Dattu

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5.

• Practical Geometry Class 7 Ex 10.1
• Practical Geometry Class 7 Ex 10.2
• Practical Geometry Class 7 Ex 10.3
• Practical Geometry Class 7 Ex 10.4

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 10
Chapter Name	Practical Geometry
Exercise	Ex 10.5
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5

Question 1.
Construct the right-angled ∆ PQR where m ∠Q = 90°, QR = 8 cm and PR = 10 cm.
Solution:
Steps of Construction:

1. Draw QR of length 8 cm.
   
2. At Q, draw QX ⊥ QR.
3. With R as centre, draw an arc of radius 10 cm.
4. Mark the meeting point of these two as P.

∆ PQR is now obtained.

Question 2.
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Solution:
Steps of Construction

1. Draw QR of length 4 cm.
2. At Q, draw QX ⊥ QR.
   
3. With R as centre, draw an arc of radius 6 cm.
4. Mark the meeting point of arc and QX as P.

∆ PQR is now obtained.

Question 3.
Construct an isosceles right-angled triangle ABC where m ∠ACB = 90° and AC = 6 cm.
Solution:
Steps of Construction

1. Draw AC of length 6 cm.
2. At C, draw CX ⊥ CA.
   
3. With C as centre, draw an arc of radius 6 cm to intersect CX at B.
4. Join AB. ∆ACB is now obtained.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2.

• Rational Numbers Class 7 Ex 9.1
• Rational Numbers Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 9
Chapter Name	Rational Numbers
Exercise	Ex 9.2
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2

Question 1.
Find the sum:

Solution:

Question 2.
Find :

Solution:

Question 3.
Find the product :

Solution:

Question 4.
Find the value of :

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3.

• Comparing Quantities Class 7 Ex 8.1
• Comparing Quantities Class 7 Ex 8.2
• Comparing Quantities Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 8
Chapter Name	Comparing Quantities
Exercise	Ex 8.3
Number of Questions Solved	11
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

Question 1.
Tell what is the profit or loss in the following transactions. Also find profit percent or loss per cent in each case.

(a) Gardening shears bought for ₹ 250 and sold for ₹ 325.
(b) A refrigerater bought for ₹ 12,000 and sold at ₹ 13,500.
(c) Acupboard bought for ₹ 2,500 and sold at ₹ 3,000.
(d) A skirt bought for ₹ 250 and sold at ₹ 150.

Solution:

Question 2.
Convert each part of the ratio to percentage

(a) 3 : 1
(b) 2 : 3 : 5
(c) 1 : 4
(d) 1 : 2 : 5

Solution:

Question 3.
The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:

Question 4.
Arun bought a car for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the percentage of the price increase?
Solution:

Question 5.
1 buy a T.V. for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it?
Solution:
C.P. = ₹ 10,000
Profit = 20% of C.P.
= 20% of ₹ 10,000
= ₹ \(\frac { 20 }{ 100 } \) × 10,000 = ₹ 2,000
∴ S.P. = C.P. + Profit
= ₹ 10,000 + ₹ 2,000 = ₹ 12,000
Hence, I get ₹ 12,000 for it.

Question 6.
Juki sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she brought it?
Solution:

Question 7.
(i) Chalk contains calcium, carbon, and oxygen in the ratio of 10 : 3 : 12. Find the percentage of carbon in chalk.
(ii) If in a stick of chalk, carbon is 3 g, what is the weight of the chalk stick?
Solution:

Question 8.
Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?
Solution:
C.P. = ₹ 275
Loss = 15% of C.P = \(\frac { 15 }{ 100 } \) × C.P
= \(\frac { 15 }{ 100 } \) × 275
= ₹ 41.25
S.P. = C.P. – Loss = 275 – 41.25 = ₹ 233.75
hence, she se11 it for ₹ 233.75.

Question 9.
Find the amount to be paid at the end of 3 years in each case:

(a) Principal = ₹ 1,200 at 12% p.a.
(b) Principal = ₹ 7,500 aat 5% p.a.

Solution:

Question 10.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years?
Solution:

Question 11.
If Meena gives an interest of ₹ 45 for one year at 9% rate p.a. What is the sum she has borrowed?
Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2.

• Comparing Quantities Class 7 Ex 8.1
• Comparing Quantities Class 7 Ex 8.3
• Comparing Quantities Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 8
Chapter Name	Comparing Quantities
Exercise	Ex 8.2
Number of Questions Solved	10
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

EXERCISE 8.2

Question 1.
Convert the given fractional numbers to per cents.

