Current Affairs updated daily
KalviGavaniKalviGavani
Class 7 · Tamil Nadu State Board

Samacheer Kalvi 7th Maths (Term 1) Measurements Exercise Solutions

Complete TN Class 7 Maths (Term 1) Measurements exercise solutions with step-by-step workings, all exercise answers, and exam tips.

Updated 12 June 202624 min read

How to Use This Page

These are complete solutions for Samacheer Kalvi 7th Maths (Term 1) Chapter 2 — Measurements. For every problem, show all steps: Given → Formula → Substitution → Calculation → Answer with units. Solve each exercise on your own first, then check the method below.

Key Concepts (Quick Recap)

  • Parallelogram: Area = base × height; Perimeter = 2 × (sum of adjacent sides).
  • Rhombus: Area = ½ × d₁ × d₂ (using diagonals) or Area = base × height. Diagonals bisect each other at 90°.
  • Trapezium: Area = ½ × h × (a + b), where a and b are the parallel sides and h is the perpendicular distance between them.
  • Circle: Circumference = 2πr = πd; Area = πr². Use π ≈ 22/7 or 3.14 as given.
  • Perimeter: Sum of all sides of a closed figure.
  • Units: Always write the correct unit (cm², m², mm²) for area and (cm, m, mm) for length/perimeter.

Exercise Solutions

Exercise 2.1 — Parallelogram

Q1. Find the area and perimeter of the following parallelograms.
Answer

(i) Given base b = 11 cm; height h = 3 cm

Area = b × h = 11 × 3 = 33 cm² Perimeter = Sum of all 4 sides = 11 + 4 + 11 + 4 = 30 cm

Area = 33 cm², Perimeter = 30 cm

(ii) Given base b = 7 cm; height h = 10 cm

Area = b × h = 7 × 10 = 70 cm² Perimeter = Sum of four sides = 13 + 7 + 13 + 7 = 40 cm

Area = 70 cm², Perimeter = 40 cm

Q2. Find the missing values.
Answer

(i) Base b = 18 cm; Height h = 5 cm — find Area

Area = b × h = 18 × 5 = 90 cm²

Area = 90 cm²

(ii) Base b = 8 m; Area = 56 sq. m — find Height

b × h = 56 8 × h = 56 h = 56 / 8 h = 7 m

Height = 7 m

(iii) Height h = 17 mm; Area = 221 sq. mm — find Base

b × h = 221 b × 17 = 221 b = 221 / 17 b = 13 mm

Base = 13 mm

Q3. Suresh won a parallelogram shaped trophy. Area = 735 sq. cm, base = 21 cm. Find the height.
Given: Area = 735 cm², base = 21 cm b × h = 735 21 × h = 735 h = 735 / 21 h = 35 cm
Answer

Height = 35 cm

Q4. Janaki has a parallelogram fabric with height 12 m and base 18 m. She cuts it into four equal parallelograms by cutting through mid-points. Find the area of each new parallelogram.
Given: base = 18 m, height = 12 m When cut through mid-points: New base = 18 / 2 = 9 m New height = 12 / 2 = 6 m Area of each new parallelogram = new base × new height = 9 × 6 = 54 m²
Answer

Area of each piece = 54 m²

Q5. A ground is in the shape of a parallelogram. Height = 14 m, base is 8 m longer than its height. Find the cost of levelling at ₹15 per sq. m.
Given: height = 14 m base = height + 8 = 14 + 8 = 22 m Area = base × height = 22 × 14 = 308 m² Cost = Area × Rate = 308 × 15 = 4620
Answer

Cost of levelling = ₹4,620

Q6 (MCQ). The perimeter of a parallelogram whose adjacent sides are 6 cm and 5 cm is:
Answer

(i) 12 cm (ii) 10 cm (iii) 24 cm (iv) 22 cm

Perimeter = 2 × (side₁ + side₂) = 2 × (6 + 5) = 2 × 11 = 22 cm
Answer(iv) 22 cm
Q7 (MCQ). The area of a parallelogram with base 10 m and height 7 m is:
Answer

(i) 70 sq. m (ii) 35 sq. m (iii) 7 sq. m (iv) 10 sq. m

Area = base × height = 10 × 7 = 70 sq. m
Answer(i) 70 sq. m
Q8 (MCQ). The base of a parallelogram with area 52 sq. cm and height 4 cm is:
Answer

