M.C.Q

MCQ 1011 Mark

In Quadrilateral $\angle\text{A} = 38^\circ, \ \angle\text{C} = 3\angle\text{A},\ \angle\text{D} = 4\angle\text{A}.$ Find the value of $\angle\text{B}=\ ?$

A
57º
B
56º
C
55º
D
80º

Answer

56º
Solution:
$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ (angle sum property)
$\angle\text{C} = 3 ( 38) = 114^\circ$
$\angle\text{D} = 4 ( 38) = 152^\circ$
So, $38^\circ + \angle\text{B} + 114^\circ + 152^\circ = 360^\circ$
$\angle\text{B} = 360^\circ - 304^\circ = 56^\circ$

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MCQ 1021 Mark

The consecutive sides of a quadrilateral have:

A
No common point.
B
One common point.
C
Two common points.
D
Infinitely many common points.

Answer

One common point.
Solution:

Consecutive sides of a Quadrilateral ABCD are
AB and BC,
BC and CD,
CD and AD,
AD and AB,
Which have only one point in common
i.e the joint point of their ends.

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MCQ 1031 Mark

In the figure, ABCD is a Rectangle. Find the values of x and y?

A
x = 55º and y = 110º
B
x = 100º and y = 100º
C
x = 50º and y = 100º
D
x = 60º and y = 120º

Answer

x = 55º and y = 110º
Solution:
ABCO is a rectangle The diagonals of a rectangle are congruent and bisect each other. Therefore, in
$\triangle\text{AOB},$ we have:
OA = OB
$\angle\text{OAB} = \angle\text{OBA} = 35^\circ$
$\text{x} = 90^\circ - 35^\circ = 55^\circ$ and $\angle\text{AOB} = 180^\circ - (35^\circ + 35^\circ) = 110^\circ$
$\text{y} = \angle\text{AOB} = 110^\circ$ [Vertically opposite angles]
Hence, x = 55° and y = 110°

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MCQ 1041 Mark

In $\triangle\text{ABC},\ \angle\text{A} = 30^\circ, \ \angle\text{B} = 40^\circ$ and $\angle\text{C} = 110^\circ.$ The angles of the triangle formed by joining the mid-points of the sides of this triangle are:

A
30º, 40º, 110º
B
70º, 70º, 40º
C
60º, 40º, 80º
D
60º, 70º, 50º

Answer

30º, 40º, 110º
Solution:
Given,
In $\angle\text{ABC}$

$\angle\text{A} = 30^\circ,\ \angle\text{B} = 40^\circ,\ \angle\text{C} = 110^\circ$
In figure, BDEF, DCEF, DEAF are parallelograms so,
$\angle\text{B} = \angle\text{E} = 40^\circ$ [$∵ \angle\text{B}$ & $\angle\text{E}$ are opposite angles of parallelogram BDEF]
$\angle\text{C} = \angle\text{F} = 110^\circ$ [$∵ \angle\text{C}$ & $\angle\text{F}$ are opposite angles of parallelogram DCEF]
$\angle\text{A} = \angle\text{D} = 30^\circ$ [$∵ \angle\text{A}$ & $\angle\text{D}$ are opposite angles of parallelogram DEAF]
Hence, $\angle\text{D} = 30^\circ$
$\angle\text{E} = 40^\circ$
$\angle\text{F} = 110^\circ$

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MCQ 1051 Mark

ABCD is a parallelogram and E and F are the centroids of triangles ABD and BCD respectively, then EF =

A
AE.
B
BE.
C
CE.
D
DE.

Answer

AE.
Solution:

Centroid is the point where all medians of a meet.
In $\triangle\text{ABD},$ E is the centroid,
And in $\triangle\text{BCD},$ F is the centroid.
By the property of centroid, centroid divides a median in 2 : 1
So from figure,
$\frac{\text{AE}}{\text{EO}}=\frac{2}{1}\Rightarrow\text{EO}=\frac{\text{AE}}{2}\ ...(1)$
Also $\frac{\text{CF}}{\text{FO}}=\frac{2}{1}\Rightarrow\text{FO}=\frac{\text{CF}}{2}\ ...(2)$
Because AC is a digonal of a parallelogram, O is its midpoint.
⇒ OA = OC
⇒ AE = CF
Adding equations (1) & (2),
$\text{EO + FO} =\frac{\text{AE}+\text{CF}}{2}=\frac{\not2\text{AE}}{\not2}$
$\Rightarrow\text{EF}=\frac{\not2\text{AE}}{\not2}$
$\Rightarrow\text{EF}=\text{AE}$

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MCQ 1061 Mark

In a parallelogram ABCD if $\angle\text{A = (3x - 20)}, \ \angle\text{ B = (Y + 15)}, \ \angle\text{C = (x + 40)}$ then find the value of x and y?

