Perimeter and Area of Plane Figures

Q1: Find the perimeter, area and length of diagonal of a rectangle, having:

i. length = 15 cm, breadth = 8 cm

Step 1: Perimeter of rectangle \(P = 2 \times (length + breadth)\)
\(P = 2 \times (15 + 8) = 2 \times 23 = 46 \, cm\)
Step 2: Area of rectangle \(A = length \times breadth\)
\(A = 15 \times 8 = 120 \, cm^2\)
Step 3: Length of diagonal \(d = \sqrt{length^2 + breadth^2}\)
\(d = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \, cm\)
Answer: Perimeter = 46 cm, Area = 120 cm², Diagonal = 17 cm

ii. length = 20 m, breadth = 15 m

Step 1: Perimeter \(P = 2 \times (20 + 15) = 2 \times 35 = 70 \, m\)
Step 2: Area \(A = 20 \times 15 = 300 \, m^2\)
Step 3: Diagonal \(d = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25 \, m\)
Answer: Perimeter = 70 m, Area = 300 m², Diagonal = 25 m

iii. length = 3.2 m, breadth = 2.4 m

Step 1: Perimeter \(P = 2 \times (3.2 + 2.4) = 2 \times 5.6 = 11.2 \, m\)
Step 2: Area \(A = 3.2 \times 2.4 = 7.68 \, m^2\)
Step 3: Diagonal \(d = \sqrt{3.2^2 + 2.4^2} = \sqrt{10.24 + 5.76} = \sqrt{16} = 4 \, m\)
Answer: Perimeter = 11.2 m, Area = 7.68 m², Diagonal = 4 m

Q2: The perimeter of a rectangle is 51.8 m and its length is 16.5 m. Find the breadth and the area of the rectangle.

Step 1: Formula for perimeter of rectangle:
\(P = 2 \times (length + breadth)\)
Step 2: Substitute given values:
\(51.8 = 2 \times (16.5 + breadth)\)
Step 3: Divide both sides by 2:
\(\frac{51.8}{2} = 16.5 + breadth\)
\(25.9 = 16.5 + breadth\)
Step 4: Subtract 16.5 from both sides:
\(breadth = 25.9 – 16.5 = 9.4 \, m\)

ii. Find the area

Step 5: Area of rectangle \(A = length \times breadth\)
\(A = 16.5 \times 9.4 = 155.1 \, m^2\)
Answer: Breadth = 9.4 m, Area = 155.1 m²

Q3: The perimeter of a rectangle is 42 m and its breadth is 7.4 m. Find the length and the area of the rectangle.

Step 1: Formula for perimeter of rectangle:
\(P = 2 \times (length + breadth)\)
Step 2: Substitute given values:
\(42 = 2 \times (length + 7.4)\)
Step 3: Divide both sides by 2:
\(\frac{42}{2} = length + 7.4\)
\(21 = length + 7.4\)
Step 4: Subtract 7.4 from both sides:
\(length = 21 – 7.4 = 13.6 \, m\)
Step 5: Area of rectangle \(A = length \times breadth\)
\(A = 13.6 \times 7.4 = 100.64 \, m^2\)
Answer: Length = 13.6 m, Area = 100.64 m²

Q4: The perimeter of a rectangle is 68 m and its length is 24 m. Find its breadth, area and diagonal.

Step 1: Formula for perimeter of rectangle:
\(P = 2 \times (length + breadth)\)
Step 2: Substitute given values:
\(68 = 2 \times (24 + breadth)\)
Step 3: Divide both sides by 2:
\(\frac{68}{2} = 24 + breadth\)
\(34 = 24 + breadth\)
Step 4: Subtract 24 from both sides:
\(breadth = 34 – 24 = 10 \, m\)
Step 5: Area of rectangle \(A = length \times breadth\)
\(A = 24 \times 10 = 240 \, m^2\)
Step 6: Length of diagonal \(d = \sqrt{length^2 + breadth^2}\)
\(d = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26 \, m\)
Answer: Breadth = 10 m, Area = 240 m², Diagonal = 26 m

Q5: The length of a rectangle is 30 cm and one of its diagonals measures 34 cm. Find the breadth, perimeter and area of the rectangle.

