M.C.Q

Question 11 Mark

The Quadrilateral forms by joining the mid-points of the sides of a Quadrilateral PQRS, taken in order, is a Rhombus if:

Answer

Diagonals of PQRS are equal.
Solution:
A quadrilateral formed by joining the mid points of the sides of the Rectangle is a rhombus. In rectangle, diagonals are equal.

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Question 21 Mark

Two parallelograms stand on equal bases and between the same parallels. The ratio of their areas is:

Answer

1 : 1
Solution:
Area of a parallelogram = base ⨯ height
Since two parallelogram stand on equal bases and between the same parallel lines,
their heights are same.
$\therefore$ Areas are also same.
$\therefore$ The ratio of their area is 1 : 1.

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Question 31 Mark

In which of the following figures are the diagonals equal?

Answer

Rectangle.
Solution:
The diagonals are equal in a rectangle.
The diagonals in a parallelogram, rhombus or trapezium need not be equal.

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Question 41 Mark

If $\angle\text{A},\ \angle\text{B},\ \angle\text{C}$ and $\angle\text{D}$ of a quadrilateral ABCD, taken in order, are in the ratio 3 : 7 : 6 : 4 then ABCD is a:

Answer

Trapezium
Solution:
Let the angles be 3x, 7x, 6x, 4x
Then 3x + 7x + 6x + 4x = 360
$\text{x}=\frac{360}{20}=18$

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Question 51 Mark

If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is:

Answer

72°
Solution:
Let one of the angle of the || gm be x°.
According to the given condition,
$\therefore$ the adjacent angle $=\frac{2}{3}\text{x}^{\circ}$
Now,
$\text{x}+\frac{2}{3}\text{x}=180^{\circ}$ ...(Sum of the adjacent angles of || gm is 180°.)
$\Rightarrow\frac{3\text{x}+2\text{x}}{3}=180^{\circ}$
$\Rightarrow\frac{5\text{x}}{3}=180^{\circ}$
$\Rightarrow5\text{x}=540^{\circ}$
$\Rightarrow\text{x}=108^{\circ}$
⇒ the adjacent angles $=\frac{2}{3}(108)=36\times2=72^{\circ}$
Hence, the smallest angle is 72°.

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Question 61 Mark

In fig if DE = 8cm, $\text{DE || BC}$ and D is the mid-Point of AB, then the true statement is:

Answer

E is the mid-Point of AC.
Solution:
By the converse of Mid-Point Theorem, which states that," If a line segment is drawn passing through the midpoint of any one side of a triangle and parallel to another side, then the line segment bisects the remaining third side.

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Question 71 Mark

The lengths of the diagonals of a rhombus are $16\ cm$ and $12\ cm.$ The length of each side of the rhombus is:

Answer

Let us assume a rhombus $\text{ABCD}$ where,
$AB = BC = CD = DA$
Now, in triangle $\text{OBC}$ by using Pythagoras theorem we get:
$BC^2 = OB^2 + OC^2$
$BC^2 = 6^2 + 8^2$
$BC^2 = 36 + 64$
$BC^2 = 100$
$\text{BC}=\sqrt{100}$
$BC = 10\ cm$
$\therefore AB = BC = CD = DA = 10\ cm$

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Question 81 Mark

In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects $\angle\text{B}$ as well as $\angle\text{D}.$ Then, $\angle\text{AMB}= ?$

Answer

90°,
Solution:
$\angle\text{ABC}=\angle\text{ADC}$ ...(Opposite angles of a parallelogram are equal)
$\Rightarrow\frac{1}{2}\angle\text{ABC}=\frac{1}{2}\angle\text{ADC}$
$\Rightarrow\angle\text{ABD}=\angle\text{ADB}$
So, $\text{AD = AB}$ ...(Sides opposite equal angles are equal.)
$\therefore\triangle\text{ABD}$ is isosceles
Also, M is the mid-point of BD.
$\therefore\text{AM}\perp\text{BD}$
$\therefore\angle\text{AMB}=90^{\circ}$

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Question 91 Mark

If a diagonal AC and BD of a quadrilateral ABCD bisect each other, then ABCD is a:

Answer

Parallelogram
Solution:
Two diagonals of quadrilateral form four triangles. Out of these four triangles two triangles of opposite to each other are congruent by SAS. By using CPCT property we can prove that both pair of opposite sides in a quadrilateral are parallel. A quadrilateral with both pair of opposite sides parallel is called parallelogram.