A
x = 30º and y = 65º
B
x = 30º and y = 95º
C
x = 32º and y = 95º
D
x = 38º and y = 85º

Answer

x = 30º and y = 95º
Solution:
Given, ABCD is a parallelogram.
So,
$\angle\text{A} =\angle\text{C}$ (Opposite angles of parallelogram are equal in size)
⇒ 3x - 20 = x + 40
⇒ 3x - x = 40 + 20
⇒ 2x = 60
⇒ x = 30º
Thus, $\angle\text{A} = 3 \times 30 - 20 = 90 - 20 = 70^\circ$
Now, $\angle\text{A}+\angle\text{B}=180^\circ$ (Sum of interior angles of parallelogram is 180°)
$⇒ 70^\circ + \angle\text{B} = 180^\circ$
$⇒ \angle\text{B}= 180^\circ - 70^\circ$
$⇒ \angle\text{B}= 110^\circ$
⇒ y + 15 = 110º
⇒ y = 95º
Hence, x = 30º and y = 95º

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MCQ 1071 Mark

Is || gm ABCD a square?

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

A
If the question can be answered by one of the given statements alone and not by the other;
B
If the question can be answered by either statement alone;
C
If the question can be answered by both the statements together but not by any one of the two;
D
If the question cannot be answered by using both the statements together.

Answer

If the question can be answered by both the statements together but not by any one of the two;
Solution:
If the diagonals of a || gm ABCD are equal, then ‖gm ABCD could either be a rectangle or a square.
If the diagonals of the ‖gm ABCD intersect at right angles, then the ‖gm ABCD could be a square or a rhombus.
However, if both the statements are combined, then ‖gm ABCD will be a square.

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MCQ 1081 Mark

Is quadrilateral ABCD a || gm?

Diagonals AC and BD bisect each other.
Diagonals AC and BD are equal.

A
If the question can be answered by one of the given statements alone and not by the other;
B
If the question can be answered by either statement alone;
C
If the question can be answered by both the statements together but not by any one of the two;
D
If the question cannot be answered by using both the statements together.

Answer

If the question can be answered by one of the given statements alone and not by the other;
Solution:
If the diagonals of a quad. ABCD bisect each other, then the quad. ABCD is a parallelogram.
So, I gives the answer.
If the diagonals are equal, then the quad. ABCD is a parallelogram.
So, II gives the answer.

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MCQ 1091 Mark

In a quadrilateral ABCD, if AO and BO are the bisectors of $\angle\text{A}$ and $\angle\text{B}$ respectively, $\angle\text{C}=70^{\circ}$ and $\angle\text{D}=30^{\circ}.$ Then, $\angle\text{AOB}=?$

A
40°
B
50°
C
80°
D
100°

Answer

50°
Solution:

We know that, sum of the angles of a quadrilateral is 360°.
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^{\circ}$
$\Rightarrow\angle\text{A}+\angle\text{B}+70^{\circ}+30^{\circ}=360^{\circ}$
$\Rightarrow\angle\text{A}+\angle\text{B}=260^{\circ}$
$\Rightarrow\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}=\frac{1}{2}(260)^{\circ}$
$\Rightarrow\angle\text{BAO}+\angle\text{ABO}=130^{\circ}...(\text{i})$
In $\triangle\text{AOB},$
$\angle\text{BAO}+\angle\text{ABO}+\angle\text{AOB}=180^{\circ}$ ...(Angle sum Property)
$\Rightarrow130^{\circ}+\angle\text{AOB}=180^{\circ}$ ...(from (i))
$\Rightarrow\angle\text{AOB}=50^{\circ}$

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MCQ 1101 Mark

PQRS is a quadrilateral. PR and QS intersect each other at O. in which of the following cases, PQRS is a parallelogram?

A
$\angle\text{P}=100^\circ,\angle\text{Q}=80^\circ,\angle\text{R}=100^\circ$
B
$\angle\text{P}=85^\circ,\angle\text{Q}=85^\circ,\angle\text{R}=95^\circ$
C
$\text{PQ}=7\text{cm},\text{QR}=7\text{cm},\text{RS}=8\text{cm},\text{SP}=8\text{cm}$
D
$\text{OP}=6.5\text{cm},\text{OQ}=6.5\text{cm},\text{OR}=5.2\text{cm},\text{OS}=5.2\text{cm}$

Answer

$\angle\text{P}=100^\circ,\angle\text{Q}=80^\circ,\angle\text{R}=100^\circ$
Solution:
In a parallelogram, opposite corner angles are equal and sum of adjacent angles = 108°
Hence, in quadrrilateral PQRS,
$\Rightarrow\angle\text{P}=\angle\text{R}$ and $\angle\text{Q}=\angle\text{S}$
Also, $\angle\text{P}+\angle\text{Q}=\angle\text{Q}+\text{R}=180^\circ$
Hence, if $\angle\text{P}=100^\circ$ and $\angle\text{Q}=80^\circ,$ then
$\angle\text{P}+\angle\text{Q}=100^\circ+80^\circ=180^\circ$
Also, if $\angle\text{Q}+=80^\circ$ and $\angle\text{R}=100^\circ$ then
$\angle\text{Q}+\angle\text{R}=80^\circ+100^\circ=180^\circ$

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MCQ 1111 Mark

In a Trapezium ABCD, if $\text{AB || CD},$ then (AC² + BD²) =?