(a) \(\frac { 1 }{ 8 } \)
(b) \(\frac { 5 }{ 4 } \)
(c) \(\frac { 3 }{ 40 } \)
(d) \(\frac { 2 }{ 7 } \)

Solution:

Question 2.
Convert the given decimal fractions to per cents.

(a) 0.65
(b) 2.1
(c) 0.02
(d) 12.35

Solution:

Question 3.
Estimate what part of the figures is coloured and hence find the percent which is coloured.

Solution:

Question 4.
Find:

(a) 15% of 250
(b) 1% of 1 hour
(c) 20% of ₹ 2500
(d) 75% of 1 kg.

Solution:

Question 5.
Find the whole quantity if

(a) 5% of it is 600.
(b) 12% of it is ₹ 1080.
(c) 40% of it is 500 km.
(d) 70% of it is 14 minutes.
(e) 8% of it is 40 liters.

Solution:

Question 6.
Convert given percents to decimal fractions and also to fractions in the simplest forms.

(a) 25%
(b) 150%
(c) 20%
(d) 5%

Solution:

Question 7.
In a city, 30% are females, 40% are males, and the remaining are children. What percent are children?
Solution:
[100% (30% + 40%)] are children
⇒ [100% – 70% ] are children
⇒ 30% are children.

Question 8.
Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
Solution:
Percentage of voters who voted = 60%
Percentage of voters who did not vote = (100 – 60)% = 40%
Total number of voters = 15000
Number of voters who did not vote = 40% of 15000
= ( \(\frac { 40 }{ 100 } \) × 15000 )
= 6000

Question 9.
Meeta saves ₹ 400 from her salary. If this is 10% of her salary. What is her salary?
Solution:
Let her salary be ₹ P. Then, 10% of P = 400
⇒ \(\frac { 10 }{ 100 } \) × p = 400
⇒ \(\frac { P }{ 10 } \) = 400
⇒ P = 400 × 10
⇒ P = 4000
Hence, her salary is ₹ 4000.

Question 10.
A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Solution:
Out of 100, 25% matches are won. Then, out of 20, the number of matches that the team won
= \(\frac { 25 }{ 100 } \) × 20
= \(\frac { 1 }{ 4 } \) × 20
= 5

We hope the NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2.

• Congruence of Triangles Class 7 Ex 7.1
• Congruence of Triangles Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 7
Chapter Name	Congruence of Triangles
Exercise	Ex 7.2
Number of Questions Solved	10
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2

Question 1.
Which congruence criterion do you use in the following?

Given:
So.
AC = DF
AB = DE
BC = EF
so ∆ ABC = ∆ DEF

Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ∆ PQR ≅ ∆ XYZ

Given : ∠MLN = ∠ FGH
∠NML = ∠GFH
ML = GF
So, ∆ LMN ≅ ∆ GFH

Given : EB = DB
AE = BC
∠A = ∠C = 90°
So, ∆ ABE ≅ ∆ CDB
Solution:

(a) SSS congruence criterion
(b) SAS congruence criterion
(c) ASA congruence criterion
(d) RHS congruence criterion.

Question 2.
You want to show that ∆ ART ≅ ∆ PEN,

(а) If you have to use SSS criterion, then you need to show
(i)AR = (ii) RT = (iii) AT =

(b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have
(i) RT = and (ii) PN =

(c) If it is given that AT = PN and you are to use ASA criterion, you need to have
(i) ? (ii) ?
Solution:

Question 3.
You have to show that ∆ AMP = ∆ AMQ.
In the following proof, supply the missing reasons.


Solution:

Question 4.
In ∆ ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ∆ PQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A student says that ∆ ABC = ∆ PQR? by AAA congruence criterion. Is he justified’? Why or why not?
Solution:
No! he is not justified because AAA is not a criterion for congruence of triangles.

Question 5.
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ∆ RAT = ?

Solution:
∆ RAT ≅ ∆ WON

Question 6.
Complete the congruence statement:


Solution:
∆ BCA = ∆ BTA
∆ QRS = ∆ TPQ

Question 7.
In a squared sheet, draw two triangles of equal areas such that

1. the triangles are congruent
2. the triangles are not congruent. What can you say about their perimeters?

Solution:

Question 8.
Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Solution:

In ∆ ABC and ∆ DEF,
AB = DF (= 2 cm)
BC = ED (= 4 cm)
CA = EF (= 3 cm)
∠BAC = ∠EDF
∠ABC = ∠DEF
But ∆ ABC is not congruent to ∆ DEF.

