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Home · Schooling · CBSE · Class 9 · Mathematics · Measuring Space: Perimeter and Area · End-of-chapter exercises

NCERT Class 9 Mathematics Measuring Space: Perimeter and Area: Exercise Solutions

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 9 Mathematics book, chapter “Measuring Space: Perimeter and Area”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

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How these were checked: the solutions were drafted with AI and then read and corrected on 23 Oct 2026. Every numerical final answer is recomputed from the question's own numbers by a calculator check that runs before each release; proofs and explanations were read, not machine-checked. The NCERT textbook is the authority — if something here disagrees with it, follow the book.

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Exercises Q1
Fig. 6.41 shows (a + b)² = a² + 2ab + b² as areas. Draw area figures for (a + b)(a − b) = a² − b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. (a + b)(a − b): draw a square of side a (area a²) and cut a small square of side b out of one corner (take a > b). The L-shaped region left has area a² − b².
  why?
  Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.
2. Cut the L-shape along the inner edge into two rectangles: one a × (a − b) and one b × (a − b).
3. Place the b × (a − b) rectangle against the end of the a × (a − b) rectangle so that their sides of length a − b match.
4. Together they form one rectangle with sides (a + b) and (a − b), so (a + b)(a − b) = a² − b².
  why?
  Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.
5. For (a + b + c)², draw a large square with side length (a + b + c).
6. Divide each side of the large square into three segments of lengths a, b, and c.
7. Draw lines parallel to the sides through these division points to create a 3 × 3 grid of smaller rectangles.
8. The grid contains three squares with areas a², b², and c².
9. It also contains six rectangles: two of area ab, two of area bc, and two of area ca.
10. Summing all nine areas gives a² + b² + c² + 2ab + 2bc + 2ca, which equals the total area (a + b + c)².
Answer: The area figures demonstrate that (a + b)(a − b) = a² − b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
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Exercises Q2
An isosceles triangle has perimeter 40 cm and equal sides of 15 cm. Find its area.
1. The perimeter of the triangle is 40 cm and the two equal sides are 15 cm each.
2. The length of the third side is 40 − (15 + 15) = 10 cm.
3. The semi-perimeter s is (15 + 15 + 10) ÷ 2 = 20 cm.
4. Using Heron's formula, Area = √[s(s−a)(s−b)(s−c)] = √[20(20−15)(20−15)(20−10)].
5. Substituting the values gives Area = √[20 × 5 × 5 × 10].
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
6. Simplifying the product inside the square root gives √5000.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
7. Factorizing 5000 as 2500 × 2 allows us to write √5000 = √2500 × √2.
8. Since √2500 = 50, the area is 50√2 cm².
9. Approximating √2 as 1.4142, the area is 50 × 1.4142 = 70.71 cm².
Answer: 50√2 cm², 70.71 cm²
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Exercises Q3
An isosceles triangle has base 10 cm and area 60 cm². Find the equal sides.
1. Let the equal sides of the isosceles triangle be x cm.
2. The base is 10 cm, so the altitude bisects the base into two segments of 5 cm each.
3. Using the Pythagorean theorem in the right-angled triangle formed, x² = 5² + h², where h is the height.
4. The area of the triangle is given by (1/2) × base × height = 60 cm².
5. Substituting the base value, (1/2) × 10 × h = 60.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
6. Solving for h gives 5h = 60, so h = 12 cm.
7. Now substitute h = 12 into the Pythagorean equation: x² = 5² + 12².
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
8. Calculate x² = 25 + 144 = 169.
9. Taking the square root, x = √169 = 13 cm.
Answer: 13 cm
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Exercises Q4
A right-angled triangle has area 54 cm² and one leg 12 cm. Find its perimeter.