A
BC² + AD² + 2BC × AD
B
AB² + CD² + 2AB × CD
C
AB² + CD² + 2AD × BC
D
BC² + AD² + 2AB × CD

Answer

BC² + AD² + 2AB × CD
Solution:
Given: ABCD is a trapezium with $\text{AB || CD}$
Construction: Draw DE and CF $\bot$ to AB.
Then in $\triangle\text{ABC}$
$\angle\text{BAC}$ is acute
$\therefore$ BC² = AC² + AB² - 2AF : AB ...(1)
and in $\triangle\text{BDA}$
$\angle\text{DBA}$ is acute
$\therefore$ AD² = BD² + AB² - 2BE : AB ...(2)
Adding (1) and (2) we get
BC² + AD² = AC² + BD² + 2AB² - 2AF·AB - 2BE·AB
⇒ AC² + BD² = BC² + AD² - 2AB [AB - AF - BE)
= BC² + AD² - 2AB [AB - (AE + EF) - (BF+ EF)]
= BC² + AD² - 2AB [AB - (AE + EF +BF+ EF)]
= BC² + AD² - 2AB [AB - (AB+ CD)] ($\therefore$ EF = DC)
= BC² + AD² - 2AB [- (CD)]
= AD² + BC² + 2AB × CD

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MCQ 1121 Mark

If bisectors of $\angle\text{A}$ and $\angle\text{B}$ of a quadrilateral ABCD intersect each other at P, of $\angle\text{B}$ and $\angle\text{C}$ at Q, of $\angle\text{C}$ and $\angle\text{D}$ at R and of $\angle\text{D}$ and $\angle\text{A}$ at S, then PQRS is a:

A
Quadrilateral whose opposite angles are supplementary.
B
Parallelogram.
C
Rhombus.
D
Rectangle.

Answer

Quadrilateral whose opposite angles are supplementary.
Solution:
Let the half of the $\angle\text{A} , \angle\text{B}, \angle\text{C}$ and $\angle\text{D}$ are denoted by a, b, c and d respectively.
$\text{a + b} + \angle\text{APB} = 180$ and $\text{c + d} + \angle\text{DRC} = 180$ (angle sum property).
$\angle\text{APB} = \angle\text{QPS} $ and $\angle\text{DRC} = \angle\text{QRS}$ (vertically opposite angles).
So, $\angle\text{QPS} + \angle\text{QRS} = (180 - \text{a} - \text{b)} + (180 - \text{c} - \text{d)}$
$= 360\ -\ (\text{a + b + c + d)}$
$= 360 - \frac{1}{2} $ of $(\angle\text{ A} + \angle\text{B} + \angle\text{C} + \angle\text{D})$
$= 360 - \frac{1}{2} $ 0f $360 = 180$
Therefore, $\angle\text{QPS}$ and $\angle\text{QRS}$ are supplementary.

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MCQ 1131 Mark

In a quadrilateral ABCD, $\angle\text{A}+\angle\text{C}$ is 2 times $\angle\text{B}+\angle\text{D}$ If $\angle\text{A}=140^\circ$ and $\angle\text{D}=60^\circ$ then $\angle\text{B}=$

A
60°
B
80°
C
120°
D
None of these.

Answer

60°
Solution:
$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ\dots(1)$
Now, $\angle\text{A}+\angle\text{C}=2(\angle\text{B}+\angle\text{D})$ (given) ...(2)
Also, $\angle\text{A}=140^\circ,\ \angle\text{D}=60^\circ$
Putting value of $(\angle\text{A}+\angle\text{C})$ from eq. (2) in eq. (1)
$2(\angle\text{B}+\angle\text{D})+\angle\text{B}+\angle\text{D}=360^\circ$
$3(\angle\text{B}+\angle\text{D})=360^\circ$
$\Rightarrow\angle\text{B}+\angle\text{D}=120^\circ$
$\Rightarrow\angle\text{B}+60^\circ=120^\circ$
$\Rightarrow\angle\text{B}=60^\circ$

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MCQ 1141 Mark

Three Statements are given below:

In a, Parallelogram the angle bisectors of 2 adjacent angles enclose a right angle.
The angle bisector of a Parallelogram form a Rectangle.
The Triangle formed by joining the mid-points of the sides of an isosceles triangle is not necessarily an isosceles triangle.
Which of the statement/ statements is/ are True?

A
II
B
I
C
I and III
D
I and II

Answer

I and II
Solution:

The adjacent angles of a parallelogram are supplementary. Their halves add up to 90º. So the angle bisectors enclose a right angle.
All the adjacent angle bisectors enclose right angles. So we have a rectangle being enclosed by the angle bisectors of a parallelogram.
The triangle formed by joining the mid-points of the sides of an isosceles triangle is always an isosceles triangle, because halves of equal sides are also equal.

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MCQ 1151 Mark

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rhombus, if:

A
ABCD is a Parallelogram.
B
ABCD is rhombus.
C
Diagonals of ABCD are equal.
D
Diagonals of ABCD are perpendicular to each other.