Step 1: Use the diagonal formula:
\(d = \sqrt{length^2 + breadth^2}\)
Step 2: Substitute given values:
\(34 = \sqrt{30^2 + breadth^2}\)
Step 3: Square both sides:
\(34^2 = 30^2 + breadth^2\)
\(1156 = 900 + breadth^2\)
Step 4: Subtract 900 from both sides:
\(breadth^2 = 1156 – 900 = 256\)
Step 5: Find the square root:
\(breadth = \sqrt{256} = 16 \, cm\)
Step 6: Perimeter \(P = 2 \times (length + breadth)\)
\(P = 2 \times (30 + 16) = 2 \times 46 = 92 \, cm\)
Step 7: Area \(A = length \times breadth\)
\(A = 30 \times 16 = 480 \, cm^2\)
Answer: Breadth = 16 cm, Perimeter = 92 cm, Area = 480 cm²

Q6: The area of a rectangle is 19.6 m² and its length is 5.6 m. Find the breadth and perimeter of the rectangle.

Step 1: Formula for area of rectangle:
\(A = length \times breadth\)
Step 2: Substitute the known values:
\(19.6 = 5.6 \times breadth\)
Step 3: Divide both sides by 5.6:
\(breadth = \frac{19.6}{5.6} = 3.5 \, m\)
Step 4: Formula for perimeter:
\(P = 2 \times (length + breadth)\)
Step 5: Substitute values:
\(P = 2 \times (5.6 + 3.5) = 2 \times 9.1 = 18.2 \, m\)
Answer: Breadth = 3.5 m, Perimeter = 18.2 m

Q7: The area of a rectangle is 52 m² and its breadth is 6.5 m. Find the length and perimeter of the rectangle.

Step 1: Formula for area of rectangle:
\(A = length \times breadth\)
Step 2: Substitute the known values:
\(52 = length \times 6.5\)
Step 3: Divide both sides by 6.5:
\(length = \frac{52}{6.5} = 8 \, m\)
Step 4: Formula for perimeter:
\(P = 2 \times (length + breadth)\)
Step 5: Substitute the values:
\(P = 2 \times (8 + 6.5) = 2 \times 14.5 = 29 \, m\)
Answer: Length = 8 m, Perimeter = 29 m

Q8: The sides of a rectangular park are in the ratio 3 : 2. If its area is 1536 m², find the cost of fencing it at ₹23.50 per metre.

Step 1: Let the sides be \(3x\) and \(2x\).
Step 2: Area of rectangle \(A = length \times breadth\)
\(1536 = 3x \times 2x = 6x^2\)
Step 3: Solve for \(x^2\):
\(x^2 = \frac{1536}{6} = 256\)
Step 4: Find \(x\):
\(x = \sqrt{256} = 16\)
Step 5: Length \(= 3x = 3 \times 16 = 48 \, m\)
Breadth \(= 2x = 2 \times 16 = 32 \, m\)
Step 6: Perimeter \(P = 2 \times (length + breadth)\)
\(P = 2 \times (48 + 32) = 2 \times 80 = 160 \, m\)

iii. Find the cost of fencing

Step 7: Cost \(= \text{Perimeter} \times \text{cost per metre}\)
\(= 160 \times 23.50 = ₹ 3760\)
Answer: Cost of fencing = ₹3760

Q9: Find the cost of carpeting a room 12 m long and 8 m broad with a carpet 75 cm broad at the rate ₹116.50 per metre.

Step 1: Calculate the Area of the Room Floor.
Length of the room \( (L) = 12 \text{ m} \)
Breadth of the room \( (B) = 8 \text{ m} \)
Area of the floor \( = L \times B \)
Area \( = 12 \text{ m} \times 8 \text{ m} = 96 \text{ m}^2 \)

Step 2: Convert Carpet Breadth to Metres.
Breadth of the carpet \( = 75 \text{ cm} \)
Since \( 100 \text{ cm} = 1 \text{ m} \), we divide by 100:
Breadth in metres \( = \frac{75}{100} = 0.75 \text{ m} \)

