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

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.

Calculation:

∴ Area =(4 cos θ + 2 sin θ)(2 cos θ + 4 sin θ)

= 8 cos² θ + 16 sin θ cos θ + 4 sin θ cos θ + 8 sin²θ

= 8 + 20 sin θ cos θ

= 8 + 10 sin 2θ

When sin 2θ = 1 

⇒ θ = π/4

∴ Area = Area_max = 8 + 10 = 18

(a + b)² = (4 cos θ + 2 sin θ + 2 cos θ + 4 sin θ)²

= (6 cos θ + 6 sin θ)²

= 36(sin θ + cos θ)²

= \(36(\sqrt{2})^2 \)

= 72

∴ The value of (a + b)2 is equal to 72.

The correct answer is Option 1.

Concept:

• Distance = Speed × Time
• Perimeter of rectangle = 2(Length + Breadth)
• Area of rectangle = length × breadth
• 1 km = 1000 m
• 1 hour = 60 minutes

Explanation:

According to the question,

A man cycling along the boundary of the park at the speed of 12 km/hr (12 km/hr = \(\frac{12000}{60}\) m/minutes ) completes one round in 8 minutes

i.e. Perimeter = Distance Travelled in 8 minutes

⇒ Perimeter = \(\frac{12000}{60}× 8 \)  (As  Distance = Speed × Time)

⇒ Perimeter = 1600 m

Now According to the question,

We know,

The ratio between the length and the breadth of a rectangular park is 3 ∶ 2

Let Length = 3x and Breadth = 2x.

Now,

Perimeter of rectangle = 2(Length + Breadth)

So, Perimeter = 2(3x + 2x)

⇒ Perimeter = 2 × 5x

⇒ 1600 = 10x

⇒ x = 160

So, Length = 3x = 3 × 160 = 480 m

Breadth = 2x = 2 × 160 = 320 m

Therefore,

Area of rectangle = length × breadth

⇒ Area = 480 × 320 m²

⇒ Area = 153600 m2

Concept:

Area of Trapezium 

Area of the trapezium = {\(\frac{1}{2}\) × (a + b) × h}

Where a and b are two parallel sides of trapezium and h = perpendicular distance between a and b. 

Calculation:

Let one of the parallel sides be of length x cm. Later, the length of the other parallel side is (x + 6) cm.

Therefore, area of the trapezium = {\(\frac{1}{2}\) × (2x + 6) × 9} cm² = {\(\frac{1}{2}\) × (2x + 6) × 9} cm² = (9x + 27) cm²

But, the area of the trapezium is given as 180 cm².

⇒ 9x + 27=180

⇒ 9x = 180 - 27 = 153

⇒ x = \(\frac{153}{9}\) = 17

The two parallel sides are of lengths 17 cm and (17 + 6) cm = 23 cm.

∴ Sum of the parallel sides are (17 + 23) = 40 cm.

Concept:

Area of Quadrilateral

Area of quadrilateral = \(\frac{1}{2}\) × diagonal × (sum of height of two triangles)

 = \(\frac{1}{2}\) × diagonal × (h₁ + h₂)

Calculation:

Diagonal = BD = 30 cm

Heights, h₁ = 10 cm & h₂ = 14 cm

Sum of the heights of the triangles = h₁ + h₂ = 10 + 14 = 24 cm

Thus, area of quadrilateral ABCD = \(\frac{1}{2}\) × diagonal × (sum of height of two triangles)

= \(\frac{(30 \times 24)}{2}\) = 360 cm²

∴ Area of the quadrilateral ABCD is 360 cm2

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 (8, 0), C (3, 4), D (11, 4) 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 (8, 0), C (3, 4) 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 = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{0}}&{{0}}&1\\ {{8}}&{{0}}&1\\ {{3}}&{{4}}&1 \end{array}} \right|\)

Calculation:

Mid point of BD is same as mid point of AC

Mid point of AC = \(\rm (\frac{1 + 3}{2} , \frac{3 + 5}{2})\) = (2, 4)

Equation of BD 

y - 2 = \(\rm \frac{2 - 4}{- 1 - 2}\) (x + 1)

⇒ y - 2 = \(\rm \frac{2}{3}\)(x + 1)

⇒ 3y - 6 = 2x + 2

⇒ 2x - 3y + 2 + 6 = 0

⇒ 2x - 3y + 8 = 0

∴ The equation of the diagonal BD is 2x - 3y + 8 = 0

Concept:

The diagonal of a parallelogram divides it into two parts of equal areas.