Answer

Diagonals of ABCD are equal.
Solution:

In $\triangle\text{ABC},$ P and Q are the mid-points of sides AB and BC respectively.
$\therefore\text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC ...(i)}$
In $\triangle\text{BCD},$ Q and R are the mid-points of sides BC and CD respectively.
$\therefore\text{QR || BD}$ and $\text{QR}=\frac{1}{2}\text{BD ...(ii)}$
In $\triangle\text{ADC},$ S and R are the mid-points of sides AD and CD respectively.
$\therefore\text{RS || AC}$ and $\text{RS}=\frac{1}{2}\text{AC ...(iii)}$
In $\triangle\text{ABD},$ P and S are the mid-points of sides AB and AD respectively.
$\therefore\text{SP || BD}$ and $\text{SP}=\frac{1}{2}\text{BD ... (iv)}$
$\Rightarrow\text{PQ ∥ RS}$ and $\text{QR ∥ SP}$ [From (i), (ii), (iii) and (iv)]
Thus, PQRS is a parallelogram.
Now, AC = BD (given)
$\Rightarrow\frac{1}{2}\text{AC}=\frac{1}{2}\text{BD}$
$\Rightarrow\text{PQ = QR = RS = SP }$ [From (i), (ii), (iii) and (iv)]
Hence, PQRS is a rhombus if diagonals of ABCD are equal.

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MCQ 1161 Mark

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

A
Rectangle
B
Trapezium
C
Square
D
None of these

Answer

Rectangle
Solution:

Let ABCD be a rhombus and P,Q,R and S be the mid-points of sides AB, BC, CD and DA respectively.
In $\triangle\text{ABD} $ and $\triangle\text{BDC} $
we have,
$\text{SP || BD}$ and $\text{SP}=\frac{1}{2}\text{BD}\ ...\ \text{(1)}$ [By mid-point and theorem]
$\text{RQ || BD}$ and $\text{RQ}=\frac{1}{2}\text{BD}\ ...\ \text{(2)}$ [By mid-point and theorem]
rom (1) and (2) we get,
$\text{SP || RQ}$
PQRS is a parallelogram
As diagonals of a rhombus bisect each other at right angles.
$∴ \text{AC⊥BD}$
Since, $\text{SP || BD, PQ || AC}$ and $ \text{AC⊥BD}$
$∴ \text{SP⊥PQ}$
$∴ \angle\text{QPS} = 90^\circ$
$∴ \text{PQRS}$ is a rectangle.

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MCQ 1171 Mark

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If $\angle\text{DAC}=32^\circ$ and $\angle\text{AOB}=70^\circ$ then, $\angle\text{DBC}$ is equal to:

A
24º
B
86º
C
38º
D
40º

Answer

38º
Solution:

Let ABCD is the parallelogram, in which AC and BD are diagonals that intersect at O.
We have, $\angle\text{DAC} = 32^\circ; \ \angle\text{AOB} = 70^\circ$ [given]
Since ABCD is a ||gm, so $\text{AB || CD}$ and $\text{AD || BC}.$
Since $\text{AD || BC},$ and AC is a transversal,
Then $\angle\text{ACB} = \angle\text{DAC}$ [alternate interior angles]
So, $\angle\text{ACB} = 32^\circ ⇒ \angle\text{OCB} = 32^\circ$
Since, $\angle\text{AOB}$ and $\angle\text{OBC}$ form a linear pair, then $\angle\text{AOB} + \angle\text{BOC}=180^\circ$
$⇒70^\circ+\angle\text{BOC}=180^\circ⇒\angle\text{BOC} = 180^\circ − 70^\circ ⇒ \angle\text{BOC}=110^\circ$
In $\angle\text{OBC},$
We have, $\angle\text{BOC} + \angle\text{OCB} + \angle\text{OBC} =180^\circ$ [angle sum property]
$⇒110^\circ + 32^\circ + \angle\text{OBC} = 180^\circ$
$⇒142^\circ + \angle\text{OBC}=180^\circ$
$⇒\angle\text{OBC} = 180^\circ−142^\circ$
$⇒\angle\text{OBC} = 38^\circ$
$⇒\angle\text{DBC} = 38^\circ$

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MCQ 1181 Mark

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

A
E is the mid-Point of AC.
B
AB = BC.
C
DE = BC.
D
DE and BC meet at some point ifwe extend both of them indefinitely.

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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MCQ 1191 Mark

What is the length of PQ in a trapezium ABCD in which $\text{AB || DC}$ and P and Q are mid-points on AD and BC respectively?

A
$\frac{1}{2}\text{(AB + CD)}$
B
$\frac{1}{2}\text{(AB + BD)}$
C
$\frac{1}{2}\text{AB}$
D
$\frac{1}{2}\text{CD}$ Join PD and Produce it to meet BA at G.

