MBApyq

Area of Triangle with Trigonometric Relations

XAT 2015 · Quantitative Ability & Data Interpretation · Hard · Quantitative Ability

All XAT 2015 questionsMore Quantitative Ability questionsAll XAT QADI

In the diagram below, CD=BF=10 unitsCD = BF = 10\text{ units} and ∠CED=∠BAF=30∘\angle CED = \angle BAF = 30^\circ. What would be the area of triangle AED? (Note: Diagram below may not be proportional to scale.)

  1. A.

    100(2+3)100 (\sqrt{2} + 3)

  2. B.

    100(3+4)100 (\sqrt{3} + 4)

  3. C.

    50(2+4)50 (\sqrt{2} + 4)

  4. D.

    50(3+4)50 (\sqrt{3} + 4)

  5. E.

    None of the above

Answer

D

Explanation

From right triangle ABFABF with ∠BAF=30∘\angle BAF = 30^\circ and BF=10BF = 10: AB=103AB = 10 \sqrt{3}. Since ED=AB=103ED = AB = 10 \sqrt{3}.

From right triangle BCFBCF with ∠BCF=60∘\angle BCF = 60^\circ: BC=103BC = \frac{10}{\sqrt{3}}.

Total height AD=AB+BC+CD=103+103+10=40+1033AD = AB + BC + CD = 10\sqrt{3} + \frac{10}{\sqrt{3}} + 10 = \frac{40 + 10\sqrt{3}}{\sqrt{3}}.

Area of △AED=12×AD×ED=12×40+1033×103=5(40+103)=50(3+4)\triangle AED = \frac{1}{2} \times AD \times ED = \frac{1}{2} \times \frac{40 + 10\sqrt{3}}{\sqrt{3}} \times 10\sqrt{3} = 5(40 + 10\sqrt{3}) = 50(\sqrt{3} + 4).

Related Quantitative Ability questions

Practise this under exam conditions

Sign in to solve it with a live timer, the on-screen CAT calculator, and streak and accuracy tracking across every question you attempt.

Solve in the workspace