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In triangle ABC, medians BD and CE intersect at F such that m∠BFE = 30°. If the lengths of BD and CE are 18 cm and 15 cm respectively, find the area, in square cm, of ∆ABC. Enter your response (as an integer)

Question

In triangle ABC, medians BD and CE intersect at F such that m∠BFE = 30°. If the lengths of BD and CE are 18 cm and 15 cm respectively, find the area, in square cm, of ∆ABC. Enter your response (as an integer)

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Solution

The area of a triangle can be found using the formula 1/2 * base * height. In this case, the base and height of the triangle are the lengths of the medians BD and CE, respectively.

However, since the medians intersect at a 30° angle, we can use the formula for the area of a triangle when two sides and the included angle are known. This formula is 1/2 * a * b * sin(C), where a and b are the lengths of the two sides and C is the included angle.

Here, a = BD = 18 cm, b = CE = 15 cm, and C = m∠BFE = 30°.

So, the area of ∆ABC = 1/2 * 18 cm * 15 cm * sin(30°)

= 1/2 * 18 cm * 15 cm * 1/2 (since sin(30°) = 1/2)

= 135 square cm.

So, the area of ∆ABC is 135 square cm.

This problem has been solved

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