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Question 101 Mark

The bisectors of the angles of a parallelogram enclose a:

Answer

Rectangle
Solution:
The bisectors of the angles of a parallelogram encloses a rectangle.

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Question 111 Mark

D and E are the mid-points of the sides AB and AC of $\triangle\text{ABC}$ and O is any point on the side BC, O is joined to A. If P and Q are the mid-points of OB and OC res, Then DEQP is:

Answer

A Parallelogram
Solution:
By mid-point theorem, DE is parallel to BC. In triangle BOA, DP parallel to OA and OA is parallel to QE in triangle AOC (mid-point theorem) because D and P are mid-points in triangle BOA and E and Q are mid-points in triangle AOC.
So, DP is parallel to EQ. In quadrilateral DPQE, both pair of opposite sides are parallel. So, it become parallelogram.

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Question 121 Mark

In a Trapezium $\text{ABCD}$, if $\text{AB || CD},$ then $(AC^2 + BD^2) =$ ?

Answer

Given: $\text{ABCD}$ is a trapezium with $\text{AB || CD}$
Construction: Draw $DE$ and $CF \perp$ to $AB$

Then in $\triangle\text{ABC}$
$\angle \text{BAC}$ is acute
$\therefore BC^2 = AC^2 + AB^2 - 2AF : AB ...(1)$
and In $\triangle \text{BDA}$
$\angle \text{DBA}$ is acute
$\therefore AD^2 = BD^2 + AB^2 - 2BE.AB ...(2)$
Adding $(1)$ and $(2)$ we get
$BC^2 + AD^2 = AC^2 + BD^2 + 2AB^2 - 2AF.AB - 2BE.AB$
$\Rightarrow AC^2 + BD^2 = BC^2 + AD^2 - 2AB [AB - AF - BE]$
$= BC^2 + AD^2 - 2AB [AB - (AE + EF) - (BF + EF)]$
$= BC^2 + AD^2 - 2AB [AB - (AE + EF + BF + EF)]$
$= BC^2 + AD^2 - 2AB [AB - (AB + CD)] (\therefore EF = DC)$
$= BC^2 + AD^2 - 2AB [(-CD)]$
$= AD^2 + BC^2 + 2AB \times CD$

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Question 131 Mark

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a:

Answer

Prallelogram.
Solution:

P, Q, R & S are the mid-points of AB, BC, CD & AD respectively.
Consider $\triangle\text{ADB},$
If in a triangle, the mid-points of two sides are joint by a line then the line is parallel to the third side.
$\Rightarrow\text{PS}||\text{DB}$ in $\triangle\text{ADB}$
Similarly in $\triangle\text{CDB},$
RQ || DB
Hence PS || RQ ...(1)
Similarly in $\triangle\text{ABC}$ and $\triangle\text{ADC}$
SR || AC, PQ || AC
⇒ SR || PQ ...(2)
From eq. (1) and (2), PQRS is a parallelogram.

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Question 141 Mark

P is the mid-point of side BC of a parallelogram ABCD such that $\angle\text{BAP}=\angle\text{DAP}.$ If AD = 10cm, then CD =

Answer

5cm.
Solution:

Let a line parallel to AB is drawn from P to meet AD at Q.
PQ || AB || DC
Q is also mid-point of AD.
Now, consider parallelogram ABPQ.
$\angle\text{PAQ}=\angle\text{APB}$ (Alternate angles)
Also $\angle\text{PAQ}=\angle\text{BAP}$ (Given)
$\Rightarrow\angle\text{APB}=\angle\text{BAP}$
So $\triangle\text{ABP}$ is isoseceles triangle.
$\Rightarrow\text{BP}=\text{AB}$
i.e. $\text{AB}=\frac{10}{2}=5\text{cm}$

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Question 151 Mark

In a Quadrilateral ABCD, $\angle\text{A} = \angle\text{C},\ \angle\text{B} = 2\angle\text{A}, \ \angle\text{D} = \frac{1}{21}\angle\text{A}.$ Then $\angle\text{A}, \ \angle\text{B}, \ \angle\text{C} $ and $\angle\text{D}$ respectively are:

Answer

$80^\circ,\ 160^\circ,\ 80^\circ,\ \text{and}\ 40^\circ$
Solution:
$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ (angle sum property)
$\angle\text{A} + 2\angle\text{A} + \angle\text{A} + \frac{1}{2}$ of $\angle\text{A} = 360^\circ$
$\frac{9}{2}$ of $\angle\text{A} = 360^\circ$
$\angle\text{A} = 80^\circ$
So, $\angle\text{B}=2(80^\circ)=160^\circ,\ \angle\text{C}=80^\circ$ and $\angle\text{D} = \frac{1}{2}$ of $80^\circ=40^\circ$

