Class 7

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 55 Answers Solutions Chapter 15 Statistics.

Statistics Class 7 Practice Set 55 Answers Solutions Chapter 15

Question 1.
The height of 30 children in a class is given in centimeters. Draw up a frequency table of this data.
131, 135, 140, 138, 132, 133, 135, 133, 134, 135, 132, 133, 140, 139, 132, 131, 134, 133, 140, 140, 139, 136, 137, 136, 139, 137, 133, 134, 131, 140
Solution:

Question 2.
In a certain colony, there are 50 families. The number of people in every family is given below. Draw up the frequency table.
5, 4, 5, 4, 5, 3, 3, 3, 4, 3, 4, 2, 3, 4, 2, 2, 2, 2, 4, 5, 1, 3, 2, 4, 5, 3, 3, 2, 4, 4, 2, 3, 4, 3, 4, 2, 3, 4, 5, 3, 2, 3, 2, 3, 4, 5, 3, 2, 3, 2
Solution:

Question 3.
A dice was cast 40 times and each score noted is given below. Draw up a frequency table for this data.
3, 2, 5, 6, 4, 2, 3, 1, 6, 6, 2, 3, 5, 3, 5, 3, 4, 2, 4, 5, 4, 2, 6, 3, 3, 2, 4, 3, 3, 4, 1, 4, 3, 3, 2, 2, 5, 3, 3, 4.
Solution:

Question 4.
The number of chapatis that 30 children in a hostel need at every meal is given below. Make a frequency table for these scores.
3, 2, 2, 3, 4, 5, 4, 3, 4, 5, 2, 3, 4, 3, 2, 5, 4, 4, 4, 3, 3, 2, 2, 2, 3, 4, 3, 2, 3, 2.
Solution:

Maharashtra Board Class 7 Maths Chapter 15 Statistics Practice Set 55 Intext Questions and Activities

Question 1.
Make groups of 10 children in your class. Find the average height of the children in each group. (Textbook pg. no. 96)
Solution:
(Students should attempt the above activities on their own.)

Question 2.
With the help of your class teacher, note the daily attendance for a week and find the average attendance. (Textbook pg. no. 96)
Solution:
(Students should attempt the above activities on their own.)

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 54 Answers Solutions Chapter 15 Statistics.

Statistics Class 7 Practice Set 54 Answers Solutions Chapter 15

Question 1.
The daily rainfall for each day of a week in a certain city is given in millimeters. Find the average rainfall during the week.
9, 11, 8, 20, 10, 16, 12
Solution:
\(\text { Average rainfall during the week }=\frac{\text { sum of rainfall for each day of the week }}{\text { number of days }}\)
= \(\frac{9+11+8+20+10+16+12}{7}\)
= \(\frac { 86 }{ 7 }\)
= 12.285 ≈ 12.29
∴ The average rainfall during the week is 12.29 mm.

Question 2.
During the annual function of a school, a Women’s Self-help Group had set up a snacks stall. Their sales every hour were worth Rs 960, Rs 830, Rs 945, Rs 800, Rs 847, Rs 970 respectively. What was the average of the hourly sales?
Solution:
\(\text { Average hourly sales }=\frac{\text { sum of sales every hour }}{\text { number of hours }}\)
= \(\frac{960+830+945+800+847+970}{6}\)
= \(\frac { 5352 }{ 6 }\)
= Rs 892
∴ The average of the hourly sales was Rs 892.

Question 3.
The annual rainfall in Vidarbha in five years is given below. What is the average rainfall for those 5 years?
900 mm, 650 mm, 450 mm, 733 mm, 400 mm.
Solution:
\(\text { Average rainfall for } 5 \text { years }=\frac{\text { sum of annual rainfall in five years }}{\text { number of years }}\)
= \(\frac{900+650+450+733+400}{5}\)
= \(\frac { 3133 }{ 5 }\)
= 626.6
∴ The average rainfall in Vidarbha for 5 years was 626.6 mm.

Question 4.
A farmer bought some sacks of animal feed. The weights of the sacks are given below in kilograms. What is the average weight of the sacks?
49.8, 49.7, 49.5, 49.3, 50,48.9, 49.2, 48.8.
Solution:
\(\text { Average weight of the sacks }=\frac{\text { sum of weight of each sack }}{\text { number of sacks }}\)
= \(\frac{49.8+49.7+49.5+49.3+50+48.9+49.2+48.8}{8}\)
= \(\frac { 395.2 }{ 8 }\)
= \(\frac { 3952 }{ 80 }\)
= 49.4
∴ The average weight of the sacks is 49.4 kg.