1. Let the two legs of the right-angled triangle be 12 cm and b cm.
2. The area of a right-angled triangle is given by (1/2) × base × height.
3. Substituting the known values, we get 54 = (1/2) × 12 × b.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
4. Solving for b, we find b = 9 cm.
5. Now we use the Pythagorean theorem to find the hypotenuse c, where c² = 12² + 9².
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
6. Calculating the squares, c² = 144 + 81 = 225.
7. Taking the square root, c = 15 cm.
8. The perimeter is the sum of all three sides: 12 + 9 + 15.
9. Adding these values gives 36 cm.
Answer: 36 cm
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Exercises Q5
A triangle has sides in the ratio 2 : 3 : 4 and perimeter 45 cm. Find its area.
1. Let the sides of the triangle be 2x, 3x, and 4x.
2. The perimeter is 2x + 3x + 4x = 9x.
3. Given the perimeter is 45 cm, so 9x = 45.
4. Solving for x gives x = 5.
5. The sides of the triangle are 2(5) = 10 cm, 3(5) = 15 cm, and 4(5) = 20 cm.
6. The semi-perimeter s is 45 ÷ 2 = 22.5 cm.
7. Using Heron's formula, Area = √[s(s − a)(s − b)(s − c)].
8. Substitute the values: Area = √[22.5(22.5 − 10)(22.5 − 15)(22.5 − 20)].
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
9. Area = √[22.5 × 12.5 × 7.5 × 2.5].
10. As fractions s = 45/2, s − a = 25/2, s − b = 15/2, s − c = 5/2, so the product is (45 × 25 × 15 × 5)/16 = 84375/16 (= 5273.4375).
11. 84375 = 5625 × 15 = 75² × 15, so area = √84375/4 = 75√15/4 cm².
12. Area ≈ 72.62 cm².
Answer: 75√15/4 cm², 72.62
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Exercises Q6
Find the area of the triangle with sides 7 cm, 24 cm, 25 cm in two different ways.
1. Identify the sides of the triangle as a = 7 cm, b = 24 cm, and c = 25 cm.
2. Check if it is a right-angled triangle by verifying if 7² + 24² = 25².
3. Calculate 49 + 576 = 625, which equals 25², confirming it is a right-angled triangle with legs 7 cm and 24 cm.
4. Method 1: Use the formula Area = (1/2) × base × height for a right-angled triangle.
5. Substitute the values to get Area = (1/2) × 7 × 24.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
6. Calculate (1/2) × 168 = 84 cm².
7. Method 2: Use Heron's formula where s is the semi-perimeter.
8. Calculate the semi-perimeter s = (7 + 24 + 25) / 2 = 56 / 2 = 28 cm.
9. Apply Heron's formula: Area = √(s(s-a)(s-b)(s-c)).
10. Substitute the values: Area = √(28(28-7)(28-24)(28-25)).
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
11. Simplify the terms inside the square root: √(28 × 21 × 4 × 3).
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
12. Calculate the product: 28 × 21 = 588 and 4 × 3 = 12, so √(588 × 12).
13. Multiply 588 × 12 = 7056.
14. Find the square root of 7056, which is 84 cm².
15. Both methods yield the same area.
Answer: 84 cm²
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Exercises Q7
A bicycle wheel has diameter 60 cm (π ≈ 22/7). Distance covered in 100 turns?
1. The diameter of the bicycle wheel is 60 cm.
2. The radius of the wheel is half of the diameter, so r = 60 ÷ 2 = 30 cm.
3. The distance covered in one turn is the circumference of the wheel, which is 2πr.
  why?
  Circumference = 2πr = πd: π is the same ratio (circumference ÷ diameter) for every circle. Take π = 22/7 or 3.14 as the question says.
4. Substituting the values, circumference = 2 × (22/7) × 30 cm.
  why?
  Circumference = 2πr = πd: π is the same ratio (circumference ÷ diameter) for every circle. Take π = 22/7 or 3.14 as the question says.
5. Calculating the circumference gives (1320/7) cm.
6. The total distance covered in 100 turns is 100 times the circumference.
7. Total distance = 100 × (1320/7) cm = 132000/7 cm.
8. Converting the fraction to a decimal, 132000 ÷ 7 ≈ 18857.14 cm.
9. To convert the distance into meters, divide by 100.
10. 18857.14 cm ÷ 100 = 188.57 m.
Answer: 18857.14 cm, 188.57 m
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Exercises Q8
Area of a quadrant of a circle with circumference 66 cm (π ≈ 22/7).
1. Let the radius of the circle be r cm.
2. The circumference of the circle is given by the formula 2πr.
