Question

What will be the area of triangle PQR formed by joining the midpoints of the sides of the triangle whose vertices are (3, 3), (9, 9) and (4, 6)?

(a) \(\frac {1}{2}\) units

(b) \(\frac {3}{2}\) units

(c) \(\frac {3}{5}\) units

(d) \(\frac {5}{2}\) units

I had been asked this question in quiz.

This intriguing question comes from Geometry in portion Coordinate Geometry of Mathematics – Class 10

Answer 1

Correct choice is (b) \(\frac {3}{2}\) units

Easiest explanation: ABC is the triangle formed by the coordinates A (3, 3), B (9, 9) and C (4, 6) and DEF is the triangle formed by joining the midpoints of AC, BC, AB.

D is the midpoint of AC. Coordinates of D = \( ( \frac {x_1+x_2}{2}, \frac {y_1+y_2}{2} ) = ( \frac {3+4}{2}, \frac {3+6}{2} ) = ( \frac {7}{2}, \frac {9}{2} ) \)

E is the midpoint of AB. Coordinates of E = \( ( \frac {x_1+x_2}{2}, \frac {y_1+y_2}{2} ) = ( \frac {3+9}{2}, \frac {3+9}{2} ) = ( \frac {12}{2}, \frac {12}{2} ) \) = (6, 6)

F is the midpoint of BC. Coordinates of F = \( ( \frac {x_1+x_2}{2}, \frac {y_1+y_2}{2} ) = ( \frac {9+4}{2}, \frac {9+6}{2} ) = ( \frac {13}{2}, \frac {15}{2} ) \)

The area of triangle formed by D \( ( \frac {7}{2}, \frac {9}{2} ) \), E (6, 6) and F\( ( \frac {13}{2}, \frac {15}{2} ) \)

We know that, area of triangle = \(\frac {1}{2}\){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 )}

The coordinates of vertices of the triangle are \( ( \frac {7}{2}, \frac {9}{2} ) \), (6, 6) and \( ( \frac {13}{2}, \frac {15}{2} ) \).

The area of triangle = \(\frac {1}{2} \{ \frac {7}{2}\) (6 – \(\frac {15}{2}\)) + 6(\(\frac {15}{2} – \frac {9}{2}\)) + \(\frac {13}{2} ( \frac {9}{2} – 6 ) \} \) = \(\frac {1}{2} \{ – \frac {21}{4}\) + 18 – \(\frac {39}{4} \} = \frac {3}{2} \) units