Question

find the area of a parallelogram ABCD in which AB = 14 cm BC = 10 cm and AC = 16 cm(using herons formula and the answer should be 138.4cm square)

Answer 1

Hello Dear.

Refers to the attachment for the Answer.

From the attachment.

In Δ ABC,
Using Heron's Formula,
S = (a + b + c)/2
⇒ S = (10 + 14 +16)/2
⇒ S = 40/2
⇒ S = 20 

Area = $\sqrt{S(S - a)(S - b)(S - c)}$
∴ Area of the triangles = $\sqrt{20(20 - 10)(20 - 14)(20 - 16)}$
⇒ Area of Δ ABC = $\sqrt{20(10)(6)(4)}$
⇒ 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.


Hence, the Area of the Parallelogram is 138.56 cm².


[ Note ⇒ The answer is different in points because I have taken the Exact value of √3. Although, it does not effect, such errors should be neglected.]



Hope it helps.

Answer 2

Answer:

ello Dear.

Refers to the attachment for the Answer.

From the attachment.

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.