Prasanna

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 10.1 8th Std Maths Answers Solutions Chapter 10 Division of Polynomials.

Practice Set 10.1 8th Std Maths Answers Chapter 10 Division of Polynomials

Question 1.
Divide and write the quotient and the remainder.
i. 21m² ÷ 7m
ii. 40a³ ÷ (-10a)
iii. (- 48p⁴) ÷ (- 9p²)
iv. 40m⁵ ÷ 30m³
v. (5x³ – 3x²) ÷ x²
vi. (8p³ – 4p²) ÷ 2p²
vii. (2y³ + 4y² + 3 ) ÷ 2y²
viii. (21x⁴ – 14x² + 7x) ÷ 7x³
ix. (6x⁵ – 4x⁴ + 8x³ + 2x²) ÷ 2x²
x. (25m⁴ – 15m³ + 10m + 8) ÷ 5m³
Solution:
i. 21m² ÷ 7m

∴ Quotient = 3m
Remainder = 0

ii. 40a³ ÷ (-10a)

∴ Quotient = -4a²
Remainder = 0

iii. (- 48p⁴) ÷ (- 9p²)

∴ Quotient = \(\frac { 16 }{ 3 }\) p²
Remainder = 0

iv. 40m⁵ ÷ 30m³

∴ Quotient = \(\frac { 4 }{ 3 }\) m²
Remainder = 0

v. (5x³ – 3x²) ÷ x²

∴ Quotient = 5x – 3
Remainder = 0

vi. (8p³ – 4p²) ÷ 2p²

∴ Quotient = 4p – 2
Remainder = 0

vii. (2y³ + 4y² + 3 ) ÷ 2y²

∴ Quotient = y + 2
Remainder = 3

viii. (21x⁴ – 14x² + 7x) ÷ 7x³

∴ Quotient = 3x
Remainder = -14x² + 7x

ix. (6x⁵ – 4x⁴ + 8x³ + 2x²) ÷ 2x²

∴ Quotient = 3x³ – 2x² + 4x + 1
Remainder = 0

x. (25m⁴ – 15m³ + 10m + 8) ÷ 5m³

∴ Quotient = 5m – 3
Remainder = 10m + 8

Maharashtra Board Class 8 Maths Chapter 10 Division of Polynomials Practice Set 10.1 Intext Questions and Activities

Question 1.
Fill in the blanks in the following examples. (Textbook pg. no. 61)

1. 2a + 3a = __
2. 7b – 4b = __
3. 3p × p² = __
4. 5m² × 3m² = __
5. (2x + 5y) × \(\frac { 3 }{ x }\) = __
6. (3x² + 4y) × (2x + 3y) = __

Solution:

1. 2a + 3a = 5a
2. 7b – 4b = 3b
3. 3p × p² = 3p³
4. 5m² × 3m² = 15m⁴
5. (2x + 5y) × \(\frac { 3 }{ x }\) = \(6+\frac { 15y }{ x }\)
6. (3x² + 4y) × (2x + 3y) = 6x³ + 9x²y + 8xy + 12y²

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 11.1 8th Std Maths Answers Solutions Chapter 11 Statistics.

Practice Set 11.1 8th Std Maths Answers Chapter 11 Statistics

Question 1.
The following table shows the number of saplings planted by 30 students. Fill in the boxes and find the average number of saplings planted by each student.

Solution:

Question 2.
The following table shows the electricity (in units) used by 25 families of Eklara village in a month of May. Complete the table and answer the following questions.

i. How many families use 45 units electricity?
ii. State the score, the frequency of which is 5.
iii. Find N, and ∑ f_i × x_i .
iv. Find the mean of electricity used by each family in the month of May.
Solution:

i. 2 families used 45 units of electricity.
ii. The score for which the frequency is 5 is 75
iii. N = 25 and ∑ f_i × x_i = 1425
iv.

The mean of electricity used by each.

Question 3.
The number of members in the 40 families in Bhilar are as follows:
1, 6, 5, 4, 3, 2, 7, 2, 3, 4, 5, 6, 4, 6, 2, 3, 2, 1, 4, 5, 6, 7, 3, 4, 5, 2, 4, 3, 2, 3, 5, 5, 4, 6, 2,3, 5, 6, 4, 2. Prepare a frequency table and And the mean of members of 40 families.
Solution:


∴ The mean of the members of 40 families is 3.9.

