Area MCQ Quiz in தமிழ் - Objective Question with Answer for Area - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Apr 4, 2025
Latest Area MCQ Objective Questions
Top Area MCQ Objective Questions
Area Question 1:
Find the area of the quadrilateral whose vertices are A (5, 3), B (6, - 4), C (- 3, - 2) and D (- 4, 7) ?
Answer (Detailed Solution Below)
Area Question 1 Detailed Solution
CONCEPT:
Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC =
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 =
⇒ Area of Δ ABC =
Area Question 2:
Find the area of the quadilateral whose vertices are A (0, 0), B (0, 10), C (8, 2) and D (8, 7) ?
Answer (Detailed Solution Below)
Area Question 2 Detailed Solution
CONCEPT:
Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC =
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 =
⇒ Area of Δ ABC =
Area Question 3:
If the points A (- 2, - 1), B (1, 0), C (4, 3) and D (1, 2) are the vertices of a quadilateral ABCD then find the area of the quadilateral ABCD ?
Answer (Detailed Solution Below)
Area Question 3 Detailed Solution
CONCEPT:
Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC =
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 =
⇒ Area of Δ ABC =
Area Question 4:
Find the are of the quadilateral whose vertices are A (2, - 2), B (8, 4), C (5, 7) and D (-1, 1) ?
Answer (Detailed Solution Below)
Area Question 4 Detailed Solution
CONCEPT:
Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC =
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 =
⇒ Area of Δ ABC =
Area Question 5:
A(1,2,−1) , B(2,5,−2) and C(4,4,−3) are three vertices of a rectangle. What is the area of the rectangle?
Answer (Detailed Solution Below)
Area Question 5 Detailed Solution
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
Length of vector
Length of vector
Area of the rectangle is the product of the lengths of vectors
∴ The area of the rectangle is
Hence, the correct answer is Option 3.
Area Question 6:
One of the diameter of the circle circumscribing the rectangle
Answer (Detailed Solution Below)
Area Question 6 Detailed Solution
Let the equation of the circle circumscribing the rectangle
Since
and,
Also, centre of (1) lies on
Subtracting (2) from (3), we get
Solving (4) and (5), we get
Substituting the values of
Radius of the circle is
Since the rectangle is inscribed in the circle, Its diagonal will be the diameter of the circle.
Also, the length of the rectangle
hence, the area of rectangle
Area Question 7:
In square ABCD, P is the midpoint of DC and Q is midpoint of AD. If the area of the quadrilateral QBCP is 15. What is the area of square ABCD?
Answer (Detailed Solution Below)
Area Question 7 Detailed Solution
Concept:
Area of triangle =
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 =
Area of QPD =
Also given area of QBCP = 15
Now, Area of ABQ + area of QPD + area of QBCP = area of ABCD
⇒
⇒
⇒ 3a2 + 120 = 8a2
⇒ 5a2 = 120 a2 = 24
Hence area of square ABCD = 24
(2) is correct
Area Question 8:
A lawn 30 m long and 16 m wide is surrounded by a path 2 m wide. What is the area of the path?
Answer (Detailed Solution Below)
Area Question 8 Detailed Solution
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.
Area Question 9:
A wall is of the form of a trapezium with height 4 m and parallel sides being 3 m and 5 m. What is the cost of painting the wall, if the rate of painting is Rs. 25/- per square metre?
Answer (Detailed Solution Below)
Area Question 9 Detailed Solution
Formula used:
Area of trapezium =
Calculation:
By using the above formula,
Area of the wall =
The cost of painting 1 m2 = Rs. 25
The cost of painting 16 m2 = 16 × 25 = Rs. 400
∴ The Required cost is Rs. 400.
Area Question 10:
The area of a rectangle whose length is 5 more than twice its width is 75 square units. What is the perimeter of the rectangle?
Answer (Detailed Solution Below)
Area Question 10 Detailed Solution
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.