NCERT Solutions for Class 9th Maths Chapter 6 In-text Questions — Area of a Triangle
Book page 132 Updated on2026-09-08
Q1.
You may wonder, like earlier, is there a gap in our argument? What would we do if angle EFG is obtuse and the triangle were shaped like triangle EFG in Fig. 6.20B? Please work out the answer to this question.
Answer
There is a gap, and it is closed by subtracting instead of adding.
The proof on page 132 drops the perpendicular from the apex E to the base FG, splitting the base as b = b₁ + b₂ and the triangle into two right triangles. If ∠EFG is obtuse, the foot of the perpendicular falls outside the segment FG, on FG produced backwards. There is no b₁ + b₂ to write.
Let the foot of the perpendicular from E be J, with F between J and G. Let JF = p and FG = b, so JG = p + b, and let the height be h.
ar(ΔEJG) = ½ (p + b) h ar(ΔEJF) = ½ p h
ΔEFG = ΔEJG − ΔEJF = ½ (p + b) h − ½ p h = ½ p h + ½ b h − ½ p h = ½ b h
When ∠EFG is obtuse the foot J lies outside FG, so ΔEFG is a difference of two right triangles, not a sum.
Why it happens: the same expression ½bh comes out because p cancels. Whether the foot of the perpendicular lands inside the base, on an endpoint, or outside it, the area depends only on the base and the perpendicular distance from the opposite vertex to the line of that base. The parallelogram route (two congruent copies of the triangle make a parallelogram) avoids the case-splitting altogether — which is why the book calls it "a nicer way".