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Question 161 Mark

P is any point on the side BC of a $\triangle\text{ABC}.$ P is joined to A. If D and E are the midpoints of the sides AB and AC respectively and M and N are the midpoints of BP and CP respectively then quadrilateral DENM is:

Answer

A parallelogram.
Solution:

In $\triangle\text{ABC},$ D and E are the mid-points of sides AB and AC respectively.
$\Rightarrow\text{DE || BC}$ and $\text{DE}=\frac{1}{2}\text{BC }...(\text{i})$
Now, $\text{MN = MP + PN}=\frac{1}{2}\text{BP}+\frac{1}{2}\text{CP}=\frac{1}{2}(\text{BP + CP})$
$\therefore\text{MN}=\frac{1}{2}\text{BC ...(ii)}$
From (i) and (ii),
$\text{DE || MN}$ and $\text{DE = MN}$
Hence, DENM is a parallelogram.

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Question 171 Mark

Which of the following is not true for a parallelogram?

Answer

Opposite angles are bisected by the diagonals.
Solution:
We know that, in a || gm opposite sides are equal, opposite angles are equal and also the diagonals bisect each other.
So, opposite angles are bisected by the diagonals is not true.

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Question 181 Mark

In a triangle ABC, P, Q and R are the mid-points of the sides BC, CA and AB respectively. If AC = 21cm, BC = 29cm and AB = 30cm, find the perimeter of the quadrilateral ARPQ?

Answer

51cm
Solution:
Given:
A, B, C is a triangle .
P, Q, R are the mid-points of sides BC , CA and AB .
AC = 21cm
BC = 29cm
AB = 30cm
To find:
Perimeter of quadrilateral ARQP?
Q is the mid-point of AC
P is the mid-point of BC
QP is parallel to AB
QP = half of AB (according to mid point theorem)
AB = 30cm, QP = 15CM (QP is half of BA) (proved above)
R is the mid-point of side AB.
QP is also parallel to AR (half of side AB)
PR is parallel to AC
PR = half of AC (according to mid point theorem)
AC = 21cm, PR = 10.5cm (PR is half of AC) (proved above)
PR is parallel to AQ (AQ is half of AC)
Since, in quadrilateral ARQP both the opposite sides are parallel it is a parallelogram.
Therefore, ARQP is a parallelogram.
We know that
In parallelogram, opp sides are equal.
Therefore,
PR = AQ = 10.5cm
QP = AR = 15cm
10.5cm + 10.5cm + 15cm + 15cm = 51cm.
Therefore the perimeter of quadrilateral,
ARQP = 51cm.

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Question 191 Mark

The figure formed by joining the mid-points of the adjacent sides of a rectangle is a:

Answer

Rhombus.
Solution:
The figure formed by joining the mid points of the adjacent sides of a rectangle is a rhombus.

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Question 201 Mark

The Quadrilateral formed by joining the mid-points of the sides of a Quadrilateral PQRS, taken in order, is a rectangle if:

Answer

Diagonals of PQRS are at right angles.
Solution:
Diagonals of PQRS are at right angles form all the internal angles as right angles. [according to angle property of rectangle, i.e., all the angles of a rectangle are right angle (90º)].

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Question 211 Mark

The two diagonals are equal in a:

Answer

Rectangle
Solution:

ABCD is a rectangle with AC and BD as its diagonals.
ABCD is a rectangle
$∴ \angle\text{A} = 90^\circ, \ \text{AD = BC}$
$\text{AD || BC}$ and AB is a transversal
$∴ \angle\text{A} + \angle\text{B} = 180^\circ$
$⇒ \angle\text{B} = 90^\circ$
In $\triangle\text{ABD}$ and $\triangle\text{BAC}$
AB = BA
$\angle\text{A} = \angle\text{B}$
AD = BC
$∴ \triangle\text{ABD}≅\triangle\text{BAC}$ (SAS)
⇒ BD = AC (c.p.c.t)
Hence, the diagonals of a rectangle are equal.

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Question 221 Mark

The opposite sides of a quadrilateral have.

Answer

No common point.
Solution:
We can look at a quadrilateral as:

The opposite sides of the above quadrilateral AB and CD have no point in common.

