How many tiles will cover a floor? How much paint does a wall need? How big is a plot of land? These ques­tions ask for area, the amount of sur­face a shape cov­ers. In this les­son you will find the area of rec­tan­gles and squares, work back­wards from an area to a side, and solve prob­lems about tiles, car­pets and paths round a gar­den.

Area of a rec­tan­gle

Draw a rec­tan­gle 5 cm long and 3 cm wide on squared paper. It holds 3 rows of squares, with 5 squares of 1 cm² in each row. That is 5×3=155 \times 3 = 15 squares, so its area is 15 cm².

This works for any rec­tan­gle:

area of a rectangle=length×breadth=l×b\text{area of a rectangle} = \text{length} \times \text{breadth} = l \times b

Both lengths must be in the same unit. If they are in cm, the area is in cm². If they are in m, the area is in m².

Area of a square

A square is a rec­tan­gle whose length and breadth are equal. If its side is aa,

area of a square=a×a=a2\text{area of a square} = a \times a = a^2

Exam­ple 1

Find the area of a rec­tan­gle 12 m long and 7.5 m wide.

12×7.5=9012 \times 7.5 = 90

The area is 90 m².

Exam­ple 2

Find the area of a square of side 15 cm.

15×15=22515 \times 15 = 225

The area is 225 cm².

Work­ing back­wards

Exam­ple 3

The area of a rec­tan­gle is 96 cm² and its length is 12 cm. Find its breadth and its perime­ter.

b=96÷12=8b = 96 \div 12 = 8

The breadth is 8 cm, and the perime­ter is 2(12+8)=402(12 + 8) = 40 cm.

Exam­ple 4

The area of a square field is 1600 m². Find its side and the cost of fenc­ing it at ₹20 per metre.

The side is the num­ber that mul­ti­plied by itself gives 1600. Since 40×40=160040 \times 40 = 1600, the side is 40 m.

Perime­ter =4×40=160= 4 \times 40 = 160 m. Cost =160×20=3200= 160 \times 20 = 3200 rupees.

Same perime­ter, dif­fer­ent area

A rec­tan­gle 6 cm by 4 cm and a square of side 5 cm both have perime­ter 20 cm. But their areas are 6×4=246 \times 4 = 24 cm² and 5×5=255 \times 5 = 25 cm². Shapes with the same perime­ter need not have the same area. Of all rec­tan­gles with a given perime­ter, the square has the largest area.

Tiles and costs

To find how many tiles cover a floor, divide the area of the floor by the area of one tile. Work in the same units.

Exam­ple 5

A floor is 6 m long and 4 m wide. How many square tiles of side 25 cm are needed to cover it?

Area of the floor =600×400=240000= 600 \times 400 = 240000 cm² (using 6 m = 600 cm and 4 m = 400 cm).

Area of one tile =25×25=625= 25 \times 25 = 625 cm².

number of tiles=240000625=384\displaystyle \text{number of tiles} = \frac{240000}{625} = 384

You can also count along the edges: 600÷25=24600 \div 25 = 24 tiles fit along the length and 400÷25=16400 \div 25 = 16 along the breadth, and 24×16=38424 \times 16 = 384.

Exam­ple 6

Find the cost of car­pet­ing a room 5 m by 4.5 m at ₹350 per square metre.

Area =5×4.5=22.5= 5 \times 4.5 = 22.5 m². Cost =22.5×350=7875= 22.5 \times 350 = 7875 rupees.

Paths round a gar­den

A path round a rec­tan­gle adds a strip all the way round. Find the area of the big rec­tan­gle and take away the area of the small one.

Exam­ple 7: Path out­side

A rec­tan­gu­lar gar­den is 20 m long and 15 m wide. A path 2 m wide runs all round it, out­side. Find the area of the path.

The path adds 2 m on each side, so the outer rec­tan­gle is 20+2+2=2420 + 2 + 2 = 24 m long and 15+2+2=1915 + 2 + 2 = 19 m wide.

area of path=24×19−20×15=456−300=156\text{area of path} = 24 \times 19 - 20 \times 15 = 456 - 300 = 156

The path cov­ers 156 m².

Exam­ple 8: Path inside

The same gar­den has instead a path 2 m wide run­ning all round it, inside the edge. Find the area of the path.

Now the inner rec­tan­gle is 20−4=1620 - 4 = 16 m long and 15−4=1115 - 4 = 11 m wide.

area of path=20×15−16×11=300−176=124\text{area of path} = 20 \times 15 - 16 \times 11 = 300 - 176 = 124

The path cov­ers 124 m².

Exam­ple 9: Two paths cross­ing

Two paths, each 2 m wide, cross through the mid­dle of the same 20 m by 15 m gar­den, one par­al­lel to the length and one par­al­lel to the breadth. Find the area of the paths.

The path along the length is a rec­tan­gle 20×2=4020 \times 2 = 40 m². The path along the breadth is 15×2=3015 \times 2 = 30 m². The square in the mid­dle where they cross, 2×2=42 \times 2 = 4 m², has been counted twice, so take it away once.

40+30−4=6640 + 30 - 4 = 66

The paths cover 66 m².

Prac­tice

  1. Find the area of a rec­tan­gle 18 cm long and 11 cm wide.
  2. Find the area of a square of side 2.5 m.
  3. The area of a rec­tan­gle is 144 m² and its breadth is 9 m. Find its length and perime­ter.
  4. The area of a square is 625 cm². Find its side.
  5. How many tiles of 20 cm by 20 cm are needed for a floor 5 m by 4 m?
  6. Find the cost of paint­ing a wall 8 m long and 3.5 m high at ₹25 per square metre.
  7. A gar­den is 30 m by 20 m. A path 2.5 m wide runs round it out­side. Find the area of the path.
  8. A room is 9 m by 7 m. A car­pet is laid leav­ing a bor­der 1 m wide all round, inside the walls. Find the area of the car­pet and of the bor­der.
  9. Two paths 3 m wide cross through the mid­dle of a field 40 m by 25 m, par­al­lel to its sides. Find the area of the paths.

Answers

Show answers
  1. 18×11=19818 \times 11 = 198 cm².
  2. 2.5×2.5=6.252.5 \times 2.5 = 6.25 m².
  3. Length =144÷9=16= 144 \div 9 = 16 m. Perime­ter =2(16+9)=50= 2(16 + 9) = 50 m.
  4. 25×25=62525 \times 25 = 625, so the side is 25 cm.
  5. 25×20=50025 \times 20 = 500 tiles (25 along the length and 20 along the breadth).
  6. Area =28= 28 m². Cost =28×25=700= 28 \times 25 = 700 rupees.
  7. Outer rec­tan­gle 35×25=87535 \times 25 = 875 m². Path =875−600=275= 875 - 600 = 275 m².
  8. Car­pet =7×5=35= 7 \times 5 = 35 m². Bor­der =63−35=28= 63 - 35 = 28 m².
  9. 40×3+25×3−3×3=120+75−9=18640 \times 3 + 25 \times 3 - 3 \times 3 = 120 + 75 - 9 = 186 m².