equal to 180⁰]

Hence, the exterior arms OA and OC are not in the same straight line

5. A line segment AP stands at point P of a straight line BC such that ∠APB = 5x – 40⁰ and ∠APC = x + 10⁰; find the value of x and angle APB.

Solution:

Given

A line segment AP stands at P and

∠APB = 5x – 40⁰

∠APC = x + 10⁰

(i) BPC is a straight line

∠APB + ∠APC = 180⁰

5x – 40⁰ + x + 10⁰ = 180⁰

6x – 30⁰ = 180⁰

6x = 180⁰ + 30⁰

We get,

6x = 210⁰

x = 210⁰ / 6

x = 35⁰

(ii) ∠APB = 5x – 40⁰

= 5 × 35⁰ – 40⁰

We get,

= 175⁰ – 40⁰

= 135⁰