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Ratio of areas of ∆MNO, ∆MOP and ∆MPQ in Fig. 9.18 is
a. 2 : 1 : 3
b. 1 : 3 : 2
c. 2 : 3 : 1
d. 1 : 2 : 3

Ratio of areas of ∆MNO, ∆MOP and ∆MPQ in Fig. 9.18 is a. 2 : 1 : 3, b. 1 : 3 : 2, c. 2 : 3 : 1, d. 1 : 2 : 3

Solution:

Using the figure

Area of ∆ MNO = 1/2 × NO × MO

Substituting the values

= 1/2 × 4 × 5

= 10 cm²

Area of ∆ MOP = 1/2 × OP × MO

Substituting the values

= 1/2 × 2 × 5

= 5 cm²

Area of ∆ MPQ = 1/2 × PQ × MO

Substituting the values

= 1/2 × 6 × 5

= 15 cm²

We know that

Required ratio = 10 : 5 : 15 = 2 : 1 : 3

Therefore, the ratio of areas is 2 : 1 : 3.

✦ Try This: 12 unit squares are joined to form a rectangle with the least perimeter. Perimeter of the rectangle is

☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 11

NCERT Exemplar Class 7 Maths Chapter 9 Problem 11

Ratio of areas of ∆MNO, ∆MOP and ∆MPQ in Fig. 9.18 is a. 2 : 1 : 3, b. 1 : 3 : 2, c. 2 : 3 : 1, d. 1 : 2 : 3

Summary:

Ratio of areas of ∆ MNO, ∆MOP and ∆MPQ in Fig. 9.18 is 2 : 1 : 3

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