XAT 2015 Q10 : Area of right triangle

Decode the given data

Given Data:

1. CD = BF = 10 units
2. ∠CED = ∠BAF = 30^o

The triangle AED is a right triangle.

The area of the triangle can be computed if we know the measure of AD and ED, the two perpendicular sides.

1. Compute length of ED

In right triangle CDE
∠CED = 30^o

Therefore, tan ∠CED = tan 30^o = \\frac{1}{\sqrt3}\\) = \\frac{CD}{ED}\\)

Therefore ED = \\sqrt{3}\\) x CD = 10\\sqrt{3}\\)

2. Compute length of AB

In right triangle ABF
∠BAF = 30^o
Therefore, tan ∠BAF = tan 30^o = \\frac{1}{\sqrt3}\\) = \\frac{BF}{AB}\\)
Hence, AB = \\sqrt{3}\\) x BF = 10\\sqrt{3}\\)

3. Compute length of BC

Triangles EDC and FBC are right triangles. ∠EDC = ∠FBC. Both are right angles.
∠DCE = ∠BCF. Vertically opposite angles are equal.

Because two angles of the two triangles EDC and FBC are equal these two triangles are similar triangles.

So, ∠BFC = ∠CED = 30^o
tan ∠BFC = tan 30^o = \\frac{1}{\sqrt3}\\) = \\frac{BC}{BF}\\)
We know BF = 10
So, BC = \\frac{10}{\sqrt3}\\).

Compute AD (Add 1 + 2 + 3 from step 1)

AD = AB + BC + CD = 10\\sqrt{3}\\) + \\frac{10}{\sqrt3}\\) + 10

Compute Area

Area of triangle AED = \\frac{1}{2}\\) ED×AD = \\frac{1}{2}\\) × 10 \\sqrt{3}\\) × (10 + \\frac{10}{\sqrt3}\\) + 10\\sqrt{3}\\)) = 50 × ( \\sqrt{3}\\) + 4)

  The correct answer is Choice D.