(i) 48 cm (ii) 104 cm (iii) 13 cm (iv) 26 cm

b × h = 52 b × 4 = 52 b = 52 / 4 b = 13 cm
Answer(iii) 13 cm
Q9 (MCQ). What happens to the area if the base is increased 2 times and the height is halved?
Answer

(i) Decreases to half (ii) Remains the same (iii) Increases two times (iv) None

Original Area = b × h New Area = (2b) × (h/2) = 2b × h/2 = b × h = Original Area
Answer(ii) Remains the same
Q10 (MCQ). In a parallelogram, the base is three times its height. If height = 8 cm, find the area.
Answer

(i) 64 sq. cm (ii) 192 sq. cm (iii) 32 sq. cm (iv) 72 sq. cm

Given: h = 8 cm, b = 3h = 3 × 8 = 24 cm Area = b × h = 24 × 8 = 192 cm²
Answer(ii) 192 sq. cm

Exercise 2.2 — Rhombus

Q1. Find the area of rhombus PQRS shown in the following figures.
Answer

(i) Diagonals given: d₁ = 16 cm, d₂ = 8 cm

Area of rhombus = ½ × d₁ × d₂ = ½ × 16 × 8 = ½ × 128 = 64 cm²

Area = 64 cm²

(ii) Base b = 15 cm, Height h = 11 cm

Area of rhombus = base × height = 15 × 11 = 165 cm²

Area = 165 cm²

Q2. Find the area of a rhombus whose base is 14 cm and height is 9 cm.
Given: base = 14 cm, height = 9 cm Area = base × height = 14 × 9 = 126 cm²
Answer

Area = 126 cm²

Q3. Find the missing values.
Answer

(i) d₁ = 19 cm, d₂ = 16 cm — find Area

Area = ½ × d₁ × d₂ = ½ × 19 × 16 = ½ × 304 = 152 cm²

Area = 152 cm²

(ii) d₁ = 26 m, Area = 468 sq. m — find d₂

½ × d₁ × d₂ = 468 ½ × 26 × d₂ = 468 26 × d₂ = 468 × 2 26 × d₂ = 936 d₂ = 936 / 26 d₂ = 36 m

d₂ = 36 m

(iii) d₂ = 12 mm, Area = 180 sq. mm — find d₁

½ × d₁ × d₂ = 180 ½ × d₁ × 12 = 180 d₁ × 12 = 180 × 2 d₁ × 12 = 360 d₁ = 360 / 12 d₁ = 30 mm

d₁ = 30 mm

Q4. The area of a rhombus is 100 sq. cm and the length of one of its diagonals is 8 cm. Find the length of the other diagonal.
Given: Area = 100 cm², d₁ = 8 cm ½ × d₁ × d₂ = 100 ½ × 8 × d₂ = 100 4 × d₂ = 100 d₂ = 100 / 4 d₂ = 25 cm
Answer

Other diagonal = 25 cm

Q5. A sweet is shaped like a rhombus with diagonals 4 cm and 5 cm. The surface should be covered by aluminium foil. Find the cost of foil for 400 such sweets at ₹7 per 100 sq. cm.
Given: d₁ = 4 cm, d₂ = 5 cm Step 1: Area of one sweet Area = ½ × d₁ × d₂ = ½ × 4 × 5 = 10 cm² Step 2: Total area for 400 sweets Total area = 400 × 10 = 4000 cm² Step 3: Cost of foil Rate = ₹7 per 100 cm² Cost = (4000 / 100) × 7 = 40 × 7 = 280
Answer

Total cost = ₹280

Q6 (MCQ). The area of the rhombus with side 4 cm and height 3 cm is:
Answer

(i) 7 sq. cm (ii) 24 sq. cm (iii) 12 sq. cm (iv) 10 sq. cm

Area = base × height = 4 × 3 = 12 cm²
Answer(iii) 12 sq. cm
Q7 (MCQ). The area of the rhombus when both diagonals measure 8 cm is:
Answer

