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Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 5 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Practice Set 5 Answers Solutions Chapter 1

Construct triangles of the measures given below:

Question 1.
In ∆MAN, m∠MAN = 90°, l(AN) = 8 cm, l(MN) = 10 cm.
Solution:

Question 2.
In the right-angled ∆STU, hypotenuse SU = 5cm and l(ST) = 4cm.
Solution:

Question 3.
In ∆ABC, l(AC) = 7.5 cm, m∠ABC = 90°, l(BC) = 5.5cm.
Solution:

Question 4.
In ∆PQR, l(PQ) = 4.5 cm, l(PR) = 11.7cm, m∠PQR = 90°.
Solution:

Question 5.
Students should take examples of their own and practice construction of triangles.
i. In ∆PQR, l(PQ) = 5 cm, l(QR) = 6.8 cm, l(PR) = 5.5 cm.
ii. In ∆XYZ, l(XY) = 5.7 cm, m∠Y = 120°, l(YZ) = 7 cm.
iii. In ∆RST, l(ST) = 6.7 cm, m∠S = 60°, m∠T = 40°.
iv. In ∆UVW, m∠U = 90°, l(UV) = 5 cm, l(VW) = 6 cm.
Solution:
i. In ∆PQR, l(PQ) = 5 cm, l(QR) = 6.8 cm, l(PR) = 5.5 cm.

ii. In ∆XYZ, l(XY) = 5.7 cm, m∠Y = 120°, l(YZ) = 7 cm.

iii. In ∆RST, l(ST) = 6.7 cm, m∠S = 60°, m∠T = 40°.

iv. In ∆UVW, m∠U = 90°, l(UV) = 5 cm, l(VW) = 6 cm.

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 4 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Practice Set 4 Answers Solutions Chapter 1

Construct triangles of the measures given below:

Question 1.
In ∆SAT, l(AT) = 6.4 cm, m∠A = 45°, m∠T = 105°.
Solution:

Question 2.
In ∆MNP, l(NP) = 5.2 cm, m∠N = 70°, m∠P = 40°
Solution:

Question 3.
In ∆EFG, l(EG) = 6 cm, m∠F = 65°, m∠G = 45°.
Solution:
In ∆EFG,
m∠E + m∠F + m∠G = 180° …(sum of measures of angles of a triangle)
m∠E + 65° + 45° = 180°
m∠E + 110° = 180°
m∠E = 180° – 110°
m∠E = 70°

Question 4.
In ∆XYZ, l(XY) = 7.3 cm, m∠X = 34°, m∠Y = 95°.
Solution:

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 4 Intext Questions and Activities

Question 1.
In ∆ABC, m∠A = 60°, m∠B = 40°, l(AC) = 6 cm. (Textbook pg. no. 5)
1. Can you draw ∆ABC?
2. What further information is required before it can be drawn?
3. Which property can be used to get it?
4. Draw the rough figure to find out.
Solution:
1. ∆ABC cannot be drawn using the given information.
Seg AC is included inside the angles ∠A and ∠C. Since measure of ∠C is not known, the triangle cannot be drawn.
2. To draw the triangle, measure of ∠C is required.
3. The property of sum of the measures of the angles of a triangle can be used to find out m∠C.
4. In ∆ABC,
m∠A + m∠B + m∠C = 180°
∴ 60° + 40° + m∠C = 180°
∴ 100° + m∠C = 180°
m∠C = 180°- 100°
∴ m∠C = 80°

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 3 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Practice Set 3 Answers Solutions Chapter 1

Draw triangles with the measures given below:

Question 1.
In ∆MAT, l(MA) = 5.2 cm, m∠A = 80°, l(AT) = 6 cm.
Solution:

Question 2.
In ∆NTS, m∠T = 40°, l(NT) = l(TS) = 5 cm.
Solution:

Question 3.
In ∆FUN, l(FU) = 5 cm, l(UN) = 4.6 cm, m∠U = 110°.
Solution:

