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

Given:

A triangle with sides represented by prime numbers and the perimeter of the triangle as 24 cm.

Concept Used:

A triangle is valid if the sum of its two sides is greater than the third side. P = a + b + c {where a,b,c are the sides of triangle} and Area of triangle = \(\frac{1}{2}. base . height\)

Solution:

We have prime numbers as 2,3,5,7,11,13,17 and 19.

Since the perimeter of the triangle is 24cm so we have taken prime numbers till 19 only.

Now let us consider pairs of sides of triangles represented by these prime numbers.

We have, (11,11,2), (19,3,2), and (17,5,2) these three pairs can be obtained since the perimeter of the triangle is 24cm.

As we know the perimeter of the triangle is equal to the sum of its sides.

⇒ P = a + b + c              {where a,b,c are the sides of triangle}

Only one pair from the pairs (11,11,2), (19,3,2), (17,5,2) satisfied the triangle property which states that the sum of two sides of a triangle is greater than the third side.

In (11,11,2): a + b > c 

⇒ 11 + 11 > 2 , 11 + 2 > 11

In (19,3,2): a + b > c this condition does not satisfied.

⇒ 3 + 2 < 19 similarly for (17,5,2): 5 + 2 < 17

⇒ a = 11, b = 11 and c = 2 

For this P = 24cm 

Area of triangle = \(\frac{1}{2}. base . height\)

We have,

height of isosceles triangle = \( \sqrt{a^2 - \frac{b^2}{4}} \)      {where a = side and b is the base}

height = \(\sqrt{(11)^2 - \frac{(2)^2}{4}} \)

⇒ height = \(\sqrt{121 -1} \)

⇒ height = \(\sqrt{120}\)

Area of triangle = \(\frac{1}{2}.2. \sqrt{120}\)

⇒ area of triangle = \(\sqrt{120}\)

⇒ area of triangle = 2 \(\sqrt{30}\) cm²

\(\therefore\) Option 2 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 

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 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:

• 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.