# Mathematics Questions and Answers – Area of a Triangle

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This set of Mathematics 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=$$\frac{1}{2}\begin{Vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{Vmatrix}$$

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

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

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

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

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.

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

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) –$$\frac{8}{5}$$

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

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

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

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.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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