(i) 64 sq. cm (ii) 32 sq. cm (iii) 30 sq. cm (iv) 16 sq. cm

Area = ½ × d₁ × d₂ = ½ × 8 × 8 = 32 cm²
Answer(ii) 32 sq. cm
Q8 (MCQ). The area of the rhombus is 128 sq. cm and the length of one diagonal is 32 cm. Find the other diagonal.
Answer

(i) 12 cm (ii) 8 cm (iii) 4 cm (iv) 20 cm

½ × d₁ × d₂ = 128 ½ × 32 × d₂ = 128 16 × d₂ = 128 d₂ = 128 / 16 d₂ = 8 cm
Answer(ii) 8 cm
Q9 (MCQ). The height of the rhombus whose area is 96 sq. m and side is 24 m is:
Answer

(i) 8 m (ii) 10 m (iii) 2 m (iv) 4 m

Area = base × height 96 = 24 × h h = 96 / 24 h = 4 m
Answer(iv) 4 m
Q10 (MCQ). The angle between the diagonals of a rhombus is:
Answer

(i) 120° (ii) 180° (iii) 90° (iv) 100°

The diagonals of a rhombus bisect each other at right angles, so the angle between them is 90°.

Answer(iii) 90°

Exercise 2.3 — Trapezium

Q1. Find the missing values in the trapezium table.
Answer

(i) h = 10 m, a = 12 m, b = 20 m — find Area

Area = ½ × h × (a + b) = ½ × 10 × (12 + 20) = ½ × 10 × 32 = 5 × 32 = 160 m²

Area = 160 m²

(ii) a = 13 cm, b = 28 cm, Area = 492 cm² — find h

½ × h × (a + b) = 492 ½ × h × (13 + 28) = 492 ½ × h × 41 = 492 h × 41 = 492 × 2 h × 41 = 984 h = 984 / 41 h = 24 cm

h = 24 cm

(iii) h = 19 m, b = 16 m, Area = 323 m² — find a

½ × h × (a + b) = 323 ½ × 19 × (a + 16) = 323 ½ × 19 × (a + 16) = 323 a + 16 = (323 × 2) / 19 a + 16 = 646 / 19 a + 16 = 34 a = 34 - 16 a = 18 m

a = 18 m

(iv) h = 16 cm, a = 15 cm, Area = 360 cm² — find b

½ × h × (a + b) = 360 ½ × 16 × (15 + b) = 360 8 × (15 + b) = 360 15 + b = 360 / 8 15 + b = 45 b = 45 - 15 b = 30 cm

b = 30 cm

Q2. Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.
Given: a = 24 cm, b = 20 cm, h = 15 cm Area = ½ × h × (a + b) = ½ × 15 × (24 + 20) = ½ × 15 × 44 = 330 cm²
Answer

Area = 330 cm²

Q3. The area of a trapezium is 1586 sq. cm. The distance between its parallel sides is 26 cm. If one of the parallel sides is 84 cm, find the other side.
Given: Area = 1586 cm², h = 26 cm, a = 84 cm ½ × h × (a + b) = 1586 ½ × 26 × (84 + b) = 1586 13 × (84 + b) = 1586 84 + b = 1586 / 13 84 + b = 122 b = 122 - 84 b = 38 cm
Answer

Other side = 38 cm

Q4. The area of a trapezium is 1080 sq. cm. If the lengths of its parallel sides are 55.6 cm and 34.4 cm, find the distance between them.
Given: Area = 1080 cm², a = 55.6 cm, b = 34.4 cm ½ × h × (a + b) = 1080 ½ × h × (55.6 + 34.4) = 1080 ½ × h × 90 = 1080 45h = 1080 h = 1080 / 45 h = 24 cm
Answer

Height = 24 cm

Q5. The area of a trapezium is 180 sq. cm and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the length of the parallel sides.
Given: Area = 180 cm², h = 9 cm Let one side = a cm, then other side b = a + 6 cm ½ × h × (a + b) = 180 ½ × 9 × (a + a + 6) = 180 9/2 × (2a + 6) = 180 9 × (2a + 6) = 360 2a + 6 = 360 / 9 2a + 6 = 40 2a = 40 - 6 2a = 34 a = 17 cm b = 17 + 6 = 23 cm
Answer

