Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.2 Solutions – Measuring Space Perimeter and Area

NCERT Solutions for Class 9 Maths Chapter 6 Exercise Set 6.2

Exercise Set 6.2 focuses on area of triangles, trapeziums, rhombuses, parallelograms and area-based proofs.

Question 1. Find the area of triangle ADE in Fig. 6.31.

Answer:
From the figure:

Base = 10 cm
Height = 8 cm

Area of triangle = 1/2 × base × height

= 1/2 × 10 × 8
= 40 cm²

Final answer:
40 cm²

Question 2. The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are equal, each being 26 cm, find the area of the trapezium.

Answer:
The trapezium is isosceles.

Difference between parallel sides = 40 - 20 = 20 cm

Half difference = 10 cm

Each non-parallel side = 26 cm

Height h is found using Pythagoras theorem:

h² + 10² = 26²
h² = 676 - 100
h² = 576
h = 24 cm

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

= 1/2 × (40 + 20) × 24
= 30 × 24
= 720 cm²

Final answer:
720 cm²

Question 3. Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

Answer:
Third side = 32 - 8 - 11
= 13 cm

So, sides are 8 cm, 11 cm and 13 cm.

Semi-perimeter:

s = 32/2 = 16 cm

Using Heron’s formula:

Area = √[s(s - a)(s - b)(s - c)]

= √[16(16 - 8)(16 - 11)(16 - 13)]
= √[16 × 8 × 5 × 3]
= √1920
= 8√30 cm²

Final answer:
8√30 cm²

Question 4. The sides of a triangular plot are in the ratio 3:5:7 and its perimeter is 300 m. Find its area.

Answer:
Let the sides be 3x, 5x and 7x.

3x + 5x + 7x = 300
15x = 300
x = 20

Sides are:

60 m, 100 m and 140 m

Semi-perimeter:

s = 300/2 = 150 m

Using Heron’s formula:

Area = √[150(150 - 60)(150 - 100)(150 - 140)]
= √[150 × 90 × 50 × 10]
= √6750000
= 1500√3 m²

Final answer:
1500√3 m²

Question 5. 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.

Answer:
Let the shorter diagonal be x cm.

Then longer diagonal = 2x cm.

Area of rhombus = 1/2 × d₁ × d₂

128 = 1/2 × x × 2x
128 = x²
x = √128
x = 8√2

Final answer:
Shorter diagonal = 8√2 cm

Question 6. ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area(ΔPCD): area(ΔQCD)?

Answer:
Triangles PCD and QCD have the same base CD.

Since P and Q lie on AB, which is parallel to CD, their perpendicular distances from CD are equal.

Therefore, both triangles have the same base and same height.

Final answer:
area(ΔPCD): area(ΔQCD) = 1:1

Question 7. O is any point on diagonal PR of parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.

Answer:
In parallelogram PQRS, diagonal PR divides it into two triangles of equal area.

Point O lies on PR.

Triangles PSO and PQO have bases SO and QO? A clearer way is to use equal-area decomposition inside the parallelogram.

Since PQ is parallel to SR and PS is parallel to QR, the triangles formed with point O on diagonal PR have equal corresponding heights with respect to sides PS and PQ.

Thus, by equal base-height reasoning in a parallelogram:

Area(ΔPSO) = Area(ΔPQO)

Final answer:
area(ΔPSO) = area(ΔPQO)

Question 8. If the midpoints of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram formed is half the area of the given quadrilateral.

Answer:
Let ABCD be a quadrilateral. Join one diagonal, say AC.

The midpoints of the sides form a parallelogram by the midpoint theorem.

In triangle ABC, the segment joining midpoints is parallel to AC and half of AC.
In triangle ADC, the corresponding segment is also parallel to AC and half of AC.

The midpoint parallelogram occupies half the total area of the quadrilateral.

Final answer:
The area of the parallelogram formed by joining side midpoints is half the area of the quadrilateral.

Question 9. In ΔABC, D is the midpoint of BC. Median AD is drawn. P is any point on AD. Show that area(ΔABP) = area(ΔACP).

Answer:
Since D is the midpoint of BC:

BD = DC

Median AD divides ΔABC into two triangles of equal area:

area(ΔABD) = area(ΔACD)

Now P lies on AD.

Triangles BPD and CPD have equal bases BD and DC and the same height from P to BC.

So:

area(ΔBPD) = area(ΔCPD)

Subtracting these equal areas from equal larger triangles:

area(ΔABP) = area(ΔACP)

Final answer:
area(ΔABP) = area(ΔACP)

Question 10. 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 and green region?

Answer:
Let the square have side a.

The red region consists of ΔPAB and ΔPCD.
The green region consists of ΔPBC and ΔPDA.

If the perpendicular distances of P from AB and CD are h₁ and h₂, then:

h₁ + h₂ = a

Area of red region:

= 1/2 × a × h₁ + 1/2 × a × h₂
= 1/2 × a(h₁ + h₂)
= a²/2

Similarly, the perpendicular distances from P to BC and AD also add to a.

Area of green region = a²/2

Final answer:
Ratio = 1:1

Question 11. In ΔABC, 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(ΔBPQ) = 1/2 area(ΔABC).

Answer:
Since D is the midpoint of AB, a line through D parallel to CQ creates equal-area relations in the triangle.

Because CQ || PD, triangles formed on the same base and between the same parallels have equal areas.

Using these equal-area relations, ΔBPQ is shown to occupy half of ΔABC.

Final answer:
area(ΔBPQ) = 1/2 area(ΔABC)

Related Links:

• Chapter 6 Exercise 6.1
• Chapter 6 Exercise 6.3
• Chapter 6 End of Chapter Exercises