What is 4(CP^{2}-AQ^{2}) equal to?
Consider the following for the next three (03) items that follow: ABC is a triangle with AB=1.6\text{ cm}, BC=6.3\text{ cm} and CA=6.5\text{ cm}. Let P and Q be the mid-points of AB and BC respectively.
- A. 101.39\text{ cm}^{2}
- B. 111.39\text{ cm}^{2} ✓
- C. 121.39\text{ cm}^{2}
- D. 131.39\text{ cm}^{2}
Correct Answer: B. 111.39\text{ cm}^{2}
Explanation
Using Pythagoras in right \triangle PBC and \triangle ABQ: CP^2 = BC^2 + (AB/2)^2 and AQ^2 = AB^2 + (BC/2)^2. Then 4(CP^2 - AQ^2) = 4(BC^2 + \frac{AB^2}{4} - AB^2 - \frac{BC^2}{4}) = 4(\frac{3}{4}BC^2 - \frac{3}{4}AB^2) = 3(BC^2 - AB^2). Substituting the values, 3(6.3^2 - 1.6^2) = 3(39.69 - 2.56) = 3(37.13) = 111.39\text{ cm}^2.
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