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=\(\frac{1}{2}\begin{Vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{Vmatrix}\)
View Answer

Answer: d
Explanation: The area of a triangle with vertices (x1,y1), (x2,y2), (x3,y3) is given by
Δ=\(\frac{1}{2}\begin{Vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{Vmatrix}\).
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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.
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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=-\(\frac{8}{3}\).
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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.
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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.

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

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter