Area of Triangles (Using Integration) PYQs
Practice Class 12 CBSE Board Previous Year Questions (2008-2026)
Q1
2013 Board
00:00
Using integration, find the area of the triangle whose vertices are $(-2, 2)$, $(0, 5)$, and $(3, 2)$. 5 Marks
Step 1: Find Equations of the Sides
Let $A=(-2,2), B=(0,5), C=(3,2)$.
Equation of AB: $y - 2 = \frac{5-2}{0-(-2)}(x+2) \Rightarrow y = \frac{3}{2}x + 5$.
Equation of BC: $y - 5 = \frac{2-5}{3-0}(x-0) \Rightarrow y = -x + 5$.
Equation of AC: $y = 2$ (Horizontal line).
Step 2: Set up the Integrals
Area $= \int_{-2}^0 (y_{AB} - y_{AC}) \, dx + \int_0^3 (y_{BC} - y_{AC}) \, dx$.
Area $= \int_{-2}^0 (\frac{3}{2}x + 5 - 2) \, dx + \int_0^3 (-x + 5 - 2) \, dx$.
Area $= \int_{-2}^0 (\frac{3}{2}x + 3) \, dx + \int_0^3 (-x + 3) \, dx$.
Step 3: Integrate and Evaluate
$= [\frac{3x^2}{4} + 3x]_{-2}^0 + [-\frac{x^2}{2} + 3x]_0^3$.
$= (0) - (3 - 6) + (-4.5 + 9) - (0) = 3 + 4.5 = 7.5$ sq. units.
7.5 sq. units