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Answer to Q194:

From the diagram above, find the area of the green triangle.

Area of the green triangle = ½ x BC x BD

But, BD² = AD² + BA²
BD² = 5² + 10²
BD = √(25 + 100) = √125
BD = 11.18

But also, ∠ DBA = ∠ BDF = δ
sin δ = ⁵⁄₁₁.₁₈
δ = sin⁻¹ (0.4472)
δ = 26.57°

So, ∠ BDF = 26.57°
that makes ∠ DCB = 90° - 26.57°
∠ DCB = 63.43°

and, DA = FB = 5
sin (DCB) = 5 ÷ BC
BC = 5 ÷ sin 63.43°
BC = 5.59

Area of the green triangle = ½ x BC x BD
= ½ x 5.59 x 11.18
= 31.25

Answer: 31.25 sq. unit



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Answer to Q194:

From the diagram above, find the area of the green triangle.

Area of the green triangle = ½ x BC x BD

But, BD² = AD² + BA²
BD² = 5² + 10²
BD = √(25 + 100) = √125
BD = 11.18

But also, ∠ DBA = ∠ BDF = δ
sin δ = ⁵⁄₁₁.₁₈
δ = sin⁻¹ (0.4472)
δ = 26.57°

So, ∠ BDF = 26.57°
that makes ∠ DCB = 90° - 26.57°
∠ DCB = 63.43°

and, DA = FB = 5
sin (DCB) = 5 ÷ BC
BC = 5 ÷ sin 63.43°
BC = 5.59

Area of the green triangle = ½ x BC x BD
= ½ x 5.59 x 11.18
= 31.25

Answer: 31.25 sq. unit

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