Parallel sides are 17 cm and 23 cm

Q6. The sunshade of a window is in the form of isosceles trapezium whose parallel sides are 81 cm and 64 cm and the distance between them is 6 cm. Find the cost of painting the surface at ₹2 per sq. cm.
Given: a = 81 cm, b = 64 cm, h = 6 cm Step 1: Area of trapezium Area = ½ × h × (a + b) = ½ × 6 × (81 + 64) = 3 × 145 = 435 cm² Step 2: Cost of painting Cost = Area × Rate = 435 × 2 = 870
Answer

Cost of painting = ₹870

Q7. A window is in the form of a trapezium whose parallel sides are 105 cm and 50 cm and the distance between the parallel sides is 60 cm. Find the cost of the glass used to cover the window at ₹15 per 100 sq. cm.
Given: a = 105 cm, b = 50 cm, h = 60 cm Step 1: Area of trapezium Area = ½ × h × (a + b) = ½ × 60 × (105 + 50) = 30 × 155 = 4650 cm² Step 2: Cost of glass Rate = ₹15 per 100 cm² Cost = (4650 / 100) × 15 = 46.5 × 15 = 697.50
Answer

Cost of glass = ₹697.50

Q8 (MCQ). The area of the trapezium, if the parallel sides are 8 cm and 10 cm and the height is 5 cm, is:
Answer

(i) 45 sq. cm (ii) 40 sq. cm (iii) 18 sq. cm (iv) 50 sq. cm

Area = ½ × h × (a + b) = ½ × 5 × (10 + 8) = ½ × 5 × 18 = 45 cm²
Answer(i) 45 sq. cm
Q9 (MCQ). In a trapezium, if the sum of the parallel sides is 10 m and the area is 140 sq. m, then the height is:
Answer

(i) 7 cm (ii) 40 cm (iii) 14 cm (iv) 28 cm

½ × h × (a + b) = 140 ½ × h × 10 = 140 5h = 140 h = 140 / 5 h = 28 m
Answer(iv) 28 cm (Note: The question states height in cm though the area is in sq. m — the calculated value is 28.)
Q10 (MCQ). When the non-parallel sides of a trapezium are equal, it is known as:
Answer

(i) a square (ii) a rectangle (iii) an isosceles trapezium (iv) a parallelogram

Answer(iii) an isosceles trapezium

Exercise 2.4 — Miscellaneous and Challenge Problems

Q1. The base of the parallelogram is 16 cm and the height is 7 cm less than its base. Find the area of the parallelogram.
Given: base = 16 cm height = base - 7 = 16 - 7 = 9 cm Area = base × height = 16 × 9 = 144 cm²
Answer

Area = 144 cm²

Q2. An agricultural field is in the form of a parallelogram, whose area is 68.75 sq. hm. The distance between the parallel sides is 6.25 hm. Find the length of the base.
Given: Area = 68.75 sq. hm, height = 6.25 hm b × h = 68.75 b × 6.25 = 68.75 b = 68.75 / 6.25 b = 6875 / 625 b = 11 hm
Answer

Length of base = 11 hm (i.e., 1.1 km)

Q3. A square and a parallelogram have the same area. If the side of the square is 48 m and the height of the parallelogram is 18 m, find the length of the base of the parallelogram.
Given: Side of square = 48 m, height of parallelogram = 18 m Step 1: Area of square Area of square = 48 × 48 = 2304 m² Step 2: Area of parallelogram = Area of square b × h = 2304 b × 18 = 2304 b = 2304 / 18 b = 128 m
Answer

Base of parallelogram = 128 m

Q4. The height of the parallelogram is one-fourth of its base. If the area of the parallelogram is 676 sq. cm, find the height and the base.
Given: height = ¼ × base, Area = 676 cm² Let base = b cm Then height = b/4 cm b × h = 676 b × (b/4) = 676 b²/4 = 676 b² = 676 × 4 b² = 2704 b = √2704 = 52 cm height = b/4 = 52/4 = 13 cm
Answer

Base = 52 cm, Height = 13 cm

Q5. The area of a rhombus is 576 sq. cm and the length of one diagonal is half of the length of the other diagonal. Find the lengths of the diagonals.
Given: Area = 576 cm², d₂ = ½ × d₁ ½ × d₁ × d₂ = 576 ½ × d₁ × (d₁/2) = 576 d₁²/4 = 576 d₁² = 576 × 4 d₁² = 2304 d₁ = √2304 = 48 cm d₂ = 48/2 = 24 cm
Answer