Step 3: Calculate the Required Length of the Carpet.
Area of the carpet \( = \) Area of the floor
Length of carpet \( \times \) Breadth of carpet \( = 96 \text{ m}^2 \)
Length of carpet \( \times 0.75 \text{ m} = 96 \text{ m}^2 \)
Length of carpet \( = \frac{96}{0.75} \)
Length of carpet \( = 128 \text{ m} \)

Step 4: Calculate the Total Cost of Carpeting.
Rate of carpeting \( = ₹116.50 \text{ per metre} \)
Total Cost \( = \text{Total Length} \times \text{Rate} \)
Total Cost \( = 128 \times 116.50 \)
Total Cost \( = ₹14,912 \)

Answer: The total cost of carpeting the room is ₹14,912.

Q10: A verandah 50 m long and 12 m broad is to be paved with tiles, each measuring 6 dm by 5 dm. Find the number of tiles needed.

Step 1: Area of verandah \(= length \times breadth\)
\(= 50 \times 12 = 600 \, m^2\)
Step 2: Convert 6 dm to meters:
\(6 \, dm = \frac{6}{10} = 0.6 \, m\)
Step 3: Convert 5 dm to meters:
\(5 \, dm = \frac{5}{10} = 0.5 \, m\)
Step 4: Area of one tile \(= 0.6 \times 0.5 = 0.3 \, m^2\)
Step 5: Number of tiles \(= \frac{\text{Area of verandah}}{\text{Area of one tile}}\)
\(= \frac{600}{0.3} = 2000\)
Answer: Number of tiles needed = 2000

Q11: The length and breadth of a rectangular field are in the ratio 7 : 5 and its perimeter is 384. Find the cost of reaping the field at ₹1.25 per sq. metre.

Step 1: Let length = \(7x\) and breadth = \(5x\)
Step 2: Formula for perimeter:
\(P = 2 \times (length + breadth)\)
Step 3: Substitute given values:
\(384 = 2 \times (7x + 5x) = 2 \times 12x = 24x\)
Step 4: Solve for \(x\):
\(x = \frac{384}{24} = 16\)
Step 5: Length \(= 7x = 7 \times 16 = 112 \, m\)
Breadth \(= 5x = 5 \times 16 = 80 \, m\)
Step 6: Area \(A = length \times breadth = 112 \times 80 = 8960 \, m^2\)
Step 7: Cost \(= \text{Area} \times \text{rate}\)
\(= 8960 \times 1.25 = ₹ 11200\)
Answer: Cost of reaping the field = ₹11200

Q12: A room 9.5 m long and 6 m wide is surrounded by a verandah 1.25 m wide. Calculate the cost of cementing the floor of this verandah at ₹28 per sq. metre.

Step 1: Length of room = 9.5 m
Breadth of room = 6 m
Width of verandah = 1.25 m
Step 2: Outer length \(= 9.5 + 2 \times 1.25 = 9.5 + 2.5 = 12 \, m\)
Outer breadth \(= 6 + 2 \times 1.25 = 6 + 2.5 = 8.5 \, m\)
Step 3: Area of outer rectangle \(= 12 \times 8.5 = 102 \, m^2\)
Area of room \(= 9.5 \times 6 = 57 \, m^2\)
Step 4: Area of verandah \(= 102 – 57 = 45 \, m^2\)
Step 5: Cost \(= \text{Area of verandah} \times \text{rate}\)
\(= 45 \times 28 = ₹ 1260\)
Answer: Cost of cementing the verandah floor = ₹1260

Q13: A rectangular grassy plot is 125 m long and 74 m broad. It has a path 2.5 m wide all round it on the inside. Find the cost of levelling the path at ₹6.80 per m².

Step 1: Width of path = 2.5 m
Length of inner rectangle \(= 125 – 2 \times 2.5 = 125 – 5 = 120 \, m\)
Breadth of inner rectangle \(= 74 – 2 \times 2.5 = 74 – 5 = 69 \, m\)
Step 2: Area of outer rectangle \(= 125 \times 74 = 9250 \, m^2\)
Area of inner rectangle \(= 120 \times 69 = 8280 \, m^2\)
Step 3: Area of path \(= \text{Area of outer rectangle} – \text{Area of inner rectangle}\)
\(= 9250 – 8280 = 970 \, m^2\)
Step 4: Cost \(= \text{Area of path} \times \text{rate}\)
\(= 970 \times 6.80 = ₹ 6596\)
Answer: Cost of levelling the path = ₹6596

Q14: A rectangular plot of land measures 95 m by 72 m. Inside the plot, a path of uniform width 3.5 m is to be constructed all around. The rest of the plot is to be laid with grass. Find the total expense involved in constructing the path at ₹46.50 per m² and laying the grass at ₹3.75 per m².

