RS Aggarwal Solutions for Class 6 Chapter 20 Two Dimensional Reflection Symmetry

Exercise 20.1

Question 1. Which of the following are closed curves? Which of them are simple?
(i) A straight line segment
(ii) A triangle
(iii) A hexagon
(iv) A wavy closed shape
(v) A crescent shape (open on one side)
(vi) A crescent shape (closed)
(vii) A curved arrow pointing upward
Answer: Figures (ii), (iii), (iv), (vi), and (vii) are closed curves. Among these, figures (ii), (iii), (iv), and (vi) are simple closed curves.

Exam Tip: A closed curve has no breaks or gaps; a simple closed curve does not cross itself. Remember that straight or curved paths that loop back on themselves without intersecting qualify as simple closed curves.

Question 2. Define perimeter of a closed figure.
Answer: The perimeter refers to the total distance around the outer boundary of a closed shape. In other words, it is the sum of all the side lengths that make up the edge of that figure.

Exam Tip: Always add every side of a polygon to find perimeter; use the formula 2(length + width) for rectangles and 4 × side for squares to save time on calculations.

Question 3. Find the perimeter of each of the following shapes:
(i) A trapezoid with sides 4 cm, 2 cm, 1 cm, and 5 cm
(ii) A trapezoid with sides 23 cm, 35 cm, 40 cm, and 35 cm
(iii) A parallelogram with all sides 15 cm
(iv) A pentagon with all sides 3 cm
Answer:
(i) Perimeter = sum of all side lengths = 4 + 2 + 1 + 5 = 12 cm
(ii) Perimeter = 23 + 35 + 40 + 35 = 133 cm
(iii) Perimeter = 15 + 15 + 15 + 15 = 60 cm
(iv) Perimeter = 3 + 3 + 3 + 3 + 3 = 15 cm
In simple words: To find how far around a shape, just add up the length of every edge. For shapes with equal sides, multiply the side length by the number of sides.

Exam Tip: Ensure you identify and add every single side, especially in irregular figures. Check your addition twice to avoid arithmetic errors.

Exercise 20.2

Question 1. Find the perimeter of the rectangle whose lengths and breadths are given below:
(i) 7 cm, 5 cm
(ii) 5 cm, 4 cm
(iii) 7.5 cm, 4.5 cm
Answer:
(i) The formula for a rectangle's perimeter is 2 × (Length + Breadth). With length 7 cm and breadth 5 cm: Perimeter = 2 × (7 + 5) = 2 × 12 = 24 cm
(ii) With length 5 cm and breadth 4 cm: Perimeter = 2 × (5 + 4) = 2 × 9 = 18 cm
(iii) With length 7.5 cm and breadth 4.5 cm: Perimeter = 2 × (7.5 + 4.5) = 2 × 12 = 24 cm
In simple words: To find the perimeter of a rectangle, add the length and width together, then multiply the result by 2. This works because a rectangle has two pairs of equal sides.

Exam Tip: Remember the formula 2(L + B) for rectangles - it is faster than adding all four sides separately and reduces calculation mistakes.

Question 2. Find the perimeter of the squares whose sides are given below:
(i) 10 cm
(ii) 5 m
(iii) 115.5 cm
Answer:
(i) The perimeter of a square equals 4 times the length of one side. For a side of 10 cm: Perimeter = 4 × 10 = 40 cm
(ii) For a side of 5 m: Perimeter = 4 × 5 = 20 m
(iii) For a side of 115.5 cm: Perimeter = 4 × 115.5 = 462 cm
In simple words: Since all four sides of a square are equal, just multiply one side by 4 to get the total distance around it.

Exam Tip: Always use the 4 × side formula for squares - it is the quickest method and reduces errors in calculation.

Question 3. Find the side of the square whose perimeter is:
(i) 16 m
(ii) 40 cm
(iii) 22 cm
Answer:
(i) The side of a square can be found by dividing the perimeter by 4. When perimeter is 16 m: Side = 16 ÷ 4 = 4 m
(ii) When perimeter is 40 cm: Side = 40 ÷ 4 = 10 cm
(iii) When perimeter is 22 cm: Side = 22 ÷ 4 = 5.5 cm
In simple words: If you know how far around a square is, divide that distance by 4 to find the length of one side.

Exam Tip: This is the reverse of the perimeter formula - always divide by 4 when finding the side from the perimeter of a square.

