Question

find the area of a parallelogram ABCD in which ab 14 cm BC 10 cm and 16 cm​

Answer 1

Answer:

138.56 cm³

Step-by-step explanation:

In Δ ABC,

Using Heron's Formula,

S = (a + b + c)/2

⇒ S = (10 + 14 +16)/2

⇒ S = 40/2

⇒ S = 20 

Area = \sqrt{S(S - a)(S - b)(S - c)}S(S−a)(S−b)(S−c) 

∴ Area of the triangles = \sqrt{20(20 - 10)(20 - 14)(20 - 16)}20(20−10)(20−14)(20−16) 

⇒ Area of Δ ABC = \sqrt{20(10)(6)(4)}20(10)(6)(4) 

⇒ Area = √4800

⇒ Area = 10√48

⇒ Area = 10√(2 × 2 × 2 × 2 × 3)

⇒ Area = 10 × 2 × 2√3

⇒ Area of Δ ABC = 40√3 cm²

∴ Area of Δ ABC = 69.28 cm².

We know,

Area of the Δ ADC = Area of the ΔABC.

[∵ Both the triangles are congruent)

∴ Area of the Parallelogram = 2 × Area of Δ ABC.

⇒ Area of the Parallelogram = 2 × 69.28 cm²

∴ Area of the Parallelogram = 138.56 cm².

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