Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.2 Solutions

Area is the amount of space covered by a two-dimensional shape, measured in square units.
Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.2 connects triangle area, Heron’s formula, trapezium area, rhombus diagonals and equal-area proofs.

Exercise 6.2 from Chapter 6, Measuring Space Perimeter and Area Class 9, shifts from perimeter to area. Students use area formulas for triangles, trapeziums, parallelograms and rhombuses, along with Heron formula Class 9 and equal-area reasoning. Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.2 Solutions cover 11 textbook questions, including numerical problems and proof-based questions. The chapter explains that a triangle’s area is half of base × height, and a median of a triangle divides it into two triangles with equal area.

Key Takeaways

• Triangle Area: Area of a triangle is 1/2 × base × height.
• Heron’s Formula: A triangle’s area can be found from its three side lengths.
• Trapezium Area: Area of a trapezium is 1/2 × sum of parallel sides × height.
• Median Property: A median of a triangle divides it into two equal-area triangles.

Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.2 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.2 Solutions for Area of Triangle

Exercise 6.2 begins with area calculations using base, height and Heron’s formula. Fig. 6.31 shows triangle ADE inside a rectangle of length 10 cm and height 8 cm.

Q1. Find the area of triangle ADE in Fig. 6.31.

The area of triangle ADE is 40 cm².

Given:

Rectangle length = 10 cm

Rectangle height = 8 cm

In Fig. 6.31:

AD = 8 cm

Distance of E from AD = 10 cm

Use:

Area of triangle = 1/2 × base × height

Substitute values:

Area of triangle ADE = 1/2 × AD × 10

Area of triangle ADE = 1/2 × 8 × 10

Area of triangle ADE = 40 cm²

Answer:

The area of triangle ADE is 40 cm².

Ganita Manjari Class 9 Chapter 6 Exercise 6.2: Area of Trapezium Class 9

A trapezium with equal non-parallel sides can be split into a rectangle and two equal right triangles. This helps find its height before using the trapezium area formula.

Q2. The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.

The area of the trapezium is 720 cm².

Given:

Parallel sides = 40 cm and 20 cm

Equal non-parallel sides = 26 cm each

Difference of parallel sides:

40 - 20 = 20 cm

Since the non-parallel sides are equal, the extra length is split equally.

Each right triangle base:

20 ÷ 2 = 10 cm

Let height be h.

Using Baudhāyana-Pythagoras theorem:

26² = h² + 10²

676 = h² + 100

h² = 676 - 100

h² = 576

h = 24 cm

Now use:

Area of trapezium = 1/2 × sum of parallel sides × height

Area = 1/2 × (40 + 20) × 24

Area = 1/2 × 60 × 24

Area = 720 cm²

Answer:

The area of the trapezium is 720 cm².

Class 9 Maths Chapter 6 Exercise 6.2 Solutions Using Heron Formula Class 9

Heron’s formula is used when all three sides of a triangle are known. The semi-perimeter is found first, then the area is calculated from the side lengths.

Q3. Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

The area of the triangle is 8√30 cm².

Given:

Two sides = 8 cm and 11 cm

Perimeter = 32 cm

Third side:

Third side = 32 - 8 - 11

Third side = 13 cm

So the sides are:

8 cm, 11 cm, 13 cm

Semi-perimeter:

s = 32/2

s = 16 cm

Use Heron’s formula:

Area = √[s(s - a)(s - b)(s - c)]

Substitute values:

Area = √[16(16 - 8)(16 - 11)(16 - 13)]

Area = √[16 × 8 × 5 × 3]

Area = √1920

Area = √(64 × 30)

Area = 8√30 cm²

Answer:

The area of the triangle is 8√30 cm².

Q4. The sides of a triangular plot are in the ratio 3:5:7; its perimeter is 300 m. Find its area.

The area of the triangular plot is 1500√3 m².

Given:

Ratio of sides = 3:5:7

Perimeter = 300 m

Sum of ratio parts:

3 + 5 + 7 = 15

One part:

300 ÷ 15 = 20 m

So the sides are:

3 × 20 = 60 m

5 × 20 = 100 m

7 × 20 = 140 m

Semi-perimeter:

s = 300/2

s = 150 m

Use Heron’s formula:

Area = √[s(s - a)(s - b)(s - c)]

Area = √[150(150 - 60)(150 - 100)(150 - 140)]

Area = √[150 × 90 × 50 × 10]

Area = √6750000

Area = 1500√3 m²

Answer:

The area of the triangular plot is 1500√3 m².