Question 4. Find the breadth of the rectangle whose perimeter is 360 cm and whose length is
(i) 116 cm
(ii) 140 cm
(iii) 102 cm
Answer:
The formula for a rectangle's perimeter is 2 × (Length + Breadth). Rearranging: Breadth = (Perimeter ÷ 2) - Length
(i) When perimeter is 360 cm and length is 116 cm: Breadth = (360 ÷ 2) - 116 = 180 - 116 = 64 cm
(ii) When perimeter is 360 cm and length is 140 cm: Breadth = (360 ÷ 2) - 140 = 180 - 140 = 40 cm
(iii) When perimeter is 360 cm and length is 102 cm: Breadth = (360 ÷ 2) - 102 = 180 - 102 = 78 cm
In simple words: Start by dividing the perimeter in half, then subtract the length. What remains is the breadth.

Exam Tip: Always rearrange the perimeter formula carefully - dividing the perimeter by 2 first avoids confusion and simplifies the calculation.

Question 5. A rectangular piece of lawn is 55 m wide and 98 m long. Find the length of the fence around it.
Answer: The length of lawn is 98 m and its width is 55 m. The fence encircles the lawn, so its length equals the perimeter of the rectangular lawn. Using the perimeter formula: Perimeter = 2 × (Length + Width) = 2 × (98 + 55) = 2 × 153 = 306 m. Therefore, the fence needed is 306 m long.

Exam Tip: Always recognize that fencing "around" something means calculating the perimeter - use the standard rectangle formula and double-check your addition before multiplying by 2.

Question 6. The side of a square field is 65m. What is the length of the fence required all around it?
Answer: The square field has a side measuring 65 m. The fence required to enclose it equals the perimeter of the square. Perimeter of the square = 4 × Side of the square = 4 × 65 = 260 m. Thus, the fence length needed is 260 m.

Exam Tip: For square fields, always apply 4 × side directly - this gives the fence length needed instantly without extra steps.

Question 7. Two sides of a triangle are 15 cm and 20 cm. The perimeter of the triangles is 50 cm. What is the third side?
Answer: Given: Perimeter of the triangle is 50 cm. The first side measures 15 cm and the second side measures 20 cm. We need to find the third side. The perimeter of a triangle equals the sum of all three sides. Therefore, the third side = (Perimeter of the triangle) - (Sum of the other two sides) = 50 - (15 + 20) = 50 - 35 = 15 cm.

Exam Tip: Always subtract the sum of known sides from the total perimeter to find the missing side - this avoids calculation errors in triangle problems.

Question 8. A wire of length 20 m is to be folded in the form of a rectangle. How many rectangles can be formed by folding the wire if the sides are positive integers in metres?
Answer: The wire length is 20 m, which becomes the perimeter when folded into a rectangle. Thus, perimeter of the rectangle = 20 m. Using the perimeter formula: 2 × (Length + Breadth) = 20 m, which gives (Length + Breadth) = 20 ÷ 2 = 10 m. Since both length and breadth must be positive integers in metres, the possible pairs are: (1m, 9m), (2m, 8m), (3m, 7m), (4m, 6m), and (5m, 5m). Therefore, five different rectangles can be formed with this wire.

Exam Tip: List all possible integer pairs systematically - ensure each pair adds up to half the perimeter and count all distinct rectangles to avoid missing any.

Question 9. A square piece of land has each side equal to 100 m. If 3 layers of metal wire have to be used to fence it, what is the length of the wire?
Answer: The square field has each side measuring 100 m. To fence one layer around it requires a length equal to the perimeter: Perimeter = 4 × 100 = 400 m. This perimeter is the wire length needed for a single layer. Since three layers must be applied: Total wire length = 3 × 400 = 1200 m.

Exam Tip: When multiple layers of fencing are needed, calculate the single perimeter first, then multiply by the number of layers - this ensures accuracy and clarity.

Question 10. Shikha runs around a square of side 75 m. Priya runs around a rectangle with length 60 m and breadth 45 m. Who covers the smaller distance?
Answer: When Shikha and Priya run around their respective shapes, they each cover a distance equal to the perimeters of those shapes. Distance covered by Shikha = Perimeter of the square = 4 × 75 = 300 m. Distance covered by Priya = Perimeter of the rectangle = 2 × (60 + 45) = 2 × 105 = 210 m. Comparing the two distances: 210 m is less than 300 m. Therefore, Priya covers the shorter distance.

Exam Tip: Always compute each perimeter separately before comparing - avoid the temptation to estimate, as even slightly different dimensions can yield very different results.

