Area of a Triangle MCQ Quiz in मल्याळम - Objective Question with Answer for Area of a Triangle - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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:

Formula:

Area of triangle = \(|\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1\\x_3 & y_3 & 1 \end{vmatrix}|\)

Calculation:

Given:

The area of the triangle formed by vertices (-2 , 0), (-2 , 8) and (t , 4) is 12 square units.

Area = \(|\frac{1}{2}\begin{vmatrix} -2 & 0& 1 \\ -2 & 8 & 1\\t & 4 & 1 \end{vmatrix}|\)

⇒ |-2(8 × 1 - 4 × 1) -0(-2 × 1 - t × 1) + 1(-2 × 4 - t × 8)| = 12 × 2

⇒ |-16 - 8t| = 24

⇒ |- 2 - t| = 3

⇒ - 2 - t = 3 or  -2 - t = -3

⇒ t = -5, 1

So, the correct answer is option 2.

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

Given;

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

Concept:

Use concept of area of triangle using matrix.

Calculation:

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

\(\frac{1}{2}\left|\begin{array}{rrr} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{array}\right|=9\)

\(\frac{1}{2}[-3(0-k)-0+1(3k-0)]=9\)

3k + 3k = 18

k = 3

Hence the option (1) is correct.

Concept:

Formula:

Area of triangle = \(|\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1\\x_3 & y_3 & 1 \end{vmatrix}|\)

Calculation:

Given:

The area of the triangle formed by vertices (-2 , 0), (-2 , 8) and (t , 4) is 12 square units.

Area = \(|\frac{1}{2}\begin{vmatrix} -2 & 0& 1 \\ -2 & 8 & 1\\t & 4 & 1 \end{vmatrix}|\)

⇒ |-2(8 × 1 - 4 × 1) -0(-2 × 1 - t × 1) + 1(-2 × 4 - t × 8)| = 12 × 2

⇒ |-16 - 8t| = 24

⇒ |- 2 - t| = 3

⇒ - 2 - t = 3 or  -2 - t = -3

⇒ t = -5, 1

So, the correct answer is option 2.

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 

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:

• 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.