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XAT 2015 Q10 : Area of right triangle

This aptitude question appeared as a part of the quantitative reasoning and data interpretation section of the XAT 2015. Question 10 of a total of 33 questions that appeared in this section in XAT 2015. This one is a medium level difficulty question and tests the following concepts: Similar triangles, trigonometric identities, and area of triangles.

Question

In the diagram below, CD = BF = 10 units and ∠CED = ∠BAF = 30o. What would be the area of triangle AED? (Note: Diagram below may not be drawn to scale.)

XAT 2015 geometry triangles

  1. 100 x (\\sqrt{2}\\) + 3)
  2. \\frac{100}{(\sqrt3 + 4)}\\)
  3. \\frac{50}{(\sqrt3 + 4)}\\)
  4. 50 x (\\sqrt{3}\\) + 4)
  5. None of the above

  Correct Answer      Choice D. 50 x (\\sqrt{3}\\) + 4)


Explanatory Answer

  • Decode the given data

    Given Data:

    1. CD = BF = 10 units
    2. ∠CED = ∠BAF = 30o
    XAT 2015 geometry triangles

    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 = 30o

    Therefore, tan ∠CED = tan 30o = \\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 = 30o
    Therefore, tan ∠BAF = tan 30o = \\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 = 30o
    tan ∠BFC = tan 30o = \\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.

Video Explanation

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