Area of Triangle PYQs
Practice Class 12 CBSE Board Previous Year Questions (2014-2026)
Q1
2024 (MCQ)
00:00
The area of a triangle with vertices $(2, 7), (1, 1)$ and $(10, 8)$ is: 1 Mark
Area $= \frac{1}{2} |\begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix}|$.
$= \frac{1}{2} |2(1-8) - 7(1-10) + 1(8-10)|$.
$= \frac{1}{2} |-14 + 63 - 2| = \frac{47}{2} = 23.5$ sq. units.
23.5 sq units
Q2
2019
00:00
Find the value of $k$ if the area of the triangle with vertices $(k, 0), (4, 0)$ and $(0, 2)$ is 4 square units. 2 Marks
Area $= \frac{1}{2} |\begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix}| = 4 \Rightarrow |\Delta| = 8$.
Expanding along $C_2$: $|-2(k - 4)| = 8 \Rightarrow |k - 4| = 4$.
Case 1: $k - 4 = 4 \Rightarrow k = 8$.
Case 2: $k - 4 = -4 \Rightarrow k = 0$.
So $k = 0, 8$.
0, 8
Q3
2023
00:00
Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants. 2 Marks
Let $P(x, y)$ be any point on the line $AB$. Since $A, B, P$ are collinear, area $= 0$.
$\frac{1}{2} \begin{vmatrix} x & y & 1 \\ 1 & 3 & 1 \\ 0 & 0 & 1 \end{vmatrix} = 0$.
Expanding along $R_3$: $1(3x - y) = 0 \Rightarrow y = 3x$.
y = 3x
Q4
2025 Board
00:00
If the area of the triangle with vertices $(x, 4), (2, -6)$ and $(5, 4)$ is 35 sq. units, find the value(s) of $x$ using determinants. 3 Marks
Area $= \frac{1}{2} |\begin{vmatrix} x & 4 & 1 \\ 2 & -6 & 1 \\ 5 & 4 & 1 \end{vmatrix}| = 35 \Rightarrow |\Delta| = 70$.
Expanding along $R_1$: $|x(-6-4) - 4(2-5) + 1(8+30)| = 70$.
$|-10x + 12 + 38| = 70 \Rightarrow |-10x + 50| = 70$.
Case 1: $-10x + 50 = 70 \Rightarrow -10x = 20 \Rightarrow x = -2$.
Case 2: $-10x + 50 = -70 \Rightarrow -10x = -120 \Rightarrow x = 12$.
So $x = -2, 12$.
x = -2, 12