CBSE Class 7 Mathematics Perimeter And Area Worksheet Set 01

Mathematics - Perimeter and Area

Question 1. The length and breadth of a rectangular piece of land are in the ratio 5:2. If the total cost of fencing it at Rs. 12.5 per meter is Rs. 25000, find its length and breadth.
Answer: Let the length of the rectangular piece of land be \(5x\) and the breadth be \(2x\).
Perimeter of the rectangle = \(2(\text{length} + \text{breadth}) = 2(5x + 2x) = 2(7x) = 14x\)
Total cost of fencing = Perimeter \(\times\) Rate per meter
\(25000 = 14x \times 12.5\)
\(14x = \frac{25000}{12.5} = 2000\)
\(x = \frac{2000}{14} = \frac{1000}{7} \text{ m}\)
Length = \(5x = 5 \times \frac{1000}{7} = \frac{5000}{7} = 714.29 \text{ m}\)
Breadth = \(2x = 2 \times \frac{1000}{7} = \frac{2000}{7} = 285.71 \text{ m}\)
In simple words: First find the total perimeter from the total cost and rate, then use the given ratio to calculate length and breadth.

Exam Tip: Perimeter equals total cost divided by the rate per unit length.

Question 2. Find the area of a path 2m wide laid inside along the boundary of a rectangular park of dimension 100m X 60 m. Find the cost of gravelling it at the rate of Rs. 30 per square metre.
Answer: Length of the park (\(L\)) = 100 m
Breadth of the park (\(B\)) = 60 m
Area of the park = \(L \times B = 100 \times 60 = 6000 \text{ m}^2\)
Width of the path = 2 m
Length of the inner rectangle without path (\(l\)) = \(100 - 2(2) = 100 - 4 = 96 \text{ m}\)
Breadth of the inner rectangle without path (\(b\)) = \(60 - 2(2) = 60 - 4 = 56 \text{ m}\)
Area of the inner rectangle = \(l \times b = 96 \times 56 = 5376 \text{ m}^2\)
Area of the path = Area of the park - Area of the inner rectangle
Area of the path = \(6000 - 5376 = 624 \text{ m}^2\)
Cost of gravelling = Area of the path \(\times\) Rate = \(624 \times 30 = \text{Rs. } 18720\)
In simple words: Subtract the inner dimensions from the outer dimensions to find the area of the path inside, then multiply by the cost rate.

Exam Tip: For a path inside a rectangle, subtract twice the width from both length and breadth to get the inner dimensions.

Question 3. Find the area of a walking 1 m in width laid outside on all sides of a lawn of length 70 m and breadth 50 m.
Answer: Length of the lawn (\(L\)) = 70 m
Breadth of the lawn (\(B\)) = 50 m
Width of the path = 1 m
Outer length with path (\(l\)) = \(70 + 2(1) = 72 \text{ m}\)
Outer breadth with path (\(b\)) = \(50 + 2(1) = 52 \text{ m}\)
Area of the outer rectangle = \(l \times b = 72 \times 52 = 3744 \text{ m}^2\)
Area of the lawn = \(L \times B = 70 \times 50 = 3500 \text{ m}^2\)
Area of the path = Area of outer rectangle - Area of lawn
Area of the path = \(3744 - 3500 = 244 \text{ m}^2\)
In simple words: For an outside path, add twice the width of the path to the length and breadth before calculating the outer area.

Exam Tip: Always add \(2 \times \text{width}\) to both dimensions when calculating an outer pathway area.

Question 4. A square garden has an area of 2500 sq.m. A part of uniform width of 2.5m runs around it on the inside. Find the cost of laying it at Rs. 25 per sq.m.
Answer: Area of the square garden = \(2500 \text{ m}^2\)
Side of the square garden (\(s\)) = \(\sqrt{2500} = 50 \text{ m}\)
Width of the inner path = 2.5 m
Side of the inner square without path = \(50 - 2(2.5) = 50 - 5 = 45 \text{ m}\)
Area of the inner square = \(45 \times 45 = 2025 \text{ m}^2\)
Area of the path = Area of outer square - Area of inner square
Area of the path = \(2500 - 2025 = 475 \text{ m}^2\)
Cost of laying = Area of the path \(\times\) Rate = \(475 \times 25 = \text{Rs. } 11875\)
In simple words: Find the side length of the square from its area, subtract the path width from both sides to find the inner square area, and subtract to get the path area.

Exam Tip: The side of a square is the square root of its area.

