WBBSE Solutions For Class 7 Maths Arithmetic Chapter 7 Areas Of Rectangles And Square

Arithmetic Chapter 7 Areas Of Rectangles And Square

Question 1. Choose the correct answer:

1.  If the perimeter of a square is xom then its area is.

1. 4x² sq.m.
2. \(\frac{x^2}{16}\) sqcm
3. \(\frac{x^2}{4}\) sqm.
4. \(\frac{x^2}{2}\) sq.cm.

The perimeter of the square = 4 × length of each side.

x = 4 × length of each side

length of each side = \(\frac{x}{4}\)

Area of square = (length of side)²

⇒ \(\left(\frac{x}{4}\right)^2\)

Area of square = \(\frac{x^2}{16}\) sq. cm.

∴ The option (2) \(\frac{x^2}{16}\) 22 sq,cm is correct answer.

Class 7 Arithmetic Chapter 7 Questions

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2. 36000 sq. mm =?

1. 3.65qcm.
2. 36 sq.cm.
3. 360 sq.cm.
4. 3600sq.cm.

1cm = 10mm

? = 36000

⇒ \(\frac{36000}{10} \times 1\) = 3600 sqcm

1.sq mm = 0.01sq centimeter

⇒ \(\frac{36000}{1} \times 0.01\)

= 3605qcm.

∴ The option (3) 360 sq cm is the correct answer

3. How many tiles each \(1 \frac{1}{3}\) meter square will cover a floor 20 metres square?

1. 200
2. 225
3. 15
4. None of these.

Area of the Flood = 20m x 20m = 400m²

Each tile has dimensions of \(\frac{4}{3} m \times \frac{4}{3} m \text {. }\)

Area of one tile = \(\frac{4}{3} m \times \frac{4}{3} m=\frac{4 \times 4}{3 \times 3}=\frac{16}{9} m^2\)

Number of tiles \(=\frac{\text { Area of the floor }}{\text { Area of one tile }}\)

⇒ \(\frac{400 m^2}{\frac{16}{9} m^2}\)

⇒ \(400 \times \frac{9}{16}\)

⇒ 400 × 0.5625

Number of tiles = 225

∴ The option (2) 225 is the correct answer

WBBSE Class 7 Maths Solutions

Question 2. write true or False:

1. 1Arc = 10 sq.m. → False

2. The length. of a rectangle is 4 times its breadth

If the breadth is 6.5 cm then its area is 169 sq. cm

length of a rectangle is l = 4xb

If breadth is 6.5

The area is 169 sq. cm

l = 4×6·5

l = 26cm

Area = l × b

=26×6.5

= 165 sq. cm.

→ True

3. If the perimeter of a square is (4a+16) com then the length of each side is (a +4) cm.

The perimeter of the square = 4 × length of each side.

4a+16 = 4(a+4)

4a+16

→ True.

Question 3. Fill in the blanks:

1. 1Sq·Hm = 10000 Sq.m

2. If the area of the square is (x²+4x+4) Cm then its perimeter is

Area of Square = (length of side)²

⇒ \(\sqrt{x^2+4 x+4}\) = length of side.

⇒ \(\sqrt{(x)^2+2 \cdot 2 x+(2)^2}\) = length of side.

⇒ \(\sqrt{(x+2)^2}\) = length of side.

x+2 = length of side.

perimeter of a square = 4x length of each side

= 4x(x+2)

perimeter of a square = 4x+8 cm.

3. The ratio of length and breadth of a rectangle is 3:2 and its area is 600 sq.m. The perimeter is _______m

l:b= 3:2 ⇒ l = 3x, b = 2x Area = 600sq.m

Area = 600 Sqm

lxb = 600 Sq.m.

3x × 2х =600

6x² = 600

x² = \(\frac{600}{6}\)

x² = 100

x = 10m

4. l = 32, b = 2x

l=3×10 , b = 2×10

l= 30m, b=20m

perimeter of a rectangle is 2(l+b)

⇒ 2(30+20)

⇒2(50)

⇒ 100m

Class 7 Arithmetic Textbook Solutions

Question 4. The area of a rectangle is (x²-64) sq m (x>8) and the length is (x+8)m then find its breadth.
Solution:

Given:

The area of the rectangle is x²-64 square meters.

The length of the rectangle is x+ 8 meters

The area of the Rectangle is

Area = Length x Breadth

Let ‘b’ be the breadth of the rectangle.

x²-64 = (x+8)×b

b= \(\frac{x^2-64}{x+8} \Rightarrow \frac{(x)^2-(8)^2}{x+8}\)

a²-b² = (a+b)(a-b)

b= \(\frac{(x+8)(x-8)}{x+8}\)

Since x>8, x+8=0 and we can cancel x+8 in the numerator and denominator

b=2-8

The breadth of the rectangle is x-8 meters.

