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

Formula Used:

Area of triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is given by 

Calculation

Given: A(K, 1), B(2, 4), C(1, 1), Area = 6

⇒ Area =

⇒ 6 =

⇒ 12 = |3K - 3|

⇒ 3K - 3 = ±12

Case 1: 3K - 3 = 12

⇒ 3K = 15

⇒ K = 5

Case 2: 3K - 3 = -12

⇒ 3K = -9

⇒ K = -3

∴ K = 5 and -3

Hence option 2 is correct

Concept Used:

Area of a triangle = |AB × AC|

Calculation:

Given:

Vertices: A(1, 1, 1), B(1, 2, 3), C(2, 3, 1)

AB = B - A = (1-1, 2-1, 3-1) = (0, 1, 2)

AC = C - A = (2-1, 3-1, 1-1) = (1, 2, 0)

 AB × AC =

⇒ AB × AC =

⇒ AB × AC =

|AB × AC| =

⇒ |AB × AC| =

Area = |AB × AC|

⇒ Area =

∴ Area of the triangle is

Hence option 2 is correct

Calculation

Given:

Lines:  and 

1) Factorize the pair of straight lines:

⇒

⇒

⇒

⇒ and

2) Find the intersection points:

Intersection of and :

⇒ ⇒

⇒ ⇒ ⇒

⇒ ⇒

⇒

Intersection of and :

⇒ ⇒

⇒ ⇒ ⇒ ⇒

⇒ ⇒

⇒

Intersection of and :

⇒ , ⇒

3) Area of triangle ABC:

⇒

⇒

⇒

Hence option 4 is correct

Calculation:

Given:

The equations of the lines are: 12x² - 20xy + 7y² = 0 and x + y - 1 = 0.

Factorize 12x² - 20xy + 7y² = 0:

12x² - 14xy - 6xy + 7y² = 0

2x(6x - 7y) - y(6x - 7y) = 0

(2x - y)(6x - 7y) = 0

So, the equations of the lines are:

2x - y = 0 => y = 2x ---(1)

6x - 7y = 0 => y = (6/7)x ---(2)

x + y - 1 = 0 ---(3)

Intersection of (1) and (3):

x + 2x - 1 = 0 => 3x = 1 => x = 1/3, y = 2/3

A = (1/3, 2/3)

Intersection of (2) and (3):

x + (6/7)x - 1 = 0 => (13/7)x = 1 => x = 7/13, y = 6/13

B = (7/13, 6/13)

Intersection of (1) and (2):

2x = (6/7)x => 14x = 6x => 8x = 0 => x = 0, y = 0

O = (0, 0)

Area of the triangle OAB:

Area =

Area =

Area =

∴ The area of the triangle is 4/39.

Hence option 4 is correct

Let be the right angled triangle inscribed in a circle of radius ,

diameter,

. It will be maximum if ,

,

,

(a) is not correct,

,

,

(d) is not correct,

,

(c) is not correct,

Hence (a), (c), (d) are not correct,

Also we know that,

,

,

So, (b) is correct

Concept:

Area of a triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃), is given by the expression

Area = 

Calculation:

Here, vertices are (3, 13), (5, -8), and (4, -2)

∴ Area of triangle = 

 sq. units 

Hence, option (4) is correct. 

Given:

Perpendicular of right angled triangle = 8 cm

Area = 20 cm²

Formula used:

Area of right angled triangle = (1/2) × perpendicular × base

Calculation:

⇒ 20 cm² = (1/2) × 8 × base

⇒ base = 20/4

⇒ 5 cm

∴ The length of base is 5 cm

Concept:

Area of a triangle =  × base × altitude

Area of ΔABC = 

Calculation:

Area of a triangle =  × base × altitude

=  × c × a sin∠CBA

=  × 10 cm × 4cm sin 30° 

= 5 × 4 ×  (as sin 30° = )

= 10 cm²

Concept:

Area of triangle =  ×(Area of the triangle formed by median as a side) 

The area of a triangle whose side lengths are a, b and c is given by:

, Where 's' is semi-perimeter of the triangle.