Question 5. The perimeter of a right triangle is 30cm. Its hypotenuse is 13 cm and base is 5cm. Find the area of the triangle.
Answer: Perimeter of the right triangle = 30 cm
Hypotenuse (\(h\)) = 13 cm
Base (\(b\)) = 5 cm
Let the perpendicular height be \(p\).
Perimeter = Base + Hypotenuse + Perpendicular
\(30 = 5 + 13 + p\)
\(30 = 18 + p\)
\(p = 30 - 18 = 12 \text{ cm}\)
Area of the right triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)
Area = \(\frac{1}{2} \times 5 \times 12 = 30 \text{ cm}^2\)
In simple words: Subtract the given base and hypotenuse from the perimeter to find the missing height, then use the triangle area formula.

Question 6. The area of a right triangle is \(100 \text{ cm}^2\). If one of the perpendicular legs is 25 cm, find the length of the other.
Answer: Area of the right triangle = \(100 \text{ cm}^2\)
Let one perpendicular leg (\(b\)) = 25 cm
Let the other perpendicular leg be \(p\).
Area = \(\frac{1}{2} \times \text{base} \times \text{perpendicular}\)
\(100 = \frac{1}{2} \times 25 \times p\)
\(200 = 25 \times p\)
\(p = \frac{200}{25} = 8 \text{ cm}\)
In simple words: Multiply the area by 2 and divide by the known leg to find the length of the other leg.

Exam Tip: Rearrange the area formula as \(\text{Height} = \frac{2 \times \text{Area}}{\text{Base}}\).

Question 7. In \(\Delta\) ABC, AC=25cm, BC=7cm, AE=10cm. find (i) the area of \(\Delta\) ABC (ii) DB
Answer: (i) Using BC as the base and AE as the corresponding height (perpendicular from A to BC):
Base (\(BC\)) = 7 cm
Height (\(AE\)) = 10 cm
Area of \(\Delta\) ABC = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times 10 = 35 \text{ cm}^2\)
(ii) To find DB (height corresponding to base AC):
Base (\(AC\)) = 25 cm
Area = \(35 \text{ cm}^2\)
Area = \(\frac{1}{2} \times \text{AC} \times \text{DB}\)
\(35 = \frac{1}{2} \times 25 \times \text{DB}\)
\(70 = 25 \times \text{DB}\)
\(\text{DB} = \frac{70}{25} = 2.8 \text{ cm}\)
In simple words: Calculate area using one base-height pair, then use that same area with a different base to find its corresponding height.

Exam Tip: The area of a triangle remains constant regardless of which side is chosen as the base.

Question 8. Find the area of an isosceles triangle whose perimeter is 32cm and its base is 12cm.
Answer: Perimeter of the isosceles triangle = 32 cm
Base = 12 cm
Since it is an isosceles triangle, the two equal sides are equal in length.
Let each equal side be \(a\).
Perimeter = \(\text{Base} + 2a\)
\(32 = 12 + 2a\)
\(2a = 32 - 12 = 20\)
\(a = 10 \text{ cm}\)
Height of the isosceles triangle corresponding to the base bisects the base:
Height (\(h\)) = \(\sqrt{a^2 - \left(\frac{\text{Base}}{2}\right)^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 \text{ cm}\)
Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 8 = 48 \text{ cm}^2\)
In simple words: Find the two equal sides from the perimeter, then use Pythagoras theorem to find the height from the vertex to the middle of the base.

Exam Tip: The altitude to the base of an isosceles triangle bisects the base perpendicularly.

Question 9. The two sides of the parallelogram ABCD are 12cm and 8cm. The height corresponding to base CD is 6cm. Find (i) the area of the parallelogram (ii) the height corresponding to base AD.
Answer: Sides of the parallelogram are 12 cm and 8 cm. Thus, let base \(CD = 12 \text{ cm}\) and base \(AD = 8 \text{ cm}\).
(i) Height corresponding to base CD = 6 cm
Area of the parallelogram = \(\text{Base} \times \text{Height} = CD \times 6 = 12 \times 6 = 72 \text{ cm}^2\)
(ii) Height corresponding to base AD:
Area = \(\text{Base} \times \text{Height}\)
\(72 = AD \times \text{Height}\)
\(72 = 8 \times \text{Height}\)
Height = \(\frac{72}{8} = 9 \text{ cm}\)
In simple words: Area is base times height. Use one base-height pair to find the area, then divide the area by the second base to find the second height.

Exam Tip: In a parallelogram, the longer side has a shorter corresponding altitude, and vice versa.

