Maharashtra Board Practice Set 52 Class 7 Maths Solutions Chapter 14 Algebraic Formulae – Expansion of Squares

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

Maharashtra Board Practice Set 51 Class 7 Maths Solutions Chapter 14 Algebraic Formulae – Expansion of Squares

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

Maharashtra Board Practice Set 49 Class 7 Maths Solutions Chapter 13 Pythagoras Theorem

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

Maharashtra Board Practice Set 48 Class 7 Maths Solutions Chapter 13 Pythagoras Theorem

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

Maharashtra Board Practice Set 43 Class 7 Maths Solutions Chapter 11 Circle

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 43 Answers Solutions Chapter 11 Circle.

Circle Class 7 Practice Set 43 Answers Solutions Chapter 11

Question 1.
Choose the correct option.
If arc AXB and arc AYB are corresponding arcs and m(arc AXB) = 120° then m(arc AYB) =
(A) 140°
(B) 60°
(C) 240°
(D) 160°
Solution:
(C) 240°

Hint:
Measure of major arc = 360° – measure of corresponding minor arc
∴ m (arc AYB) = 360 – m (arc AXB)
∴ m (arc AYB) = 360 – 120
∴ m (arc AYB) = 240°

Question 2.
Some arcs are shown in the circle with centre ‘O’ Write the names of the minor arcs, major arcs and semicircular arcs from among them.

Solution:
Minor arcs : arc QXP, arc PR, arc RY, arc YQ, arc QX, arc XP, arc PRY.
Major arcs : arc PYQ, arc PQR, arc RQY, arc XPQ, arc XQP, arc XQR
Semicircular arcs : arc QPR, arc QYR.

Question 3.
In a circle with centre O, the measure of a minor arc is 110°. What is the measure of the major arc PYQ?

Solution:
Measure of major arc = 360° – measure of corresponding minor arc
∴ m (arc PYQ) = 360 – 110
∴ m (arc PYQ) = 250°
∴ The measure of the major arc PYQ is 250°.

Maharashtra Board Class 7 Maths Chapter 11 Circle Practice Set 43 Intext Questions and Activities

Question 1.
Measure the circumference and diameter of the objects given below and enter the ratio of the circumference to its diameter in the table.

Examine the ratio of the circumference to the diameter. What do we see? (Textbook pg. no. 75)
Solution:
The ratio of circumference to the diameter is same and is approximately equal to 3.14.

Question 2.
Place a cylindrical bottle on a paper and trace the outline of its base. Use a thread to measure the circumference of the circle. (Textbook pg. no. 75)
Solution:
(Students should attempt the above activities on their own)

Question 3.
Measure the circumference of a bangle with the help of a thread. (Textbook pg. no. 75)
Solution:
(Students should attempt the above activities on their own)

Question 4.
Measure the circumference of any cylindrical object using a thread. (Textbook pg. no. 75)
Solution:
(Students should attempt the above activities on their own)

Maharashtra Board Practice Set 45 Class 7 Maths Solutions Chapter 12 Perimeter and Area

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 45 Answers Solutions Chapter 12 Perimeter and Area.

Perimeter and Area Class 7 Practice Set 45 Answers Solutions Chapter 12

Question 1.
If the side of a square is 12 cm, find its area.
Solution:
Area of a square = (side)² = (12)²
= 144 sq. cm.
∴ The area of the square is 144 sq. cm.

Question 2.
If the length of a rectangle is 15 cm and breadth is 5 cm, find its area.
Solution:
Area of a rectangle = length × breadth
= 15 × 5
= 75 sq. cm.
∴ The area of the rectangle is 75 sq. cm.

Question 3.
The area of a rectangle is 102 sq. cm. If its length is 17 cm, what is its perimeter?
Solution:
Area of a rectangle = length × breadth
∴ 102 = 17 × breadth
∴ breadth = $$\frac { 102 }{ 17 }$$ = 6 cm
Perimeter of rectangle = 2 (length + breadth)
= 2 (17 + 6)
= 2 × 23
= 46 cm
∴ The perimeter of rectangle is 46 cm.

Question 4.
If the side of a square is tripled, how many times will its area be as compared to the area of the original square?
Solution:
Let the side of the square be a.
∴ Area of a square = (side)² = a²
New side of the square = 3 × a = 3a
∴ New area of the square = (3a)²
= 9a²
= 9 × area of original square
∴ If the side of a square is tripled, its area will become 9 times the area of the original square.