Class 9 Maths Ganita Manjari Chapter 6 Solutions for Area of Rhombus

The area of a rhombus can be found using its diagonals. Exercise 6.2 uses the formula 1/2 × diagonal 1 × diagonal 2.

Q5. One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.

The shorter diagonal is 8√2 cm.

Let the shorter diagonal be x cm.

Then the longer diagonal is 2x cm.

Use:

Area of rhombus = 1/2 × d1 × d2

Substitute values:

128 = 1/2 × x × 2x

128 = x²

x = √128

x = √(64 × 2)

x = 8√2 cm

Answer:

The length of the shorter diagonal is 8√2 cm.

Measuring Space Perimeter and Area Class 9: Parallelogram Area Proofs

Area comparisons in parallelograms depend on equal bases or equal heights. The textbook explains that a parallelogram with base b and height h has area bh.

Q6. ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area (∆PCD): area (∆QCD)?

The ratio area (∆PCD): area (∆QCD) is 1:1.

Given:

ABCD is a parallelogram.

P and Q lie on AB.

Triangles PCD and QCD have the same base CD.

Since AB ∥ CD, points P and Q are at the same perpendicular distance from CD.

So both triangles have the same height.

Use:

Area of triangle = 1/2 × base × height

Therefore:

Area (∆PCD) = Area (∆QCD)

Ratio:

Area (∆PCD): Area (∆QCD) = 1:1

Answer:

The required ratio is 1:1.

Q7. O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.

Area (∆PSO) = Area (∆PQO).

Given:

PQRS is a parallelogram.

O lies on diagonal PR.

To prove:

Area (∆PSO) = Area (∆PQO)

Proof:

1. Triangles PSO and PQO have common base PO.
2. Points S and Q are opposite vertices of parallelogram PQRS.
3. The perpendicular distances of S and Q from diagonal PR are equal.
4. Since O lies on PR, the perpendicular distances from S and Q to line PO are equal.
5. Triangles PSO and PQO have the same base and equal heights.

Therefore:

Area (∆PSO) = Area (∆PQO)

Answer:

The areas of triangles PSO and PQO are equal.

Class 9 Maths Area Solutions for Midpoint and Median Properties

The median of a triangle divides it into two equal-area triangles. Exercise 6.2 uses this idea in proofs involving quadrilaterals and triangles.

Q8. If the mid-points of the sides of a 4-gon are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon.

The parallelogram formed by joining the midpoints has half the area of the given 4-gon.

Let ABCD be the given 4-gon.

Let E, F, G and H be the midpoints of AB, BC, CD and DA.

Join EF, FG, GH and HE.

To prove:

Area (EFGH) = 1/2 × Area (ABCD)

Proof:

1. In triangle ABC, E and F are midpoints of AB and BC.

So:

EF ∥ AC

EF = AC/2

1. In triangle ADC, H and G are midpoints of AD and DC.

So:

HG ∥ AC

HG = AC/2

1. Therefore:

EF ∥ HG

EF = HG

1. A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram.

So:

EFGH is a parallelogram.

1. The midpoint construction divides the 4-gon into four corner triangles and the central parallelogram.
2. Using equal-base and equal-height triangle area relations, the four corner triangles together occupy half the area of ABCD.

Therefore:

Area (EFGH) = 1/2 × Area (ABCD)

Answer:

The parallelogram formed by joining the midpoints has half the area of the given 4-gon.

Q9. In ∆ABC, the midpoint of BC is D. Median AD is drawn. P is any point on AD. Show that area (∆ABP) = area (∆ACP).

Area (∆ABP) = Area (∆ACP).

Given:

D is the midpoint of BC.

AD is a median.

P lies on AD.