Question 11. The dimensions of a photographs are 30 cm × 20 cm. What length of wooden frame is needed to frame the picture?
Answer: The photograph has dimensions 30 cm by 20 cm. The wooden frame goes around the entire perimeter of the photograph. The required frame length = Perimeter of the photograph = 2 × (Length + Breadth) = 2 × (30 + 20) = 2 × 50 = 100 cm.

Exam Tip: Recognise that framing a picture requires covering its entire boundary - the frame length needed equals the photograph's perimeter exactly.

Question 12. The length of a rectangular field is 100 m. If its perimeter is 300 m, what is its breadth?
Answer: The rectangular field has length 100 m and perimeter 300 m. Using the perimeter formula for a rectangle: Perimeter = 2 × (Length + Breadth). From this, Breadth = (Perimeter ÷ 2) - Length. Substituting the values: Breadth = (300 ÷ 2) - 100 = 150 - 100 = 50 m.

Exam Tip: Rearrange the rectangle perimeter formula systematically - divide the perimeter by 2 first, then subtract the length to isolate the breadth.

Question 13. To fix fence wires in a garden. 70 m long and 50 m wide, Arvind bought metal pipes for posts. He fixed a post every 5 metres apart; each post was 2 m long. What is total length og the pipersh he bought for the posts?
Answer: The garden is 70 m long and 50 m wide. First, calculate the perimeter: Perimeter = 2 × (Length + Breadth) = 2 × (70 + 50) = 2 × 120 = 240 m. Posts are placed every 5 metres around this perimeter, so the number of posts needed = 240 ÷ 5 = 48 posts. Since each post measures 2 m in length, the total pipe length required = 48 × 2 = 96 m.

Exam Tip: Break the problem into steps: find perimeter, then divide by spacing to get post count, then multiply by individual post length for the total - this prevents mixing up the calculations.

Question 14. Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs 12 per meter.
Answer: The park has length 175 m and breadth 125 m. Perimeter of the park = 2 × (Length + Breadth) = 2 × (175 + 125) = 2 × 300 = 600 m. The fencing rate is Rs 12 per metre. Total fencing cost = Rs 12 × 600 = Rs 7,200.

Exam Tip: Always find the perimeter first, then multiply by the rate per unit - ensure you maintain unit consistency throughout the calculation to avoid errors.

Question 15. The perimeter of a rectangular pentagon is 100 cm. How long is each side?
Answer: A regular pentagon is a closed polygon having five sides of equal length. The perimeter of this pentagon is 100 cm. Since the perimeter equals 5 times the length of one side: Side of the regular pentagon = Perimeter ÷ 5 = 100 ÷ 5 = 20 cm.

Exam Tip: For regular polygons (all sides equal), always divide the perimeter by the number of sides to find each individual side - this is much faster than other methods.

Question 16. Find the perimeter of a regular hexagon with each side measuring 8 m.
Answer: A regular hexagon is a closed polygon with six sides of equal length. Given that each side measures 8 m, the perimeter of the hexagon = 6 × Side of the hexagon = 6 × 8 = 48 m.

Exam Tip: For any regular polygon, multiply the side length by the number of sides - with hexagons, always use the factor 6.

Question 17. A rectangular piece of land measure 0.7 km by 0.5 km. Each side is to be fenced with four rows of wires. What length of the wire is needed?
Answer: The rectangular land measures 0.7 km by 0.5 km. Perimeter of the rectangular land = 2 × (Length + Breadth) = 2 × (0.7 + 0.5) = 2 × 1.2 = 2.4 km. This perimeter represents the wire length for a single row around the land. Since four rows must be applied: Total wire length needed = 4 × 2.4 = 9.6 km.

Exam Tip: When dealing with multiple rows of fencing, compute the perimeter once, then multiply by the row count - this method avoids repeated calculations and ensures consistency.