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Question 231 Mark

The opposite sides of a quadrilateral have:

Answer

No common point.
Solution:

ABCD is a Quadrilateral.
The opposite sides AB and DC, AD and BC have no common point.

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Question 241 Mark

If the diagonals of a rhombus of a rhombus are 18cm and 24cm respectively, then its side is equal to:

Answer

15cm.
Solution:

Let BD = 24cm and AC = 18cm (Given)
Now, $\frac{\text{AC}}{2}=\frac{18}{2}=9\text{cm}$ and $\text{BO}=\frac{\text{BD}}{2}=\frac{24}{2}=12\text{cm}$
Now, $\text{AB}=\sqrt{(\text{AO})^2+(\text{BO})^2}$ (Diagonals make 90° between them)
$=\sqrt{9^2+12^2}$
$=\sqrt{81+144}$
$=\sqrt{225}$
$\text{AB}=15\text{cm}$

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Question 251 Mark

In Quadrilateral ABCD, $\angle\text{A} = (3\text{x})^\circ, \ \angle\text{B} = (5\text{x})^\circ,\ \angle\text{C} = (20\text{x})^\circ,\ \angle\text{D} = (8\text{x})^\circ.$ Find the value of x?

Answer

10
Solution:
$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360$ (angle sum property)
3x + 5x + 20x + 8x = 360
36x = 360
x = 10

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Question 261 Mark

The diagonals AC and BD of a rectangle ABCD intersect each other at P. If $\angle\text{ABD} = 50^\circ,$ then $\angle\text{DPC} =\ ?$

Answer

80º
Solution:
Given,
ABCD is a rectangle

Diagonals AC & BD intersect each other at P
$\angle\text{ABD} = 50^\circ$
$∵$ diagonals of rectangle bisect each other and are equal in length
$⇒ \angle\text{ABD} = \angle\text{PDC}$ [alternate angles]
$⇒ \angle\text{PDC}= \angle\text{PCD} = 50^\circ$
In $\triangle\text{DPC}$
$⇒ \angle\text{DPC} + \angle\text{PCD} + \angle\text{PDC} = 180^\circ$
$⇒ \angle\text{DPC} + 50^\circ + 50^\circ= 180^\circ$
$⇒ \angle\text{DPC} = 180^\circ - 100^\circ = 80^\circ$

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Question 271 Mark

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a:

Answer

Parallelogram
Solution:
Let ABCD be a quadrilateral in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.
Join AC
In $\triangle\text{ABC},$ the points P and Q are the mid-points of sides AB and BC respectively.
$\therefore\ \text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC}\ ...\ \text{(i)}$ [By mid-point theorem]
Again, in $\triangle\text{DAC},$ the points S and R are the mid-points of AD and DC.
$\therefore\ \text{SR || AC}$ and $\text{SR}=\frac{1}{2}\text{AC}\ ...\ \text{(ii)}$

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Question 281 Mark

If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form.

Answer

A rectangle
Solution:

PNQM is a rectangle.

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Question 291 Mark

Given a triangular prism, then what can we conclude about the lateral faces?

Answer

Faces are Parallelogram.
Solution:
There are five faces in triangular prism. two triangles and three quadrilaterals in which both pair of opposite sides equal. A quadrilateral with both pair of opposite sides equal is called parallelogram. So, lateral faces are parallelogram.

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Question 301 Mark

The parallel sides of a trapezium are a and b respectively. The line joining the mid-points of its non-parallel sides will be:

Answer

$\frac{1}{2}(\text{a}+\text{b})$
Solution:

E and F are the given to be the mid-points of AD and BC respectively.
$\therefore\text{EF} = \frac{1}{2}(\text{AB + DC})$
$=\frac{1}{2}(\text{a + b})$

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Question 311 Mark

Write the correct answer in the following:
Three angles of a quadrilateral are 75º, 90º and 75º. The fourth angle is:

Answer

120º
Solution:
Fourth angle of the quadrilateral
= 360° - (75° + 90° + 75°)
= 360° - 240°
= 120°
Hence, (d) is the corrent answer.

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Question 321 Mark

Write the correct answer in the following:
Which of the following is not true for a parallelogram?

Answer

Opposite angles are bisected by the diagonals.
Solution:
We know that, in a parallelogram, opposite sides are equal, opposite angles are equal, opposite angles are not bisected by the diagonals and diagonals bisect each other.