Maharashtra Board Class 7 Maths Chapter 12 Perimeter and Area Practice Set 45 Intext Questions and Activities

Question 1.
A rectangular playground is 65m long and 30m wide. A pathway of 1.5 m width goes all around the ground, outside it. Find the area of the pathway. (Textbook pg. no. 82)

Solution:
The playground is rectangular.
₹ABCD is the playground. Around it is a pathway 1.5 m wide.
Around ₹ABCD we get the rectangle ₹PQRS
Length of new rectangle PQRS = 65 + 1.5 + 1.5 = 68 m
Breadth of new rectangle PQRS = 30 + 1.5 + 1.5 = 33m
Area of path = Area of rectangle PQRS – Area of rectangle ABCD = 68 x 33 – 65 x 30
= 2244 – 1950
= 294 sq m

Question 2.
Is there another way to find the area of the pathway in the problem above? (Textbook pg. no. 82)

Solution:
Yes. The area of the pathway can be found by dividing it into rectangles and adding the areas of these rectangles.
Length of rectangle 1 = 30 + 1.5 + 1.5 = 33 m
Breadth of rectangle 1 = 1.5 m
∴ Area of rectangle 1 = 33 x 1.5
= 49.5 sq. m
Area of rectangle 4 = Area of rectangle 1
= 49.5 sq. m.
Length of rectangle 2 = 65 m
breadth of rectangle 2 = 1.5 m
∴ Area of rectangle 2 = 65 x 1.5
= 97.5 sq. m.
Area of rectangle 3 = area of rectangle 2
= 97.5 sq. m.
∴ Area of pathway = Sum of area of the 4 rectangles = 49.5 + 49.5 + 97.5 + 97.5
= 294 sq. m.

Question 3.
The length and the width of a mobile phone are 13 cm and 7 cm respectively. It has a screen PQRS as shown in the figure. What is the area of the screen? (Textbook pg. no. 82)

Solution:
ABCD is the rectangle formed by the edges of the mobile. PQRS is the rectangle formed by leaving a 1.5 cm wide edge alongside AB, BC, and DC, and a 2 cm edge alongside DA.
Length of rectangle PQRS = 9.5 cm
Breadth of rectangle PQRS = 4 cm
Area of screen = Area of rectangle PQRS = 9.5 x 4
= 38 sq .cm

Maharashtra Board Practice Set 47 Class 7 Maths Solutions Chapter 12 Perimeter and Area

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 47 Answers Solutions Chapter 12 Perimeter and Area.

Perimeter and Area Class 7 Practice Set 47 Answers Solutions Chapter 12

Question 1.
Find the total surface area of cubes having the following sides:
i. 3 cm
ii. 5 cm
iii. 7.2 m
iv. 6.8 m
v. 5.5 m
Solution:
i. Total surface area of cube = 6l²
= 6 × (3)²
= 6 × 9
= 54 sq. cm.

ii. Total surface area of cube = 6l²
= 6 × 5²
= 6 × 25
= 150 sq. cm.

iii. Total surface area of cube = 6l²
= 6 × (7.2)²
= 6 × 51.84
= 311.04 sq. m.

iv. Total surface area of cube = 6l²
= 6 × (6.8)²
= 6 × 46.24
= 277.44 sq. m.

v. Total surface area of cube = 6l²
= 6 × (5.5)²
= 6 × 30.25
= 181.5 sq. m.

Question 2.
Find the total surface area of the cuboids of length, breadth and height as given below:
i. 12 cm, 10 cm, 5 cm
ii. 5 cm, 3.5 cm, 1.4 cm
iii. 2.5 m, 2 m, 2.4 m
iv. 8 m, 5 m, 3.5 m
Solution:
i. Total surface area of cuboid
= 2 (lb + bh + lh)
= 2 (12 × 10 + 10 × 5 + 12 × 5)
= 2 (120 + 50 + 60)
= 2 × 230
= 460 sq. cm.

ii. Total surface area of cuboid
= 2 (lb + bh + lh)
= 2 (5 × 3.5 + 3.5 × 1.4 + 5 × 1.4)
= 2(17.5 + 4.9 + 7)
= 2 × 29.4
= 58.8 sq. cm.

iii. Total surface area of cuboid = 2 (lb + bh + lh)
= 2(2.5 × 2 + 2 × 2.4 + 2.5 × 2.4)
= 2 (5 + 4.8 + 6)
= 2 × 15.8
= 31.6 sq. m.

iv. Total surface area of cuboid = 2 (lb + bh + lh)
= 2 (8 × 5+ 5 × 3.5 + 8 × 3.5)
= 2(40 + 17.5 + 28)
= 2 × 85.5
= 171 sq. m.

