Question

In triangle ABC, the lengths of the sides are 5 cm, 8 cm and 9 cm. If D, E and F are the midpoints of BC, CA and AB respectively, then the area of triangle DEF, in $cm^2$, is

• A. $\frac{3\sqrt{11}}{2}$
• B. $6\sqrt{11}$
• C. $3\sqrt{11}$
• D. $\frac{9}{2}\sqrt{11}$

Hint:

The triangle formed by joining the midpoints of the sides of a triangle always has one-fourth the area of the original triangle.

Answer 1

The Correct Option is A

Concept: The triangle formed by joining the midpoints of the sides of a triangle is called the medial triangle. Its area is one-fourth of the area of the original triangle.

Step 1: Find the area of triangle ABC using Heron's formula.
Given sides: \[ a=5,\quad b=8,\quad c=9 \] Semi-perimeter: \[ s=\frac{5+8+9}{2}=11 \] Area of triangle ABC: \[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \] \[ = \sqrt{11(11-5)(11-8)(11-9)} \] \[ = \sqrt{11\times6\times3\times2} \] \[ = \sqrt{396} = 6\sqrt{11} \]

Step 2: Use the medial triangle property.
Since D, E and F are midpoints, \[ \text{Area}(\triangle DEF) = \frac14 \text{Area}(\triangle ABC) \] \[ = \frac14(6\sqrt{11}) \] \[ = \frac{3\sqrt{11}}{2} \]