Question 4.
The number of Science and Mathematics projects submitted by Model high school, Nandpur in last 20 years at the state level science exhibition is:
2, 3 ,4, 1, 2, 3, 1, 5, 4, 2, 3, 1, 3, 5, 4, 3, 2, 2, 3, 2. Prepare a frequency table and find the mean of the data.
Solution:


∴ The mean of the given data is 2.75.

Maharashtra Board Class 8 Maths Chapter 11 Statistics Practice Set 11.1 Intext Questions and Activities

Question 1.
The number of pages of a book Ninad read for five consecutive days were 60, 50, 54, 46, 50. Find the average number of pages he read everyday. (Textbook pg. no. 67)
Solution:
\(\frac{60+[50]+[54]+[46]+50}{[5]}=\frac{260}{[5]}=[52]\)
∴ Average number of pages read daily is 52

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 35 Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Algebraic Expressions and Operations on them Class 7 Practice Set 35 Answers Solutions Chapter 8

Question 1.
Multiply:
i. 16xy × 18xy
ii. 23xy² × 4yz²
iii. (12a + 17b) × 4c
iv. (4x + 5y) × (9x + 7y)
Solution:
i. 16xy × 18xy
= 16 × 18 × xy × xy
= 288x²y²

ii. 23xy² × 4yz²
= 23 × 4 × xy² × yz²
= 92xy³z²

iii. (12a + 17b) × 4c = 12a × 4c + 17b × 4c
= 48ac + 68bc

iv. (4x + 5y) × (9x + 7y)
= 4x × (9x + 7y) + 5y × (9x + 7y)
= (4x × 9x) + (4x × 7y) + (5y × 9x) + (5y × 7y)
= 36x² + 28xy + 45xy + 35y²
= 36x² + 73xy + 35y²

Question 2.
A rectangle is (8x + 5) cm long and (5x + 3) cm broad. Find its area. Solution:
Length of the rectangle = (8x + 5) cm
Breadth of the rectangle = (5x + 3) cm
∴ Area of the rectangle = length × breadth
= (8x + 5) × (5x + 3)
= 8x × (5x + 3) + 5 × (5x + 3)
= (8x × 5x) + (8x × 3) + (5 × 5x) + (5 × 3)
= 40x² + 24x + 25x + 15
= 40x² + 49x + 15
∴ The area of the rectangle is (40x² + 49x + 15) sq. cm.

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 36 Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Algebraic Expressions and Operations on them Class 7 Practice Set 36 Answers Solutions Chapter 8

Question 1.
Simplify (3x – 11y) – (17x + 13y) and choose the right answer.
(A) 7x – 12y
(B) -14x – 54y
(C) -3(5x + 4y)
(D) -2(7x + 12y)
Solution:
(D) -2(7x + 12y)

Hints:
(3x – 11y) – (17x + 13y) = 3x – 11y – 17x – 13y
= – 14x – 24y
= – 2 × 7x – 2 × 12y
= – 2(7x + 12y)

Question 2.
The product of (23x²y³z) and (-15x³yz²) is __
(A) -34x⁵y⁴z³
(B) 34x²y³z⁵
(C) 145x³y²z
(D) 170x³y²z³
Solution:
(A) -34x⁵y⁴z³

Question 3.
Solve the following equations:
i. \(4 x+\frac{1}{2}=\frac{9}{2}\)
ii. 10 = 2y + 5
iii. 5m – 4 = 1
iv. 6x – 1 = 3x + 8
v. 2(x – 4) = 4x + 2
vi. 5(x + 1) = 74
Solution:
i. \(4 x+\frac{1}{2}=\frac{9}{2}\)

ii. 10 = 2y + 5

iii. 5m – 4 = 1

iv. 6x – 1 = 3x + 8

v. 2(x – 4) = 4x + 2

vi. 5(x + 1) = 74

Question 4.
Rakesh’s age is less than Sania’s age by 5 years. The sum of their ages is 27 years. How old are they?
Solution:
Let the age of Rakesh be x years.
∴ Sania’s age = (x + 5) years.
According to the given condition,
x + (x + 5) = 27
∴ 2x + 5 = 27
∴ 2x = 27 – 5
∴ 2x = 22
∴ \(x=\frac { 22 }{ 2 }=11\)
Sania’s age = x + 5 = 11 + 5 = 16 years
∴ The ages of Rakesh and Sania are 11 years and 16 years respectively.