Step 1: Width of path = 3.5 m
Length of inner rectangle \(= 95 – 2 \times 3.5 = 95 – 7 = 88 \, m\)
Breadth of inner rectangle \(= 72 – 2 \times 3.5 = 72 – 7 = 65 \, m\)
Step 2: Area of entire plot \(= 95 \times 72 = 6840 \, m^2\)
Area of grass area \(= 88 \times 65 = 5720 \, m^2\)
Step 3: Area of path \(= \text{Area of entire plot} – \text{Area of grass area}\)
\(= 6840 – 5720 = 1120 \, m^2\)
Step 4: Cost of path \(= 1120 \times 46.50 = ₹ 52080\)
Step 5: Cost of grass \(= 5720 \times 3.75 = ₹ 21450\)
Step 6: Total cost \(= 52080 + 21450 = ₹ 73530\)
Answer: Total expense involved = ₹73530

Q15: A rectangular hall is 22 m and 15.5 m broad. A carpet is laid inside the hall leaving all around a margin of 75 cm from the walls. Find the area of the carpet and the area of the strip left uncovered. If the width of the carpet is 82 cm, find its cost at the rate of ₹124 per metre.

Step 1: Calculate the Area of the Hall.
Length of the hall \( (L) = 22 \text{ m} \)
Breadth of the hall \( (B) = 15.5 \text{ m} \)
Area of the hall \( = L \times B \)
Area of the hall \( = 22 \times 15.5 = 341 \text{ m}^2 \)

Step 2: Find the Dimensions of the Carpet.
The margin is left on all sides, so we subtract twice the margin from both length and breadth.
Margin \( = 75 \text{ cm} = 0.75 \text{ m} \)
Length of the carpet \( (l) = 22 – (0.75 + 0.75) = 22 – 1.5 = 20.5 \text{ m} \)
Breadth of the carpet \( (b) = 15.5 – (0.75 + 0.75) = 15.5 – 1.5 = 14 \text{ m} \)

Step 3: Calculate the Area of the Carpet and the Uncovered Strip.
Area of the carpet \( = l \times b \)
Area of the carpet \( = 20.5 \times 14 = 287 \text{ m}^2 \)
Area of the uncovered strip \( = \text{Total Area} – \text{Carpet Area} \)
Area of the uncovered strip \( = 341 – 287 = 54 \text{ m}^2 \)

Step 4: Calculate the Length of the Carpet Roll Required.
Width of the carpet roll \( = 82 \text{ cm} = 0.82 \text{ m} \)
Length of carpet needed \( = \frac{\text{Area of Carpet}}{\text{Width of Carpet Roll}} \)
Length \( = \frac{287}{0.82} = 350 \text{ m} \)

Step 5: Calculate the Total Cost.
Rate \( = ₹124 \text{ per metre} \)
Total Cost \( = \text{Length} \times \text{Rate} \)
Total Cost \( = 350 \times 124 = ₹43,400 \)

Answer: The area of the carpet is \( 287 \text{ m}^2 \), the area of the uncovered strip is \( 54 \text{ m}^2 \), and the total cost is ₹43,400.

Q16: A rectangular lawn 75 m by 60 m has two roads each 4 m wide running in the middle of it, one parallel to length and the other parallel to breadth. Find the cost of gravelling the roads at ₹14.50 per sq. metre.

