Area of a Triangle MCQ Quiz - Objective Question with Answer for Area of a Triangle - Download Free PDF

Concept 

The area of a triangle given its three vertices is 

\(​A = \frac{1}{2} | x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2 ) | \)   

Explanation:

\(A = \frac{1}{2} [|1(5-3) + 2(3-2) + 4(2-5)|] \)  

\(A = \frac{1}{2} |1× 2 + 2 × 1 + 4 × (-3) | \)

\(A = \frac{1}{2} | 2 + 2 - 12| \)  

\(A = \frac{1}{2} |-8| \)  

\(A = \frac{1}{2} × 8 = 4 \)   

Hence Option(2) is the correct answer.

Concept 

The area of a triangle given its three vertices is 

\(​A = \frac{1}{2} | x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2 ) | \)   

Explanation:

\(A = \frac{1}{2} [|1(5-3) + 2(3-2) + 4(2-5)|] \)  

\(A = \frac{1}{2} |1× 2 + 2 × 1 + 4 × (-3) | \)

\(A = \frac{1}{2} | 2 + 2 - 12| \)  

\(A = \frac{1}{2} |-8| \)  

\(A = \frac{1}{2} × 8 = 4 \)   

Hence Option(2) is the correct answer.

Concept 

The area of a triangle given its three vertices is 

\(​A = \frac{1}{2} | x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2 ) | \)   

Explanation:

\(A = \frac{1}{2} [|1(5-3) + 2(3-2) + 4(2-5)|] \)  

\(A = \frac{1}{2} |1× 2 + 2 × 1 + 4 × (-3) | \)

\(A = \frac{1}{2} | 2 + 2 - 12| \)  

\(A = \frac{1}{2} |-8| \)  

\(A = \frac{1}{2} × 8 = 4 \)   

Hence Option(2) is the correct answer.

Concept:
The area of a triangle with vertices \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right) \text { and }\left(x_{3}, y_{3}\right)\) is given by:

\(\frac{1}{2}\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|\)

Calculations:

Vertices of triangle are (-2, a) (2, -6) and (5, 4) 

Area = 35sq units

⇒ \(35=\frac{1}{2}\left|\begin{array}{ccc} -2 & a & 1 \\ 2 & -6 & 1 \\ 5 & 4 & 1 \end{array}\right|\)

⇒ \(35=\frac{1}{2}\)[(-2)(-6)(1) + (a)(1)(5) + (1)(2)(4) - (1)(-6)(5) - (a)(2)(1) - (-2)(1)(4)]

⇒ \(35=\frac{1}{2}(12+5 a+8+30-2 a+8)\)

⇒ \(35=\frac{1}{2}(58+3 a)\)

⇒ 70 = 58 + 3a

⇒ 12 = 3a

⇒ \(a=\frac{12}{3}=4\)

Hence, the correct option is 1) 4.

Concept:

• Area of a Triangle using Coordinates: The area of a triangle with vertices at (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by the formula:
• Area = (1/2) × |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
• This formula uses the concept of determinant geometry and gives the absolute area regardless of vertex order.
• Given area helps in forming an equation to find unknown coordinate values.

Calculation:

Given, vertices A(−3, 0), B(3, 0), C(0, k) and area = 9 sq. units

Using the formula: Area = (1/2) × |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

⇒ (1/2) × |−3(0 − k) + 3(k − 0) + 0(0 − 0)| = 9

⇒ (1/2) × |−3(−k) + 3k| = 9

⇒ (1/2) × |3k + 3k| = 9

⇒ (1/2) × |6k| = 9

⇒ |6k| = 18

⇒ 6k = ±18

⇒ k = ±3

∴ The value of k is 3 or −3.

Here Option 3 will be the correct answer.

Concept:

Area of a triangle with vertices (x1, y1) , (x2, y2), (x3, y3) is given by

​Area = \(\rm \dfrac 12 \rm \begin{vmatrix} \rm x_1&\rm y_1 &1 \\ x_2& \rm y_2&1 \\ \rm\rm x_3 &\rm y_3&1 \end{vmatrix}\)

Calculations:

Given that, vertices of triangle are (-3, 0), (3, 0) and (0, k)

By using the above formula,

​⇒ Area = \(\rm \dfrac 12 \rm \begin{vmatrix} \rm -3&\rm 0 &1 \\ 3& 0&1 \\ 0 &k&1 \end{vmatrix}\)

⇒ Area = \(\dfrac 12\)[-3(0 - k) - 0 + 1(3k)]

⇒ Area = 3k

According to the question, area of triangle is 9 square unit,

⇒ 3k = 9

⇒ k = 3

∴ Required value of k is 3 unit.

