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(Chapter – 12) (Heron’s Formula)(Exemplar Problems)
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Question 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.
Answer 6:
Now, first determine the area of ∆ABC.
The sides of a triangle are
AB – a = 7.5cm, BC = b = 7cm and CA = c = 6.5cm
Now, Semi- Perimeter, 𝑠 =
𝑎+𝑏+𝑐
2
==
7.5+7+6.5
2
=
21
2
∴ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎 ∆ABC = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
= 6.5𝑐𝑚
[by Heron’s formula]
= √10.5(10.5 − 7.5)(10.5 − 7) (10.5 − 6.5)
= √10.5 × 3 × 3.5 × 4
= √441 = 21𝑐𝑚2
……………(i)
Now, area of parallelogram BCED = Base × Height
……………(ii)
= BC × DF =7 × DF
1
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(Chapter – 12) (Heron’s Formula)(Exemplar Problems)
According to the question,
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎 ∆ABC = Area of parallelogram BCED
⇒
21 = 7 × DF
⇒
DF =
21
4
[From Equations (i) and (ii)]
= 3cm
Hence, the height of parallelogram is 3 cm
2
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