Question 18. Avneet buys 9 square paving slabs, each with a side of 1/2 m. He lays them in the form of a square.
(i) What is the perimeter of his arrangement?
(ii) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement?
(iii) Avneet wonders, if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges)
Answer:
(i) Nine square slabs, each with side 1/2 m, are arranged to form one large square. Since 9 = 3 × 3, the large square has 3 slabs along each edge. The side of this large square = 3 × (1/2) = 3/2 = 1.5 m. Perimeter = 4 × 1.5 = 6 m.
(ii) In a cross arrangement, the outline forms a plus sign shape. This arrangement has 12 exposed edges (each edge is 1/2 m long). Total perimeter = 12 × (1/2) = 6 m.
(iii) To achieve a larger perimeter, arrange the nine slabs in a line, one after another. This creates a long rectangle measuring 9 × (1/2) m by 1 × (1/2) m. Perimeter = 2 × [9 × (1/2) + 1 × (1/2)] = 2 × [4.5 + 0.5] = 2 × 5 = 10 m. A linear arrangement gives a perimeter of 10 m, which is greater than the square or cross configurations.
In simple words: Different shapes made from the same number of pieces can have different perimeters. A long, thin shape usually has a larger perimeter than a compact, square-like shape made from the same pieces.

Exam Tip: Remember that perimeter depends on the shape's arrangement, not just the area or number of pieces - test multiple configurations to find the maximum or minimum perimeter for the same set of objects.

Exercise 20.3

Question 1. The following figures are drawn on a squared paper. Count the number of squares enclosed by each figure and find its area, taking the area of each square as 1 cm².
Answer:
(i) The shape contains 16 complete squares. Since each square has an area of 1 cm², the total area is 16 × 1 = 16 cm²
(ii) The shape has 36 complete squares. With each square measuring 1 cm², the total area equals 36 × 1 = 36 cm²
(iii) The figure holds 15 complete squares and 6 half squares. Therefore, the total area is (15 + 6 × 1/2) = 18 cm²
(iv) This shape contains 20 complete squares and 8 half squares. So the area equals (20 + 8 × 1/2) = 24 cm²
(v) The figure has 13 complete squares, 8 portions greater than half squares, and 7 portions less than half squares. The area is (13 + 8 × 1) = 21 cm²
(vi) This shape holds 8 complete squares, 6 parts more than half squares, and 4 parts less than half squares. The area is (8 + 6 × 1) = 14 cm²

Exam Tip: When counting half squares, pair them as complete units. Always neglect the "less than half" portions and count "more than half" portions as full squares for approximation.

Question 2. On a squared paper, draw (i) a rectangle, (ii) a triangle, (iii) any irregular closed figure, Find approximate area of each by counting the number of squares complete, more than half and exactly half.
Answer:
(i) A rectangle: This shape encloses 18 complete squares. Assuming each square has an area of 1 cm², the rectangle's area would be 18 cm²
(ii) A triangle: This triangle contains 4 complete squares, 6 parts greater than half squares, and 6 parts less than half squares. Taking each square as 1 cm² and neglecting the less-than-half portions while treating greater-than-half portions as complete units, the triangle's area measures approximately 10 cm²
(iii) Any irregular figure: This figure has 10 complete squares, 1 part exactly half a square, 7 parts greater than half squares, and 6 parts less than half squares. The area works out to (10 + 1 × 1/2 + 7 × 1) = 17.5 cm²

Exam Tip: When using the counting method, always include exactly half portions in your calculation and round greater-than-half regions up to 1 complete square.

Question 3. Draw any circle on the graph paper, Count the squares and use them to estimate the area the area of the circular region.
Answer: When a circle is drawn on squared paper, count the complete squares it covers, the squares that are more than half enclosed, and those that are less than half enclosed. This circle on the squared paper contains 21 complete squares, 15 parts greater than half squares, and 8 parts less than half squares. Assuming each square measures 1 cm², and disregarding the less-than-half portions while treating more-than-half portions as full squares, the circle's area is approximately (21 + 15) = 36 cm²
In simple words: Draw a circle on grid paper, count all the boxes inside it, and add half-boxes that the circle covers more than halfway. This gives you a rough estimate of the circle's area.

Exam Tip: The grid method provides only an approximation. The more grid squares used, the more accurate the estimate becomes compared to the actual formula πr².

Question 4. Using tracing paper and centimeter graph paper to compare the areas of the following pairs of figures:
Answer: When using tracing paper, trace both figures onto graph paper. For the first figure: it contains 4 complete squares, 9 parts greater than half squares, and 9 parts less than half squares. Assuming 1 square = 1 cm² and disregarding the less-than-half portions while counting more-than-half portions as full squares, the area is (4 + 9) = 13 cm². For the second figure: it holds 8 complete squares, 11 parts greater than half squares, and 10 parts less than half squares. Using the same method where 1 square = 1 cm², the area becomes (8 + 11) = 19 cm². Comparing these two shapes, the area of the second figure exceeds that of the first figure.
In simple words: Trace the shapes onto graph paper, count complete boxes and half-boxes, then compare the totals to see which figure is larger.