  why?
  Circumference = 2πr = πd: π is the same ratio (circumference ÷ diameter) for every circle. Take π = 22/7 or 3.14 as the question says.
3. Substitute the given value: 2 × (22/7) × r = 66.
  why?
  Circumference = 2πr = πd: π is the same ratio (circumference ÷ diameter) for every circle. Take π = 22/7 or 3.14 as the question says.
4. Simplify the equation: (44/7) × r = 66.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
5. Solve for r: r = 66 × (7/44).
6. Calculate r: r = 10.5 cm.
7. The area of a full circle is given by the formula πr².
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
8. A quadrant is one-fourth of a circle, so its area is (1/4) × πr².
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
9. Substitute the values: Area = (1/4) × (22/7) × (10.5)².
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
10. Calculate the square of the radius: (10.5)² = 110.25.
11. Multiply the terms: Area = (1/4) × (22/7) × 110.25.
12. Simplify the fraction: (22/7) × 110.25 = 346.5.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
13. Divide by 4: Area = 346.5 / 4.
14. Final calculation: Area = 86.625 cm².
Answer: 86.625 cm²
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Exercises Q9
A car wheel has outer radius 28 cm (π ≈ 22/7). Distance in one turn, and number of turns in 1 km?
1. The distance covered in one turn is equal to the circumference of the wheel.
2. Circumference = 2 × π × r = 2 × (22/7) × 28 cm.
  why?
  Circumference = 2πr = πd: π is the same ratio (circumference ÷ diameter) for every circle. Take π = 22/7 or 3.14 as the question says.
3. Circumference = 176 cm.
  why?
  Circumference = 2πr = πd: π is the same ratio (circumference ÷ diameter) for every circle. Take π = 22/7 or 3.14 as the question says.
4. Convert the total distance of 1 km into centimeters: 1 km = 100000 cm.
5. Number of turns = Total distance ÷ Distance in one turn.
6. Number of turns = 100000 ÷ 176.
7. Number of turns = 568.18.
Answer: Distance in one turn = 176 cm; Number of turns = 568.18
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Exercises Q10
Two rectangles have the same area and the same perimeter. Must they be congruent?
1. Let the first rectangle have length l and width w.
2. Let the second rectangle have length L and width W.
3. Since the areas are equal, lw = LW.
4. Since the perimeters are equal, 2(l + w) = 2(L + W), which simplifies to l + w = L + W.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
5. The pair of numbers (l, w) are the roots of the quadratic equation x² − (l + w)x + lw = 0.
6. The pair of numbers (L, W) are the roots of the quadratic equation x² − (L + W)x + LW = 0.
7. Because the sums and products are equal, both pairs satisfy the same quadratic equation.
8. A quadratic equation has a unique set of roots, so {l, w} = {L, W}.
9. This means the two rectangles have the same length and width, just possibly swapped.
10. Therefore, the two rectangles are congruent.
Answer: Yes, they must be congruent.
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Exercises Q11
Using the area of a parallelogram (base × height) and Fig. 6.42 (a trapezium cut into a parallelogram and a triangle), show the area of a trapezium with parallel sides a, b and height h is ½(a + b)h.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Consider a trapezium ABCD with parallel sides AB = a and CD = b, and height h.
2. Draw a line through vertex A parallel to side BC, intersecting side CD at point E.
3. This construction divides the trapezium into a parallelogram ABCE and a triangle ADE.
4. In parallelogram ABCE, the base is AB = a and the height is h, so its area is a × h.
5. In triangle ADE, the base is DE = CD − CE = b − a and the height is h, so its area is ½(b − a)h.
6. The total area of the trapezium is the sum of the areas of the parallelogram and the triangle.
7. Area = a × h + ½(b − a)h.
8. Factor out h to get Area = h(a + ½b − ½a).
9. Simplify the expression inside the brackets to get Area = h(½a + ½b).
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
10. Factor out ½ to obtain the final formula Area = ½(a + b)h.
