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# Area of shaded portion in Fig. 9.15 is

a. 25 cm²

b. 15 cm²

c. 14 cm²

d. 10 cm²

**Solution:**

From the figure

Length of rectangle = 5 cm

Breadth of rectangle = 3 + 1 = 4 cm

We know that

Area of shaded portion = 1/2 × __Area of rectangle__

= 1/2 × (l × b)

Substituting the values

= 1/2 × (5 × 4)

= 1/2 × 20

So we get

= 10 cm²

Therefore, the area of the shaded portion is 10 cm².

**✦ Try This: **16 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 7**

## Area of shaded portion in Fig. 9.15 is a. 25 cm², b. 15 cm², c. 14 cm², d. 10 cm²

**Summary:**

Area of shaded portion in Fig. 9.15 is 10 cm²

**☛ Related Questions:**

- Area of parallelogram ABCD (Fig. 9.16) is not equal to a. DE × DC, b. BE × AD, c. BF × DC, d. BE × B . . . .
- Area of triangle MNO of Fig. 9.17 is a. 1/2 MN x NO, b. 1/2 NO x MO, c. 1/2 MN x OQ, d. 1/2 NO x OQ
- Ratio of area of ∆MNO to the area of parallelogram MNOP in the same figure 9.17 is a. 2 : 3, b. 1 : . . . .

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