Exam Tip: Always use the same grid size for both figures when comparing, and consistently apply the rule for counting more-than-half and less-than-half squares.

Exercise 20.4

Question 1. Find the area of a rectangle, whose
(i) Length = 6cm, breadth = 3 cm
(ii) Length = 8 cm, breadth = 3 cm
(iii) Length = 4.5 cm, breadth = 2 cm
Answer:
(i) The formula for rectangle area is Length × Breadth. With Length = 6 cm and Breadth = 3 cm, the area = 6 × 3 = 18 cm²
(ii) Using Length = 8 cm and Breadth = 3 cm, the area = 8 × 3 = 24 cm²
(iii) With Length = 4.5 cm and Breadth = 2 cm, the area = 4.5 × 2 = 9 cm²

Exam Tip: Always multiply the two perpendicular sides (length and breadth). Make sure both measurements are in the same unit before multiplying.

Question 2. Find the area of a square whose side is:
(i) 5 cm
(ii) 4.1 cm
(iii) 5.5 cm
(iv) 2.6 cm
Answer: The formula for square area is Side × Side.
(i) Side = 5 cm, so area = 5 × 5 = 25 cm²
(ii) Side = 4.1 cm, so area = 4.1 × 4.1 = 16.81 cm²
(iii) Side = 5.5 cm, so area = 5.5 × 5.5 = 30.25 cm²
(iv) Side = 2.6 cm, so area = 2.6 × 2.6 = 6.76 cm²

Exam Tip: Remember that for a square, both sides are equal, so you square the side length. Be careful with decimal multiplication to avoid calculation errors.

Question 3. The area of a rectangle is 49 cm² and its breadth is 2.8 cm. Find the length of the rectangle.
Answer: Given: Area = 49 cm² and Breadth = 2.8 cm. Using the formula Area = Length × Breadth, we can rearrange to find Length = Area / Breadth = 49 / 2.8 = 17.5 cm

Exam Tip: When finding one dimension from area and another dimension, always divide the area by the known dimension. Check your answer by multiplying back.

Question 4. The side of a square is 70 cm. Find its area and perimeter.
Answer: Given: Side = 70 cm. The area of the square = Side × Side = 70 × 70 = 4900 cm². The perimeter of the square = 4 × Side = 4 × 70 = 280 cm

Exam Tip: Don't confuse area (measured in square units) with perimeter (measured in linear units). Area uses multiplication of sides, while perimeter adds all four sides together.

Question 5. The area of a rectangle is 225 cm² and its one side is 25 cm, find its other side.
Answer: Given: Area = 225 cm² and one side = 25 cm. Using Area = Length × Breadth, the other side = Area / one side = 225 / 25 = 9 cm

Exam Tip: The method is straightforward - divide the total area by the known side to get the unknown side. This works for any rectangle.

Question 6. What will happen to the area of a rectangle If its
(i) Length and breadth are trebled (ii) Length is doubled and breadth is same
(iii) Length is doubled and breadth is halved
Answer:
(i) If the length and breadth are trebled: Let the original length and breadth be l and b respectively. Original area = l × b = lb. When both dimensions are tripled, they each become three times their original values. Therefore, the new length = 3l and the new breadth = 3b. The new area = 3l × 3b = 9lb. This shows that the area of the rectangle becomes 9 times its original area.
(ii) If the length is doubled and breadth remains the same: With the original area = l × b = lb. When length doubles while breadth stays unchanged, the new length = 2l and breadth = b. The new area = 2l × b = 2lb. Therefore, the area of the rectangle becomes 2 times its original area.
(iii) If the length is doubled and breadth is halved: The original area = l × b = lb. When length doubles and breadth becomes half, the new length = 2l and new breadth = b/2. The new area = 2l × b/2 = lb. The new area remains the same as the original area.

Exam Tip: Understand how scaling one or both dimensions affects area. When both dimensions scale by factor k, area scales by k². When only one dimension changes, area changes proportionally.

Question 7. What will happen to the area of a square if its side is :
(i) Tripled
(ii) Increased by half of it
Answer:
(i) If the side of a square is tripled: Let the original side of the square be s. The original area = s × s = s². When the side is tripled, the new side becomes 3s. The new area = 3s × 3s = 9s². This means that the area becomes 9 times the original area.
(ii) If the side of a square is increased by half of it: Let the original side be s. The original area = s × s = s². When the side is increased by half its length, the new side equals \( s + \frac{1}{2}s = \frac{3}{2}s \). The new area = \( \frac{3}{2}s × \frac{3}{2}s = \frac{9}{4}s² \). This shows that the area becomes 9/4 times (or 2.25 times) the original area.