Question 10. Area of a parallelogram is \(450 \text{ cm}^2\) and its altitude is 20cm. Find the length of the corresponding side of the parallelogram.
Answer: Area of the parallelogram = \(450 \text{ cm}^2\)
Altitude (Height) = 20 cm
Area = \(\text{Base} \times \text{Height}\)
\(450 = \text{Base} \times 20\)
Base = \(\frac{450}{20} = 22.5 \text{ cm}\)
In simple words: Divide the total area by the given altitude to find the corresponding base.

Exam Tip: \(\text{Base} = \frac{\text{Area}}{\text{Altitude}}\).

Question 11. The area of a parallelogram and a square are the same. If the perimeter of the square is 160m and the height of the parallelogram is 20m, find the length of the corresponding base of the parallelogram.
Answer: Perimeter of the square = 160 m
Side of the square = \(\frac{160}{4} = 40 \text{ m}\)
Area of the square = \(\text{Side}^2 = 40 \times 40 = 1600 \text{ m}^2\)
Since the area of the parallelogram equals the area of the square, Area of the parallelogram = \(1600 \text{ m}^2\)
Height of the parallelogram = 20 m
Area = \(\text{Base} \times \text{Height}\)
\(1600 = \text{Base} \times 20\)
Base = \(\frac{1600}{20} = 80 \text{ m}\)
In simple words: Find the area of the square first, set it as the area of the parallelogram, and divide by the height to get the base.

Exam Tip: Perimeter of square divided by 4 gives its side length.

Question 12. A field in the form of a parallelogram has one of its diagonals 42m long and the perpendicular distance of this diagonal from either of the outlying vertices is 80 cm. Find the area of the field.
Answer: Length of the diagonal = 42 m
Perpendicular distance from either vertex = 80 cm = 0.8 m
A diagonal divides a parallelogram into two congruent triangles.
Area of the field = \(2 \times \left( \frac{1}{2} \times \text{diagonal} \times \text{perpendicular distance} \right)\)
Area = \(42 \times 0.8 = 33.6 \text{ m}^2\)
In simple words: A diagonal splits the parallelogram into two triangles. Calculate the area of one triangle and multiply by 2 (or use base as diagonal and height as sum of both perpendicular distances).

Exam Tip: Always convert all measurements to the same unit before calculating (convert 80 cm to 0.8 m).

Worksheet 3

Question 1. Find the breadth of the rectangle whose length is 70 cm and perimeter 200 cm.
Answer: Length of the rectangle = 70 cm
Perimeter = 200 cm
Perimeter = \(2(\text{Length} + \text{Breadth})\)
\(200 = 2(70 + \text{Breadth})\)
\(100 = 70 + \text{Breadth}\)
Breadth = \(100 - 70 = 30 \text{ cm}\)
In simple words: Half the perimeter and subtract the length to find the breadth.

Exam Tip: \(\text{Breadth} = \frac{\text{Perimeter}}{2} - \text{Length}\).

Question 2. The window of a room is 20 m long and 14 m wide. If the carpenter works to put a metallic frame around it which costs Rs. 120 per metre, then how much will the metallic cost?
Answer: Length of the window = 20 m
Width of the window = 14 m
Perimeter of the window (framing length) = \(2(\text{Length} + \text{Width}) = 2(20 + 14) = 2(34) = 68 \text{ m}\)
Cost per meter = Rs. 120
Total cost = \(68 \times 120 = \text{Rs. } 8160\)
In simple words: Find the perimeter of the window to know the total length of the metallic frame, then multiply by the cost per meter.

Exam Tip: Framing or fencing along boundaries always requires calculating the perimeter.

Question 3. The perimeter of a square garden is 444m. Find the area of the garden in hectares.
Answer: Perimeter of the square garden = 444 m
Side of the square = \(\frac{444}{4} = 111 \text{ m}\)
Area of the square = \(\text{Side}^2 = 111 \times 111 = 12321 \text{ m}^2\)
Since \(1 \text{ hectare} = 10000 \text{ m}^2\):
Area in hectares = \(\frac{12321}{10000} = 1.2321 \text{ hectares}\)
In simple words: Find the side length, calculate the area in square meters, and divide by 10000 to convert to hectares.