Diagonals: d₁ = 48 cm, d₂ = 24 cm

Q6. A ground is in the form of an isosceles trapezium with parallel sides measuring 42 m and 36 m. The distance between the parallel sides is 30 m. Find the cost of levelling at ₹135 per sq. m.
Given: a = 42 m, b = 36 m, h = 30 m Step 1: Area of trapezium Area = ½ × h × (a + b) = ½ × 30 × (42 + 36) = ½ × 30 × 78 = 15 × 78 = 1170 m² Step 2: Cost of levelling Cost = Area × Rate = 1170 × 135 = 1,57,950
Answer

Cost of levelling = ₹1,57,950

Q7. In a parallelogram PQRS, PM and PN are the heights corresponding to the sides QR and RS respectively. If the area of the parallelogram is 900 sq. cm and the lengths of PM and PN are 20 cm and 36 cm respectively, find the length of the sides QR and SR.
Given: Area = 900 cm², PM = 20 cm, PN = 36 cm Considering QR as base with height PM: QR × PM = 900 QR × 20 = 900 QR = 900 / 20 QR = 45 cm Considering SR as base with height PN: SR × PN = 900 SR × 36 = 900 SR = 900 / 36 SR = 25 cm
Answer

QR = 45 cm, SR = 25 cm

Q8. If the base and height of a parallelogram are in the ratio 7:3 and the height is 45 cm, find the area of the parallelogram.
Given: base : height = 7 : 3, height = 45 cm Let base = 7x cm, height = 3x cm 3x = 45 x = 15 Base = 7 × 15 = 105 cm Height = 45 cm (given) Area = base × height = 105 × 45 = 4725 cm²
Answer

Area = 4725 cm²

Q9. Find the area of the parallelogram ABCD if AC is 24 cm and BE = DF = 8 cm.
Given: Diagonal AC = 24 cm, perpendicular heights BE = 8 cm, DF = 8 cm A parallelogram is divided by diagonal AC into two equal triangles. Area of △ABC = ½ × AC × BE = ½ × 24 × 8 = 96 cm² Area of △ADC = ½ × AC × DF = ½ × 24 × 8 = 96 cm² Area of parallelogram ABCD = Area of △ABC + Area of △ADC = 96 + 96 = 192 cm²
Answer

Area = 192 cm²

Q10. The area of the parallelogram ABCD is 1470 sq. cm. If AB = 49 cm and AD = 35 cm, find the heights DF and BE.
Given: Area = 1470 cm², AB = 49 cm, AD = 35 cm Taking AB as base, height = DF: AB × DF = 1470 49 × DF = 1470 DF = 1470 / 49 DF = 30 cm Taking AD as base, height = BE: AD × BE = 1470 35 × BE = 1470 BE = 1470 / 35 BE = 42 cm
Answer

DF = 30 cm, BE = 42 cm

Q11. One of the diagonals of a rhombus is thrice the other. If the sum of the lengths of the diagonals is 24 cm, find the area of the rhombus.
Given: d₁ = 3 × d₂, d₁ + d₂ = 24 Substituting: 3d₂ + d₂ = 24 4d₂ = 24 d₂ = 6 cm d₁ = 3 × 6 = 18 cm Area = ½ × d₁ × d₂ = ½ × 18 × 6 = 54 cm²
Answer

Area = 54 cm²

Q12. A man has to build a rhombus shaped swimming pool with one diagonal 13 m and the other diagonal is twice the first. Find the area of the swimming pool and also find the cost of cementing the floor at ₹15 per sq. m.
Given: d₁ = 13 m, d₂ = 2 × 13 = 26 m Step 1: Area of rhombus Area = ½ × d₁ × d₂ = ½ × 13 × 26 = 169 m² Step 2: Cost of cementing Cost = Area × Rate = 169 × 15 = 2535
Answer

Area = 169 m², Cost of cementing = ₹2,535

Q13. Find the height of the parallelogram whose base is four times the height and whose area is 576 sq. cm.
Given: base = 4 × height, Area = 576 cm² Let height = h cm Then base = 4h cm Area = base × height 4h × h = 576 4h² = 576 h² = 576 / 4 h² = 144 h = √144 h = 12 cm Base = 4 × 12 = 48 cm
Answer