Maharashtra Board Class 7 Maths Chapter 15 Statistics Practice Set 54 Intext Questions and Activities

Question 1.
Rutuja practised skipping with a rope all seven days of a week. The number of times she jumped the rope in one minute every day is given below. Find the average number of jumps per minute.
60, 62, 61, 60, 59, 63, 58. (Textbook pg. no. 96)
Solution:
\(\text { Average }=\frac{\text { Sum of the number of jumps ons even days }}{\text { Total number of days }}\)
= \(\frac{[60]+[62]+[61]+[60]+[59]+[63]+[58]}{7}\)
= \(\frac { 423 }{ 7 }\)
= 60.42
∴ Average number of jumps per minute = 60.4

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 53 Answers Solutions Chapter 14 Algebraic Formulae – Expansion of Squares.

Algebraic Formulae – Expansion of Squares Class 7 Practice Set 53 Answers Solutions Chapter 14

Question 1.
Factorize the following expressions:
i. p² – q²
ii. 4x² – 25y²
iii. y² – 4
iv. \(\mathrm{p}^{2}-\frac{1}{25}\)
v. \(9 x^{2}-\frac{1}{16} y^{2}\)
vi. \(x^{2}-\frac{1}{x^{2}}\)
vii. a²b – ab
viii. 4x²y – 6x²
ix. \(\frac{1}{2} y^{2}-8 z^{2}\)
x. 2x² – 8y²
Solution:
i. p² – q²
Here, a = p, b = q
∴ p² – q² = (p + q)(p – q)
….[(a² – b²) = (a + b)(a – b)]

ii. 4x² – 25y²
= (2x)² – (5y)²
Here, a = 2x, b = 5y
∴ (2x)² – (5y)² = (2x + 5y)(2x – 5y)
….[(a² – b²) = (a + b)(a – b)]

iii. y² – 4
= y² – 2²
Here, a = y, b = 2
∴ y² – 2² = (y + 2)(y – 2)
….[(a² – b²) = (a + b)(a – b)]

iv. \(\mathrm{p}^{2}-\frac{1}{25}\)
Here a = \(\frac { 1 }{ 25 }\), b = \(\frac { 1 }{ 5 }\)
\(p^{2}-\left(\frac{1}{5}\right)^{2}=\left(p+\frac{1}{5}\right)\left(p-\frac{1}{5}\right)\)
….[(a² – b²) = (a + b)(a – b)]

v. \(9 x^{2}-\frac{1}{16} y^{2}\)
Here a = 3x, b = \(\frac { 1 }{ 4 }y\)
∴\((3 x)^{2}-\left(\frac{1}{4} y\right)^{2}=\left(3 x+\frac{1}{4} y\right)\left(3 x-\frac{1}{4} y\right)\)
….[(a² – b²) = (a + b)(a – b)]

vi. \(x^{2}-\frac{1}{x^{2}}\)
Here a = x, b = \(\frac { 1 }{ x }\)
\(x^{2}-\left(\frac{1}{x}\right)^{2}=\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)\)
….[(a² – b²) = (a + b)(a – b)]

vii. a²b – ab
= a (ab – b)
= ab (a – 1)

viii. 4x²y – 6x²
= 2 (2x²y – 3x²)
= 2x² (2y – 3)

ix. \(\frac{1}{2} y^{2}-8 z^{2}\)

x. 2x² – 8y²
= 2 (x² – 4y²)
= 2 [x² – (2y)²]
= 2(x + 2y)(x – 2y)
….[(a² – b²) = (a + b)(a – b)]

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

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 48 Answers Solutions Chapter 13 Pythagoras Theorem.

Pythagoras Theorem Class 7 Practice Set 48 Answers Solutions Chapter 13

Question 1.
In the figures below, find the value of ‘x’.

Solution:
i. In ∆LMN, ∠M = 90°.
Hence, side LN is the hypotenuse.
According to Pythagoras’ theorem,
l(LN)² = l(LM)² + l(MN)²
∴ x² = 72 + 24²
∴ x² = 49 + 576
∴ x² = 625
∴ x² = 25²
∴ x = 25 units

ii. In ∆PQR, ∠Q = 90°.
Hence, side PR is the hypotenuse.
According to Pythagoras’ theorem,
l(PR)² = l(PQ)² + l(QR)²
∴ 412 = 92 + x²
∴ 1681 = 81 + x²
∴ 1681 – 81 = x²
∴ 1600 = x²
∴ x² = 1600
∴ x² = 40²
∴ x = 40 units

iii. In AEDF, ∠D = 90°.
Hence, side EF is the hypotenuse.
According to Pythagoras’ theorem,
l(EF)² = l(ED)² + l(DF)²
∴ 17² = x² + 8²
∴ 289 = x² + 64
∴ 289 – 64 = x²
∴ 225 = x²
∴ x² = 225
∴ x² = 15²
∴ x = 15 units

Question 2.
In the right-angled ∆PQR, ∠P = 90°. If l(PQ) = 24 cm and l(PR) = 10 cm, find the length of seg QR.
Solution:

In ∆PQR, ∠P = 90°.
Hence, side QR is the hypotenuse.
According to Pythagoras’ theorem,
l(QR)² = l(PR)² + l(PQ)²
∴ l(QR)² = 10² + 24²
∴ l(QR)² = 100 + 576
∴ l(QR)² =676
∴ l(QR)² = 26²
∴ l(QR) = 26 cm
∴ The length of seg QR is 26 cm.