Semi-perimeter of the triangle = s = 

Calculation:

Given: length of three medians of a triangle are 9 cm, 12 cm and 15 cm

Let s be semi-perimeter of the triangle formed by median as a side

∴ s = 

Now, Area of the triangle formed by median as a side = 

As we know,

Area of triangle =  ×(Area of the triangle formed by median as a side) 

= 

CONCEPT:

If three points A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are collinear then area of Δ ABC is zero i.e 

CALCULATION:

Given: The points (x, -1), (2, 1) and (4, 5) are collinear

Let A = (x, - 1), B = (2, 1) and C = (4, 5)

Let's find the area of the Δ ABC

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of the Δ ABC then area of Δ ABC is given by: 

Here, x₁ = x, y₁ = - 1, x₂ = 2, y₂ = 1, x₃ = 4 and y₃ = 5

So, area of  Δ ABC = 

⇒ Area of Δ ABC = 2 - 2x

∵ The points A, B and C are collinear  ⇒ Area of ΔABC = 0

⇒ 2 - 2x = 0

⇒ x = 1

Hence, option C is the correct answer.

Concept:

The area of triangle ABC with vertices A = (x1, y1), B = (x2, y2) and C = (x3, y3) is given by, 

Area = 

Calculations:

Consider, the vertices of a triangle ABC are A = (1, -3) = (x₁, y₁), B = (4, -3) = (x2, y2) and C = (-9, k) = (x3, y3)

The area of triangle ABC with vertices A = (x1, y1), B = (x2, y2) and C = (x3, y3) is given by, 

Area = 

⇒ 15 = 

⇒ 30 = 

⇒ 30 = 3k + 9

⇒ k = 7

Hence, If the vertices of a triangle are (1, -3), (4, -3) and (-9, k) and its area is 15 sq. units then value of k is 7

Given:

Height (perpendicular) of the triangle = 5 cm

Area of the triangle = 20 cm²

Formula used:

Area of triangle = (1/2) × b × h

b = Base of the triangle

h = Height of the triangle

Calculation:

According to the question,

20 cm² = (1/2) × b × 5cm

⇒ b = 40/5 = 8 cm

∴ The required length of the base of the triangle is 8 cm.

CONCEPT:

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

Note: If the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

CALCULATION: 

Here, we have to find the value of k for  which the points A (- 2, 3), B (1, 2) and C (k, 0) are collinear

Let x1 = - 2, y1 = 3, x2 = 1, y2 = 2, x3 = k and y3 = 0.

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the  vertices of a Δ ABC then area of Δ ABC = 

⇒ A = (k - 7)/2

∵ The given points are collinear.

As we know that, if the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

⇒ A = (k - 7)/2 = 0

⇒ k = 7

Hence, option C is the correct answer.

Formula used - 

area of triangle ABC = (1/2) × base × height

In triangle ABC, If two sides are AB and BC and angle between two sides is θ 

∴ Area of triangle ABC = (1/2) × AB × AC × sinθ 

Given - 

AB = 14 cm, BC = 15 cm, ∠B = 30°,

In another Triangle ABC, BC = 20 cm 

Solution - 

⇒ area of triangle is same in both the cases. so,

⇒ (1/2) × AB × BC × sin30° = (1/2) × BC × AD

⇒ 15 × 14 × (1/2) = 20 × AD

⇒ AD = 5.25 cm

∴ AD = 5.25 cm

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: P(0,0), Q(2,5) and R(-3,4) are the vertices of the triangle PQR

Let x1 = 0, y1 = 0, x2 = 2, y2 = 5, x3 = - 3 and y3 = 4.

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the  vertices of a Δ ABC then area of Δ ABC = 

⇒ A = 23/2

Hence, option C is the correct answer.