Area of triangle having three coordinates (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area = (1/2) | [x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)] |

Calculation:

Here in the figure diagonal AC divides ABCD into two equal parts

Area(ABCD) = 2 × Area (ΔABC)    ----(i)

Area (ΔABC) = (1/2) | [1 (2 - 5) + (-1) (5 - 3) + 3 (3 - 2)] |

⇒ Area (ΔABC) = (1/2) | [ -3 - 2 + 3 ] |

⇒ Area (ΔABC) = (1/2) | - 2  | = 1

Now, From (i), we get 

Area(ABCD) = 2 × 1 = 2

∴ The area of parallelogram is 2 square unit.

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 (- 4, 5), B (0, 7), C (5, - 5) and D (- 4, - 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 (- 4, 5), B (0, 7), C (5, - 5) 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 = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{- 4}}&{{5}}&1\\ {{0}}&{{7}}&1\\ {{5}}&{{-5}}&1 \end{array}} \right|\)

Given:

Diameter = 14 cm

Concept:

TSA of hemisphere = 3πr² 

Calculation:

Radius, r = 14/2 = 7 cm

∴ TSA of solid hemisphere = 3 × (22/7) × 7² 

⇒ 462 cm² 

Hence, the total surface area of a solid hemisphere is 462 cm².

Concept:

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

2. Distance formula:  A(x, y) and B(a, b) be any two points, by distance formula,

 \(\rm AB = \sqrt {(x - a)^2 + (y - b)^2}\)

3. (a + b)2 = a2 + 2ab + b2

4. (a - b)2 = a2 - 2ab + b2

Calculation:

Given that, coordinates of point A, C and D are (3, 2), (- 3, 2) and (k, -1) respectively. 

Area of Δ ABC = 9 sq. unit

According to the question, |AD| = |CD| 

Using distance formula,

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

Taking square of both side

⇒ (3 - k)² + 9 = (3 + k)2 + 9

⇒ 9 + k² - 6k = 9 + k² + 6k

⇒ 12k = 0 ⇒ k = 0

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 (1, - 2), B (3, 6), C (5, 10) and D (3, 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 (1, - 2), B (3, 6), C (5, 10) 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 = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{1}}&{{-2}}&1\\ {{3}}&{{6}}&1\\ {{5}}&{{10}}&1 \end{array}} \right|\)

Concept:

Area of rectangle = Length × Width

Calculation:

Given Figure:

As we can see side length of the figure is (a + b) and width is (a + b) 

Area of figure = Length × width = (a + b) × (a + b) = (a + b)²         ..... (1)

Now,

Area of figure = Area of section 1 + Area of section 2 + Area of section 3 + Area of section 4

= a × a + a × b + b × a + b × b

=  a² + 2ab + b²

∴ (a + b)² = a2 + 2ab + b2

Concept:

• Distance = Speed × Time
• Perimeter of rectangle = 2(Length + Breadth)
• Area of rectangle = length × breadth
• 1 km = 1000 m
• 1 hour = 60 minutes

Explanation:

According to the question,

A man cycling along the boundary of the park at the speed of 12 km/hr (12 km/hr = \(\frac{12000}{60}\) m/minutes ) completes one round in 8 minutes

i.e. Perimeter = Distance Travelled in 8 minutes

⇒ Perimeter = \(\frac{12000}{60}× 8 \)  (As  Distance = Speed × Time)

⇒ Perimeter = 1600 m

Now According to the question,

We know,

The ratio between the length and the breadth of a rectangular park is 3 ∶ 2

Let Length = 3x and Breadth = 2x.

Now,

Perimeter of rectangle = 2(Length + Breadth)

So, Perimeter = 2(3x + 2x)

⇒ Perimeter = 2 × 5x

⇒ 1600 = 10x

⇒ x = 160

So, Length = 3x = 3 × 160 = 480 m

Breadth = 2x = 2 × 160 = 320 m

Therefore,

Area of rectangle = length × breadth

⇒ Area = 480 × 320 m²

⇒ Area = 153600 m2

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