Answer

$\frac{1}{2}\text{(AB + CD)}$
Solution:
In $\triangle\text{PCD}$ and $\triangle\text{APG},$
$\angle\text{DPC} = \angle\text{GPA},$
$\angle\text{PDC} = \angle\text{AGP}$
$∴\triangle\text{PCD}≅\triangle\text{APG}$
CD = AG and PD = PG
In $\triangle\text{BGD},$
P is the mid-point of GD
Q is the mid-point of BD
Therefore, By mid-point theorem, $\text{PQ || AB}$
and $\text{PQ} = \frac{1}{2}\text{(GB)}$
but GB = GA + AB = CD + AB
$∴\text{PQ}=\big(\frac{1}{2}\big)\text{(AB + CD)}$

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MCQ 1201 Mark

Three angles of a quadrilateral are 80º, 95º and 112º. Its fourth angle is:

A
100°
B
73º
C
85º
D
78º

Answer

73º
Solution:
Let the fourth angle be x
80º + 95º + 112º + xº = 360º (Sum of angles of quadrilateral)
287º + xº = 360º
x = 360º – 287º
= 73º
Hence, 73º is correct.

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MCQ 1211 Mark

Three statements are given below:

In a || gm, the angle bisectors of two adjacent angles enclose a right angle.
The angle bisectors of a || gm form a rectangle.
The triangle formed by joining the mid-points of the sides of an isosceles triangle is not necessarily an isosceles triangle.

Which is true?

A
I only
B
II only
C
I and II
D
II and III

Answer

I and II
Solution:
However, the triangle formed by joining the mid-point of the sides of an isosceles triangle is surely an isosceles triangle.
So, III is false
Thus, I and II are true.

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MCQ 1221 Mark

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

A
Rectangle.
B
Parallelogram.
C
Rhombus.
D
Square.

Answer

Parallelogram.
Solution:

PQ || SR || AC
QR || PS || BD
{Because line joining the mid-points of two sides of triangle is || to third side}
Now because AC is not prependicular to BD in parallelogram,
⇒ SR is not perpendicular to QR
Also $\triangle\text{ASP}\not\cong\triangle\text{DRS}$
$⇒ \text{PS} \neq \text{SR}$
⇒ PQRS is just a parallelogram.

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MCQ 1231 Mark

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

A
None of these.
B
PQRS is a Parallelogram.
C
Diagonals of PQRS are equal.
D
PQRS is a Rhombus.

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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MCQ 1241 Mark

In the given figure, ABCD is a || gm and E is the mid-point of BC. Also, DE and AB when produced meet at F. Then:

A
$\text{AF}=\frac{3}{2}\text{AB}$
B
AF = 2AB
C
AF = 3AB
D
AF² = 2AB²

Answer

AF = 2AB
Solution:
In $\triangle\text{CDE}$ and $\triangle\text{BFE,}$
$\angle\text{DEC}=\angle\text{FEB}$ ...(Vertically opposite angles)
$\angle\text{DCE}=\angle\text{FBE}$ ...(Alternate angles)
and $\text{CE = BE}$ ...(E is the mid-point.)
$\therefore\triangle\text{CDE}\cong\triangle\text{BFE}$ ...(By AA congruence criterion)
$\therefore\text{CD = BF}$ ...(C.P.C.T.)
Now,
$\text{AF = AB + BF}$
$\Rightarrow\text{AF + AB + CD}$ ...(from (i))
$\Rightarrow\text{AF = 2AB}$

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MCQ 1251 Mark

Is quadrilateral ABCD a rhombus?

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

A
If the question can be answered by one of the given statements alone and not by the other;
B
If the question can be answered by either statement alone;
C
If the question can be answered by both the statements together but not by any one of the two;
D
If the question cannot be answered by using both the statements together.

Answer

If the question can be answered by both the statements together but not by any one of the two;
Solution:
If the quad. ABCD is a ‖gm, it could be a rectangle or square or rhombus.
So, statement I is not sufficient to answer the question.
If the diagonals AC and BD are perpendicular to each other, then the ‖gm could be a square or rhombus.
So, statement II is not sufficient to answer the question.
However, if the statements are combined, then the quad. ABCD is a rhombus.

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MCQ 1261 Mark

In E is the mid-point of median AD such that BE produced meets AC at F. If AC = 10.5cm, then AF =

A
3cm.
B
3.5cm.
C
2.5cm.
D
5cm.

Answer

3.5cm.
Solution:

A line DG is drawn parallel to EF to meet AC.
FE || DG and FE || GH
Now, consider $\triangle\text{ADG}.$
E is the mid-point of AD and EF is line from E || to Base DG.
So by property, it will meet AG at its midpoint
i.e. F is midpoint of AG.
⇒ AF = FG ...(1)
Now, consider $\triangle\text{FBC}\ \&\ \triangle\text{GDC}$
FE || GH and FE || GD
D is mid-point of BC.
$\Rightarrow\frac{\text{DC}}{\text{BC}}=\frac{1}{2}\dots(2)$
Because $\triangle\text{FBC}\sim\triangle\text{GDC},$
$\Rightarrow\frac{\text{GC}}{\text{FC}}=\frac{1}{2}$
⇒ FC = 2GC
or FG = GC ...(3)
From equation (1) and (3)
AF = FG = GC
$\Rightarrow\text{AF}=\frac{\text{AC}}{3}=\frac{10.5}{3}=3.5\text{cm}$

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MCQ 1271 Mark

In a $\triangle\text{ABC},$ 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?