Concept:

If \(\rm (x_1, y_1), (x_2, y_2)\) and \(\rm (x_3, y_3)\) are the vertices of a triangle then, 

Area = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

Properties of Determinants

\(\rm \begin{vmatrix} \rm x_1 & \rm y_1 & kx \\\ \rm x_2 & \rm y_2 & ky \\\ \rm x_3 & \rm y_3 & kz \end{vmatrix} = k \begin{vmatrix} \rm x_1 & \rm y_1 & x \\\ \rm x_2 & \rm y_2 & y \\\ \rm x_3 & \rm y_3 & z \end{vmatrix}\)

Calculations:

If \(\rm (x_1, y_1), (x_2, y_2)\) and \(\rm (x_3, y_3)\) are the vertices of a triangle then,

Area = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

Given: the area is ‘k’ square units

⇒ k = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

⇒ \(\rm \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix} = 2k\)

Now,  \(\begin{vmatrix} \rm x_1 & \rm y_1 & 4 \\\ \rm x_2 & \rm y_2 & 4 \\\ \rm x_3 & \rm y_3 & 4 \end{vmatrix}^2\)

= \(\begin{vmatrix} \rm x_1 & \rm y_1 & 4 \\\ \rm x_2 & \rm y_2 & 4 \\\ \rm x_3 & \rm y_3 & 4 \end{vmatrix}\begin{vmatrix} \rm x_1 & \rm y_1 & 4 \\\ \rm x_2 & \rm y_2 & 4 \\\ \rm x_3 & \rm y_3 & 4 \end{vmatrix}\)

= \(4\begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}.4\begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

= 16.(2k)(2k)

= 64k²

Concept:

If \(\rm (x_1, y_1), (x_2, y_2)\) and \(\rm (x_3, y_3)\) are the vertices of a triangle then, 

Area = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

Note: Area is always a positive quantity, therefore we always take the absolute value of the determinant for the area. 

Calculation:

Given vertices are A (1, 1) ,B ( 6, 0) and C ( 3, 2).

We know that area of triangle ABC is given by, 

Δ = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

⇒ Δ = \(\rm\frac{1}{2}\begin{vmatrix} 1& 1 & 1\\ 6& 0 & 1\\ 3& 2 & 1 \end{vmatrix}\) 

⇒ Δ = \(\frac{1}{2}\left [ 1\left ( 0-2 \right )-1 (6-3)+1 (12-0)\right ]\) 

⇒ Δ =  \(\frac{7}{2}\)  

The correct option is 1.

Concept:

The area of a triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃), is given by the expression

\(\rm Area = \frac{1}{2}\begin{vmatrix} x_{1} &x_{2} & 1\\ y_{1}& y_{2} &1 \\ z_{1}&z_{2} & 1 \end{vmatrix}\)  

Calculation:

Given, Area of triangle, A = 4 sq. unit and vertices (K, 0), (4, 0), (0, 2) 

since the area is always Positive But the determinant can be both positive and negative.  

∴ Δ = ± 4 . 

⇒ ± 4 = \(\rm = \frac{1}{2}\begin{vmatrix} k &0 & 1\\ 4& 0 &1 \\ 0&2 & 1 \end{vmatrix}\)  

⇒ ± 4  \(\rm = \frac{1}{2} \left [ k(0-2) - 0 (4-0)+1(8-0)\right ]\) 

⇒ ± 8 = -2k + 8 

So, 8 = -2k + 8 or  -8 = -2k +8 

⇒ k = 0 or  8  .

The correct option is 2.

Concept:

The area of the triangle using determinant is given by-

\(\Delta = \frac{1}{2}\begin{vmatrix} x_{1} &y_{1} &1 \\ x_{2} &y_{2} &1 \\ x_{3} &y_{3} &1 \end{vmatrix}\)      ----(1)

Formula used:

The n^th term of a GP is given by;

T_n = a.r^{(n - 1)}      ----(2)

where a = first term and r = common ratio.

Calculation:

Since a, x₁, x₂ are in GP with common ratio r,

⇒ x₁ = a.r, x₂ = a.r²      [using (2)]

Since b, y1, y2 are in GP with common ratio s,

⇒ y1 = b.s, y2 = b.s²      [using (2)]

∴ The area of triangle is given by-

\(\Delta = \frac{1}{2}\begin{vmatrix} a &b &1 \\ x_{1} &y_{1} &1 \\ x_{2} &y_{2} &1 \end{vmatrix}\)      [using (1)]

\(\Rightarrow \Delta = \frac{1}{2}\begin{vmatrix} a &b &1 \\ ar &bs &1 \\ ar^{2} &bs^{2} &1 \end{vmatrix}\)

\(\Rightarrow \Delta = \frac{1}{2}ab\begin{vmatrix} 1 &1 &1 \\ r &s &1 \\ r^{2} &s^{2} &1 \end{vmatrix}\)

