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rss GMAT Practice Question (One of Hundreds)
Figure 1
In the figure above, which is not necessarily drawn to scale, triangle ABC is a right triangle. AB = 5, BC = 12. AD = AC, and angle ADC = 45 degrees. CDEF is a square. What is the area of the circle inscribed in the square CDEF?
Correct Answer: B
Find the area of each individual shaded region and add them up.
  1. Look at triangle ABC: We already know that it is a right triangle, and AB = 5, BC = 12. Using the Pythagorean Theorem, we can find the length of the Hypotenuse AC
    (12)2 + (5)2 = c2
    144 + 25 = c22
    169 = c2
    c = 13
    AC = 13
  2. Since we know that AC = 13 and we are given that AD = AC, triangle ADC is an Isosceles triangle. Recall that in an Isosceles triangle the two angles opposite the sides that are congruent are also congruent. This means that angle ADC = angle ACD = 45 degrees. Let us find angle DAC: we know that in any triangle the sum of all its three angles equals 180 degrees. So, angle DAC + angle ACD + angle ADC = 180
    angle DAC + 45 + 45 = 180
    angle DAC + 90 = 180
    angle DAC = 180-90 = 90 degrees.
  3. This means that triangle ADC is also a right triangle, and it is a special right 45-45-90 degree triangle. We know that in a 45-45-90 triangle, the hypotenuse = (21/2) leg . We found out that leg AC = 13, so the Hypotenuse DC = 13(21/2)
  4. Now, we arrive at the circle inscribed within the square. Note that the diameter of the circle is equal to the length of one of sides of the square. So, the radius = 1/2 diameter . We know that the length of one side in the square, which is side DC = 13(21/2) . This means that the diameter of the circle = 13(21/2) so the radius of the circle = (13/2)(21/2)
  5. Now that we found the radius of the circle, we calculate the area of the circle using A = πr2
    Area of circle = {[(13/2)(21/2)]2}π = (169/2) / (4)π = (169/2)π, which is correct answer B
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