Question 3.
A matchbox is 4 cm long, 2.5 cm broad and 1.5 cm in height. Its outer sides are to be covered exactly with craft paper. How much paper will be required to do so?
Solution:
Length of the matchbox (l) = 4 cm, breadth (b) = 2.5 cm, height (h) = 1.5 cm
∴ Total surface area of the matchbox = 2 (lb + bh + lh)
= 2 (4 × 2.5 + 2.5 × 1.5 + 4 × 1.5)
= 2 (10 + 3.75 + 6)
= 2 × 19.75
= 39.5 sq. cm.
∴ 39.5 sq. cm paper will be required.

Question 4.
An open box of length 1.5 m, breadth 1 m, and height 1 m is to be made for use on a trolley for carrying garden waste. How much sheet metal will be required to make this box? The inside and outside surface of the box is to be painted with rust proof paint. At a rate of Rs 150 per sq. m, how much will it cost to paint the box?
Solution:
Length of the box (l) = 1.5 m, breadth (b) = 1 m, height (h) = 1 m
Since, the box is open at top,
∴ Sheet required to make the box = total surface area of the box – area of the top
= 2 (lb + bh + lh) – lb
= 2lb + 2bh + 2lh – lb
= lb + 2bh + 2lh
= 1.5 × 1 + 2 × 1 × 1 + 2 × 1.5 × 1
= 1.5 + 2 + 3
= 6.5 sq. m.
Since, the inside and outside surface of the box are to be painted.
∴ Area to be painted = 2 × Area of the box = 2 × 6.5 = 13 sq. m.
Total cost of painting = area to be painted × rate per sq. m.
= 13 × 150
= Rs 1950
∴ 6.5 sq. m. sheet of metal will be required and the cost of painting the box will be Rs 1950.

Maharashtra Board Class 7 Maths Chapter 12 Perimeter and Area Practice Set 47 Intext Questions and Activities

Question 1.
Measure the length and breadth of the courts laid out for games such as kho-kho, kabaddi, tennis, badminton, etc. Find out their perimeters and areas. (Textbook pg. no. 81)
Solution:
(Students should attempt the above activities on their own)

Question 2.
Take mobile handsets of different sizes and find the area of their screens. (Textbook pg. no. 82)
Solution:
(Students should attempt the above activities on their own)

Maharashtra Board Practice Set 46 Class 7 Maths Solutions Chapter 12 Perimeter and Area

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 46 Answers Solutions Chapter 12 Perimeter and Area.

Perimeter and Area Class 7 Practice Set 46 Answers Solutions Chapter 12

Question 1.
A page of a calendar is 45 cm long and 26 cm wide. What is its area?
Solution:
Area of page of a calendar = length × breadth
= 45 × 26
= 1170 sq. cm.
∴ The area of the page of the calendar is 1170 sq. cm.

Question 2.
What is the area of a triangle with base 4.8 cm and height 3.6 cm?
Solution:
Area of triangle = $$\frac { 1 }{ 2 }$$ × base × height
= $$\frac { 1 }{ 2 }$$ × 4.8 × 3.6
= $$\frac { 1 }{ 2 }$$ × 17.28
= 8.64 sq. cm.
∴ The area of the triangle is 8.64 sq. cm.

Question 3.
What is the value of a rectangular plot of land 75.5 m long and 30.5 m broad at the rate of Rs 1000 per square metre?
Solution:
Area of the rectangular plot = length × breadth
= 75.5 × 30.5
= 2302.75 sq. m.
Value of the plot = area of the plot × rate per square metre = 2302.75 × 1000
= Rs 230275
∴ The value of the plot is Rs 23,02,750.

Question 4.
A rectangular hall is 12 m long and 6 m broad. Its flooring is to be made of square tiles of side 30 cm. How many tiles will fit in the entire hall? How many would be required if tiles of side 15 cm were used?
Solution:
Area of the rectangular hall = length × breadth
= 12 × 6
= 72 sq. m.
Side of the square shaped tile = 30 cm
= $$\frac { 30 }{ 100 }$$ m …[1cm = $$\frac { 1 }{ 100 }$$m]
= $$\frac { 3 }{ 10 }$$ m
Area of the tile = (side)²
= $$\left(\frac{3}{10}\right)^{2}$$
= $$\frac{9}{100}$$ sq.m
Number of tiles required = $$\frac{\text { Area of the hall }}{\text { Area of each tile }}$$
= $$72 \div \frac{9}{100}$$
= $$72 \times \frac{100}{9}$$
= 800
∴ 800 square shaped tiles of 30 cm side will be required.
If the side of the square is reduced to half, its area will become $$\frac { 1 }{ 4 }$$ times the original.
i. e. number of tiles required will become 4 times the original tiles.
∴ Number of tiles required = 4 × number of tiles of side 30 cm
= 4 × 800
= 3200
∴ 3200 square shaped tiles of 15 cm side will be required.

Question 5.
Find the perimeter and area of a garden with measures as shown in the figure alongside.

