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# Find the area of a parallelogram given in Fig. 12.2. Also find the length of the altitude from vertex A on the side DC.

**Solution:**

Given, ABCD is a parallelogram

We have to find the area of the parallelogram

We know that area of parallelogram = 2 × area of triangle BCD

In triangle BCD,

a = 25 cm

b = 12 cm

c = 17 cm

By Heron’s formula,

Area of triangle = √s(s - a)(s - b)(s - c)

Where s = semiperimeter

s = (a + b + c)/2

So, s = (25 + 12 + 17)/2

= 54/2

s = 27 cm

Area of triangle BCD = √27(27 - 25)(27 - 17)(27 - 12)

= √27(2)(10)(15)

= √9 × 3 × 2 × 5 × 2 × 5 × 3

= 3 × 3 × 5 × 2

= 9 × 10

Area of triangle BCD = 90 cm²

Area of ABCD = 2(90)

Area of parallelogram = 180 cm²

We have to find the length of the altitude from vertex A on the side DC.

Let the length of the altitude be h cm

Area of parallelogram = base × height

180 = 12 × h

h = 180/12

h = 90/6

h = 30/2

h = 15 cm

Therefore, the altitude is 15 cm

**✦ Try This: **The sides of a triangle are 11 m, 60 m and 61 m. The altitude to the smallest side is

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 12

**NCERT Exemplar Class 9 Maths Exercise 12.3 Problem 5**

## Find the area of a parallelogram given in Fig. 12.2. Also find the length of the altitude from vertex A on the side DC.

**Summary:**

In Fig. 12.2, the area of a parallelogram is 180 cm². Also the length of the altitude from vertex A on the side DC is 15 cm

**☛ Related Questions:**

- A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long . . . .
- The perimeter of a triangular field is 420 m and its sides are in the ratio 6 : 7 : 8. Find the area . . . .
- The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively, and . . . .

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