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Question 331 Mark

In figure, ABCD and AEFG are both parallelograms if $\angle\text{C} = 80^\circ,$ then $\angle\text{DGF}$ is:

Answer

80º
Solution:
As $\angle\text{A} = 80^\circ$ opposite angles of a parallelogram are equal and $\angle\text{DGF} = 80^\circ$ as GF is parallel to AB, corresponding angles are equal.

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Question 341 Mark

The two digonals are equal in a:

Answer

Rectangle.
Solution:
The two diagonals are equal in a rectangle (property).

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Question 351 Mark

In a Quadrilateral ABCD, AB = BC and CD = DA, then the quadrilateral is a:

Answer

Kite
Solution:
A quadrilateral having equal adjacent sides is called Kite. AB and BC are adjacent sides and AD and DC are also adjacent sides which are equal.

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Question 361 Mark

In given figure, ABCD is a parallelogram and E is the mid-point of BC. DE and AB when produced meet at F. Then, AF = ?

Answer

AF = 2AB
Solution:
By the congruency of triangles,
BEF and CED (AAS Rule) are congruent
So, DC = BF (CPCT) ...(1)
But DC = AB ...(2)
So, AB = BF
but AF = AB + BF
So, AF = 2AB

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Question 371 Mark

Three angles of a quadrilateral are $80^\circ , 95^\circ$ and $112^\circ .$ Its fourth angle is:

Answer

Let the measure of the fourth angle be $x^\circ.$
We know that, the sum of the angles of a quadrilateral is $360^\circ$
So, $80^\circ + 95^\circ + 112^\circ + x = 360^\circ$
$\Rightarrow 287^\circ + x = 360^\circ$
$\Rightarrow x = 73^\circ$
$\therefore$ Its fourth angle is $73^\circ$

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Question 381 Mark

In the fig, ABCD is a Parallelogram. The values of x and y are:

Answer

45º, 30º
Solution:
3x - 10º = x + 80º [opposite angles of a parallelogram are equal]
3x - x = 80º + 10º
2x = 90º
$\text{x} = \frac{90^\circ}{2}$
x = 45º
3x - 10º + y + 25º = 180º [In a parallelogram co-interior angles are supplementary]
3 × 45º - 10º + y + 25º = 180º
135º + 25º - 10º + y = 180º
150º + y = 180º
y = 180º - 150º
y = 30º

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Question 391 Mark

In which of the following figures are the diagonals equal?

Answer

Rectangle
Solution:
Rectangle is the correct answer. As we know that from all the quadrilaterals given in other options, diagonals of a rectangle are equal.

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Question 401 Mark

If ABCD is a parallelogram with two adjacent angles $\angle\text{A}=\angle\text{B}$ then the parallelogram is a:

Answer

Rectangle.
Solution:
Given that ABCD is a parallelogram.
We konw that, opposite sides of a parallelogram are parallel.
$\Rightarrow\angle\text{A}+\angle\text{B}=180^{\circ}$ ...(interior angles)
Also, $\angle\text{A}=\angle\text{B}=90^{\circ}$ ...(Given)
Since opposite angles of a parallelogram are equal,
$\angle\text{A}=\angle\text{C}$ and $\angle\text{B}=\angle\text{D}$
So, $\angle\text{A}=\angle\text{C}=\angle\text{B}=\angle\text{D}=90^{\circ}$
$\therefore$ ABCD is a rectangle.

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Question 411 Mark

Write the correct answer in the following:
If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form:

Answer

A rectangle.
Solution:
PNQM is a rectangle.

Hence, (c) is the correct answer.

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Question 421 Mark

Three statements are given below:

In a rectangle ABCD, the diagonal AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$
In a square ABCD, the diagonal AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$
In a rhombus ABCD, the diagonal AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$

Which is true?

Answer

II and III
Solution:

Consider I.
We know that, in a rectangle the diagonals are not bisectors of each other, since the adjacent side.
Thus, I is false.

Consider II.
We know that, in a square the diagonals bisect the opposite angles.
So, in a square ABCD, the diagonals AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$
Thus, II is true.

Consider III.
We know that, in a rhombus the diagonals bisect the opposite angles.
So, in a rhombus ABCD, the diagonals AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$
Thus, III is true.

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Question 431 Mark

In each of the questions one question is followed by two statements I and II. Choose the correct option.
Is quadrilateral ABCD a rhombus?

Quadrilateral ABCD is a ||gm.
Diagonals AC and BD are perpendicular to each other.