Question 4.
In ∆PRS, l(RS) = 5.5 cm, l(RP) = 4.2 cm, m∠R = 90°.
Solution:

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 2 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Practice Set 2 Answers Solutions Chapter 1

Question 1.
Draw triangles with the measures given below:
i. In ∆ABC, l(AB) = 5.5 cm, l(BC) = 4.2 cm, l(AC) = 3.5 cm.
ii. In ∆STU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm.
iii. In ∆PQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm.
Solution:
i. In ∆ABC, l(AB) = 5.5 cm, l(BC) = 4.2 cm, l(AC) = 3.5 cm.

ii. In ∆STU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm.

iii. In ∆PQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm.

Question 2.
Draw an isosceles triangle with base 5 cm and the other sides 3.5 cm each.
Solution:

Question 3.
Draw an equilateral triangle with side 6.5 cm.
Solution:

Question 4.
Choose the lengths of the sides yourself and draw one equilateral, one isosceles and one scalene triangle.
Solution:
i. Equilateral triangle LMN, l(LM) = l(MN) = l(LN) = 4 cm.

ii. Isosceles triangle STU, l(ST) = l(TU) = 4cm, l(SU) = 6 cm

iii. Scalene triangle XYZ, l(XY) = 4.5 cm, l(XY) = 6.5 cm, l(XZ) = 5.5 cm

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 2 Intext Questions and Activities

Question 1.
Draw ∆ABC such that l(AB) = 4 cm, and l(BC) = 3 cm. (Textbook pg. no. 3)

1. Can this triangle be drawn?
2. A number of triangles can be drawn to fulfill these conditions. Try it out.
3. Which further condition must be placed if we are to draw a unique triangle using the above information?

Solution:

1. ∆ABC triangle cannot be drawn as length of third side is not given.
2. For ∆ABC to draw l(AC) > l(AB) + l(BC)
   i.e., l(AC) > 4 + 3
   i.e., l(AC) > 7 cm
   ∴ number of triangles can be drawn if l(AC) > 7 cm
3. l(AC) > l(AB) + l(BC) is the required condition to draw a unique triangle.

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 44 Answers Solutions Chapter 12 Perimeter and Area.

Perimeter and Area Class 7 Practice Set 44 Answers Solutions Chapter 12

Question 1.
If the length and breadth of a rectangle are doubled, how many times the perimeter of the old rectangle will that of the new rectangle be?
Solution:
Let the length of the old rectangle be l and breadth be b.
∴ Perimeter of old rectangle = 2(l + b)
Length of new rectangle = 2l and breadth = 2b
∴ Perimeter of new rectangle = 2(2l + 2b)
= 2 x 2 (l + b)
= 2 x perimeter of old rectangle
∴ The perimeter of new rectangle will be twice the perimeter of old rectangle.

Question 2.
If the side of a square is tripled, how many times the perimeter of the first square will that of the new square be?
Solution:
Let the length of the square be a.
Perimeter of square = 4 x side
= 4 x a = 4a
Side of new square = 3 x a = 3a
Perimeter of new square = 4 x side
= 4 x 3a = 3 x 4a = 3x perimeter of original square.
∴ The perimeter of new square will be three times the perimeter of original square.

Question 3.
Given alongside is the diagram of a playground. It shows the length of its sides. Find the perimeter of the playground.

Solution:

Side AF = side BC + side DE
∴ Side AF = 15 + 15 = 30 m
Side FE = side AB + side CD
∴ Side FE = 10 + 5 = 15 m
∴ Perimeter of the playground = side AB + side BC + side CD + side DE + side FE + side AF
= 10 + 15 + 5 + 15 + 15 + 30
= 90 m.
∴ The perimeter of the playground is 90 m.

Question 4.
As shown in the figure, four napkins all of the same size were made from a square piece of cloth of length 1 m. What length of lace will be required to trim all four sides of all the napkins?

