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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