Answer: Area of trapezium = ½(a + b)h
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Exercises Q12
By cutting a trapezium into two triangles, show its area is ½ × (sum of parallel sides) × height.
1. Consider a trapezium ABCD with parallel sides AB and CD, and height h.
2. Draw the diagonal AC to divide the trapezium into two triangles, triangle ABC and triangle ACD.
3. The area of triangle ABC is ½ × base × height, which is ½ × AB × h.
4. The area of triangle ACD is ½ × base × height, which is ½ × CD × h.
5. The total area of the trapezium is the sum of the areas of these two triangles.
6. Area = ½ × AB × h + ½ × CD × h.
7. Factor out the common term ½ × h to get ½ × h × (AB + CD).
8. Thus, the area is ½ × (sum of parallel sides) × height.
Answer: Area of trapezium = ½ × (sum of parallel sides) × height
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Exercises Q13
Show how two identical trapeziums make a parallelogram, and get the trapezium area formula from it.
1. Take two identical trapeziums with parallel sides a and b and height h.
2. Place the second trapezium upside down next to the first one so that their non-parallel sides align.
3. The combined shape forms a parallelogram because opposite sides are equal and parallel.
4. The base of this parallelogram is the sum of the parallel sides of the trapezium, which is a + b.
5. The height of the parallelogram remains h, the same as the height of the trapezium.
6. The area of the parallelogram is calculated as base × height, which is (a + b) × h.
7. Since the parallelogram is made of two identical trapeziums, the area of one trapezium is half of the parallelogram's area.
8. Therefore, the area of the trapezium is 1/2 × (a + b) × h.
Answer: Area of trapezium = 1/2 × (sum of parallel sides) × height
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Exercises Q14
Show that the area of a kite is half the product of its diagonals, (i) by algebra, (ii) by geometry.
1. Let the kite be ABCD with AB = AD and CB = CD. Its diagonals AC and BD cross at O at right angles (AC is the line of symmetry).
2. (i) The kite is made of four right triangles AOB, BOC, COD, DOA with legs along the diagonals.
3. (i) Area = ½·AO·BO + ½·CO·BO + ½·CO·DO + ½·AO·DO = ½·BO·(AO + CO) + ½·DO·(AO + CO) = ½·(AO + CO)·(BO + DO).
4. (i) Since AO + CO = AC and BO + DO = BD, area = ½ × AC × BD.
5. (ii) Draw the rectangle around the kite whose sides pass through A, B, C, D parallel to the diagonals; its sides are AC and BD, so its area is AC × BD.
6. (ii) The diagonals cut this rectangle into four small rectangles, and each side of the kite is a diagonal of one of them, cutting it into two equal halves.
7. (ii) The kite contains exactly one half of each small rectangle, so its area is ½ × AC × BD.
Answer: Area of a kite = ½ × AC × BD, half the product of its diagonals
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Exercises Q15
(i) Rectangle with sides 2a, 2b vs rectangle with sides a, b; (ii) triangle with sides 2a, 2b, 2c vs triangle with sides a, b, c; (iii) triangle with sides 3a, 3b, 3c vs a, b, c. Show the areas are 4, 4 and 9 times, and check whether 4, 4 and 9 copies of the small shape fit into the big one.
1. (i) Area(ABCD) = ab and area(PQRS) = 2a × 2b = 4ab, which is 4 times as much.
2. (i) Yes: the lines joining the midpoints of opposite sides of PQRS cut it into four rectangles, each a × b, so 4 copies of ABCD fill PQRS exactly.
3. (ii) For △ABC, with s = (a + b + c)/2, Heron's formula gives area = √(s(s − a)(s − b)(s − c)).
4. (ii) For △PQR every side doubles, so its semi-perimeter is 2s and each of s − a, s − b, s − c doubles as well.
5. (ii) Its area = √(2s × 2(s − a) × 2(s − b) × 2(s − c)) = √16 × √(s(s − a)(s − b)(s − c)) = 4 × area(△ABC).
6. (ii) Yes, 4 copies fit: join the midpoints of the sides of △PQR. By the midpoint theorem each joining segment is half the side it is parallel to, so the four small triangles all have sides a, b, c and are congruent to △ABC.
7. (iii) With sides 3a, 3b, 3c the semi-perimeter is 3s and each factor triples, so the area = √81 × √(s(s − a)(s − b)(s − c)) = 9 × area(△ABC).
