Moderate Level Worksheet

Part A: Subjective Questions - Very Short Answer (1 Mark Each)

Building on basic area concepts, let's explore more complex problems involving ratio, conversions, and reverse calculations.

These moderate-level questions will strengthen your understanding of area formulas and their applications.

1. Write the formula for the area of a trapezium.

Area = 12a+bh12a+bh (Note: where a, b are parallel sides and h is height)

Perfect! Area of trapezium = 1/2 × (sum of parallel sides) × height.

2. A rhombus has diagonals 12 cm and 9 cm. Find its area.

Answer: cm2

Excellent! Area = 1/2 × d₁ × d₂ = 12 × 12 × 9 = 54 cm2.

3. Find the base of a triangle whose area = 60 cm2 and height = 10 cm.

Answer: cm

Great! Using Area = 12 × b × h, so 60 = 12 × b × 10, therefore b = 12 cm.

4. What is the height of a parallelogram whose base = 20 cm and area = 160 cm2?

Answer: cm

Correct! Using Area = base × height, so 160 = 20 × h, therefore h = 8 cm.

5. Convert 4 m² to cm2.

Answer: cm2

Perfect! 1 m² = 10000 cm2, so 4 m² = 4 × 10000 = 40000 cm2.

Drag each problem type to its appropriate solution approach:

Part A: Section B – Short Answer Questions (2 Marks Each)

1. The length and breadth of a rectangle are in the ratio 3 : 2 and its area is 150 cm2. Find length and breadth.

Let length = and breadth =

Area = length × breadth = 3x × 2x =

Given area = 150 cm2, so 6x² =

x² = , therefore x =

Length = 3x = cm

Breadth = 2x = cm

Excellent! Length = 15 cm and Breadth = 10 cm.

2. The area of a parallelogram is 180 cm2. If its base is 15 cm, find the height.

Formula: Area = ×

Substituting: 180 = × height

Height = 180 ÷ 15 = cm

Perfect! The height is 12 cm.

3. A field in the shape of a trapezium has parallel sides 10 m and 20 m and height 12 m. Find its area.

Formula: Area = 12 × (a + b) × h

Sum of parallel sides = + = m

Area = 12 × ×

Area = 12 × 360 = m2

Great work! The area is 180 m².

4. Find the area of a triangle whose sides are 8 cm, 6 cm and 10 cm (using Heron's formula).

Given: a = 8 cm, b = 6 cm, c = 10 cm

Semi-perimeter s = (a + b + c)/2 = ( + + )/2

s = 24/2 = cm

Area = √[s(s-a)(s-b)(s-c)]

Area = √[12 × × × ]

Area = √[576] = cm2

Excellent! The area is 24 cm2.

Part A: Section C – Long Answer Questions (4 Marks Each)

1. The sides of a triangle are 7 cm, 8 cm, and 9 cm.

(a) Find its area using Heron's formula.

Semi-perimeter s = (7 + 8 + 9) ÷ 2 = ÷ 2 = cm

s − a = 12 − 7 =

s − b = 12 − 8 =

s − c = 12 − 9 =

Area = √[12 × 5 × 4 × 3] = √

Area ≈ cm2 (round to 2 decimal places)

Great! The area is approximately 26.83 cm2.

(b) Find its height corresponding to the largest side.

The largest side = cm (this will be the base)

Using Area = 12 × base × height

26.83 = 12 × 9 × h

Height = (26.83 × 2) ÷ 9 ≈ cm

Perfect! The height is approximately 5.96 cm.

2. A field is in the shape of a rhombus whose perimeter is 400 m and one of its diagonals is 160 m.

(a) Find its area.

Perimeter = 400 m, so each side = 400 ÷ 4 = m

Given diagonal d₁ = 160 m

Diagonals of rhombus bisect at right angles

Half of d₁ = m

Using Pythagoras: (side)² = d122 + d222

100² = 80² + d222

10000 = 6400 + d222

d222 =

d₂/2 = m, so d₂ = m

(b) Find its other diagonal.

The other diagonal d₂ = 120 m (calculated above).

Area = 12 × d₁ × d₂ = 12 × 160 × 120 = m²

Excellent! Area = 9600 m² and other diagonal = 120 m.

3. A plot of land is in the shape of a trapezium whose parallel sides are 50 m and 30 m, and the distance between them is 20 m.

(a) Find its area.

Area = 12 × (a + b) × h = 12 × ( + ) ×

Area = 12 × × 20 = m²

Perfect! The area is 800 m².

(b) Find the cost of fencing the plot at ₹15 per metre.

For fencing, we need all four sides. Given: two parallel sides 50 m and 30 m

Note: Non-parallel sides are not given, so we assume equal non-parallel sides.

Using Pythagoras: non-parallel side = √[(50−30)²/4 + 20²] = √[100 + 400] = √500 ≈ m

Perimeter ≈ 50 + 30 + 22.36 + 22.36 ≈ m

Cost = 124.72 × 15 ≈ ₹ (round to nearest rupee)

Great! The approximate cost is ₹1871.

4. A triangle has sides 26 m, 28 m, and 30 m.

(a) Find its area using Heron's formula.

s = (26 + 28 + 30) ÷ 2 = ÷ 2 = m

s − a = 42 − 26 =

s − b = 42 − 28 =

s − c = 42 − 30 =

Area = √[42 × 16 × 14 × 12] = √ = m2

Excellent! The area is 336 m².

(b) Find the height corresponding to the largest side.

Largest side = m (this is the base)

Area = 12 × base × height

336 = 12 × 30 × h

Height = (336 × 2) ÷ 30 = m

Perfect! The height is 22.4 m.

Part B: Objective Questions - Test Your Understanding!

Answer these multiple choice questions:

6. The base of a triangle is doubled, height remains same. Area becomes:

(a) Half (b) Same (c) Double (d) Four times

Correct! Area = 12 × base × height. If base doubles, area also doubles.

7. The diagonals of a rhombus are 16 cm and 12 cm. Area = ?

(a) 48 cm2 (b) 96 cm2 (c) 192 cm2 (d) 144 cm2

Perfect! Area = 12 × 16 × 12 = 96 cm2.

8. The area of a triangle and a rectangle having same base and height are in ratio:

(a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 1 : 3

Excellent! Triangle area = 12bh and Rectangle area = bh, so ratio is 1:2.

9. 1 m² = ? dm²

(a) 10 (b) 100 (c) 1000 (d) 10000

Correct! 1 m = 10 dm, so 1 m² = 10 × 10 = 100 dm².

10. The area of a rhombus with diagonals d₁, d₂ is:

(a) d₁ × d₂ (b) 12 d₁ × d₂ (c) 2 × d₁ × d₂ (d) d₁ + d₂

Perfect! Rhombus area = 1/2 × product of diagonals.

🎉 Outstanding Work! You've Mastered Moderate Area Concepts!

Here's what you learned:

• Reverse Calculations: Finding missing dimensions when area is known

• Heron's Formula: Calculating triangle area when three sides are given: √[s(s-a)(s-b)(s-c)]

• Ratio Problems: Solving for dimensions when given in ratio form

• Unit Conversions: Converting between m², cm2, dm², and hectares

• Area Relationships: Understanding how area changes when dimensions are modified

• Comparative Analysis: Relating areas of different shapes with same base and height

These advanced problem-solving skills will prepare you for complex real-world applications in construction, agriculture, and design!