Solution:
Side of the square piece of cloth = 1 m
∴ Side of each napkin = 0.5 m
Length of lace that will be required for 1 napkin = perimeter of the napkin
= 4 x side = 4 x 0.5 = 2 m
∴ Perimeter of 4 napkins = 4 x 2 = 8 m
∴ 8 metre long lace will be required to trim all four napkins.

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 32  Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Algebraic Expressions and Operations on them Class 7 Practice Set 32 Answers Solutions Chapter 8

Question 1.
Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.
i. 7x
ii. 5y – 7z
iii. 3x³ – 5x² – 11
iv. 1 – 8a – 7a² – 7a³
v. 5m – 3
vi. a
vii. 4
viii. 3y² – 7y + 5
Solution:
i. Monomial
ii. Binomial
iii. Trinomial
iv. Polynomial
v. Binomial
vi. Monomial
vii. Monomial
viii. Trinomial

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 9.2 8th Std Maths Answers Solutions Chapter 9 Discount and Commission.

Practice Set 9.2 8th Std Maths Answers Chapter 9 Discount and Commission

Question 1.
John sold books worth Rs 4500 for a publisher. For this he received 15% commission. Complete the following activity to find the total commission John obtained.
Solution:
Selling price of the books = Rs 4500
Rate of commission = 15%
Commission obtained = 15% of selling price
\(=\frac{[15]}{[100]} \times[4500]\)
= 15 × 45
∴ Commission obtained = 675 Rupees.
∴ The total commission obtained by John is Rs 675.

Question 2.
Rafique sold flowers worth Rs 15,000 by giving 4% commission to the agent. Find the commission he paid. Find the amount received by Rafique.
Solution:
Here, selling price of flowers = Rs 15,000,
Rate of commission = 4%
i. Commission = 4% of selling price
= \(\frac { 4 }{ 100 }\) × 15,000
= 4 x 150
∴ Commission = Rs 600

ii. Amount received by Rafique = selling price – commission
= 15,000 – 600
= Rs 14,400
∴ Rafique paid Rs 600 as commission and the amount received by him was Rs 14,400.

Question 3.
A farmer sold food grains for Rs 9200 through an agent. The rate of commission was 2%. How much amount did the agent get ?
Solution:
Here, selling price of food grains = Rs 9200,
Rate of commission = 2%
Commission = 2% of selling price
= \(\frac { 2 }{ 100 }\) × 9200
= 2 × 92
= Rs 184
∴ The agent got a commission of Rs 184.

Question 4.
Umatai purchased following items from a Khadi – Bhandar.
i. 3 sarees for Rs 560 each.
ii. 6 bottles of honey for Rs 90 each.
On the purchase, she received a rebate of 12%. How much total amount did Umatai pay?
Solution:
Here, number of sarees = 3,
Price of each saree = Rs 560
∴ Cost of 3 sarees = 560 × 3
= Rs 1680 …(i)
Also, number of honey bottles = 6,
Price of each bottle = Rs 90
∴ Cost of 6 honey bottles = 90 × 6
= Rs 540
Total amount of purchase
= cost of 3 sarees + cost of 6 honey bottles
= 1680 + 540 … [From (i) and (ii)]
= Rs 2220 …(iii)
Rate of rebate = 12%
Rebate = 12% of total amount of purchase
= \(\frac { 12 }{ 100 }\) × 2220
= 12 × 22.20
= Rs 266.40 ..(iv)
Amount paid by Umatai
= Total amount of purchase – Rebate
= 2,220 – 266.40 … [From (iii) and (iv)]
= Rs 1953.60
∴ The total amount paid by Umatai is Rs 1953.60.