8. (iii) Yes, 9 copies fit: divide each side of △PQR into three equal parts and draw lines through these points parallel to the sides. This makes 9 small triangles (6 pointing one way, 3 the other way), each with sides a, b, c.
Answer: (i) 4 times, and 4 copies fit; (ii) 4 times, and 4 copies fit; (iii) 9 times, and 9 copies fit
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Exercises Q16
What fraction is shaded? (i) Fig. 6.43: in a triangle, the shaded 4-gon has corners at one vertex, the midpoint of the side from that vertex, and the two trisection points of the side opposite it; (ii) Fig. 6.44: in a square, the small square formed by the four lines from each corner to the midpoint of a side.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. (i) Call the triangle PTR, with P the vertex at the shaded corner, M the midpoint of PT, and X, Y the points dividing TR into three equal parts (X nearer T).
2. (i) The shaded 4-gon PMXY is triangle PTY with triangle TMX removed.
3. (i) Triangles PTY and PTR have the same height from P, and TY = (2/3)TR, so area(△PTY) = (2/3) area(△PTR).
4. (i) TX = (1/3)TR, so area(△TPX) = (1/3) area(△PTR); M is the midpoint of TP, so area(△TMX) = ½ area(△TPX) = (1/6) area(△PTR).
5. (i) Shaded = 2/3 − 1/6 = 4/6 − 1/6 = 3/6 = 1/2 of the triangle.
6. (ii) Take the square with corners (0, 0), (2, 0), (2, 2), (0, 2). The four lines are y = x/2 and y = 1 + x/2 (slope ½), and y = 2 − 2x and y = 4 − 2x (slope −2).
7. (ii) The slopes ½ and −2 multiply to −1, so lines from different pairs are perpendicular, and each pair is parallel: the shaded shape is a rectangle.
8. (ii) Its corners: y = x/2 meets y = 2 − 2x at (0.8, 0.4) and y = 4 − 2x at (1.6, 0.8); y = 1 + x/2 meets y = 4 − 2x at (1.2, 1.6) and y = 2 − 2x at (0.4, 1.2).
9. (ii) Sides: from (0.8, 0.4) to (1.6, 0.8) is √(0.8² + 0.4²) = √0.8, and from (1.6, 0.8) to (1.2, 1.6) is √(0.4² + 0.8²) = √0.8. So it is a square of area 0.8.
10. (ii) Fraction = 0.8 ÷ 4 = 1/5.
Answer: (i) 1/2; (ii) 1/5
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Exercises Q17
What fraction of the rectangle is covered by the circles? Fig. 6.45: 3 equal circles in a row, touching each other and the rectangle; Fig. 6.46: the same with 4 circles.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Let each circle have radius r. The circles touch each other and the sides of the rectangle, so the rectangle is 2r tall and as long as the diameters placed in a row.
2. Fig. 6.45: 3 circles; rectangle 6r × 2r = 12r²; circles 3 × πr² = 3πr².
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
3. Fraction = 3πr² ÷ 12r² = π/4.
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
4. Fig. 6.46: 4 circles; rectangle 8r × 2r = 16r²; circles 4πr².
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
5. Fraction = 4πr² ÷ 16r² = π/4.
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
6. With π ≈ 22/7, π/4 ≈ 22/28 = 11/14, about 0.786.
Answer: In both figures the circles cover π/4 ≈ 11/14 of the rectangle
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Exercises Q18
Conjecture the fraction of a rectangle covered by n equal circles fitted in a row in the same way; test for 10, 20 and 50 circles, then prove it.
1. Let the diameter of each circle be d, so the radius is r = d/2.
2. The area of one circle is πr² = π(d/2)² = πd²/4.
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
3. The area of the rectangle containing n circles in a row is length × width = (nd) × d = nd².
4. The total area of the n circles is n × (πd²/4) = nπd²/4.
5. The fraction of the rectangle covered is (nπd²/4) / (nd²) = π/4.
6. For 10 circles, the fraction is π/4 ≈ 3.14159/4 ≈ 0.7854.
7. For 20 circles, the fraction is π/4 ≈ 0.7854.
8. For 50 circles, the fraction is π/4 ≈ 0.7854.
9. The fraction is independent of n and equals π/4.
10. The value 11/14 is approximately 0.7857, which is a close rational approximation to π/4.
Answer: The fraction is π/4; 11/14
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Exercises Q19