Step 1: Length of lawn = 75 m
Breadth of lawn = 60 m
Width of each road = 4 m
Step 2: Area of road parallel to length:
\(= length \times width = 75 \times 4 = 300 \, m^2\)
Step 3: Area of road parallel to breadth:
\(= breadth \times width = 60 \times 4 = 240 \, m^2\)
Step 4: Area of overlap (road intersection):
\(= width \times width = 4 \times 4 = 16 \, m^2\)
Step 5: Total area of roads:
\(= 300 + 240 – 16 = 524 \, m^2\)
Step 6: Cost \(= \text{area} \times \text{rate}\)
\(= 524 \times 14.50 = ₹ 7598\)
Answer: Cost of gravelling the roads = ₹7598

Q17: The length and breadth of a rectangular park are in the ratio 5 : 2. A 2.5 m wide path running all around the outside of the park has an area of 305 m². Find the dimensions of the park.

Step 1: Width of path = 2.5 m
Step 2: Length including path \(= 5x + 2 \times 2.5 = 5x + 5\)
Breadth including path \(= 2x + 2 \times 2.5 = 2x + 5\)
Step 3: Area including path \(= (5x + 5)(2x + 5)\)
Step 4: Area of park \(= 5x \times 2x = 10x^2\)
Step 5: Area of path \(= \text{Area including path} – \text{Area of park} = 305 \, m^2\)
\((5x + 5)(2x + 5) – 10x^2 = 305\)
Step 6: Expand:
\( (5x)(2x) + (5x)(5) + (5)(2x) + (5)(5) – 10x^2 = 305 \)
\( 10x^2 + 25x + 10x + 25 – 10x^2 = 305 \)
\( 35x + 25 = 305 \)
Step 7: Subtract 25 from both sides:
\(35x = 305 – 25 = 280\)
\(x = \frac{280}{35} = 8\)
Step 8: Length \(= 5x = 5 \times 8 = 40 \, m\)
Breadth \(= 2x = 2 \times 8 = 16 \, m\)
Answer: Length = 40 m, Breadth = 16 m

Q18: Find the perimeter, area and diagonal of a square each of whose sides measures:

i. Side = 16 cm

Step 1: Perimeter of square \(P = 4 \times side\)
\(P = 4 \times 16 = 64 \, cm\)
Step 2: Area of square \(A = side^2\)
\(A = 16^2 = 256 \, cm^2\)
Step 3: Diagonal of square \(d = side \times \sqrt{2}\)
\(d = 16 \times \sqrt{2} = 16 \times 1.414 = 22.62 \, cm\)
Answer: Perimeter = 64 cm, Area = 256 cm², Diagonal = 22.62 cm

ii. Side = 8.5 m

Step 1: Perimeter of square:
\(P = 4 \times 8.5 = 34 \, m\)
Step 2: Area of square:
\(A = 8.5^2 = 72.25 \, m^2\)
Step 3: Diagonal of square:
\(d = 8.5 \times \sqrt{2} = 8.5 \times 1.414 = 12.02 \, m\)
Answer: Perimeter = 34 m, Area = 72.25 m², Diagonal = 12.02 m

iii. Side = 2.5 dm

Step 1: Perimeter of square:
\(P = 4 \times 2.5 = 10 \, dm\)
Step 2: Area of square:
\(A = 2.5^2 = 6.25 \, dm^2\)
Step 3: Diagonal of square:
\(d = 2.5 \times \sqrt{2} = 2.5 \times 1.414 = 3.54 \, dm\)
Answer: Perimeter = 10 dm, Area = 6.25 dm², Diagonal = 3.54 dm

Q19: The perimeter of a square is 28 cm. Find its area and the length of its diagonal.

Step 1: Formula for perimeter of square:
\(P = 4 \times side\)
Step 2: Substitute the given perimeter:
\(28 = 4 \times side\)
Step 3: Solve for side:
\(side = \frac{28}{4} = 7 \, cm\)
Step 4: Area of square:
\(A = side^2 = 7^2 = 49 \, cm^2\)
Step 5: Diagonal \(d = side \times \sqrt{2} = 7 \times 1.414 = 9.9 \, cm\)
Answer: Area = 49 cm², Diagonal = 9.9 cm

Q20: The diagonal of a square is \(5\sqrt{2}\) m. Find its area and perimeter.