Question 5.
Use the given information and fill in the boxes with suitable numbers.
Smt. Deepanjali purchased a house for Rs 7,50,000 from Smt. Leelaben through an agent. Agent has charged 2 % brokerage from both of them.
Solution:
i. Smt. Deepanjali paid 7,50,000 × \(\frac { 2 }{ 100 }\)
= 7,500 × 2 = Rs 15,000 brokerage for purchasing the house.

ii. Smt. Leelaben paid brokerage of Rs 15,000

iii. Total brokerage received by the agent is = 15,000 + 15,000 = Rs 30,000

iv. The cost of house Smt. Deepanjali paid is = 7,50,000 + 15,000 = Rs 7,65,000

v. The selling price of house for Smt.Leelaben is = 7,50,000 – 15,000
= Rs 7,35,000

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 52 Answers Solutions Chapter 14 Algebraic Formulae – Expansion of Squares.

Algebraic Formulae – Expansion of Squares Class 7 Practice Set 52 Answers Solutions Chapter 14

Algebraic Formulae Expansion Of Squares Class 7 Question 1.
Factorise the following expressions and write them in the product form.
i. 201a³b²
ii. 91xyt²
iii. 24a²b²
iv. tr²s³
i. 201a³b²
= 3 × 67 × a³ × b²
= 3 × 67 × a × a × a × b × b

ii. 91xyt²
= 7 × 13 × x × y × t²
= 7 × 13 × x × y × t × t

iii. 24a²b²
= 2 × 2 × 2 × 3 × a² × b²
= 2 × 2 × 2 × 3 × a × a × b × b

iv. tr²s³
= t × r² × s³
= t × r × r × s × s × s

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 51 Answers Solutions Chapter 14 Algebraic Formulae – Expansion of Squares.

Algebraic Formulae – Expansion of Squares Class 7 Practice Set 51 Answers Solutions Chapter 14

Question 1.
Use the formula to multiply the following:
i. (x + y)(x – y)
ii. (3x – 5)(3x + 5)
iii. (a + 6)(a – 6)
iv. \(\left(\frac{x}{5}+6\right)\left(\frac{x}{5}-6\right)\)
Solution:
i. Here, a = x, b = y
(x + y)(x – y) = x² – y²
…. [(a + b)(a – b) = a² – b²]

ii. Here, a = 3x, b = 5
(3x – 5) (3x + 5) = (3x)² – 5²
…. [(a + b)(a – b) = a² – b²]
= 9x² – 25

iii. Here, A = a, B = 6
(a + 6) (a – 6) = a² – 6²
…. [(A + B)(A – B) = A² – B²]
= a² – 36

iv. Here, a = \(\frac { x }{ 5 }\), b = 6
\(\left(\frac{x}{5}+6\right)\left(\frac{x}{5}-6\right)=\left(\frac{x}{5}\right)^{2}-(6)^{2}\)
…. [(a + b)(a – b) = a² – b²]
= \(\frac{x^{2}}{25}-36\)

Question 2.
Use the formula to find the values:
i. 502 × 498
ii. 97 × 103
iii. 54 × 46
iv. 98 × 102
Solution:
i. 502 × 498 = (500 + 2) (500 – 2)
Here, a = 500, b = 2
∴ (500 + 2) (500 – 2) = 500² – 2²
…. [(a + b)(a – b) = a² – b²]
= 250000 – 4
= 249996
∴ 502 × 498 = 249996

ii. 97 × 103 = (100 – 3) (100 + 3)
Here, a = 100, b = 3
∴ (100 – 3) (100 + 3) = 100² – 3²
…. [(a + b)(a – b) = a² – b²]
= 10000 – 9
= 9991
∴ 97 × 103 = 9991

iii. 54 × 46 = (50 + 4) (50 – 4)
Here, a = 50, b = 4
∴ (50 + 4) (50 – 4) = 50² – 4²
…. [(a + b)(a – b) = a² – b²]
= 2500 – 16 = 2484
∴ 54 × 46 = 2484

iv. 98 × 102 = (100 – 2) (100 + 2)
Here, a = 100, b = 2
∴ (100 – 2) (100 + 2) = 100² – 2²
…. [(a + b)(a – b) = a² – b²]
= 10000 – 4
= 9996
∴ 98 × 102 = 9996

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 49 Answers Solutions Chapter 13 Pythagoras Theorem.