Fig. 6.47: nine identical rectangles (4 lying in the top row, 5 standing in the bottom row) form a large rectangle of area 72 cm². Find the perimeter of each small rectangle.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Let the length of each small rectangle be l cm and its breadth be b cm.
2. From the top row, the length of the large rectangle is 4l, and from the bottom row, it is 5b.
3. Equating the lengths gives 4l = 5b, so l = (5/4)b.
4. The width of the large rectangle is the sum of the vertical dimensions: l + b.
5. The area of the large rectangle is 4l × (l + b) = 72.
6. Substitute l = (5/4)b into the area equation: 4 × (5/4)b × ((5/4)b + b) = 72.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
7. Simplify to 5b × (9/4)b = 72, which gives (45/4)b² = 72.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
8. Solving for b² gives b² = 72 × 4 / 45 = 32/5.
9. Thus b = √(32/5) = 4√10 / 5 cm.
10. Calculate l = (5/4)b = (5/4) × (4√10 / 5) = √10 cm.
11. The perimeter of each small rectangle is 2(l + b) = 2(√10 + 4√10 / 5).
12. Simplify the perimeter: 2(9√10 / 5) = 18√10 / 5 cm.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
13. Approximate the value: 18 × 3.162 / 5 ≈ 11.38 cm.
Answer: 18√10/5 cm, 11.38
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Exercises Q20
Fig. 6.48: lines from the top vertex of a triangle to the two trisection points of the base. Show the blue triangle (left third) and the red triangle (right third) have equal areas, and find a way to cut the blue one into pieces that cover the red one.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Let the triangle be ABC with apex A and base BC, and let D and E divide BC into three equal parts, BD = DE = EC. Blue = △ABD, red = △AEC.
2. Both triangles have a base of length BC/3 on line BC and the same apex A, so both have the same height h, the distance of A from BC.
3. Area(△ABD) = ½ × BD × h = ½ × EC × h = area(△AEC).
4. One way to cut: let M and N be the midpoints of AB and AD. Cut the blue triangle along MN and turn the small triangle AMN through 180° about N, so that NA falls on ND. The blue triangle becomes a parallelogram with base BD and height h/2.
  why?
  The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.
5. Cut off a triangle at one slanted end of this parallelogram and slide it to the other end (more than one such cut if the slant is large); it becomes a rectangle of base BC/3 and height h/2.
6. The red triangle turns into the same rectangle by the same steps (cut along the line joining the midpoints of AE and AC, then straighten).
7. So the pieces of the blue triangle, rearranged into that rectangle, can be put together by reversing the red triangle's steps to cover the red triangle exactly. Other dissections also work.
Answer: The blue and red triangles each have area ½ × (BC/3) × h, so they are equal; the blue triangle can be cut into a rectangle of base BC/3 and height h/2, which rebuilds the red triangle
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Exercises Q21
Fig. 6.49: a quarter circle in a square, centred at a vertex and passing through the two adjacent vertices, and semicircles on the two sides through that vertex. Show the shaded regions A (common to both semicircles) and B (in the quarter circle, outside both semicircles) have equal area.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Let the square have side s. The quarter circle is centred at a corner O with radius s; the two semicircles stand on the two sides through O, each with radius s/2.
2. Area of the quarter circle = ¼ × π × s² = πs²/4.
3. Area of each semicircle = ½ × π × (s/2)² = πs²/8, so the two together have area πs²/4.
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
4. No point of a semicircle on a side of length s through O is farther than s from O, so both semicircles lie inside the quarter circle.
5. The two semicircles overlap exactly in region A, so together they cover an area of πs²/8 + πs²/8 − A = πs²/4 − A.
6. Region B is the part of the quarter circle outside both semicircles: B = πs²/4 − (πs²/4 − A) = A.
Answer: area(A) = area(B)
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Exercises Q22