A
52cm
B
20cm
C
51cm
D
80cm

Answer

51cm
Solution:
In $\triangle\text{ABC},$ R and P are the mid-points of AB and BC.
$\therefore\ \text{RP || AC},\ \text{RP}=\frac{1}{2\text{AC}}$ [By mid-point theorem]
In a quadrilateral RPQA,
$⇒\text{RP || AQ}⇒\text{RP || AQ}$
$\therefore\ \text{RPQA}$ is a parallelogram
$⇒\text{AR}=\frac{1}{2\text{AB}}$
$\therefore\ \text{AR}=12×30=15\text{cm}$
$⇒ \text{AR} = \text{PQ} = 15\text{cm}$ [Since, opposite sides are equal]
$⇒\text{RP}=\frac{1}{2\text{AC}}=\frac{1}{2}×21=10.5\text{cm}$ [Since, opposite sides are equal]
⇒ Perimeter of ARPQ = AR + QP + RP + AQ
= 15 + 15 + 10.5 + 10.5
= 51cm
$∴$ Perimeter of ARPQ is 51cm.

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MCQ 1281 Mark

A
Rectangle.
B
Parallelogram.
C
Rhombus.
D
Quadrilateral whose opposite angles are supplementary.

Answer

Quadrilateral whose opposite angles are supplementary.
Solution:

In $\triangle\text{APB},$ by angle sum property,
$\angle\text{APB}+\angle\text{PAB}+\angle\text{PBA}=180^{\circ}$
$\Rightarrow\angle\text{APB}+\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}=180^{\circ}$
$\Rightarrow\angle\text{APB}=180^{\circ}-\Big(\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}\Big)$
In $\triangle\text{CRD},$ by angle sum property,
$\angle\text{CRD}+\angle\text{RDC}+\angle\text{RCD}=180^{\circ}$
$\Rightarrow\angle\text{CRD}+\frac{1}{2}\angle\text{D}+\frac{1}{2}\angle\text{C}=180^{\circ}$
$\Rightarrow\angle\text{CRD}=180^{\circ}-\Big(\frac{1}{2}\angle\text{D}+\frac{1}{2}\angle\text{C}\Big)$
Now, $\angle\text{SPQ}+\angle\text{SRQ}=\angle\text{APB}+\angle\text{CRD}$
$=360^{\circ}-\Big(\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\frac{1}{2}\angle\text{D}\Big)$
$=360^{\circ}=\frac{1}{2}(\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D})$
$=360^{\circ}-\frac{1}{2}\times360^{\circ}$
$=360^{\circ}-180^{\circ}$
$=180^{\circ}$
Now, $\angle\text{PSR}+\angle\text{PQR}=360^{\circ}-(\angle\text{SPQ}+\angle\text{SRQ})$
$=360^{\circ}-180^{\circ}$
$=180^{\circ}$
Hence, PQRS is a quadrilateral whose opposite angles are supplementary.

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MCQ 1291 Mark

In the given figure, ABCD is a rhombus. Then:

A
AC² + BD² = AB²
B
AC² + BD² = 2AB²
C
AC² + BD² = 4AB²
D
2(AC² + BD²) = 3AB²

Answer

AC² + BD² = 4AB²Solution:
The diagonals of a rhombus bisect each other at right angles.
$\Rightarrow\text{OA}=\frac{1}{2}\text{AC}$ and $\text{OB}=\frac{1}{2}\text{BD}$
Also, $\angle\text{AOB}=90^{\circ}$
In right $\triangle\text{AOB},$
$\therefore\text{AB}^2=\text{OA}^2+\text{OB}^2$ ...(By Pythagoras theorem)
$\Rightarrow\text{AB}^2=\frac{1}{4}\text{AC}^2+\frac{1}{4}\text{BD}^2$
$\Rightarrow\text{AC}^2+\text{BD}^2=4\text{AB}^2$

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MCQ 1301 Mark

In Fig. the quadrilateral ABCD circumscribes a circle with centre O. If $\angle\text{AOB} = 115^\circ,$ then find $\angle\text{COD}.$

A
23º
B
24º
C
127º
D
115º

Answer

115º
Solution:
$\therefore \angle\text{AOB} = \angle\text{COD}$ (vertically opposite angle)
$\therefore \angle\text{COD}=115^\circ$

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MCQ 1311 Mark

In the given figure, ABCD is a parallelogram in which $\angle\text{BDC}=45^{\circ}$ and $\angle\text{BAD}=75^{\circ}.$ Then, $\angle\text{CBD}=?$

