Vincen wrote: ↑

Tue Apr 28, 2020 8:10 am

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In the triangle above, if \(BC=4\sqrt2,\) then what is the area of \(\triangle ABC?\)

A. \(64\)
B. \(16+16\sqrt3\)
C. \(8+8\sqrt3\)
D. \(8+4\sqrt2\)
E. \(8\sqrt2\)

[spoiler]OA=C[/spoiler]

Source: Princeton Review

If we draw an altitude from B to AC and call the intersection on AC point D, we will see that triangle ABC is split into two right triangles: ABD on the left and CBD on the right. Furthermore, triangle ABD is a 30-60-90 right triangle, and triangle CBD is a 45-45-90 right triangle. Therefore, the area of triangle ABC is the sum of the areas of the triangles ABD and CBD.

Let’s look at triangle CBD first since we are given BC = 4√2. Since BC is the hypotenuse of triangle CBD, the legs of triangle CBD, CD and BD, are each 4√2/√2 = 4. Therefore, the area of triangle CBD is ½ x 4 x 4 = 8.

Now, let’s look at triangle ABD since now we know BD = 4. Since BD, the shorter leg of triangle ABD, is 4, the longer leg, AD is 4 x √3 = 4√3. Therefore, the area of triangle ABD is ½ x 4 x 4√3 = 8√3.

Therefore, the area of triangle ABC is 8 + 8√3.

Answer: C