Fig. 6.50: a square of side 2 units with four semicircles inside on its sides, forming a 4-petal flower. Find the flower's perimeter and area.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. The square has side 2, so each semicircle has radius 1, centred at the midpoint of a side.
2. Each petal runs from a corner of the square to its centre. Its two edges are arcs of the two semicircles on the sides that meet at that corner.
3. From the centre of such a semicircle (a side's midpoint), the corner and the centre of the square are both 1 unit away and in perpendicular directions, so each petal edge is a quarter circle of radius 1.
4. Perimeter: 4 petals × 2 edges = 8 quarter arcs = 8 × ¼ × 2π × 1 = 4π units.
5. With π ≈ 22/7, 4π ≈ 88/7 ≈ 12.57 units.
6. Half a petal is a quarter circle of radius 1 minus the right triangle with legs 1 and 1: π/4 − 1/2.
7. One petal = 2 × (π/4 − 1/2) = π/2 − 1, so the four petals have area 2π − 4 square units.
8. With π ≈ 22/7, 2π − 4 ≈ 44/7 − 28/7 = 16/7 ≈ 2.29 square units.
Answer: Perimeter 4π ≈ 88/7 ≈ 12.57 units; area 2π − 4 ≈ 16/7 ≈ 2.29 square units
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Exercises Q23
Fig. 6.51: two concentric circles, centre O; chord BC of the larger circle, of length l, touches the smaller circle at A. Show the ring between the circles has area ¼πl².
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Let r be the radius of the smaller circle and R be the radius of the larger circle.
2. Since chord BC touches the smaller circle at A, OA is perpendicular to BC.
3. In a circle, a perpendicular from the centre to a chord bisects the chord, so AB = BC / 2 = l / 2.
  why?
  The perpendicular from the centre of a circle to a chord bisects the chord (the two right triangles it makes are congruent by RHS).
4. In right-angled triangle OAB, the hypotenuse is OB = R and the sides are OA = r and AB = l / 2.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
5. By the Pythagoras theorem, R² = r² + (l / 2)².
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
6. Expanding the square, R² = r² + l² / 4.
  why?
  Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.
7. Rearranging the terms gives R² − r² = l² / 4.
8. The area of the ring between the two circles is the area of the larger circle minus the area of the smaller circle.
9. Area of ring = πR² − πr² = π(R² − r²).
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
10. Substituting the value from the previous step, Area = π(l² / 4) = ¼πl².
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
Answer: The area of the ring is ¼πl².
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Exercises Q24
Fig. 6.52: semicircles on all sides of a right-angled triangle, the one on the hypotenuse drawn over the triangle. Show area(A) + area(B) = area(C), where A and B are the crescents on the two shorter sides and C is the triangle.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Let the sides of the right-angled triangle be a, b, and c, where c is the hypotenuse.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
2. The area of a semicircle with diameter x is (1/2) × π × (x/2)² = πx²/8.
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
3. Area of semicircle on side a is πa²/8, on side b is πb²/8, and on hypotenuse c is πc²/8.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
4. The circle with the hypotenuse as diameter passes through the right-angle vertex, so the semicircle on c drawn over the triangle covers the triangle C and two segments: one cut off by side a and one cut off by side b.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
5. So πc²/8 = C + (segment on a) + (segment on b).
6. Crescent A = semicircle on a minus the segment on a, and crescent B = semicircle on b minus the segment on b.
7. A + B = πa²/8 + πb²/8 − (segment on a + segment on b) = π(a² + b²)/8 − (πc²/8 − C).
8. By Pythagoras a² + b² = c², so A + B = πc²/8 − πc²/8 + C = C.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
Answer: area(A) + area(B) = area(C)
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Exercises Q25