Applying C₁ → C₁ - C₃ and C₂ → C₂ - C₃, we get,

\(\Rightarrow \Delta = \frac{1}{2}ab\begin{vmatrix} 0 &0 &1 \\ r-1 &s-1 &1 \\ r^{2}-1 &s^{2}-1 &1 \end{vmatrix}\)

\(\Rightarrow \Delta = \frac{1}{2}ab(r-1)(s-1)\begin{vmatrix} 0 &0 &1 \\ 1 &1 &1 \\ r+1 &s+1 &1 \end{vmatrix}\)      (∵ (a - b)(a + b) = a² - b²)

\(\Rightarrow \Delta = \frac{1}{2}ab(r-1)(s-1)(s-r)\)

Hence, the area of the triangle is \(\dfrac{1}{2}ab(r-1)(s-1)(s-r) \).

Given:

 vertices are A(4, 8),B(-6, 2) and C(5, 4)

Concept:

Area of triangle whose vertices are \(\rm (x_1,y_1),(x_2,y_2), and (x_3,y_3)\) is

\(\rm A = \frac{1}{2} |x_1(y_2 − y_3) + x_2(y_3 − y_1) + x_3(y_1 − y_2)|\)

Calculation:

vertices are A(4, 8),B(-6, 2) and C(5, 4)

Then the area is

\(\rm A = \frac{1}{2} |x_1(y_2 − y_3) + x_2(y_3 − y_1) + x_3(y_1 − y_2)|\)

\(\rm A=\frac{1}{2}|4(2-4)-6(4-8)+5(8-2)| \)

\(\rm A=\frac{1}{2}|-8+24+30|\)

\(\rm A=\frac{1}{2}\cdot46\)

\(\rm A=23 \ unit^2\)

Hence the option (4) is correct.

Concept:

Area of a triangle with points (x₁, y₁), (x₂, y₂) and (x₃, y₃) = \(\rm {1\over2}\begin{vmatrix} \rm x_1 & \rm y_1 & 1\\ \rm x_2 & \rm y_2& 1\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

Calculation

Given area of triangle = 10 sq. units

⇒ \({1\over2}\begin{vmatrix}4&2 & 1\\ -1 & 2& 1\\ 3 & a & 1 \end{vmatrix}\) = 10

R₁ = R₁ - R₂

⇒ \(\begin{vmatrix}5 & 0 & 0 \\ -1&2 & 1\\ 3 & a & 1 \end{vmatrix}\) = 20

⇒ |5 (2 - a) - 0 + 0| = 20

⇒ |2 - a| = 4 

⇒ a = -2 or 6

Concept: 

Area of a triangle with vertices (x₁, y₁) , (x₂, y₂), (x₃, y₃) is given by

Area = \(\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|\)

Explanation:

Given, the area of a triangle with vertices (-3, 0), (3, 0), and (0, k) is 9 square units.

∴ Area = \(\frac{1}{2}\left|\begin{array}{lll}-3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1\end{array}\right|\)

⇒ Area = \(\frac{1}{2}\)[-3(0 - k) - 0 + 1(3k - 0)]

⇒ Area = \(\frac{1}{2}\) ×  6k

⇒ Area = 3k

The area of a triangle with vertices (-3, 0), (3, 0), and (0, k) is 9 square units.

∴ 3k = 9

⇒ k = 3

Concept-

Area of triangle = \( \frac{1}{2} \{ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\}\)

Calculation-

If A(x₁, y₁) B(x2, y2) and C(x₃, y₃) be the vertices of triangle than its area will be 

Here x₁ = 3, x₂ = 4, x₃ = 0 and y₁ = -2, y2 = 0, y3 = -4

Area of triangle = \( {1} \over {2}\)[3(0 - (-4)) + 4(-4 - (-2)) + 0(-2 - 0)]

Area = \( {1} \over {2}\)(12 + (-8) + 0)

Area = 2

∴ The area of the triangle with vertices (3, -2), (4, 0), (0, -4) is 2.

Concept:

Area of a triangle with points (x1, y1), (x2, y2) and (x3, y3) = \(\rm {1\over2}\begin{vmatrix} \rm x_1 & \rm y_1 & 1\\ \rm x_2 & \rm y_2& 1\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

Calculation

Area of the triangle A = \(\rm {1\over2}\begin{vmatrix} \rm x_1 & \rm y_1 & 1\\ \rm x_2 & \rm y_2& 1\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

⇒ A = \({1\over2}\begin{vmatrix} 1 &-2 & 1\\ 3& 1& 1\\ 2 & 4 & 1 \end{vmatrix}\)

⇒ 2A = 1(1 - 4) - (-2)(3 - 2) + 1(12 - 2)

⇒ 2A  = -3 + 2 + 10

⇒ A = \(9\over2\) = 4.5 sq. unit