In Fig. 12.5, ∆ ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC a parallelogram DBCE of same area as that of ∆ ABC is constructed. Find the height DF of the parallelogram.
Solution:
Given, ABC is a triangle
The sides are AB = 7.5 cm
AC = 6.5 cm
BC = 7 cm
On base BC a parallelogram DBCE of the same area as that of triangle ABC is constructed.
We have to find the height DF of the parallelogram
In triangle ABC,
By Heron’s formula,
Area of triangle = √s(s - a)(s - b)(s - c)
Where s = semiperimeter
s = (a + b + c)/2
Here, a = 7.5 cm, b = 7 cm and c = 6.5 cm
So, s = (7.5 + 7 + 6.5)/2
= 21/2
s = 10.5 cm
Area = √10.5(10.5 - 7.5)(10.5 - 7)(10.5 - 6.5)
= √10.5(3)(3.5)(4)
= √36.75 × 12
= √441
Area of triangle ABC = 21 cm²
Area of parallelogram = base × height
Area of parallelogram DBCE = BC × DF
Given, area of parallelogram DBCE = area of triangle ABC
So, 21 = 7 × DF
DF = 21/7
DF = 3 cm
Therefore, the height of the parallelogram is 3 cm.
✦ Try This: In a quadrilateral ABCD, AB = 7 cm, BC = 6 cm, CD = 12 cm, DA = 15 cm, AC = 9 cm. Find its area.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 12
NCERT Exemplar Class 9 Maths Exercise 12.4 Problem 6
In Fig. 12.5, ∆ ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC a parallelogram DBCE of same area as that of ∆ ABC is constructed. Find the height DF of the parallelogram.
Summary:
In Fig. 12.5, ∆ ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC a parallelogram DBCE of the same area as that of ∆ ABC is constructed. The height DF of the parallelogram is 3 cm
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