Fig. 6.53: two circles of radius r, each passing through the other's centre. Find the area of the region common to both circles.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Let the two circles have centres O and O' and radius r.
2. Since each circle passes through the other's centre, the distance OO' is equal to r.
3. Let the circles intersect at points A and B.
4. Join OA, O'A, OB, and O'B.
5. In triangle OAO', OA = O'A = OO' = r, so it is an equilateral triangle.
6. Therefore, angle AOO' = 60°.
7. Similarly, triangle OBO' is equilateral, so angle BOO' = 60°.
8. The central angle subtended by the common chord AB at centre O is angle AOB = 60° + 60° = 120°.
9. The area of the sector OAB is (120/360) × πr² = (1/3)πr².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
10. The area of triangle OAB is (1/2)r² sin(120°) = (1/2)r²(√3/2) = (√3/4)r².
  why?
  In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
11. The area of the segment of the first circle cut by chord AB is Area(Sector OAB) − Area(△OAB) = (π/3 − √3/4)r².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
12. The common region is the sum of two such identical segments (one from each circle).
13. Total common area = 2 × (π/3 − √3/4)r² = (2π/3 − √3/2)r².
14. Using π ≈ 3.14, the coefficient is (2 × 3.14 / 3) − (1.732 / 2) ≈ 2.093 − 0.866 = 1.227 ≈ 1.23.
Answer: (2π/3 − √3/2)r², which is approximately 1.23 r²
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Exercises Q26
Fig. 6.54: in a rectangle, triangles of areas A, B, C as marked (C has a right angle at the inner point). Show the rectangle's area is 2(A + C)(B + C)/C.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Let the corner shared by triangles A and B be O = (0, 0), and let the rectangle be w wide and h tall, so its area is wh.
2. As in the figure, the top vertex is T = (x, h) on the top side, R = (w, y) is on the right side, and the inner point Q = (x, y) is directly below T and level with R.
3. C = △TQR has a right angle at Q with legs TQ = h − y and QR = w − x, so C = ½(h − y)(w − x).
4. A = △OTQ has base TQ = h − y on the vertical line through T, and O is x away from that line, so A = ½(h − y)x.
5. B = △OQR has base QR = w − x on the horizontal line through R, and O is y below that line, so B = ½(w − x)y.
6. A + C = ½(h − y)(x + w − x) = ½(h − y)w, and B + C = ½(w − x)(y + h − y) = ½(w − x)h.
7. 2(A + C)(B + C)/C = 2 × ½(h − y)w × ½(w − x)h ÷ [½(h − y)(w − x)] = wh, the area of the rectangle.
Answer: Area of the rectangle = 2(A + C)(B + C)/C
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Exercises Q27
Fig. 6.55: quarter circle AFB with centre O, semicircle AEB on AB as diameter, triangle AOB. Show the shaded crescent AEBF and the shaded triangle AOB have equal areas.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Let OA = OB = r. The angle AOB is 90°, so AB = √(r² + r²) = r√2.
2. Triangle AOB has area ½ × r × r = r²/2.
3. The semicircle AEB has radius AB/2 = r√2/2, so its area is ½ × π × (r√2/2)² = ½ × π × r²/2 = πr²/4.
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
4. The quarter circle AFB (centre O) has area πr²/4. Its part between chord AB and arc AFB is a segment of area πr²/4 − r²/2.
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
5. The crescent AEBF is the half-disc on AB minus this segment: πr²/4 − (πr²/4 − r²/2) = r²/2.
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
6. So the crescent AEBF and triangle AOB both have area r²/2.
Answer: area(crescent AEBF) = area(△AOB) = r²/2
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Other exercises in this chapter

Exercise Set 6.1 — 8 questions
Exercise Set 6.2 — 11 questions
Exercise Set 6.3 — 10 questions

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NCERT Class 9 Mathematics Measuring Space: Perimeter and Area: Exercise Solutions

Exercises Q1

Exercises Q2

Exercises Q3

Exercises Q4

Exercises Q5

Exercises Q6

Exercises Q7

Exercises Q8

Exercises Q9

Exercises Q10

Exercises Q11

Exercises Q12

Exercises Q13

Exercises Q14

Exercises Q15

Exercises Q16

Exercises Q17

Exercises Q18

Exercises Q19

Exercises Q20

Exercises Q21

Exercises Q22

Exercises Q23

Exercises Q24

Exercises Q25

Exercises Q26

Exercises Q27