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

By solving x = y with circle

We get

\(\mathrm{C}(\sqrt{2}, \sqrt{2}) \)

\(\mathrm{D}(-\sqrt{2},-\sqrt{2})\)

By solving x + y = 1 with

circle x² + y² = 4

we set

\(\mathrm{A}\left(\frac{1+\sqrt{7}}{2}, \frac{1-\sqrt{7}}{2}\right) \)

\(\& \mathrm{~B}\left(\frac{1-\sqrt{7}}{2}, \frac{1+\sqrt{7}}{2}\right)\)

∴ Area of Quadrilateral ACBD

= 2 × Area of ΔBCD

\(=2 \times \frac{1}{2}\left|\begin{array}{ccc} \sqrt{2} & \sqrt{2} & 1 \\ \frac{1-\sqrt{7}}{2} & \frac{1+\sqrt{7}}{2} & 1 \\ -\sqrt{2} & -\sqrt{2} & 1 \end{array}\right| \)

\(=2 \sqrt{14}\)

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

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,

\(\left( {{\rm{x}},{\rm{\;y}}} \right) = \left( {\frac{{ - 2\; + \;4}}{2},\frac{{ - 1\; + \;3}}{2}} \right) = \left( {1,{\rm{\;}}1} \right)\) 

Let the Fourth vertex be D (x_D, y_D)

P is mid-point of BD,

\(\therefore \left( {{\rm{x}},{\rm{\;y}}} \right) = \left( {1,{\rm{\;}}1} \right) = \left( {\frac{{1\; + \;{{\rm{x}}_{\rm{D}}}}}{2},\frac{{0\; + \;{{\rm{y}}_{\rm{D}}}}}{2}} \right)\) 

⇒ (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 \(y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\)

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,

\(\frac{1+7}{2},\frac{0+2}{2}\) = (4,1)

Co-ordinates of D

\(\frac{p+5}{2}=4\)

p = 3

\(\frac{q-1}{2}=1\)

q = 3

Co-ordinates of D = (3, 3)

Equation of BD

\(y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\)

Here, x₁ = 3, x₂ = 5, y₁ = 3, y₂ = -1

\((y-3)=\frac{-1-3}{5-3}\times (x-3)\)

\((y-3)=\frac{-4}{2}\times (x-3)\)

⇒ 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 =\( \frac{1}{2} \left| [x_1 (y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)] \right|\)

Calculation:

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

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

Area (ΔABC) = \( \frac{1}{2} \left| [1 (-1 – 2) + 5(2-0) + 7(0+1)] \right|\)

⇒ 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 \(\overrightarrow {{\rm{AB}}} = {\rm{\;\vec a\;and\;}}\overrightarrow {{\rm{AD}}} = {\rm{\vec b}}\)

Now,

\(\overrightarrow {{\rm{AB}}} = {\rm{\;\vec a}} = \left( {4 - 1} \right){\rm{\hat i}} + \left( {3 - 2} \right){\rm{\hat j}} = 3{\rm{\hat i}} + 1{\rm{\hat j}}\)

\(\overrightarrow {{\rm{AD}}} = {\rm{\vec b}} = \;\left( {3 - 1} \right){\rm{\hat i}} + \left( {5 - 2} \right){\rm{\hat j}} = 2{\rm{\hat i}} + 3{\rm{\hat j}}\)

Now,

\({\rm{\vec a}} \times {\rm{\vec b}} = \;\left| {\begin{array}{*{20}{c}} {{\rm{\hat i}}}&{{\rm{\hat j}}}&{{\rm{\hat k}}}\\ 3&1&0\\ 2&3&0 \end{array}} \right| = 0 - 0 + \;{\rm{\hat k}}\left( {9 - 2} \right) = 7{\rm{\hat k}}\;\)

Now,

area of the parallelogram = \(\left| {{\rm{\vec a}} \times {\rm{\vec b}}} \right|\) = 7square units

Hence, option 4 is correct.

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

Formula used:

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

\((\frac{x_1\ +\ x_2}{2}, \frac{y_1\ +\ y_2}{2})\)

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) 

= \((\frac{2\ +\ 4}{2},\ \frac{1\ +\ 3}{2})\) = (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 

\((y\ -\ y_1)\ =\ \frac{y_2\ -\ y_1}{x_2\ -\ x_1}(x\ -\ x_1)\)

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).