Question 5.
When planting a forest, the number of jambhul trees planted was greater than the number of ashoka trees by 60. If there are altogether 200 trees of these two types, how many jambhul trees were planted?
Solution:
Let the number of jambhul trees planted be x.
∴ Number of ashoka trees = x – 60
According to the given condition, x + x – 60 = 200
∴ 2x = 200 + 60
∴ 2x = 260
∴ \(x=\frac { 260 }{ 2 }=130\)
∴ 130 jambhul trees were planted.

Question 6.
Shubhangi has twice as many 20-rupee notes as she has 50-rupee notes. Altogether, she has 2700 rupees. How many 50-rupee notes does she have?
Solution:
Let the number of 50-rupee notes with shubhangi be x.
∴ Number of 20-rupee notes = 2x
∴ Total amount with Shubhangi = Number of 50-rupee notes × 50 + Number of 20-rupee notes × 20
= x × 50 + 2x × 20
= 50x + 40x
= 90x
According to the given condition,
90x = 2700
∴ \(x=\frac { 2700 }{ 90 }=30\)
∴ Shubhangi has 30 notes of 50 rupees.

Question 7.
virat made twice as many runs as Rohit. The total of their scores is 2 less than a double century. How many runs did each of them make?
Solution:
Let the runs made by Rohit be x.
∴ Runs made by Virat = 2x
According to the given condition,
x + 2x = 200 – 2
∴ 3x = 198
∴ \(x=\frac { 198 }{ 3 }=66\)
∴ Runs made by Virat = 2x = 2 × 66 = 132
∴ The runs made by Virat and Rohit are 132 and 66 respectively.

Maharashtra Board Class 7 Maths Chapter 8 Algebraic Expressions and Operations on them Practice Set 36 Intext Questions and Activities

Question 1.
Solve the following equations. (Textbook pg. no. 59)
i. x + 7 = 4
ii. 4p = 12
iii. m – 5 = 4
iv. \(\frac { t }{ 3 }=6\)
Solution:
i. x + 7 = 4
∴ x + 7 – 7 = 4 – 7 ….(Subtracting 7 from both sides)
∴ x + 0 = -3
∴ x = -3

ii. 4p = 12
∴ \(\frac{4 p}{4}=\frac{12}{4}\) ….(Dividing both sides by 4)
∴ p = 3

iii. m – 5 = 4
∴ m – 5 + 5 = 4 + 5
…. (Adding 5 to both sides)
∴ m + 0 = 9
∴ m = 9

iv. \(\frac { t }{ 3 }=6\)
∴ \(\frac { t }{ 3 }\) × 3 = 6 × 3 …. (Multiplying both sides by 3)
∴ t = 18

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 34 Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Algebraic Expressions and Operations on them Class 7 Practice Set 34 Answers Solutions Chapter 8

Question 1.
Subtract the second expression from the first.
i. (4xy – 9z); (3xy – 16z)
ii. (5x + 4y + 7z); (x + 2y + 3z)
iii. (14x² + 8xy + 3y²); (26x² – 8xy – 17y²)
iv. (6x² + 7xy + 16y²); (16x² – 17xy)
v. (4x + 16z); (19y – 14z + 16x)
Solution:
i. (4xy – 9z) – (3xy – 16z)
= 4xy – 9z – 3xy + 16z
= (4xy – 3xy) + (16z – 9z)
= xy + 7z

ii. (5x + 4y + 7z) – (x + 2y + 3z)
= 5x + 4y + 7z – x – 2y – 3z
= (5x – x) + (4y – 2y) + (7z – 3z)
= 4x + 2y + 4z

iii. (14x² + 8xy + 3y²) – (26x² – 8xy – 17y²)
= 14x² + 8xy + 3y² – 26x² + 8xy + 17y²
= (14x² – 26x²) + (8xy + 8xy) + (3y² + 17y²)
= -12x² + 16xy + 20y²

iv. (6x² + 7xy + 16y²) – (16x² – 17xy)
= 6x² + 7xy + 16y² – 16x² + 17xy
= (6x² – 16x²) + (7xy + 17xy) + 16y²
= -10x² + 24xy + 16y²

v. (4x + 16z) – (19y— 14z + 16x)
= 4x + 16z – 19y + 14z – 16x
= (4x – 16x) – 19y + (16z + 14z)
= -12x – 19y + 30z