Question 9.
If ∆ ABC and ∆ PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Solution:
BC = RQ by ASA congruence rule.

Question 10.
Explain why ∆ ABC ≅ ∆ FED

Solution:
∠ABC = ∠FED (= 90°) BC = ED
∠ACB = ∠FDE
∵ The sum of the measures of the three angles of a triangle is 180°.
∆ ABC ≅ ∆ FED (By SAS congruence criterion)

We hope the NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.2.

• Lines and Angles Class  7 Ex 5.1
• Lines and Angles Class  7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 5
Chapter Name	Lines and Angles
Exercise	Ex 5.2
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.2

Question 1.
State the property that is used in each of the following statements?

1. If a || b, then ∠ 1 = ∠ 5.
2. If ∠ 4 = ∠ 6, then a || b.
3. If ∠ 4 + ∠ 5 = 180°, then a || b.

Solution:

1. Corresponding angle property.
2. Alternate interior angle property.
3. Interior angles on the same side of the transversal are supplementary.

Question 2.
In the following figure, identify:

1. the pairs of corresponding angles.
2. the pairs of alternate interior angles.
3. the pairs of interior angles on the same side of the transversal.
4. the vertically opposite angles.

Solution:

1. ∠1, ∠5; ∠2, ∠6; ∠3, ∠7 and ∠4, ∠8 are four pairs of corresponding angles.
2. ∠2, ∠8, and ∠3, ∠5 are two pairs of alternate interior angles.
3. ∠2, ∠5, and ∠3, ∠8 are two pairs of interior angles on the same side of the transversal.
4. ∠1, ∠3; ∠2, ∠4; ∠5, ∠7 and ∠6, ∠8 are four pairs of vertically opposite angles.

Question 3.
In the adjoining figure, p || q. Find the unknown angles.
Solution:
a = 55°, b = 125°, c = 55°, d = 125°, e = 55°, f = 55°.

Question 4.
Find the value of x in each of the following figures if l || m.

Solution:
(i) Since, l || m and t is a transversal.
∴ ∠x = (180° – 110°) = 70° [Corresponding angles, Linear pair]

(ii) if l || m and a is a transversal.
Then, ∠x = 1000 [Corresponding angles]

Question 5.
In the given figure, the arms of two angles are parallel. If ∠ ABC = 70°, then find

Solution:

1. 70°
2. 70°

Question 6.
In the given figures below, decide whether l is parallel to m.


Solution:

1. l is not parallel to m
2. l is not parallel to m
3. l || m
4. l is not parallel to m

We hope the NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4.

• Simple Equations Class 7 Ex 4.1
• Simple Equations Class 7 Ex 4.2
• Simple Equations Class 7 Ex 4.3
• Simple Equations Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 4
Chapter Name	Simple Equations
Exercise	Ex 4.4
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4

Question 1.
Set up equations and solve them to find the unknown numbers in the following cases:

(a) Add 4 to eight times a number; you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from \(\frac { 5 }{ 2 } \) of the number, the result is 23.

Solution:

Question 2.
Solve the following:
(a) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest mark plus 7. The highest score is 87. What is the lowest score?
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).
(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Solution:

Question 3.
Solve the following:

1. Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
2. Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
3. The people of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees was two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?

Solution:

Question 4.
Solve the following riddle :
I am a number, Tell me my identity!
Take me seven times over And add a fifty!
To reach a triple century You still need forty!
Solution:
Let ‘x’ be the number,
Then, according to the question, we get (x × 7) + 50 = 300 – 40
7x + 50 = 260
7x = 210
x = \(\frac { 210 }{ 7 } \) = 30
So, the number is 30.

We hope the NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3.

• Simple Equations Class 7 Ex 4.1
• Simple Equations Class 7 Ex 4.2
• Simple Equations Class 7 Ex 4.4
• Simple Equations Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 4
Chapter Name	Simple Equations
Exercise	Ex 4.3
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3

Question 1.
Solve the following equations:

Solution:

Question 2.
Solve the following equations:

(a) 2 (x + 4) = 12
(6) 3 (n – 5) = 21
(c) 3 (n – 5) = -21
(d) -4 (2 + x) = 8
(e) 4 (2 – x) = 8

Solution:

Question 3.
Solve the /bilowing equations:

(a) 4 = 5 (p – 2)
(b) -4 = 5 (p – 2)
(c) 16 = 4 + 3 (t + 2)
(d) 4 + 5 (p – 1) = 34
(e) 0 = 16 + 4 (m – 6)