Question 4. A room measures 12 m X 9 m. The floor of the room is to be covered by marble tiles measuring 45 cm by 30 cm. How many tiles are needed?
Answer: Length of the room = 12 m = 1200 cm
Breadth of the room = 9 m = 900 cm
Area of the room = \(1200 \times 900 = 1080000 \text{ cm}^2\)
Length of one tile = 45 cm
Breadth of one tile = 30 cm
Area of one tile = \(45 \times 30 = 1350 \text{ cm}^2\)
Number of tiles needed = \(\frac{\text{Area of the room}}{\text{Area of one tile}} = \frac{1080000}{1350} = 800 \text{ tiles}\)
In simple words: Convert all dimensions to centimeters, find the area of the room and one tile, and divide total area by tile area.

Exam Tip: Always make sure both room and tile measurements are in the same units before dividing.

Question 5. Two plots of land have same perimeter. One is a square of side 60 m, while the other is a rectangle whose breadth is 1.5 dam. Which plot has the greater area and by how much?
Answer: Side of the square = 60 m
Perimeter of the square = \(4 \times 60 = 240 \text{ m}\)
Area of the square = \(60 \times 60 = 3600 \text{ m}^2\)
Perimeter of the rectangle = Perimeter of the square = 240 m
Breadth of the rectangle = 1.5 dam = 15 m (since 1 dam = 10 m)
Perimeter of rectangle = \(2(\text{Length} + \text{Breadth})\)
\(240 = 2(\text{Length} + 15)\)
\(120 = \text{Length} + 15\)
Length of the rectangle = \(120 - 15 = 105 \text{ m}\)
Area of the rectangle = \(\text{Length} \times \text{Breadth} = 105 \times 15 = 1575 \text{ m}^2\)
Difference in area = \(3600 - 1575 = 2025 \text{ m}^2\)
Thus, the square plot has the greater area by \(2025 \text{ m}^2\).
In simple words: Find square area and perimeter, use perimeter to find rectangle length and area, then compare both areas.

Exam Tip: Convert decameters (dam) to meters by multiplying by 10 (\(1 \text{ dam} = 10 \text{ m}\)).

Question 6. A grassy plot is 80 m X 60 m. Two cross paths each 4 m wide are constructed at right angles through the centre of the field, such that each path is parallel to one of the sides of the rectangle. Find the total area used as path.
Answer: Length of the plot (\(L\)) = 80 m
Breadth of the plot (\(B\)) = 60 m
Width of each path = 4 m
Area of the path parallel to length = \(L \times \text{width} = 80 \times 4 = 320 \text{ m}^2\)
Area of the path parallel to breadth = \(B \times \text{width} = 60 \times 4 = 240 \text{ m}^2\)
The central square common to both paths has dimensions \(4 \text{ m} \times 4 \text{ m}\), so its area = \(4 \times 4 = 16 \text{ m}^2\)
Total area of the path = (Area of horizontal path) + (Area of vertical path) - (Area of common square)
Total area = \(320 + 240 - 16 = 544 \text{ m}^2\)
In simple words: Add the areas of both crossing paths and subtract the intersection square once because it was counted twice.

Exam Tip: For cross paths intersecting at the center, always subtract the area of the central overlapping square.

Question 7. A rectangle 50 m by 30 m has 2 m wide paths joining the middle points of the opposite sides and also a path of the same breadth all round it. The remainder is covered with grass as show in the figure. Find the cost of laying out of court with grass at Rs. 25 per sq. m and paving the paths at Rs. 2 per sq. m. also, find the total cost.
Answer: Length of rectangle (\(L\)) = 50 m
Breadth of rectangle (\(B\)) = 30 m
Total area of the rectangle = \(50 \times 30 = 1500 \text{ m}^2\)
Width of paths = 2 m
The paths consist of: 1. An outer path all around of width 2 m. 2. Cross paths joining middle points of opposite sides (width 2 m each).
Let us find the total area of the paths directly or by finding the grassy areas.
Grass is divided into 4 smaller rectangles by the central paths, and enclosed by the outer boundary path.
Dimensions of each of the 4 grass rectangles:
Total length minus central path width = \(50 - 2 = 48 \text{ m}\). Half for each grass section = \(\frac{48}{2} = 24 \text{ m}\).
Total breadth minus central path width = \(30 - 2 = 28 \text{ m}\). Half for each grass section = \(\frac{28}{2} = 14 \text{ m}\).
Since there is also a 2 m outer path all around, the grass plots are actually inside the outer path.
Let's calculate total path area: - Outer path all around: Width 2 m. - Inner cross paths: Two paths of width 2 m intersecting. Total area of paths = Area of outer path + Area of cross paths - Overlaps.
Alternatively, Area of grass = Area of 4 inner rectangles = \(4 \times (22 \times 12) = 4 \times 264 = 1056 \text{ m}^2\).
Total Path Area = Total Rectangle Area - Grass Area = \(1500 - 1056 = 444 \text{ m}^2\).
Cost of laying grass = \(1056 \times 25 = \text{Rs. } 26400\)
Cost of paving paths = \(444 \times 2 = \text{Rs. } 888\)
Total Cost = \(26400 + 888 = \text{Rs. } 27288\)
In simple words: Find the area of the grass plots first, subtract from total area to get path area, then multiply each area by its respective rate and add the costs.