Exam Tip: When a square's side scales by factor k, its area scales by k². For instance, tripling the side (k=3) multiplies the area by 9 (3²). This relationship is crucial for all area scaling problems.

Question 8. Find the perimeter of a rectangle whose area is 500 cm² and breadth is 20 cm.
Answer: The area equals 500 cm² and the breadth measures 20 cm. Using the relationship Area = Length × Breadth, we can find that Length = 500 ÷ 20 = 25 cm. The perimeter is calculated using Perimeter = 2(Length + Breadth) = 2(25 + 20) = 2 × 45 = 90 cm.

Exam Tip: Always derive the missing dimension first using the area formula, then apply the perimeter formula—this two-step method avoids errors.

Question 9. A rectangle has the area equal to that of a square of side 80 cm. If the breadth of the rectangle is 20 cm, find its length.
Answer: The square has a side of 80 cm, so its area is 80 × 80 = 6400 cm². Since the rectangle's area matches this, the rectangle's area is also 6400 cm². With a breadth of 20 cm, we use Length = Area ÷ Breadth to get Length = 6400 ÷ 20 = 320 cm.

Exam Tip: When a rectangle's area equals a square's area, find the square's area first, then use division to get the rectangle's missing dimension.

Question 10. Area of a rectangle of breadth 17 cm is 340 cm². Find the perimeter of the rectangle.
Answer: The rectangle has an area of 340 cm² and a breadth of 17 cm. Using Length = Area ÷ Breadth, we get Length = 340 ÷ 17 = 20 cm. The perimeter is Perimeter = 2(Length + Breadth) = 2(20 + 17) = 2 × 37 = 74 cm.

Exam Tip: Division and addition must follow the correct order—find the length first, then add both dimensions and double.

Question 11. A marble tile measures 15 cm × 20 cm. How many tiles will be required to cover a wall of size 4 m × 6 m?
Answer: The tile has dimensions 15 cm × 20 cm, giving an area of 15 × 20 = 300 cm². The wall measures 4 m × 6 m, which converts to 400 cm × 600 cm (since 1 m = 100 cm). The wall's area is 400 × 600 = 240,000 cm². The number of tiles needed is 240,000 ÷ 300 = 800 tiles.

Exam Tip: Always convert all measurements to the same unit before dividing—mixing units leads to incorrect answers.

Question 12. A marble tile measures 10 cm × 12 cm. How many tiles will be required to cover a wall of size 3 m × 4 m? Also, find the total cost of the tiles at the rate of Rs 2 per tile.
Answer: The tile measures 10 cm × 12 cm, so its area is 10 × 12 = 120 cm². The wall is 3 m × 4 m, which equals 300 cm × 400 cm (converting using 1 m = 100 cm). The wall's area is 300 × 400 = 120,000 cm². The number of tiles required is 120,000 ÷ 120 = 1,000 tiles. At Rs 2 per tile, the total cost is 2 × 1,000 = Rs 2,000.

Exam Tip: Break the problem into steps: find tile area, find wall area, calculate tile count, then multiply by the unit cost.

Question 13. One tile of a square plot is 250 m. Find the cost of levelling it at the rate of Rs 2 per square meter.
Answer: The square plot has a side of 250 m. Its area is Side × Side = 250 × 250 = 62,500 m². The levelling rate is Rs 2 per m². The total levelling cost is 62,500 × 2 = Rs 1,25,000.

Exam Tip: For a square, always square the side length first, then multiply the area by the given rate.

Question 14. The following figures have been split into rectangles. Find the areas. (The measures are given in centimeters)
Answer:

(i) This figure is made up of two rectangles (II and IV) and two squares (I and III).
Square I: Area = Side × Side = 3 × 3 = 9 cm²
Rectangle II: Area = 2 × 1 = 2 cm²
Square III: Area = 3 × 3 = 9 cm²
Rectangle IV: Area = 2 × 4 = 8 cm²
Total area of this figure = 9 + 2 + 9 + 8 = 28 cm²

(ii) This figure consists of three rectangles (I, II, and III).
Rectangle I: Area = Length × Breadth = 3 × 1 = 3 cm²
Rectangle II: Area = 3 × 1 = 3 cm²
Rectangle III: Area = 3 × 1 = 3 cm²
Total area of this figure = 3 + 3 + 3 = 9 cm²

Exam Tip: Always label each section clearly and compute areas separately before summing—this prevents missing or double-counting regions.

