Mathematics Part i Solutions Solutions for Class 7 Math Chapter 1 Between The Lines are provided here with simple step-by-step explanations. These solutions for Between The Lines are extremely popular among class 7 students for Math Between The Lines Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part i Solutions Book of class 7 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part i Solutions Solutions. All Mathematics Part i Solutions Solutions for class 7 Math are prepared by experts and are 100% accurate.
Page No 9:
Question 1:
In each figure below, there are three angles. The measures of two are given alongside. Can you find the third?
A.
B.
C.
D.
Answer:
(A) Given ∠BAC = 30° and ∠CAD = 40°
∠BAD = ∠BAC + ∠CAD
= 30° + 40°
= 70°
(B) Given ∠BAC = 60° and ∠CAD = 45°
∠BAD = ∠BAC + ∠CAD
= 60° + 45°
= 105°
(C) Given ∠BAC = 50° and ∠BAD = 90°
∠BAD = ∠BAC + ∠CAD
⇒ ∠CAD = ∠BAD − ∠BAC
= 90° − 50°
= 40°
(D) Given ∠CAD = 50° and ∠BAD = 100°
∠BAD = ∠BAC + ∠CAD
⇒ ∠BAC = ∠BAD − ∠CAD
= 100° − 50°
= 50°
Page No 12:
Question 1:
In this figure, how much is ∠ACE?
Answer:
From the figure, we can observe that ∠ACE and ∠ECB make a linear pair.
We know that the angles in a linear pair are supplementary.
∴ ∠ACE + ∠ECB = 180°
⇒ ∠ACE + (∠ECD + ∠DCB) = 180°
Given ∠ECD = 50° and ∠DCB = 25°
⇒ ∠ACE + 50° + 25° = 180°
⇒ ∠ACE + 75° = 180°
⇒ ∠ACE = 180° − 75° = 105°
[PAGENO,12]]
Page No 13:
Question 1:
How much is the angle marked in the picture below?
Answer:
From the figure, we can see that there are two set squares ABC and CDE.
We know that each acute angle in the set square ABC measures 45°, i.e., ∠ABC = ∠ACB = 45°.
Also, the measurements of each acute angle in the set square CDE are as follows:
∠DCE = 30° and ∠DEC = 60°
Again, from the figure, we can observe that ∠ACB and ∠ACE make a linear pair.
We know that the angles in a linear pair are supplementary.
∴ ∠ACB + ∠ACE = 180°
⇒ ∠ACB + (∠ACD + ∠DCE) = 180°
⇒ 45° + ∠ACD + 30° = 180°
⇒ ∠ACD + 75° = 180°
⇒ ∠ACD = 180° − 75° = 105°
Page No 13:
Question 2:
In the figure below, ∠ACD = ∠BCE. How much is each
Answer:
From the figure, we can observe that ∠ACD and ∠DCB make a linear pair.
We know that the angles in a linear pair are supplementary.
∴ ∠ACD + ∠DCB = 180°
⇒ ∠ACD + (∠DCE + ∠ECB) = 180°
Given ∠ACD = ∠ECB and ∠DCE = 30°
⇒ ∠ACD + 30° + ∠ACD = 180°
⇒ 2∠ACD + 30° = 180°
⇒ 2∠ACD = 180° − 30°
⇒ 2∠ACD = 150°
⇒ ∠ACD = 75°
Thus, ∠ACD = ∠ECB = 75°.
Page No 15:
Question 1:
All figures below show two intersecting lines. One of the four angles formed is given. Find the other three and write them in the figure.
Answer:
(i) The given figure can be named as follows.
From the figure, we can observe that ∠AOC and ∠COB make a linear pair.
We know that the angles in a linear pair are supplementary.
∴ ∠AOC + ∠COB = 180°
Given ∠COB = 45°
⇒ ∠AOC + 45° = 180°
⇒ ∠AOC = 180° − 45° = 135°
Also, ∠AOC and ∠AOD make a linear pair.
∴ ∠AOC + ∠AOD = 180°
∠AOC = 135°, therefore
135° + ∠AOD = 180°
⇒ ∠AOD = 180° − 135° = 45°
Also, ∠AOD and ∠DOB make a linear pair.
∴ ∠AOD + ∠DOB = 180°
∠AOD = 45°, therefore
45° + ∠DOB = 180°
⇒ ∠DOB = 180° − 45° = 135°
The calculated angles can be marked in the given figure as follows.
(ii) The given figure can be named as follows.
From the figure, we can observe that ∠POR and ∠ROQ make a linear pair.
We know that the angles in a linear pair are supplementary.
∴ ∠POR + ∠ROQ = 180°
Given ∠ROQ = 120°
⇒ ∠POR + 120° = 180°
⇒ ∠POR = 180° − 120° = 60°
Also, ∠POR and ∠POS make a linear pair.