Solution:

The boundary of the garden is made of 12 sides each of length 13 m.
Perimeter of the garden = sum of the lengths of all sides
= 12 × 13
= 156 m
The garden in the given figure can be divided into 5 squares each of side 13 m.
∴ Area of the garden = 5 × area of each square part
= 5 × (side)²
= 5 × (13)²
= 5 × 169
= 845 sq. m.
∴ The perimeter and area of a garden are 156 m and 845 sq. m. respectively.

Maharashtra Board Practice Set 42 Class 7 Maths Solutions Chapter 11 Circle

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 42 Answers Solutions Chapter 11 Circle.

Circle Class 7 Practice Set 42 Answers Solutions Chapter 11

Question 1.
Complete the table below:

 Sr. No Radius (r) Diameter (d) Circumference (c) i. 7 cm ii. 28 cm iii. 616 cm iv. 72.6 cm

Solution:
i. Radius (r) = 7 cm
Diameter (d) = 2r
= 2 x 7 = 14 cm
Circumference (c) = πd
= $$\frac { 22 }{ 7 }$$ x 14
= 44 cm

ii. Diameter (d) = 28 cm
Radius (r) = $$\frac{d}{2}=\frac{28}{2}$$ = 14 cm
Circumference (c) = πd
= $$\frac { 22 }{ 7 }$$ x 28
= 88 cm

iii. Circumference (c) = 616 cm
∴ πd = 616
∴ $$\frac { 22 }{ 7 }$$ x d = 616
∴ d = 616 x $$\frac { 7 }{ 22 }$$
∴ d = 196 cm
∴ Diameter (d) = 196 cm
Radius (r) = $$\frac{\mathrm{d}}{2}=\frac{196}{2}$$ = 98 cm

iv. Circumference (c) = 72.6 cm
∴ πd = 72.6
$$\frac { 22 }{ 7 }$$ x d = 72.6
∴ $$d=72.6 \times \frac{7}{22}=\frac{726}{10} \times \frac{7}{22}=\frac{33 \times 7}{10}$$
∴ d = 23.1 cm
∴ Diameter (d) = 23.1 cm
Radius (r) = $$\frac{\mathrm{d}}{2}=\frac{23.1}{2}$$
= 11.55 cm

 Sr. No Radius (r) Diameter (d) Circumference (c) i. 7 cm 14 cm 44 cm ii. 14 cm 28 cm 88 cm iii. 98 cm 196 cm 616 cm iv. 11.55 cm 23.1 cm 72.6 cm

Question 2.
If the circumference of a circle is 176 cm, find its radius.
Solution:
Circumference (c) = 176 cm
∴ 2πr = 176
∴ 2 x $$\frac { 22 }{ 7 }$$ x r = 176
∴ $$\frac { 44 }{ 7 }$$ x r = 176
∴ r = 176 x $$\frac { 7 }{ 44 }$$ = 28 cm
∴ The radius of the circle is 28 cm.

Question 3.
The radius of a circular garden is 56 m. What would it cost to put a 4-round fence around this garden at a rate of 40 rupees per metre?
Solution:
Radius of the circular garden (r) = 56 m
∴ Circumference of the circular garden (c) = 2πr
= 2 x $$\frac { 22 }{ 7 }$$ x 56
= 352 m
∴ Length of the wire required to put 1-round fence = Circumference
∴ Length of wire required to put a 4-round fence = 4 x Circumference
= 4 x 352
= 1408 m
∴ Cost of wire per meter = Rs 40
∴ Total cost = length of wire required x cost of the wire
= 1408 x 40
= Rs 56320
∴ The cost to put a 4-round fence around the garden is Rs 56320.

Question 4.
The wheel of a bullock cart has a diameter of 1.4 m. How many rotations will the wheel complete as the cart travels 1.1 km?
Solution:
Diameter of the wheel of the bullock cart (d) = 1.4 m
Circumference of the wheel of the bullock cart (c) = πd
$$=\frac{22}{7} \times 1.4=\frac{22}{7} \times \frac{14}{10}=\frac{44}{10}=4.4 \mathrm{m}$$
Distance covered in 1 rotation = Circumference of the wheel
= 4.4 m

∴ The wheel of the bullock cart will complete 250 rotations as the cart travels 1.1 km.

Maharashtra Board Class 7 Maths Chapter 11 Circle Practice Set 42 Intext Questions and Activities

Question 1.
Identify the radii, chords and diameters in the circle alongside and write their names in the table below: (Textbook pg. no. 75)

 i. Radii ii. Chords iii. Diameters

Solution:
i. OA, OB, OC, OF
iii. AB, FC

Maharashtra Board Practice Set 44 Class 7 Maths Solutions Chapter 12 Perimeter and Area

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.