Question 15. Split the following shapes into rectangles and find the area of each. (The measures are given in centimeters)
Answer:

(i) This figure consists of two rectangles (I and II).
Rectangle I: Area = Length × Breadth = 10 × 2 = 20 cm²
Rectangle II: Area = 10 × 3/2 = 15 cm²
Total area of this figure = 20 + 15 = 35 cm²

(ii) This figure is made up of two squares (I and III) and one rectangle (II).
Square I: Area = Side × Side = 7 × 7 = 49 cm²
Square III: Area = 7 × 7 = 49 cm²
Rectangle II: Area = 21 × 7 = 147 cm²
Total area of this figure = 49 + 49 + 147 = 245 cm²

(iii) This figure is made up of two rectangles (I and II).
Rectangle I: Area = Length × Breadth = 5 × 1 = 5 cm²
Rectangle II: Area = 4 × 1 = 4 cm²
Total area of this figure = 5 + 4 = 9 cm²

Exam Tip: Use dashed lines to partition irregular shapes and label each section—methodical subdivision ensures complete and accurate calculation.

Question 16. How many tiles with dimension 5 cm and 12 cm will be needed to fit a region whose length and breadth are respectively?
(i) 100 cm and 144 cm
(ii) 70 cm and 36 cm
Answer:

(i) The tile dimensions are 5 cm × 12 cm, so its area is 5 × 12 = 60 cm². The region measures 100 cm × 144 cm, giving an area of 100 × 144 = 14,400 cm². The number of tiles needed is 14,400 ÷ 60 = 240 tiles.

(ii) The tile dimensions remain 5 cm × 12 cm, with area 5 × 12 = 60 cm². The region is 70 cm × 36 cm, with area 70 × 36 = 2,520 cm². The number of tiles needed is 2,520 ÷ 60 = 42 tiles.

Exam Tip: Always divide the total region area by a single tile area—never mix units or forget to convert dimensions.

Exercise 20.5

Question 1. The sides of a rectangle are in the ratio 5 : 4. If its perimeter is 72 cm, then its length is
(a) 40 cm
(b) 20 cm
(c) 30 cm
(d) 60 cm
Answer: (b) 20 cm
In simple words: If the sides are in ratio 5 : 4, call them 5x and 4x. Using the perimeter formula 2(5x + 4x) = 72, you get x = 4. So the length is 5 × 4 = 20 cm.

Exam Tip: When sides are in a ratio, express them using a variable—this simplifies solving for actual dimensions.

Question 2. The cost of fencing a rectangular field 34 m long and 18 m wide at As 2.25 per metre is
(a) Rs 243
(b) Rs 234
(c) Rs 240
(d) Rs 334
Answer: (b) Rs 234
In simple words: Calculate the perimeter: 2(34 + 18) = 2 × 52 = 104 m. Multiply by the rate: 104 × 2.25 = Rs 234.

Exam Tip: Fencing problems require the perimeter, not area—always use 2(Length + Breadth).

Question 3. If the cost of fencing a rectangular field at Rs. 7.50 per metre is Rs. 600, and the length of the field is 24 m, then the breadth of the field is
(a) 8 m
(b) 18 m
(c) 24 m
(d) 16 m
Answer: (d) 16 m
In simple words: From the fencing cost, find the perimeter: 600 ÷ 7.50 = 80 m. Use Breadth = Perimeter/2 - Length = 80/2 - 24 = 40 - 24 = 16 m.

Exam Tip: Work backwards from cost to find perimeter, then extract the unknown dimension using the perimeter formula.

Question 4. The cost of putting a fence around a square field at As 2.50 per metre is As 200. The length of each side of the field is
(a) 80 m
(b) 40 m
(c) 20 m
(d) None of these
Answer: (c) 20 m
In simple words: The fencing cost gives us the perimeter: 200 ÷ 2.50 = 80 m. A square has all sides equal, so Side = Perimeter ÷ 4 = 80 ÷ 4 = 20 m.

Exam Tip: For a square, always divide the perimeter by 4 to get one side—this is the key difference from rectangles.