∴ ∠POR + ∠POS = 180°
∠POR = 60°, therefore
60° + ∠POS = 180°
⇒ ∠POS = 180° − 60° = 120°
Also, ∠POS and ∠SOQ make a linear pair.
∴ ∠POS + ∠SOQ = 180°
∠POS = 120°, therefore
120° + ∠SOQ = 180°
⇒ ∠SOQ = 180° − 120° = 60°
The calculated angles can be marked in the given figure as follows.
(iii) The given figure can be named as follows.
From the figure, we can observe that ∠POM and ∠MOQ make a linear pair.
We know that the angles in a linear pair are supplementary.
∴ ∠POM + ∠MOQ = 180°
Given ∠MOQ = 90°
⇒ ∠POM + 90° = 180°
⇒ ∠POM = 180° − 90° = 90°
Also, ∠MOQ and ∠QOL make a linear pair.
∴ ∠MOQ + ∠QOL = 180°
∠MOQ = 90°, therefore
90° + ∠QOL = 180°
⇒ ∠QOL = 180° − 90° = 90°
Also, ∠LOP and ∠POM make a linear pair.
∴ ∠LOP + ∠POM = 180°
∠POM = 90°, therefore
∠LOP + 90° = 180°
⇒ ∠LOP = 180° − 90° = 90°
The calculated angles can be marked in the given figure as follows.
Page No 17:
Question 1:
In the figures below, find the measures of the angles marked.
Answer:
(i) The given figure can be named as follows.
From the figure, we can observe that ∠AOF and ∠BOE, ∠AOC and ∠DOB and ∠FOD and ∠COE are opposite angles.
We know that the opposite angles between two lines are equal.
So, we have
∠DOB = ∠AOC = 70° and ∠COE = ∠DOF = 30°
Also, ∠AOF, ∠FOD and ∠DOB make a linear set.
∴ ∠AOF + ∠FOD + ∠DOB = 180°
⇒ ∠AOF + 30° + 70° = 180°
⇒ ∠AOF + 100° = 180°
⇒ ∠AOF = 180° − 100° = 80°
Also, ∠BOE = ∠AOF = 80°
Thus, we have ∠AOF = 80°, ∠DOB = 70°, ∠BOE = 80° and ∠COE = 30°.
(ii) The given figure can be named as follows.
From the figure, we can observe that ∠AOE and ∠BOF, ∠DOE and ∠COF, and ∠AOC and ∠DOB are opposite angles.
We know that the opposite angles between two lines are equal.
So, we have
∠AOE = ∠BOF = 60° and ∠AOC = ∠DOB = 60°
Also, ∠AOE, ∠AOC and ∠COF make a linear set.
∴ ∠AOE + ∠AOC + ∠COF = 180°
⇒ 60° + 60° + ∠COF = 180°
⇒ 120° + ∠COF = 180°
⇒ ∠COF = 180° − 120° = 60°
Also, ∠DOE = ∠COF = 60°
Thus, we have ∠AOC = ∠COF = ∠BOF = ∠DOE = 60°.
(iii) The given figure can be named as follows.
From the figure, we can observe that ∠AOX and ∠BOY, as well as ∠AOY and ∠BOX, are opposite angles.
We know that the opposite angles between two lines are equal.
So, we have
∠AOX = ∠BOY = 110° and ∠AOY = ∠BOX
Also, ∠BOY = ∠BOC + ∠COY
⇒ 110° = 60° + ∠COY
⇒ ∠COY = 110° − 60° = 50°
Also, ∠AOX and ∠AOY make a linear pair.
We know that the angles in a linear pair are supplementary.
∴ ∠AOX + ∠AOY = 180°
⇒ 110° + ∠AOY = 180°
⇒ ∠AOY = 180° − 110° = 70°
So, we have ∠AOY = ∠BOX = 70°.
Thus, we have ∠AOY = ∠BOX = 70° and ∠COY = 50°.
(iv) The given figure can be named as follows.
From the figure, we can observe that ∠AOC and ∠BOD, as well as ∠AOD and ∠BOC, are opposite angles.
We know that the opposite angles between two lines are equal.
So, we have
∠AOC = ∠BOD = 130° and ∠AOD = ∠BOC
Also, ∠AOC = ∠AOE + ∠EOC
⇒ 130° = ∠AOE + 50°
⇒ ∠AOE = 130° − 50° = 80°
Also, ∠AOE and ∠EOB make a linear pair.
∴ ∠AOE + ∠EOB = 180°
⇒ ∠AOE + ∠EOC + ∠COB = 180°
⇒ 80° + 50° + ∠COB = 180°
⇒ 130° + ∠COB = 180°
⇒ ∠COB = 180° − 130° = 50°
So, we have ∠COB = ∠AOD = 50°.
Thus, we have ∠AOE = 80° and ∠COB = ∠AOD = 50°.
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