Exam Tip: Finding the remaining grassy area directly is often easier than calculating multiple path sections.

Question 8. A circular piece of thin wire is converted into a square of side 6.25 cm. if there is no loss or gain in its length, find the radius of the circular wire.
Answer: Side of the square = 6.25 cm
Perimeter of the square = \(4 \times \text{side} = 4 \times 6.25 = 25 \text{ cm}\)
Length of circular wire = Perimeter of the square = 25 cm
Circumference of the circle = \(2 \pi r = 25\)
\(2 \times \frac{22}{7} \times r = 25\)
\(\frac{44}{7} \times r = 25\)
\(r = \frac{25 \times 7}{44} = \frac{175}{44} = 3.98 \text{ cm}\)
In simple words: The perimeter of the square equals the circumference of the circle. Use the circumference formula to find the radius.

Question 9. A wire is in the form of a circle of radius 42 cm. It is bent into a square. Find the side of the square. (take \(\pi = \frac{22}{7}\))
Answer: Radius of the circle (\(r\)) = 42 cm
Circumference of the circle = \(2 \pi r = 2 \times \frac{22}{7} \times 42 = 2 \times 22 \times 6 = 264 \text{ cm}\)
Perimeter of the square = Circumference of the circle = 264 cm
Side of the square = \(\frac{\text{Perimeter}}{4} = \frac{264}{4} = 66 \text{ cm}\)
In simple words: Calculate the circle's circumference, which is equal to the square's perimeter, then divide by 4 to find the side of the square.

Exam Tip: Circumference acts as the perimeter when wire shape changes.

Question 10. The radius of a circular field is 24.5 m. Find the distance run by a boy in making 4 complete turns.
Answer: Radius of the circular field (\(r\)) = 24.5 m
Circumference of one turn = \(2 \pi r = 2 \times \frac{22}{7} \times 24.5 = \frac{44}{7} \times 24.5 = 44 \times 3.5 = 154 \text{ m}\)
Distance covered in 4 complete turns = \(4 \times 154 = 616 \text{ m}\)
In simple words: Multiply the circumference of the circle by the number of turns to get the total distance run.

Exam Tip: One complete turn equals one full circumference.

Question 11. A race track is in the form of a ring whose inner circumference is 220 m and the outer circumference is 440 m. Find the width of the track.
Answer: Inner circumference (\(2 \pi r_1\)) = 220 m
\(2 \times \frac{22}{7} \times r_1 = 220\)
\(\frac{44}{7} \times r_1 = 220\)
\(r_1 = \frac{220 \times 7}{44} = 5 \times 7 = 35 \text{ m}\)
Outer circumference (\(2 \pi r_2\)) = 440 m
\(2 \times \frac{22}{7} \times r_2 = 440\)
\(\frac{44}{7} \times r_2 = 440\)
\(r_2 = \frac{440 \times 7}{44} = 10 \times 7 = 70 \text{ m}\)
Width of the track = Outer radius - Inner radius = \(r_2 - r_1 = 70 - 35 = 35 \text{ m}\)
In simple words: Find both inner and outer radii from their respective circumferences, then subtract them to find the width of the track.

Exam Tip: Width of a circular ring = Outer radius (\(R\)) - Inner radius (\(r\)).

Question 12. Find the distance covered by the wheel of a truck in 100 rotations if the diameter of the wheel is 49 cm.
Answer: Diameter of the wheel (\(d\)) = 49 cm
Circumference of the wheel = \(\pi d = \frac{22}{7} \times 49 = 22 \times 7 = 154 \text{ cm}\)
Distance covered in 1 rotation = 154 cm
Distance covered in 100 rotations = \(100 \times 154 = 15400 \text{ cm} = 154 \text{ m}\)
In simple words: Multiply the circumference of the wheel by the number of rotations to find the total distance covered.

Exam Tip: Distance in \(n\) rotations = \(n \times (\pi \times d)\).