Area MCQ Quiz in বাংলা - Objective Question with Answer for Area - বিনামূল্যে ডাউনলোড করুন [PDF]

CONCEPT:

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = 

\(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

CALCULATION:

Given: A (5, 3), B (6, - 4), C (- 3, - 2) and D (- 4, 7) are the vertices of a quadrilateral ABCD

Here, we have to find the area of quadrilateral ABCD

Area of quadrilateral ABCD = Area of ΔABC + Area of Δ ACD

Let's find out the area of ΔABC

∵ A (5, 3), B (6, - 4), C (- 3, - 2) are the vertices of ΔABC

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

⇒ Area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{5}}&{{3}}&1\\ {{6}}&{{-4}}&1\\ {{-3}}&{{-2}}&1 \end{array}} \right|\)

CONCEPT:

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

CALCULATION:

Given: A (0, 0), B (0, 10), C (8, 2) and D (8, 7) are the vertices of a quadilateral ABCD

Here, we have to find the area of quadilateral ABCD

Area of quadilateral ABCD = Area of ΔABC + Area of Δ ACD

Let's find out the area of ΔABC

∵ A (0, 0), B (0, 10), C (8, 2) are the vertices of ΔABC

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

⇒ Area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{0}}&{{0}}&1\\ {{0}}&{{10}}&1\\ {{8}}&{{2}}&1 \end{array}} \right|\)

CONCEPT:

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

CALCULATION:

Given: A (- 2, - 1), B (1, 0), C (4, 3) and D (1, 2) are the vertices of a quadilateral ABCD

Here, we have to find the area of quadilateral ABCD

Area of quadilateral ABCD = Area of ΔABC + Area of Δ ACD

Let's find out the area of ΔABC

∵ A (- 2, - 1), B (1, 0), C (4, 3) are the vertices of ΔABC

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

⇒ Area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{- 2}}&{{-1}}&1\\ {{1}}&{{0}}&1\\ {{4}}&{{3}}&1 \end{array}} \right|\)

CONCEPT:

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

CALCULATION:

Given: A (2, - 2), B (8, 4), C (5, 7) and D (-1, 1) are the vertices of a quadilateral ABCD

Here, we have to find the area of quadilateral ABCD

Area of quadilateral ABCD = Area of ΔABC + Area of Δ ACD

Let's find out the area of ΔABC

∵ A (2, - 2), B (8, 4), C (5, 7) are the vertices of ΔABC

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

⇒ Area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{2}}&{{-2}}&1\\ {{8}}&{{4}}&1\\ {{5}}&{{7}}&1 \end{array}} \right|\)

Calculation:

Given,

Points A(1, 2, -1), B(2, 5, -2), and C(4, 4, -3) are three vertices of the rectangle.

We need to find the area of the rectangle formed by the vectors \( AB \) and \( BC \).

Length of vector \( AB \):

\( \text{Length of AB} = \sqrt{(2 - 1)^2 + (5 - 2)^2 + (-2 - (-1))^2} \)

\(= \sqrt{1^2 + 3^2 + 1^2} = \sqrt{11} \)

Length of vector \( BC \):

\( \text{Length of BC} = \sqrt{(4 - 2)^2 + (4 - 5)^2 + (-3 - (-2))^2} \)

\( = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{6} \)

Area of the rectangle is the product of the lengths of vectors \( AB \) and \( BC \):

\( \text{Area} = \sqrt{11} \times \sqrt{6} = \sqrt{66} \)

∴ The area of the rectangle is \( \sqrt{66} \) square units.

Hence, the correct answer is Option 3.

\( { x }^{ 2 }+{ y }^{ 2 }+2gx+2fy+c=0 \) ...(1)

Since \(A(-3,4)\) and \(B(5,4)\) lies on (1), we get

\( 25-6g+8f+c=0 \) ...(2)

and, \( 41+10+8f+c=0 \) ...(3)

Also, centre of (1) lies on \(4y=x+7\)

\(\therefore -g+4f+7=0\) ...(4)

Subtracting (2) from (3), we get

\( 16+16g=0\Rightarrow g=-1 \) ...(5)

Solving (4) and (5), we get \(f=-2.\)

Substituting the values of \(g\) and \(f\) in (3), we get

\( 41-10-16+c=0\Rightarrow c=-15.\)

\(\therefore\) The equation of the circle (1) becomes

\( { x }^{ 2 }+{ y }^{ 2 }-2x-4y-15=0.\)

Radius of the circle is \( =\sqrt { 1+4+15 } =2\sqrt { 5 } .\)

Since the rectangle is inscribed in the circle, Its diagonal will be the diameter of the circle.

\(\therefore\) The length of the diagonal of the rectangle \( =2\left( 2\sqrt { 5 } \right) =4\sqrt { 5 } .\)

Also, the length of the rectangle

\( =AB=\sqrt { { \left( 5+3 \right) }^{ 2 }+{ \left( 4-4 \right) }^{ 2 } } =8.\)

\(\therefore\) Its breadth \( =\sqrt { { \left( 4\sqrt { 5 } \right) }^{ 2 }-{ \left( 8 \right) }^{ 2 } } =\sqrt { 80-64 } =\sqrt { 16 } =4.\)

hence, the area of rectangle \(=8\times 4=32.\)

Concept:

Area of triangle = \(\frac12\) × base × height

Explanation:

Let the side of each square ABCD be a

Then AB = BC = CD = DA = a and AQ = DP = a/2

So area of ABCD  = a²

Area of ABQ = \(\frac12\) × a × \(\frac a2\) = \(\frac{a^2}4\)

Area of QPD = \(\frac12\) × \(\frac a2\) × \(\frac a2\) = \(\frac{a^2}8\)

Also given area of QBCP = 15

Now, Area of ABQ + area of QPD + area of QBCP = area of ABCD

⇒ \(\frac{a^2}4\) + \(\frac{a^2}8\) + 15 = a2

⇒ \(\frac{3a^2}8\) + 15 = a2

⇒ 3a2 + 120 = 8a2

⇒ 5a2 = 120 a2 = 24

Hence area of square ABCD = 24

(2) is correct

Formula used:

Area of Rectangle = length × width

Calculation:

Area of the path = 2 × 20 × 2 + 2 × 30 × 2 

⇒ 80 + 120 = 200 m²

∴ The area of the path is 200 m².

Formula used:

Area of trapezium = \(\frac{1}{2}\times(Sum\ of\ Parallel\ sides)\times Height\)

Calculation: 

By using the above formula,

Area of the wall = \(\frac{1}{2}×(5\ +\ 3)×4 \) = 16 m²

The cost of painting 1 m² = Rs. 25

The cost of painting 16 m2 = 16 × 25 = Rs. 400

∴ The Required cost is Rs. 400.

Given:

Area of Rectangle = 75 square units

Length is 5 more than twice of width

Formula used:

Area of Rectangle = length × width

Perimeter of Rectangle = length + width

Calculation:

Let the width of Rectangle be x unit

Length = (2x + 5) unit

Area of Rectangle = 75

⇒ x (2x + 5) = 75

⇒ 2x² + 5x = 75

⇒ 2x² + 5x - 75 = 0

⇒ 2x² + 15x - 10x - 75 = 0

⇒ x (2x + 15) - 5 (2x + 15) = 0

⇒ (2x + 15) (x - 5) = 0

⇒ x = 5, -15/2

neglecting the negative value of x as distance can't be negative

⇒ x = 5

Width = 5 unit

Length = 15 unit

Perimeter of Rectangle = 15 + 5 = 20 unit

∴ Perimeter of Rectangle is 20 units.