Solution:

Question 4.
(a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x = – 2.
Solution:
(a) 1. Start with x = 2
Multiply both sides by 3, 3x = 6
Subtract 2 from both sides, 3x – 2 = 4 …(1)

2. Start with x = 2
Multiply both sides by 4, 4x = 8
Add 5 to both sides, 4x + 5 = 13 …(2)

3. Start with x = 2 Multiply both sides by 5 5x = 10
Subtract 1 from both sides, 5x – 1 = 9 …(3)

(b) First equation:
Start with x = -2
Multiply both sides by 2, 2x = -4
Subtract 3 from both sides, 2x – 3 = -7

Second equation:
Start with x = – 2
Multiply both sides by – 5, – 5x = 10
Add 10 to both sides, 10 – 5x = 20

Third equation:
Start with x = -2
Divide both sides by 2, \(\frac { x }{ 2 } \) = -1
Add 3 to both sides, \(\frac { x }{ 2 } \) + 3 = 2

We hope the NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2.

• Simple Equations Class 7 Ex 4.1
• Simple Equations Class 7 Ex 4.3
• Simple Equations Class 7 Ex 4.4
• Simple Equations Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 4
Chapter Name	Simple Equations
Exercise	Ex 4.2
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2

Question 1.
Give first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(e) y – 4 = – 7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4
Solution:
(a) The given equation is x – 1 = 0
Add 1 to both sides,
x – 1 + 1 = 0 + 1 ⇒ x = 1
It is the required solution.
Check. Put the solution x = 1 back into the equation.
L.H.S. = x – 1 = 1 = 1 – 0 = R.H.S.
The solution is thus checked for its correctness.

(b) The given equation is x + 1 = 0
Subtract 1 from both sides, x + 1 – 1 = 0 – 1 ⇒ x = – 1
It is the required solution.
Check. Put the solution x = – 1 back into the equation.
L.H.S. = x + 1 = (-1)+1
= 0 = R.H.S.
The solution is thus checked for its correctness.

(c) The given equation is
x – 1 = 5
Add 1 to both sides,
x + 1 – 1 = 5 + 1 ⇒ x = 6
It is the required solution
Check. Put the solution x = 6 back into the equation.
L.H.S. = x – 1 = 6 – 1 = 5 = R.H.S.
The solution is thus checked for its correctness.

(d) The given equation is x + 6 = 2
Subtract 6 from both sides, x + 6 – 6 = 2 – 6 ⇒ x = – 4
It is the required solution.
Check. Put the solution x = – 4 back into the equation.
L.H.S. = x + 6 = – 4 + 6 = 2 = R.H.S.
The solution is thus checked for its correctness.

(e) The given equation is y – 4 = – 7
Add 4 to both sides, y – 4 + 4 = – 7 + 4 ⇒ y = – 3
It is the required solution.
Check. Put the solution
L.H.S. = y – 4 = – 3 – 4 = – 7 = R.H.S.
The solution is thus checked for its correctness.

(f) The given equation is y – 4 = 4
Add 4 to both sides,
y – 4 + 4 = 4 + 4 ⇒ y = 8
It is the required solution.
Check. Put the solution y = 8 back into the equation.
L.H.S. = y – 4 = 8 – 4 = 4 = R.H.S.
The solution is thus checked for its correctness.

(g) The given equation is y + 4 = 4
Subtract 4 from both sides, y + 4 – 4 = 4 – 4 ⇒ y = 0
It is the required solution.
Check. Put the solution y = 0 back into the equation.
L.H.S. =y + 4 = 0 + 4 = 4 = R.H.S.
The solution is thus checked for its correctness.

(h) The given equation is y + 4 = – 4
Subtract 4 from both sides, y + 4 – 4 = – 4 – 4 ⇒ y = -8
It is the required solution.
Check. Put the solution y = – 8 back into the equation.
L.H.S. = y + 4 = -8 + 4 = – 4 = R.H.S.
The solution is thus checked for its correctness.

Question 2.
Give first the step you will use to separate the variable and then solve the equation:
(a) 3l = 42
(b) \(\frac { b }{ 2 } \) = 6
(c) \(\frac { p }{ 7 } \) = 4
(d) 4x = 25
(e) 8y = 36
(f) \(\frac { z }{ 3 } \) = \(\frac { 5 }{ 4 } \)
(g) \(\frac { a }{ 5 } \) = \(\frac { 7 }{ 15 } \)
(h) 20t = – 10

Solution:

Question 3.
Give the steps you will use to separate the variable and then solve the equation :
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) \(\frac { 20p }{ 3 } \) = 40
(d) \(\frac { 3p }{ 10 } \) = 6
Solution:

Question 4.
Solve the following equations:

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4.