Step 1: Formula for diagonal of square:
\(d = side \times \sqrt{2}\)
Step 2: Substitute the given diagonal:
\(5\sqrt{2} = side \times \sqrt{2}\)
Step 3: Divide both sides by \(\sqrt{2}\):
\(side = \frac{5\sqrt{2}}{\sqrt{2}} = 5 \, m\)
Step 4: Area \(A = side^2 = 5^2 = 25 \, m^2\)
Step 5: Perimeter \(P = 4 \times side = 4 \times 5 = 20 \, m\)
Answer: Area = 25 m², Perimeter = 20 m

Q21: The diagonal of a square is 12 cm long. Find its area and perimeter.

Step 1: Calculate the Area of the Square using the Diagonal.
Diagonal \( (d) = 12 \text{ cm} \)
Formula for Area of a square when diagonal is given:
Area \( = \frac{1}{2} \times d^2 \)
Area \( = \frac{1}{2} \times (12)^2 \)
Area \( = \frac{1}{2} \times 144 \)
Area \( = 72 \text{ cm}^2 \)

Step 2: Find the Side of the Square.
Let the side of the square be \( s \).
Area \( = s^2 \)
\( s^2 = 72 \)
\( s = \sqrt{72} \)
\( s = \sqrt{36 \times 2} \)
\( s = 6\sqrt{2} \text{ cm} \)
Using the value \( \sqrt{2} \approx 1.414 \):
\( s = 6 \times 1.414 = 8.484 \text{ cm} \)

Step 3: Calculate the Perimeter of the Square.
Perimeter \( = 4 \times s \)
Perimeter \( = 4 \times 6\sqrt{2} \)
Perimeter \( = 24\sqrt{2} \text{ cm} \)
In decimal form:
Perimeter \( = 4 \times 8.484 = 33.936 \text{ cm} \)

Answer: The area of the square is \( 72 \text{ cm}^2 \) and the perimeter is \( 24\sqrt{2} \text{ cm} \) (or approximately \( 33.94 \text{ cm} \)).

Q22: The area of a square field is 32 m². Find its diagonal.

Step 1: Area of square \(A = side^2\)
\(32 = side^2\)
Step 2: Find side by taking square root:
\(side = \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \, m\)
Step 3: Diagonal \(d = side \times \sqrt{2} = 4\sqrt{2} \times \sqrt{2} = 4 \times 2 = 8 \, m\)
Answer: Diagonal = 8 m

Q23: The area of a square is 81 cm². Find its perimeter and the length of its diagonal.

Step 1: Area of square \(A = side^2\)
\(81 = side^2\)
Step 2: Find side by taking square root:
\(side = \sqrt{81} = 9 \, cm\)
Step 3: Perimeter \(P = 4 \times side = 4 \times 9 = 36 \, cm\)
Step 4: Diagonal \(d = side \times \sqrt{2} = 9 \times 1.414 = 12.73 \, cm\)
Answer: Perimeter = 36 cm, Diagonal = 12.73 cm

Q24: A square field has an area of 6.25 ares. Find the cost of putting a fence round it at ₹32.50 per metre.

Step 1: 1 are = 100 m²
Area \(= 6.25 \times 100 = 625 \, m^2\)
Step 2: Area of square \(A = side^2\)
\(625 = side^2\)
\(side = \sqrt{625} = 25 \, m\)
Step 3: Perimeter \(P = 4 \times side = 4 \times 25 = 100 \, m\)
Step 4: Cost \(= \text{Perimeter} \times \text{rate}\)
\(= 100 \times 32.50 = ₹ 3250\)
Answer: Cost of putting fence = ₹3250

Q25: The cost of ploughing a square field at ₹13.50 per square metre is ₹5400. Find the cost of fencing the field at ₹28.50 per metre.

Step 1: Cost of ploughing \(= \text{Area} \times \text{rate}\)
\(5400 = \text{Area} \times 13.50\)
Step 2: Area \(= \frac{5400}{13.50} = 400 \, m^2\)
Step 3: Area \(= side^2\)
\(side^2 = 400\)
\(side = \sqrt{400} = 20 \, m\)
Step 4: Perimeter \(P = 4 \times side = 4 \times 20 = 80 \, m\)
Step 5: Cost \(= \text{Perimeter} \times \text{rate}\)
\(= 80 \times 28.50 = ₹ 2280\)
Answer: Cost of fencing the field = ₹2280