Question 5. The length of a rectangle is three times of its width. If the length of the diagonal is 8√10 m, then the perimeter of the rectangle is
(a) 15√10 m
(b) 16√10 m
(c) 24√10 m
(d) 64 m
Answer: (d) 64 m
In simple words: Let the width be x m, so the length is 3x m. Using the Pythagorean theorem for the diagonal: \( (3x)^2 + x^2 = (8\sqrt{10})^2 \)
\( 9x^2 + x^2 = 640 \)
\( 10x^2 = 640 \)
\( x^2 = 64 \)
\( x = 8 \)
So width = 8 m and length = 24 m. Perimeter = 2(24 + 8) = 64 m.

Exam Tip: Use the diagonal and the length-width relationship together with the Pythagorean theorem to find actual dimensions before calculating perimeter.

Question 6. If a diagonal of a rectangle is thrice its smaller side, then its length and breadth are in the ratio
(a) 3:1
(b) √3:1
(c) √2:1
(d) 2√2:1
Answer: (d) 2√2:1
In simple words: When the diagonal is three times the shorter side, you can set up an equation using the Pythagorean theorem. Solving it gives the ratio of the longer side to the shorter side as 2√2:1.

Exam Tip: Always apply the Pythagorean theorem directly when working with diagonals and sides of rectangles — it's the fastest path to the ratio.

Question 7. The ratio of the areas of two squares, one having its diagonal double than the other, is
(a) 1 : 2
(b) 2:3
(c) 3 : 1
(d) 4 : 1
Answer: (d) 4 : 1
In simple words: If one square has a diagonal that is twice as long as another square's diagonal, then the area of the first square will be four times the area of the second square. This is because area depends on the square of the side length.

Exam Tip: Remember that when linear dimensions (like diagonals) double, areas increase by a factor of 4 — this relationship holds for all similar shapes.

Question 8. If the ratio of areas of two squares is 225:256, then the ratio of their perimeters is
(a) 225 : 256
(b) 256 : 225
(c) 15:16
(d) 16 : 15
Answer: (c) 15:16
In simple words: The area ratio is the square of the side ratio. To find the side ratio, take the square root of 225:256, which gives 15:16. Since perimeter is directly proportional to side length, the perimeter ratio is also 15:16.

Exam Tip: When going from area ratios to perimeter ratios, always take the square root of the area ratio first to get the side ratio — perimeters follow the same proportion as sides.

Question 9. If the sides of a square are halved, then its area
(a) remains same
(b) becomes half
(c) becomes one fourth
(d) becomes double
Answer: (c) becomes one fourth
In simple words: When you cut the side length in half, the area gets multiplied by (1/2)² = 1/4. So the new area is one-fourth of the original area.

Exam Tip: Keep in mind that area scales with the square of the linear dimension — if you halve the side, the area reduces by a factor of 4, not 2.

Question 10. A rectangular carpet has area 120 m² and perimeter 46 metres. The length of its diagonal is
(a) 15 m
(b) 16 m
(c) 17 m
(d) 20 m
Answer: (c) 17 m
In simple words: Use the area and perimeter to find the two side lengths, then apply the Pythagorean theorem to calculate the diagonal length. The diagonal comes out to 17 metres.

Exam Tip: For rectangle problems with both area and perimeter given, always set up two equations — one for area (l × b) and one for perimeter (2(l + b)) — then solve simultaneously.

Question 11. If the ratio between the length and the perimeter of a rectangular plot is 1: 3, then the ratio between the length and breadth of the plot is
(a) 1 : 2
(b) 2 : 1
(c) 3 : 2
(d) 2 : 3
Answer: (b) 2 : 1
In simple words: If the length to perimeter ratio is 1:3, then the length equals one-third of the perimeter. Setting up the equation and simplifying reveals that the length is twice the breadth, giving a 2:1 ratio.

Exam Tip: When given a ratio involving perimeter, express perimeter as 2(l + b) and substitute carefully — this algebraic step is essential for finding the side ratio.

Question 12. If the length of the diagonal of a square is 20 cm, then its perimeter is
(a) 10√2 cm
(b) 40 cm
(c) 40√2 cm
(d) 200 cm
Answer: (c) 40√2 cm
In simple words: The diagonal of a square with side s equals s√2. If the diagonal is 20 cm, then the side is 20/√2 = 10√2 cm. The perimeter is 4 times the side, which equals 40√2 cm.

Exam Tip: For a square, always remember the diagonal-side relationship: diagonal = side × √2. Use this to convert between diagonal and side length quickly.