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 7 Maths Solutions covers the 7th Std Maths Practice Set 33 Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Algebraic Expressions and Operations on them Class 7 Practice Set 33 Answers Solutions Chapter 8

Question 1.
Add:
i. 9p + 16q; 13p + 2q
ii. 2a + 6b + 8c; 16a + 13c + 18b
iii. 13x² – 12y²; 6x² – 8y²
iv. 17a²b² + 16c; 28c – 28a²b²
v. 3y² – 10y + 16; 2y – 7
vi. – 3y² + 10y – 16; 7y² + 8
Solution:
i. (9p + 16q) + (13p + 2q)
= (9p + 13p) + (16q + 2q)
= 22p + 18q

ii. (2a + 6b + 8c) + (16a + 13c + 18b)
= (2a + 16a) + (6b + 18b) + (8c + 13c)
= 18a + 24b + 21c

iii. (13x² – 12y²) + (6x² – 8y²)
= (13x² + 6x²) + [(-12y²) + (-8y²)]
= 19x² + (-20y²)
= 19x² – 20y²

iv. (17a²b² + 16c) + (28c – 28a²b²)
= [17a²b² + (-28a²b²)] + (16c + 28c)
= -11a²b² + 44c

v. (3y² – 10y + 16) + (2y – 7)
= 3y² + (-10y + 2y) + (16 – 7)
= 3y² – 8y + 9

vi. (-3y² + 10y – 16) + (7y² + 8)
= (-3y² + 7y²) + (10y) + (-16 + 8)
= 4y² + 10y – 8

Maharashtra Board Class 7 Maths Chapter 8 Algebraic Expressions and Operations on them Practice Set 33 Intext Questions and Activities

Question 1.
Answer the following questions. (Textbook pg. no. 57)

1. 3x + 4y = How many?
2. 3 guavas + 4 mangoes = 7 guavas.
3. 7m – 2n = 5m.

Solution:

1. 3x and 4y are unlike terms. Hence, they cannot be added, further to get a single term.
2. No. Guava and mango are different fruits. Hence, 3 guavas + 4 mangoes & 7 guavas.
3. No. 7m and 2n are unlike terms. Hence, 7m – 2n ≠ 5m.

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 8 Maths Solutions covers the Practice Set 8.1 8th Std Maths Answers Solutions Chapter 8 Quadrilateral: Constructions and Types.

Practice Set 8.1 8th Std Maths Answers Chapter 8 Quadrilateral: Constructions and Types

Construct the following quadrilaterals of given measures.

Question 1.
In ∆MORE, l(MO) = 5.8 cm, l(OR) = 4.4 cm, m∠M = 58°, m∠O = 105°, m∠R = 90°.
Solution:

Question 2.
Construct ∆DEFG such that l(DE) = 4.5 cm, l(EF) = 6.5 cm, l(DG) = 5.5 cm, l(DF) = 7.2 cm, l(EG) = 7.8 cm.
Solution:

Question 3.
In ∆ABCD, l(AB) = 6.4 cm, l(BC) = 4.8 cm, m∠A = 70°, m∠B = 50°, m∠C = 140°.
Solution:

Question 4.
Construct ₹LMNO such that
l(LM) = l(LO) = 6 cm,
l(ON) = l(NM) = 4.5 cm, l(OM) = 7.5 cm.
Solution:

Maharashtra Board Class 8 Maths Chapter 8 Quadrilateral: Constructions and Types Practice Set 8.1 Intext Questions and Activities

Question 1.
Construction of a triangle:
Construct the triangles with given measures. (Textbook pg. no. 41)
i. ∆ABC: l(AB) = 5 cm, l(BC) = 5.5, l(AC) = 6 cm.
Solution:

Steps of construction:
Step 1 : As shown in the rough figure, draw seg BC of length 5.5 cm as the base.
Step 2 : By taking a distance of 5 cm on the compass and placing the metal tip of the compass on point B, draw an arc on one side of BC.
Step 3 : By taking a distance 6 cm on the compass and placing the metal tip of the t compass on point C and draw an arc ’ such that it intersects the previous arc. Name the point as A.
Step 4 : Draw segments AB and AC to get the triangle. ∆ABC is the required triangle.

