चौकोन MCQ Quiz in मराठी - Objective Question with Answer for Quadrilaterals - मोफत PDF डाउनलोड करा

By solving x = y with circle

We get

By solving x + y = 1 with

circle x² + y² = 4

we set

∴ Area of Quadrilateral ACBD

= 2 × Area of ΔBCD

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 = 

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 = 

⇒ Area of Δ ABC = 

Concept:

Properties of a parallelogram

• The diagonals of a parallelogram bisect each other.

• Opposite sides of a parallelogram are congruent.

• Opposite angles of a parallelogram are congruent.

Calculation:

Diagonals of a parallelogram bisect each other,

Let the point of intersection of diagonals be P(x, y)

So, P is mid-point of AC,

Let the Fourth vertex be D (x_D, y_D)

P is mid-point of BD,

⇒ (x_D, y_D) = (1, 2)

Fourth vertex of the parallelogram is D = (1, 2)

Concept:

The equation of a line passing through two points (x₁, y₁) and (x₂, y₂) is 

In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. That is each diagonal cuts the other into two equal parts.

Calculation:

The midpoint of BD is the same as the midpoint of AC

In parallelogram ABCD, A (1, 0), B (5, -1), C (7, 2), D (p, q)

The diagonals of a parallelogram bisect each other.

O is the point of intersection of AC and BD.

Since, O is the midpoint of BD, its coordinates will be,

 = (4,1)

Co-ordinates of D

p = 3

q = 3

Co-ordinates of D = (3, 3)

Equation of BD

Here, x₁ = 3, x₂ = 5, y₁ = 3, y₂ = -1

⇒ y - 3 = -2(x - 3)

⇒ 2x + y - 9 = 0

∴ The equation of the diagonal BD is 2x + y - 9 = 0

Concept:

Area of the parallelogram

• 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 =

Calculation:

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

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

Area (ΔABC) =

⇒ Area (ΔABC) = (1/2) | [ -3 + 10 + 7 ] |

⇒ Area (ΔABC) = (1/2) | 14 | = 7

Now, From (i), we get

Area(ABCD) = 2 × 7 = 14

∴ The area of parallelogram is 14 square unit.

Concept:

• The parallelogram whose adjacent sides are the vectors a and b then the area of the parallelogram is given by  

Calculation:

Let 

Now,

Now,

Now,

area of the parallelogram =  = 7square units

Hence, option 4 is correct.

Concept:

Area of Quadrilateral

Area of quadrilateral =  × diagonal × (sum of height of two triangles)

 =  × 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 =  × diagonal × (sum of height of two triangles)

=  = 360 cm²

∴ Area of the quadrilateral ABCD is 360 cm2

Formula used:

The midpoint of two-point (x1, y1) and (x2, y2) is given by

Calculation:

Given that, point (2, 1) and (4, 3) are opposite vertices of rectangle 

The mid-point of A(2, 1) and C(4, 3) 

=  = (3, 2)

We know that the intersecting point of the diagonal of a rectangle is the same or at the midpoint. 

Therefore, the point (3, 2) will satisfy the equation 2x - y = k

⇒ 2(3) - 2 = k

⇒ k = 4

Hence, option 3 is correct.

Additional Information1. Equation of line of slope m passing through the point (x₁, y₁) is

(y - y₁) = m(x - x₁)

2. Equation of line passing through (x1, y1) and (x2, y2) is 

Concept:

If the diagonals of a quadrilateral are perpendicular, then it is a rhombus

Slope of line ax + by + c = 0 is -a/b

If m₁ and m₂ are the slope of two perpendicular lines then 

m₁ × m₂ = - 1

Calculation:

Given, x + 3y = 4       __(i)

Slope of (i) = m₁ = -1/3

6x - 2y = 7              ___(ii)

Slope of (ii) = m₂ = 6/2 = 3

Since, m1. m2 = -1 

⇒ The diagonals of a quadrilateral are perpendicular

⇒ ABCD must be rhombus.

∴ The correct answer is option (1).