Area of a Triangle - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 4 Multiple Choice Questions & Answers (MCQs) focuses on “Area of a Triangle”.

1. Which of the following is the formula for finding the area of a triangle with the vertices (x₁,y₁), (x₂,y₂), (x₃,y₃).
a) Δ=\(\begin{Vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{Vmatrix}\)
b) Δ=\(\frac{1}{2}\begin{Vmatrix}x_1&y_1&1\\x_2&y_2&0\\x_3&y_3&1\end{Vmatrix}\)
c) Δ=\(\frac{1}{2}\begin{Vmatrix}x_1&y_1&1\\x_2&1&1\\x_3&y_3&1\end{Vmatrix}\)
d) Δ=\(\frac{1}{2}\begin{Vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{Vmatrix}\)
View Answer

Answer: d
Explanation: The area of a triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃) is given by
Δ=\(\frac{1}{2}\begin{Vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{Vmatrix}\).

2. What is the area of the triangle whose vertices are (0,1), (0,2), (1,5)?
a) 1 sq.unit
b) 2 sq.units
c) \(\frac{1}{3}\) sq.units
d) \(\frac{1}{2}\) sq.units
View Answer

Answer: d
Explanation: The area of the triangle with vertices (0,1), (0,2), (1,5) is given by
Δ=\(\frac{1}{2}\begin{Vmatrix}0&1&1\\0&2&1\\1&5&1\end{Vmatrix}\)
Expanding along C₁, we get
Δ=\(\frac{1}{2}\){(0-0+1(1-2)}=\(\frac{1}{2}\) |-1|=\(\frac{1}{2}\) sq.units.

3. The area of the triangle formed by three collinear points is zero.
a) True
b) False
View Answer

Answer: a
Explanation: The given statement is true. If the three points are collinear then they will be lying in a single line. Therefore, the area of the triangle formed by collinear points is zero.

4. Find the value of k for which (1,2), (3,0), (2,k) are collinear.
a) 0
b) -1
c) 2
d) 1
View Answer

Answer: d
Explanation: The area of triangle formed by collinear points is zero.
Δ=\(\frac{1}{2}\) \(\begin{Vmatrix}1&2&1\\3&0&1\\2&k&1\end{Vmatrix}\)=0
Expanding along C₂, we get
\(\frac{1}{2}\){-2(3-2)+0-k(1-3)}=0
\(\frac{1}{2}\) {-2+2k}=0
∴k=1

5. What is the area of the triangle if the vertices are (0,2), (0, 0), (3, 0)?
a) 1 sq.unit
b) 5 sq.units
c) 2 sq.units
d) 3 sq.units
View Answer

Answer: d
Explanation: The area of the triangle (0,2), (0, 0), (3, 0) with vertices is given by
Δ=\(\frac{1}{2}\begin{Vmatrix}0&2&1\\0&0&1\\3&0&1\end{Vmatrix}\)
Expanding along R₃, we get
Δ=\(\frac{1}{2}\) {0-0+3(2-0)}
Δ=3 sq.units.

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6. Find the equation of the line joining A(2,1) and B(6,3) using determinants.
a) 2y-x=0
b) 2y-x=0
c) y-x=0
d) y-2x=0
View Answer

Answer: a
Explanation: Let C(x,y) be a point on the line AB. Thus, the points A(2,1), B(6,3), C(x,y) are collinear. Hence, the area of the triangle formed by these points will be 0.
⇒Δ=\(\frac{1}{2}\begin{Vmatrix}2&1&1\\6&3&1\\x&y&1\end{Vmatrix}\)=0
Expanding along C₃, we get
\(\frac{1}{2}\) {1(6y-3x)-1(2y-x)+1(6-6)}=0
\(\frac{1}{2}\) {6y-3x-2y+x}=\(\frac{1}{2}\) {4y-2x}=0
⇒2y-x=0

7. Find the value of k if the area is \(\frac{7}{2}\) sq. units and the vertices are (1,2), (3,5), (k,0).
a) \(\frac{8}{3}\)
b) –\(\frac{8}{3}\)
c) –\(\frac{7}{3}\)
d) –\(\frac{8}{5}\)
View Answer

Answer: b
Explanation: Given that the vertices are (1,2), (3,5), (k,0)
Therefore, the area of the triangle with vertices (1,2), (3,5), (k,0) is given by
Δ=\(\frac{1}{2}\begin{Vmatrix}1&2&1\\3&5&1\\k&0&1\end{Vmatrix}\)=\(\frac{7}{2}\)
Expanding along R₃, we get
\(\frac{1}{2}\) {k(2-5)-0+1(5-6)}=\(\frac{1}{2}\) {-3k-1}=\(\frac{7}{2}\)
⇒-\(\frac{1}{2}\) (3k+1)=\(\frac{7}{2}\)
3k=-8
k=-\(\frac{8}{3}\).

8. Find the area of the triangle with the vertices (2,3), (4,1), (5,0).
a) 3 sq.units
b) 2 sq.units
c) 0
d) 1 sq.unit
View Answer

Answer: c
Explanation: The area of the triangle with vertices (2,3), (4,1), (5,0) is given by
Δ=\(\frac{1}{2}\begin{Vmatrix}2&3&1\\4&1&1\\5&0&1\end{Vmatrix}\)
Applying R₂→R₂-R₃
Δ=\(\frac{1}{2}\begin{Vmatrix}2&3&1\\-1&1&0\\5&0&1\end{Vmatrix}\)
Expanding along R₂, we get
Δ=\(\frac{1}{2}\) {-(-1)(3-0)+1(2-5)}
Δ=\(\frac{1}{2}\) (0-0)=0.

9. Find the equation of the line joining A(5,1), B(4,0) using determinants.
a) 4x-y=4
b) x-4y=4
c) x-y=4
d) x-y=0
View Answer

Answer: c
Explanation: Let C(x,y) be a point on the line AB. Thus, the points A(5,1), B(4,0), C(x,y) are collinear. Hence, the area of the triangle formed by these points will be 0.
⇒Δ=\(\frac{1}{2}\begin{Vmatrix}5&1&1\\4&0&1\\x&y&1\end{Vmatrix}\)=0
Applying R₁→R₁-R₂
\(\frac{1}{2}\begin{Vmatrix}1&1&0\\4&0&1\\x&y&1\end{Vmatrix}\)=0
Expanding along R₁, we get
=\(\frac{1}{2}\) {1(0-y)-1(4-x)}=0
=\(\frac{1}{2}\) {-y-4+x}=0
⇒x-y=4.

10. Find the value of k for which the points (3,2), (1,2), (5,k) are collinear.
a) 2
b) 5
c) 4
d) 9
View Answer

Answer: a
Explanation: Given that the vertices are (3,2), (1,2), (5,k)
Therefore, the area of the triangle with vertices (3,2), (1,2), (5,k) is given by
Δ=\(\frac{1}{2}\) \(\begin{Vmatrix}3&2&1\\1&2&1\\5&k&1\end{Vmatrix}\)=0
Applying R₁→R₁-R₂, we get
\(\frac{1}{2}\begin{Vmatrix}2&0&0\\1&2&1\\5&k&1\end{Vmatrix}\)=0
Expanding along R₁, we get
\(\frac{1}{2}\) {2(2-k)-0+0}=0
2-k=0
k=2.

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