Question 5. The cost of cultivation of a 45m long piece of land is ₹200. If the breadth of land were less the cost would have been ₹150 calculate the breadth of the land.
Solution:

Let’s denote the breadth of the piece of land as ‘b’ meters

The length of the land is 45 meters.

The area of the land with breadth b is 45 × b sq.m.

The cost of cultivation is ₹200

If the breadth is reduced by 9m

breadth b-9 meters

The new Area is 45×(b-9)

The cost of cultivation is ₹150.

Cost per square meter = \(\frac{200}{45 b}\)

Reduced Breadth

Cost per square meter = \(\frac{150}{45(b-9)}\)

⇒ \(\frac{200}{45 b}=\frac{150}{45(b-9)}\)

⇒ \(\frac{200}{b}=\frac{150}{b-q}\)

⇒ 200(b-9) = b(150)

⇒ 2006-1800 = 150b.

Subtract 150b from both sides.

50b -1800 = 0

506 =1800

b = \(\frac{1800}{50}\)

b = 36

The breadth of the piece of land is 36 meters.

Class 7 Maths Arithmetic Problems

Question 6. A square plot of land of side 26m. A path 2m wide runs all around it from outside. Find the area of the path.
Solution:

Given

The Square Plot land side is 26m.

1. Calculate the Area of the large square.

The path runs all around it and is 2 meters wide.

The side length of the larger square is

26m+ 2m+2m = 30m

The area of the larger square is:

Area_{large sequence} = 30mx30m = 900m²

2. Calculate the Area of the original square. Path

Area_{Origina sequence} = 26m × 26m = 676m²

3. Calculate the Area of the path.

Area_Path = Area_{large sequence} – Area_{Original sequence}

= 900m²-676m² = 224m²

∴ The area of the path around the square plot of land is 224 Sq.m

Question 7. The length of the rectangle floor of a room is thrice of its breadth. It cost ₹4761 to cover the whole floor with a carpet. If one sq.m of carpet costs ₹3 then find the perimeter of the room.
Solution:

Given:

The cost to cover the whole floor ₹4761

Cost per square meter = ₹3m2

Area of the Floor = \(=\frac{\text { Total cost }}{\text { cost per square meter }}\)

= \(\frac{4761}{3}\)

= 1587Sqm²

Let the breadth of the room be ‘b’ meters.

Since the length is thrice the breadth, the length ‘l’ is 3b meters.

Area = l×b

= 3b×b

= 3b²

The area of the Floor is 1587Sq.m²

3b² = 1587

b² = \(\)

= 529

Class 7 Maths Chapter 7 Notes

Question 8. \(b^2=\frac{1587}{3}=529\)

b = \(\sqrt{529}\) = 23m

b = 23m

l is three times the breadth:-

l = 3b

= 3×23

1=69m

The perimeter p of a rectangle is given by

P = 2 × (l+b)

P = 2 × (69+23)

= 2 × 92

= 184m

∴ The perimeter of the room is 184 meters.

Question 9. A room 24 meters long and 12 meters high, costs 48960 For whitewashing its walls at 700 per square meter. Find the breadth of the room.
Solution:

Let the breadth of the room be m

Length of the room = 24m, and 12m high

The area of the room is (24×x) sq. m ⇒ 24x sq.m

Cost per square meter = ₹60

Total cost = ₹48960

Total Area of walls = Total cost/cost per square meter

Total Area of walls = \(\frac{48960}{60}\)
་
=816 sq.m.

Two longer walls each have an area of 24m × 12m

Two Shorter walls each have an area of bm × 12m

Total Area of the four walls is = 2× (24×12)+2×(b×12)

= 2×288+2×12b

576+246

The total area of the walls whitewashed is 816 sqm

576 + 246 = 816

24b = 816-576

24b = 240 = \(\frac{240}{24}\)

b ⇒ b=10

∴ The breadth of the room is 10 meters.

Class 7 Arithmetic Formulas

Question 10. what biggest size of square stones may be used to pave the floor of a room 20m long and 15m wide? Also, find the number of stones needed.
Solution:

Given the length of the room = 20m

width of room = 15m

prime factorization.

20= 2×2×5

15=3×5

The GCD of 20, 15 is 5.

So, the largest square Stone that can be used has a side length of 5 meters.

Area of room = 20m x 15m = 300 Sqm.

Area of one square stone = 5m×5m= 25 sq.m

Number of Stones = \(=\frac{\text { Total Area }}{\text { Area of one Stone }}\)

= \(\frac{300}{25}\)

= 12

The biggest size of square stones that may be used to pave the Floor is 5 meters.

The number of such stones needed to cover the floor is 12.