A
45°
B
55°
C
60°
D
75°

Answer

60°.
Solution:
Since ABCD is a parallelogram, AB || CD since opposite angles of a parallelogram are equal.
$\Rightarrow\angle\text{ABD}=\angle\text{BCD}=45^{\circ}$ ...(Alternate angles)
In $\triangle\text{ADB},$
$\angle\text{ABD}+\angle\text{BDA}+\angle\text{DAB}=180^{\circ}$ ...(Angle sum property)
$\Rightarrow45+\angle\text{BDA}+75=180$
$\Rightarrow\angle\text{BDA}+120=180$
$\Rightarrow\angle\text{BDA}=60^{\circ}$
$\Rightarrow\angle\text{CBD}=\angle\text{BDA}=60^{\circ}$ ...(Alternate angles)
$\Rightarrow\angle\text{CBD}=60^{\circ}$

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MCQ 1321 Mark

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

A
$\frac{3}{2}\text{AB}$
B
$2\text{AB}$
C
$3\text{AB}$
D
$\frac{5}{4}\text{AB}$

Answer

$2\text{AB}$
Solution:

BE || AD
⇒ BE || AD
Now, consider $\triangle\text{FAD}$
BE || AD
Also $\frac{\text{BE}}{\text{AD}}=\frac{1}{2}$
In $\triangle\text{FBE}$ and $\triangle\text{FAD},$
$\angle\text{FAD}=\angle\text{FBE}$ {Corresponding angles}
$\angle\text{ADF}=\angle\text{BEF}$ {Corresponding angles}
$\angle\text{F} =\angle\text{F}$ {Common}
Hence, $\triangle\text{FBE}\sim\triangle\text{FAD}$
$\Rightarrow\frac{\text{BF}}{\text{AF}}=\frac{\text{BE}}{\text{AD}}=\frac{1}{2}$
$\Rightarrow1-\frac{\text{BF}}{\text{AF}}=1-\frac{1}{2}$
$\Rightarrow\frac{\text{AF}-\text{BF}}{\text{AF}}=\frac{1}{2}$
$\Rightarrow\frac{\text{AB}}{\text{AF}}=\frac{1}{2}$
$\Rightarrow\text{AF}=\text{2AB}$

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MCQ 1331 Mark

In Quadrilateral ABCD, $\angle\text{A}+\angle\text{C}=140^\circ, \ \angle\text{A}:\angle\text{C}=1:3$ and $\angle\text{B}:\angle\text{D}=5:6.$ Find the values of $\angle\text{A}, \ \angle\text{B}, \ \angle\text{C}$ and $\angle\text{D}\ ?$

A
35º, 100º, 105º, 120º
B
100º, 102º, 120º, 10º
C
10º, 20º, 100º, 260º
D
90º, 90º, 100º, 80º

Answer

35º, 100º, 105º, 120º
Solution:
Given $\angle\text{A}+ \angle\text{C}=140^\circ$
And $\angle\text{A}: \angle\text{C} = 1:3$
And $\angle\text{B}:\angle\text{D}=5:6$
$\Rightarrow\ \angle\text{A} = \frac{1}{4}×1400=35^\circ$
$\Rightarrow\ \angle\text{C} = \frac{3}{4}×1400=105^\circ$
Now according to angle sum property of quadrilateral,
$\angle\text{A} +\angle\text{B}+ \angle\text{C}+ \angle\text{D} = 360^\circ$
$\Rightarrow\ 350^\circ+\angle\text{B}+ 1050^\circ +\angle\text{D} = 360^\circ$
$\Rightarrow\ \angle\text{B}+\angle\text{D} = 360^\circ−140^\circ=220^\circ$
$\Rightarrow\ 5\text{x} + 6\text{x} = 220^\circ$
$\Rightarrow\ \text{x} = 20^\circ$
So, $\angle\text{B} = 5×20^\circ=100^\circ$
And $\angle\text{D}=6×20^\circ=120^\circ$

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MCQ 1341 Mark

We get a rhombus by joining the mid-points of the sides of a:

A
Parallelogram.
B
Rhombus.
C
Rectangle.
D
Triangle.

Answer

Rectangle.
Solution:

$\text{PR}||\text{AD}\Rightarrow\text{AB}\not\bot\text{AD}$
$\text{QS}||\text{AB}\Rightarrow\text{PR}\not\bot\text{QS}$
Since diagonals of PQRS are not making 90° between them,
PQRS is not a Rhombus.

P, Q, R and S are the mid-points,
PR and QS are diagonals of quadrilateral PQRS.
PR || AD, QS || AB
Because they are Formed by joning of mid-points of sides of Rhombus ABCD.
AD is not $\bot$ to AB
⇒ PR will not be $\bot$ to QS
i.e angle between diagonals PR & QS is not 90°.
So, PQRS is not a Rhombus.

PR and QS are making 90° with each - other.
Because PR || AD, QS || AB and $\text{AD}\perp\text{AB}$
So PR and QS are diagonals of PQRS and are $\perp$ to each other.
Hence , PQRS is a Rhombus.

By joining the mid-points of sides of a triangle, no quadrilateral is formed.