Height = 12 cm, Base = 48 cm

Q14. A table top is in the shape of a trapezium with parallel sides 200 cm and 150 cm and height 50 cm. Find the cost of the glass used to cover the table at ₹6 per 10 sq. cm.
Given: a = 200 cm, b = 150 cm, h = 50 cm Step 1: Area of trapezium Area = ½ × h × (a + b) = ½ × 50 × (200 + 150) = ½ × 50 × 350 = 25 × 350 = 8750 cm² Step 2: Cost of glass Rate = ₹6 per 10 cm² Cost = (8750 / 10) × 6 = 875 × 6 = 5250
Answer

Cost of glass = ₹5,250

Q15. Arivu has a land ABCD with measurements given. A portion ABED is used for cultivation (where E is the midpoint of DC). Find the cultivated area.
Given: ABED is a trapezium with: height h = 18 m One parallel side a = 24 m E is midpoint of DC, so other parallel side b = 24/2 = 12 m Area of trapezium ABED = ½ × h × (a + b) = ½ × 18 × (24 + 12) = 9 × 36 = 324 m²
Answer

Cultivated area = 324 m²

Common Mistakes

1. Confusing base with side in parallelogram: The height is the perpendicular distance, not the length of a slanted side. Always use the perpendicular height in Area = base × height.

2. Forgetting the ½ in rhombus and trapezium formulas: Area of rhombus = ½ × d₁ × d₂ (not d₁ × d₂). Area of trapezium = ½ × h × (a + b) (not h × (a + b)). Missing the ½ doubles your answer.

3. Mixing up diagonals with sides in a rhombus: When the problem gives diagonals, use ½ × d₁ × d₂. When it gives base and height, use base × height. Do not mix these formulas.

4. Wrong unit conversions in cost problems: If the rate is given per 100 sq. cm or per sq. m, divide the total area by the rate unit first, then multiply by the price. For example, ₹6 per 10 sq. cm means: Cost = (Area / 10) × 6.

5. Using diameter instead of radius in circle formulas: When the diameter is given, first convert: r = d/2. Circumference = 2πr (not 2πd) and Area = πr² (not πd²).

6. Incorrectly finding the missing dimension: When asked to find base or height, set up the equation: b × h = Area, then solve for the unknown. Do not divide the area by 2 — the ½ only appears in rhombus (diagonal formula) and trapezium formulas.

7. Adding parallel sides instead of multiplying in parallelogram area: Area of a parallelogram = base × height (multiply). Perimeter = 2 × (sum of adjacent sides) — do not confuse the two.

Exam Checklist

  • [ ] I can find the area and perimeter of a parallelogram using A = b × h and P = 2(a + b).
  • [ ] I can find the area of a rhombus using both formulas: A = ½ × d₁ × d₂ and A = base × height.
  • [ ] I can find the area of a trapezium using A = ½ × h × (a + b).
  • [ ] I can solve for a missing dimension (base, height, diagonal) by setting up an equation.
  • [ ] I can handle unit conversions and cost-based word problems correctly.
  • [ ] I remember that diagonals of a rhombus bisect each other at 90°.
  • [ ] I never forget the ½ factor in rhombus (diagonal formula) and trapezium formulas.
  • [ ] I can identify whether a problem needs the diagonal formula or the base × height formula for a rhombus.

Internal Links

  • [Class 7 Maths (Term 1) Measurements — Chapter Guide](/school/tamil-nadu-state-board/class/7/maths/term-1-measurements)
  • [All Class 7 Maths Solutions](/school/tamil-nadu-state-board/class/7/maths/solutions)
  • [Class 7 Maths Subject Guide](/school/tamil-nadu-state-board/class/7/maths)

Frequently Asked Questions

Area of a parallelogram = base × height. The height must be the perpendicular distance from the base to the opposite side.
← Browse all Class 7 Mathematics Exercise Solutions

Other question types — same chapter

Chapter Notes

Tamil Nadu Class 7 Maths: Measurements

Exercise Solutions

Samacheer Kalvi 7th Maths (Term 1) Direct and Inverse Proportion Exercise Solutions

Exercise Solutions

Samacheer Kalvi 7th Maths (Term 3) Percentage and Simple Interest Exercise Solutions