Answer

If the question can be answered by both the statements together but not by any one of the two.
Solution:
Here, we can observe that neither I not II can alone justify the answer to the given question. But if we consider both I and II together then they completely satisfies the answer.
So the question can be answered by both the statement together but not by any one of the two.

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Question 441 Mark

Three statements are given below:

In a Rectangle ABCD, the diagonals AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$
In a Square ABCD, the diagonals AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$
In rhombus ABCD, the diagonals AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$

Which is True?

Answer

II and III
Solution:
In square and rhombus, all sides are equal. By joining the points A and C diagonal AC formed. we get two triangles ABC and ADC which are congruent (SAS congruence). Also, opposite sides are parallel. So, by using alternate angle property we can prove that angle $\text{BAC}= \angle\text{DCA}$ and angle $\text{DAC}= \angle\text{ACB}.$ But by CPCT $\angle\text{DAC}= \angle\text{BAC}.$
So, all four angles made by diagonal AC with end points A and C are equal which proves that diagonal AC bisects $\angle\text{A}$ and $\angle\text{C}$ both.

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Question 451 Mark

In a trapezium ABCD, if E and F be the mid-point of the diagonals AC and BD respectively. Then, EF = ?

Answer

$\frac{1}{2}(\text{AB}-\text{CD})$
Solution:

Construction: join CF and extend it to meet AB at G.
In $\triangle\text{CDF}$ and $\triangle\text{GFB},$
$\angle\text{CDF}=\angle\text{GFB}$ ...(Vertically opposite angles)
$\text{AB || CD},$
So, $\angle\text{DCF}=\angle\text{GFB,}$ ...(alternate angles)
$\text{DF = FB}$ ...(F is the mid-point of BD)
$\Rightarrow\triangle\text{CDF}\cong\triangle\text{GFB}$ ...(ASA congruence criterion)
So, $\text{CD = GB}$ and $\text{CF = GF}$ ...(C.P.C.T.)
Since E and F are the mid-points of CA and CG respectively,
we have $\text{EF}=\frac{1}{2}\text{AG}$
$=\frac{1}{2}(\text{AB}-\text{GB})$
$=\frac{1}{2}(\text{AB}-\text{CD})$

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Question 461 Mark

In each of the questions one question is followed by two statements I and II. Choose the correct option.
Is || gm ABCD a square?

Diagonals of ||gm ABCD are equal.
Diagonals of ||gm ABCD intersect at right angles.

Answer

If the question can be answered by both the statements together but not by any one of the two.
Solution:
We know that when the diagonals of a parallelogram are equal, it might be a square or a rectangle. But if the diagonals of that parallelogram intersect at a right angle, then it is definitely the a square. Thus, it can be concluded that both I and II together will give answer.
So, the question can be answered by both the statements together but not by any one of the two.

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Question 471 Mark

In each of the questions one question is followed by two statements I and II. Choose the correct option.
Is quadrilateral ABCD a parallelogram?

Its opposite sides are equal.
Its opposite angles are equal.

Answer

If the question can be answered by either statement alone.
Solution:
We know that a quadrilateral is a parallelogram when either I or II holds true.
So, the question can be answered by either statement alone.

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Question 481 Mark

Is quad. ABCD a parallelogram?

Its opposite sides are equal.
Its opposite angles are equal.

Answer

If the question can be answered by either statement alone;
Solution:
If the opposite sides of a quad. ABCD are equal, the quadrilateral is a parallelogram.
If the opposite angles are equal, then the quad. ABCD is a parallelogram.

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Question 491 Mark

In the given figure, ABCD is a parallelogram in which $\angle\text{BAD}=75{^\circ}$ and $\angle\text{CBD}=60^{\circ}.$ Then, $\angle\text{BDC}=?$

Answer

45°
Solution:
We know that, the opposite angles of a parallelogram are equal.
$\therefore\angle\text{C}=\angle\text{A}=75^{\circ}$
In $\triangle\text{BCD},$
$\angle\text{BCD}+\angle\text{BDC}+\angle\text{CBD}=180^{\circ}$ ...(Angle sum property)
$\Rightarrow75^{\circ}+\angle\text{BDC}+60^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{BDC}=45^{\circ}$

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Question 501 Mark

Which of the following is not true for the Parallelogram?

Answer

Opposite angles are bisected by the diagonals.
Solution:
If opposite angles are bisected by diagonals in parallelogram, all four bisected angles become equal which leads to equal adjacent side. That is not true in case of parallelogram.

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MATHS M.C.Q

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