ii. ∆DEF: m∠D = 35°, m∠F = 100°, l(DF) = 4.8 cm.
Solution:

Steps of construction:
Step 1 : As shown in the rough figure, draw seg DF of length 4.8 cm as the base.
Step 2 : Placing the centre of the protractor at point D, mark point P such that m∠PDF = 35°.
Step 3 : Placing the centre of the protractor at point F, mark point Q such that m∠QFD = 100°.
Step 4 : Draw ray DP and ray FQ. Name their point of intersection as E.
∆DEF is required triangle.

iii. ∆MNP: l(MP) = 6.2 cm, l(NP) = 4.5 cm, m∠P = 75°.
Solution:

Steps of construction:
Step 1 : As shown in the rough figure, draw seg PN of length 4.5 cm as the base.
Step 2 : Placing the centre of the protractor at point P, mark point Q such that m∠QPN = 75°.
Step 3 : By taking a distance of 6.2 cm on the compass and placing the metal tip at point P, draw an arc on ray PQ. Name the point as M.
Step 4 : Draw seg MN to get the triangle. ∆MNP is the required triangle.

iv. ∆XYZ: m∠Y = 90°, l(XY) = 4.2 cm, l(XZ) = 7 cm.
Solution:

Steps of construction:
Step 1 : As shown in the rough figure, draw seg XY of 4.2 cm as the base.
Step 2 : Placing the centre of the protractor at point Y, mark point Q such that m∠QYX = 90°.
Step 3 : By taking a distance of 7 cm on the compass and placing the metal tip on point X, draw an arc on ray YQ. Name the point as Z.
Step 4 : Draw seg XZ to get the triangle. ∆XYZ is the required triangle.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 8.3 8th Std Maths Answers Solutions Chapter 8 Quadrilateral: Constructions and Types.

Practice Set 8.3 8th Std Maths Answers Chapter 8 Quadrilateral: Constructions and Types

Question 1.
Measures of opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of its each angle.
Solution:
Let ₹PQRS be the parallelogram.
m∠Q = (3x – 2)° and m∠S = (50 – x)°

m∠Q = m∠S
…..(i)
[Opposite angles of a parallelogram are congruent]
∴ 3x – 2 = 50 – x
∴ 3x + x = 50 + 2
∴ 4x = 52
∴ x = \(\frac { 52 }{ 4 }\)
∴ x = 13
Now, m∠Q = (3x – 2)°
= (3 x 13 – 2)° = (39 – 2)° = 37°
∴ m∠S = m∠Q = 37° …[From(i)]
m∠P + m∠Q = 180°
….[Adjacent angles of a parallelogram are supplementary]
∴ m∠P + 37° = 180°
∴ m∠P = 180° – 37° = 143°
∴ m∠R = m∠P = 143°
…..[Opposite angles of a parallelogram are congruent]
∴ The measures of the angles of the parallelogram are 37°, 143°, 37° and 143°.

Question 2.
Referring the given figure of a parallelogram, write the answers of questions given below.
i. If l(WZ) = 4.5 cm, then l(XY) = ?
ii. If l(YZ) = 8.2 cm, then l(XW) = ?
iii. If l(OX) = 2.5 cm, then l(OZ) = ?
iv. If l(WO) = 3.3 cm, then l(WY) = ?
v. If m∠WZY = 120°, then m∠WXY = ? and m∠XWZ = ?

Solution:
i. l(WZ) = 4.5 cm … [Given]
l(X Y) = l(WZ) ….[Opposite sides of a parallelogram are congrument ]
∴ l(X Y) = 4.5cm

ii. l(YZ) = 8.2 cm …[Given]
l(XW) = l(YZ)
…[Opposite sides of a parallelogram are congruent]
∴ l(XW) = 8.2cm … [Given]

iii. l(OX) = 2.5 cm …[Given]
l(OZ) = l(OX)
….[Diagonals of a parallelogram bisect each other]
∴ l(OZ) = 2.5cm

