Measuring Space | Class 9 Mathematics Notes

6.8 AREA OF A TRIANGLE

This section derives the formula for the area of a triangle as half the product of its base and height (1/2 × base × height). It begins with right-angled triangles and extends to all triangles by enclosing them in rectangles and parallelograms. The section discusses the case of obtuse triangles and how the formula still applies. It also provides a geometric proof by showing that two congruent triangles can be combined to form a parallelogram, whose area is base × height, hence each triangle has half that area. The section introduces the property that a median divides a triangle into two triangles of equal area, a surprising and important result. It encourages reflection on whether these two triangles can be cut and rearranged to cover each other. The section concludes with Heron's formula, which allows calculation of the area of a triangle using only the lengths of its sides, without needing the height. Several examples demonstrate Heron's formula for equilateral, isosceles, and scalene triangles, verifying results with the base-height formula.

📊 Diagram: Fig. 6.20A; Fig. 6.20B; Fig. 6.21A; Fig. 6.21B; Fig. 6.22; Fig. 6.23; Fig. 6.24; Fig. 6.25

🧪 Activity: Think and Reflect: Explore whether two triangles formed by a median can be cut and rearranged to cover each other.

🔗 Connection: Leads to the next section on Brahmagupta's formula for cyclic quadrilaterals and generalizations.