• The Triangle and its Properties Class 7 Ex 6.1
• The Triangle and its Properties Class 7 Ex 6.2
• The Triangle and its Properties Class 7 Ex 6.3
• The Triangle and its Properties Class 7 Ex 6.5
• The Triangle and its Properties Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 6
Chapter Name	The Triangle and its Properties
Exercise	Ex 6.4
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4

Question 1.
Is it possible to have a triangle with the following sides?

1. 2 cm, 3 cm, 5 cm
2. 3 cm, 6 cm, 7 cm
3. 6 cm, 3 cm, 2 cm.

Solution:

1. Since, 2 + 3 > 5
   So the given side lengths cannot form a triangle.
2. We have, 3 + 6 > 7, 3 + 7 > 6 and 6 + 7 > 3
   i. e., the sum of any two sides is greater than the third side.
   So, these side lengths form a triangle.
3. We have, 6 + 3 > 2, 3 + 2 \(\ngtr \) 6
   So, the given side lengths cannot form a triangle.

Question 2.
Take any point O in the interior of a triangle PQR. Is

1. OP + OQ > PQ ?
2. OQ + OR > QR?
3. OR + OP > RP ?

Solution:

1. Yes ! OP + OQ > PQ …(1)
   Sum of the lengths of any two sides of a triangle is greater than the length of the third side
2. Yes! OQ + OR > QR …(2)
   Sum of the lengths of any two sides of a triangle is greater than the length of the third side
3. Yes! OR + OP > RP …(3)
   Sum of the lengths of any two sides of a triangle is greater than the length of the third side

Question 3.
AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM?
(Consider the sides of triangles ∆ ABM and ∆ AMC.)

Solution:
Using triangle inequality property in triangles ABM and AMC, we have
AB + BM > AM …(1) and, AC + MC > AM …(2)
Adding (1) and (2) on both sides, we get
AB + (BM + MC) + AC > AM + AM
⇒ AB + BC + AC > 2AM

Question 4.
ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD ?

Solution:
In ∆ ABC, AB + BC > AC …(1)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
In ∆ ACD, CD + DA > AC …(2)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
Adding (1) and (2),
AB + BC + CD + DA > 2AC …(3)
In ∆ ABD, AB + DA > BD …(4)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
In ∆ BCD, BC + CD > BD …(5)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
Adding (4) and (5),
AB + BC + CD + DA > 2BD …(6)
Adding (3) and (6),
2 [AB + BC + CD + DA] > 2 (AC + BD)
⇒ AB + BC + CD + DA > AC + BD.

Question 5.
ABCD is a quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)?

Solution:
In ∆ OAB, OA + OB > AB ….(1)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side.
In ∆ OBC, OB + OC > BC ….(2)
Sum of the lengths of any two sides of a triangle la greater than the length of the third side.
In ∆ OCA,OC + OA > CA ….(3)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
In ∆ OAD, OA + OD > AD ….(4)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
Adding (1), (2), (3) and (4),
2(OA + OB + OC + OD) > AB + BC + CD + DA
⇒ AB + BC + CD + DA < 2 (OA + OB + OC + OD)
⇒ AB + BC + CD + DA < 2(OA + OC + OB + OD)
⇒ AB + BC + CD + DA < 2 (AC + BD).

Question 6.
The lengths of the two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Solution:
Let x cm be the length of the third side.
∴ Sum of the lengths of any two sides of a triangle is greater than the length of the third side.
∴ We should have

∴ The length of the third side should be any length between 3 cm and 27 cm.

We hope the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and it’s Properties Ex 6.4 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3.

• The Triangle and its Properties Class 7 Ex 6.1
• The Triangle and its Properties Class 7 Ex 6.2
• The Triangle and its Properties Class 7 Ex 6.4
• The Triangle and its Properties Class 7 Ex 6.5
• The Triangle and its Properties Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 6
Chapter Name	The Triangle and its Properties
Exercise	Ex 6.3
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3

Question 1.
Find the value of the unknown x in the following diagrams:

Solution:

Question 2.
Find the values of the unknowns x and y in the following diagrams:

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3, drop a comment below and we will get back to you at the earliest.