Pythagoras Theorem Class 7 Practice Set 49 Answers Solutions Chapter 13

Question 1.
Find the Pythagorean triplets from among the following sets of numbers:
i. 3,4,5
ii. 2,4,5
iii. 4,5,6
iv. 2,6,7
v. 9,40,41
vi. 4,7,8
Solution:
i. 3² = 9,4² = 16, 5² = 25
Now, 9 + 16 = 25
∴ 3² + 42 = 5²
∴ 3, 4 and 5 is a Pythagorean triplet.

ii. 2² = 4, 4² = 16, 5² = 25
But, 4 + 16 ≠ 25
∴ 2² + 4² ≠ 5²
∴ 2, 4 and 5 is not a Pythagorean triplet.

iii. 4² = 16, 5² = 25, 6² = 36
But 16 + 25 ≠ 36
∴ 4² + 5² ≠ 6²
∴ 4, 5 and 6 is not a Pythagorean triplet.

iv. 2² = 4, 6² = 36, 7² = 49
But, 4 + 36 ≠ 49
∴ 2² + 6² ≠ 7²
∴ 2, 6 and 7 is not a Pythagorean triplet.

v. 9² = 81, 40² = 1600,41² = 1681
Now, 81 + 1600 = 1681
∴ 9² + 40² = 41²
∴ 9,40 and 41 is a Pythagorean triplet.

vi. 4² = 16, 7² = 49, 8² = 64
But, 16 + 49 ≠ 64
∴ 4² + 7² ≠ 8²
∴ 4, 7 and 8 is not a Pythagorean triplet.

Question 2.
The sides of some triangles are given below. Find out which ones are right-angled triangles?
i. 8,15,17
ii. 11,12,15
iii. 11,60,61
iv. 1.5, 1.6, 1.7
v. 40, 20, 30
Solution:
i. 8² = 64, 15² = 225, 17² = 289
Now, 64 + 225 = 289
∴ 8² + 15² = 17²
The above expression is of the from (hypotenuse)² = (base)² + (height)²
∴ The sides of lengths 8,15,17 will form a right-angled triangle.

ii. 11² = 121, 12² = 144, 15² = 225
But, 121 + 144 ≠ 225
∴ 11² + 12² ≠ 25²
∴ The above expression is not of the from
(hypotenuse)² = (base)² + (height)²
∴ The sides of lengths 11, 12, 15 will not form a right-angled triangle.

iii. 11² = 121, 60² = 3600, 61² = 3721
Now, 121 +3600 = 3721
∴ 11² + 60² = 61²
∴ The above expression is of the from
(hypotenuse)² = (base)² + (height)²
∴ The sides of lengths 11, 60, 61 will form a right-angled triangle.

iv. 1.5² = 2.25, 1.6² = 2.56, 1.7² = 2.89
But, 2.25 + 2.56 ≠ 2.89
∴ 1.5² + 1.6² ≠ 1.7²
∴ The above expression is not of the from
(hypotenuse)² = (base)² + (height)²
∴ The sides of lengths 1.5, 1.6, 1.7 will not form a right-angled triangle.

v. 40² = 1600, 20² = 400, 30² = 900
But, 400 + 900 ≠ 1600
∴ 20² + 30² ≠ 40²
∴ The above expression is not of the from (hypotenuse)² = (base)² + (height)²
∴ The sides of lengths 40, 20, 30 will not form a right-angled triangle.

Maharashtra Board Class 7 Maths Chapter 13 Pythagoras’ Theorem Practice Set 49 Intext Questions and Activities

Question 1.
From the numbers 1 to 50, pick out the Pythagorean triplets. (Textbook pg. no. 90)
Solution:

1. 3,4,5
2. 5,12,13
3. 7,24,25
4. 8,15,17
5. 9,40,41
6. 12,35,37
7. 20,21,29