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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

Solved CBSE Board Question: Area of Triangle Using Integration

Problem: Using integration, find the area of the region bounded by the triangle whose vertices are A(1, 0), B(2, 2), and C(3, 1). (CBSE Board Exam)

Step 1: Equations of the Boundary Lines
  • Equation of line AB: y - 0 = ((2 - 0)/(2 - 1))(x - 1) => y = 2(x - 1)
  • Equation of line BC: y - 2 = ((1 - 2)/(3 - 2))(x - 2) => y = - (x - 2) + 2 => y = 4 - x
  • Equation of line AC: y - 0 = ((1 - 0)/(3 - 1))(x - 1) => y = (1/2)(x - 1)
Step 2: Area Integration Working

Area(Triangle ABC) = Area under AB (x from 1 to 2) + Area under BC (x from 2 to 3) - Area under AC (x from 1 to 3).
= integral from 1 to 2 of 2(x - 1) dx + integral from 2 to 3 of (4 - x) dx - integral from 1 to 3 of (1/2)(x - 1) dx.
= [ (x - 1)^2 ]_1^2 + [ 4x - x^2/2 ]_2^3 - (1/2)[ (x - 1)^2 / 2 ]_1^3.
= (1 - 0) + [ (12 - 9/2) - (8 - 2) ] - (1/4)(4 - 0) = 1 + [ 7.5 - 6 ] - 1 = 1.5 sq units (or 3/2 sq units).

Examiner Tip: In CBSE evaluation criteria, you can verify your answer using the coordinate geometry determinant formula: (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| = (1/2)|1(2 - 1) + 2(1 - 0) + 3(0 - 2)| = (1/2)|1 + 2 - 6| = 3/2 sq units. Always show full integration steps to earn full marks.