iv. l(WO) = 3.3 cm … [Given]
l(WO) = \(\frac { 1 }{ 2 }\) l(WY)
….[Diagonals of a parallelogram bisect each other]
∴ 3.3 = \(\frac { 1 }{ 2 }\) l(WY)
∴ 3.3 x 2 = l(WY)
∴ l(WY) = 6.6cm

v. m∠WZY =120° … [Given]
m∠WXY = m∠WZY
…..[Opposite angles of a parallelogram are congrument]
∴ m∠WXY = 120° …(i)
m∠XWZ + m∠WXY = 180°
….[Adjacent angles of a parallelogram are supplementary]
∴ m∠XWZ + 120° = 180° … [From (i)]
∴ m∠XWZ = 180°- 120°
∴ m∠XWZ = 60°

Question 3.
Construct a parallelogram ABCD such that l(BC) = 7 cm, m∠ABC = 40°, l(AB) = 3 cm
Solution:

Opposite sides of a parallelogram are congruent.
∴ l(AB) = l(CD) = 3cm
l(BC) = l(AD) = 7 cm

Question 4.
Ratio of consecutive angles of a quadrilateral is 1 : 2 : 3 : 4. Find the measure of its each angle. Write with reason, what type of a quadrilateral it is.
Solution:
Let ₹PQRS be the quadrilateral.
Ratio of consecutive angles of a quadrilateral is 1 : 2 : 3 : 4.
Let the common multiple be x.
∴m∠P = x°, m∠Q = 2x°, m∠R = 3x° and m∠S = 4x°
In ₹PQRS,
m∠P + m∠Q + m∠R + m∠S = 360°
…[Sum of the measures of the angles of a quadrilateral is 360°]
∴x° + 2x° + 3x° + 4x° = 360°
∴10 x° = 360°
∴x° = \(\frac { 360 }{ 10 }\)
∴x° = 36°
∴m∠P = x° = 36°

m∠Q = 2x° = 2 × 36° = 72°
m∠R = 3x° = 3 × 36° = 108° and
m∠S = 4x° = 4 × 36° = 144°
∴The measures of the angles of the quadrilateral are 36°, 72°, 108°, 144°.
Here, m∠P + m∠S = 36° + 144° = 180°
Since, interior angles are supplementary,
∴side PQ || side SR
m∠P + m∠Q = 36° + 72° = 108° ≠ 180°
∴side PS is not parallel to side QR.
Since, one pair of opposite sides of the given quadrilateral is parallel.
∴The given quadrilateral is a trapezium.

Question 5.
Construct ₹BARC such that
l(BA) = l(BC) = 4.2 cm, l(AC) = 6.0 cm, l(AR) = l(CR) = 5.6 cm
Solution:

Question 6.
Construct ₹PQRS, such that l(PQ) = 3.5 cm, l(QR) = 5.6 cm, l(RS) = 3.5 cm, m∠Q = 110°, m∠R = 70°. If it is given that ₹PQRS is a parallelogram, which of the given information is unnecessary?
Solution:

1. Since, the opposite sides of a parallelogram are congruent.
   ∴ Either l(PQ) or l(SR) is required.
2. To construct a parallelogram lengths of adjacent sides and measure of one angle is required.
   ∴ Either l(PQ) and m∠Q or l(SR) and m∠R is the unnecessary information given in the question.

Maharashtra Board Class 8 Maths Chapter 8 Quadrilateral: Constructions and Types Practice Set 8.3 Intext Questions and Activities

Question 1.
Draw a parallelogram PQRS. Take two rulers of different widths, place one ruler horizontally and draw lines along its edges. Now place the other ruler in slant position over the lines drawn and draw lines along its edges. We get a parallelogram. Draw the diagonals of it and name the point of intersection as T.

1. Measure the opposite angles of the parallelogram.
2. Measure the lengths of opposite sides.
3. Measure the lengths of diagonals.
4. Measure the lengths of parts of the diagonals made by point T. (Textbook pg. no. 47)

Solution:
[Students should attempt the above activities on their own.]

Question 2.
In the given figure of ₹ABCD, verify with a divider that seg AB ≅ seg CB and seg AD ≅ seg CD. Similarly measure ∠BAD and ∠BCD and verify that they are congruent. (Textbook pg. no. 48)

Solution:
[Students should attempt the above activities on their own.]