Question 3.
In the right-angled ∆LMN, ∠M = 90°. If l(LM) = 12 cm and l(LN) = 20 cm, find the length of seg MN.
Solution:

In ∆LMN, ∠M = 90°.
Hence, side LN is the hypotenuse.
According to Pythagoras’ theorem,
l(LN)² = l(LM)² + l(MN)²
∴ 20² = 12² + l(MN)²
∴ l(MN)² = 20² – 12²
∴ l(MN)² = 400 – 144
∴ l(MN)² = 256
∴ l(MN)² = 16²
∴ l(MN)= 16 cm
∴ The length of seg MN is 16 cm.

Question 4.
The top of a ladder of length 15 m reaches a window 9 m above the ground. What is the distance between the base of the wall and that of the ladder?
Solution:

The wall and the ground are perpendicular to each other. Hence, the ladder leaning against the wall forms a right-angled triangle.
In ∆ABC, ∠B = 90°
According to Pythagoras’ theorem,
l(AC)² = l(AB)² + l(BC)²
∴ 15² = l(BC)² + 9²
∴ 225 = l(BC)² + 81
∴ 225 – 81 = l(BC)²
∴ 144 = l(BC)²
∴ 12² = l(BC)²
∴ l(BC) = 12
∴ The distance between the base of the wall and that of the ladder is 12 m.

Maharashtra Board Class 7 Maths Chapter 13 Pythagoras’ Theorem Practice Set 48 Intext Questions and Activities

Question 1.
Write the name of the hypotenuse of each of the right angled triangles shown below.
i.

The hypotenuse of ∆ABC is__
ii.

The hypotenuse of ∆LMN is__
iii.

The hypotenuse of ∆XYZ is__
Solution:
i. AC
ii. MN
iii. XZ

Question 2.
Draw right-angled triangles with the lengths of hypotenuse and one side as shown in the rough figures below. Measure the third side. Verify the Pythagoras’ theorem. (Textbook pg. no. 87)

Solution:

i. From the figure, by measurement,
l(AB) = 4 cm
Now, in right-angled triangle ABC,
l(AB)² + l(BC)² = (4)² + (3)²
= 16 + 9
∴ l(AB)² + l(BC)² = 25 …. (i)
l(AC)² = (5)² = 25 ….(ii)
∴ From (i) and (ii),
l(AC)² = l(AB)² + l(BC)²
∴ Pythagoras’ theorem is verified.
(Students should draw the triangles PQR and XYZ and verify the Pythagoras ’ theorem)

Question 3.
Without using a protractor, can you verify that every angle of the vacant quadrilateral in the adjacent figure is a right angle? (Textbook pg. no. 89)
Solution:

In the square ABCD the shaded triangles are right-angled and are the same.
In ∆LBM,
m∠BLM + m∠BML + m∠LBM = 180° …. (Sum of the measures of the angles of a triangles is 180° )
∴ m∠BLM + m∠BML + 90° = 180°
∴ m∠BLM + m∠BML = 90° …. (i)
Now, ∆LBM and ∆LAP are same.
∴ m∠BML = m∠ALP …. (ii)
∴ m∠BLM + m∠ALP = 90° …. IFrom (i) and (ii)l
Now, m∠ALP + m∠PLM + m∠BLM = 180° …. (The measure of a straight angle is 180°)
∴ m∠ALP + m∠BLM + m∠PLM = 180°
∴ 90° + m∠PLM = 180°
∴ m∠PLM = 180°- 90° = 90°
∴ m∠PLM is a right angle.
Similarly, we can prove that the other angles of the vacant quadrilateral are right angles.

Question 4.
On a sheet of card paper, draw a right-angled triangle of sides 3 cm, 4 cm and 5 cm. Construct a square on each of the sides. Find the area of each of the squares and verify Pythagoras’ theorem. (Textbook pg. no. 89)
Solution:

Area of square ABLM = l(AB)² = 32 = 9 sq.cm
Area of square BCPN = l(BC)²= 42 = 16 sq.cm
Area of square ACQR = l(AC)² = 52 = 25 sq.cm
Now, 25 = 16 + 9
i.e. 5² = 4² + 3²
∴ l(AC)² = l(BC)² + l(AB)²
∴ (hypotenuse)² = (base)² + (height)²