To prove:

Area (∆ABP) = Area (∆ACP)

Proof:

1. Since D is the midpoint of BC:

BD = DC

1. Line AD passes through D, so B and C are at equal perpendicular distances from AD.
2. Since P lies on AD, AP is a common base for triangles ABP and ACP.
3. Triangles ABP and ACP have the same base AP and equal heights.

Therefore:

Area (∆ABP) = Area (∆ACP)

Answer:

The areas of triangles ABP and ACP are equal.

Class 9 Maths Chapter 6 Exercise 6.2 Solutions for Square Area Ratio

A point inside a square divides the square into four triangles. Opposite triangle pairs have equal total area because their heights add up to the side of the square.

Q10. Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD. What is the ratio of the areas of the red region (∆PAB and ∆PCD) and the green region (∆PBC and ∆PDA)?

The ratio of the red region to the green region is 1:1.

Let the side of the square be a.

For the red region:

Area (∆PAB) + Area (∆PCD)

Both triangles have bases AB and CD.

Since AB = CD = a

The sum of their heights is a.

So:

Red area = 1/2 × a × a

Red area = a²/2

For the green region:

Area (∆PBC) + Area (∆PDA)

Both triangles have bases BC and DA.

Since BC = DA = a

The sum of their heights is a.

So:

Green area = 1/2 × a × a

Green area = a²/2

Therefore:

Red area : Green area = a²/2 : a²/2

Red area : Green area = 1:1

Answer:

The ratio of the red region to the green region is 1:1.

Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.2 Solutions for Final Proof

Question 11 uses midpoint and parallel-line ideas together. The construction makes triangle BPQ occupy half the area of triangle ABC.

Q11. In ∆ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ ∥ PD. PQ is joined. Prove that Area (∆BPQ) = 1/2 Area (∆ABC).

Area (∆BPQ) = 1/2 Area (∆ABC).

Let BC be the base of ∆ABC.

Let the perpendicular height of A from BC be h.

Since D is the midpoint of AB, D is halfway from A to B.

So the perpendicular height of D from BC is:

h/2

Since CQ ∥ PD, the parallel construction places Q so that triangle BPQ has half the product needed for ∆ABC.

Using coordinate-style area comparison:

Let:

B = (0, 0)

C = (c, 0)

A = (0, h)

Then:

Area (∆ABC) = 1/2 × c × h

Area (∆ABC) = ch/2

Since D is the midpoint of AB:

D = (0, h/2)

Let:

P = (p, 0)

Line PD is parallel to CQ.

From the parallel-line condition, Q lies on AB at height:

Height of Q = ch/(2p)

Now find area of ∆BPQ:

Area (∆BPQ) = 1/2 × BP × height of Q

Area (∆BPQ) = 1/2 × p × ch/(2p)

Area (∆BPQ) = ch/4

Compare with ∆ABC:

Area (∆ABC) = ch/2

So:

Area (∆BPQ) = 1/2 × Area (∆ABC)

Answer:

Area (∆BPQ) = 1/2 Area (∆ABC).

Measuring Space Perimeter and Area Class 9 Concepts Used in Exercise 6.2

Exercise 6.2 is built around area formulas and equal-area proofs. The main tools are triangle area, Heron’s formula, trapezium area, rhombus diagonal area and the median property.

Area of Triangle Class 9

The area of a triangle is half the product of its base and height.

Copy-friendly formula:

Area of triangle = 1/2 × base × height

Using symbols:

Area = 1/2 × b × h

Heron Formula Class 9

Heron’s formula finds the area of a triangle when all three sides are known.

Copy-friendly formula:

s = (a + b + c)/2

Area = √[s(s - a)(s - b)(s - c)]

Area of Trapezium Class 9

The area of a trapezium uses the sum of parallel sides and the height.

Copy-friendly formula:

Area of trapezium = 1/2 × (a + b) × h

Here:

a and b = parallel sides

h = height

Area of Rhombus Class 9

The area of a rhombus can be found from its diagonals.

Copy-friendly formula:

Area of rhombus = 1/2 × d1 × d2

Median Divides Triangle into Equal Areas

A median joins a vertex to the midpoint of the opposite side.

Copy-friendly result:

If AD is a median of ∆ABC, then:

Area (∆ABD) = Area (∆ACD)

Quick Formula Table for Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.2 Solutions

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6