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MCQ 1351 Mark

In a rhombus ABCD, if $\angle\text{ACB} = 40^\circ,$ then $\angle\text{ADB} =\ ?$

A
55º
B
60º
C
50º
D
25º

Answer

50º
Solution:

Given ABCD is a rhombus. Diagonals bisect each other perpendicularly.
Hence $\angle\text{BOC} = 90^\circ$
Given $\angle\text{OCB} = 40^\circ$
$\text{AD || BC}$ and BD is the transversal
$∴ \angle\text{ADB} = \angle\text{DBC}$ (Alternate angles)
Hence in right angled $\triangle\text{BOC},$
$\angle\text{BOC} + \angle\text{OCB} + \angle\text{OBC} = 180^\circ$
$⇒ 90^\circ + 40^\circ + \angle\text{OBC} = 180^\circ$
$⇒130^\circ + \angle\text{OBC} = 180^\circ$
$⇒ \angle\text{OBC} = 180^\circ - 130^\circ = 50^\circ$
But $\angle\text{OBC} = \angle\text{DBC}$
Therefore, $\angle\text{ADB} = 50^\circ.$

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MCQ 1361 Mark

If bisector of $\angle\text{A}$ and $\angle\text{B}$ of a quadrilateral ABCD intersect each other at p, $\angle\text{B}$ and $\angle\text{C}$ at Q, $\angle\text{C}$ and $\angle\text{D}$ at R and, $\angle\text{D}$ and $\angle\text{A}$ at S then PQRS is a:

A
Rhombus.
B
Quadrilateral whose opposite angles are supplementary.
C
Rectangle.
D
Parallelogram.

Answer

Rectangle.
Solution:
Let's assume our quadrilateral ABCD as a parallelogram:

we know
$\angle\text{DCB} + \angle\text{ABC} = 180^\circ$ (Co-interior angles of parallelogram are supplementary)
$⇒\frac{1}{2}\angle\text{DCB} + \frac{1}{2}\angle\text{ABC} = 90^\circ$ (Both sides divide by 2)
$⇒\angle\text{1} + \angle\text{2} =90^\circ... (1)$
In, $\triangle\text{CQB}$ we know
$ ⇒\angle\text{1} + \angle\text{2} +\angle\text{CQB} = 180^\circ ... (2)$
From eq (1) and eq (2), We get
$⇒ \angle\text{CQB} = 180^\circ − 90^\circ$
$⇒\angle\text{CQB} = 90^\circ$
$⇒\angle\text{PQR} = 90^\circ$ (because $\angle\text{CQB} = \angle\text{PQR},$ vertically opposite angles)
Similarly, it can be shown
$\angle\text{QPS} = \angle\text{PSR} =\angle\text{SRQ} = 90^\circ$
So, quadrilateral PQRS is a rectangle.

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MCQ 1371 Mark

Diagonals of a quadrilateral ABCD bisect each other. If $\angle\text{A}=45^\circ,$ then $\angle\text{B}=$

A
115°
B
120°
C
125°
D
135°

Answer

135°
Solution:

Consider $\triangle\text{AOD}\ \&\ \triangle\text{COB},$
AO = CO {Diagonals bisects each other}
OD = OB {Diagonals bisects each other}
$\angle\text{AOD}=\angle\text{COB}$ (Opposite angles)
So by SAS property, $\triangle\text{AOD}\cong\triangle\text{COB},$
$\Rightarrow\angle\text{ADO}=\angle\text{CBO}\dots(1)$
$\angle\text{ABD}=180^\circ-\angle\text{A}-\angle\text{ADO}$ $($in $\triangle\text{ADB})$
$=180^\circ-45^\circ-\angle\text{ADO}$
$\angle\text{ABD}=135^\circ-\angle\text{ADO}\dots(2)$
$\angle\text{B}=\angle\text{ABD}+\angle\text{CBO}$
Putting values From eq (1) and (2)
$\angle\text{B}=135^\circ-\angle\text{ADO}+\angle\text{ADO}$
$\angle\text{B}=135^\circ$

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MCQ 1381 Mark

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

A
1 : 2
B
2 : 1
C
1 : 3
D
1 : 1

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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MCQ 1391 Mark

In a Trapezium ABCD, if $\text{AB || CD},$ then (AC² + BD²) = ?

A
BC² + AD² + 2AB.CD
B
BC² + AD² + 2BC.AD
C
AB² + CD² + 2AD.BC
D
AB² + CD²+ 2AB.CD

Answer

BC² + AD² + 2AB.CD
Solution:
Given: ABCD is a trapezium with $\text{AB || CD}$
Construction: Draw DE and CF $⊥$ to AB

Then in $\triangle\text{ABC}$
$\angle\text{BAC}$ is acute
$∴$ BC² = AC² + AB² - 2AF : AB ...(1)
and In $\triangle\text{BDA}$
$\angle\text{DBA}$ is acute
$∴$ AD² = BD² + AB² - 2BE.AB ...(2)
Adding (1) and (2) we get
BC² + AD² = AC² + BD² + 2AB² - 2AF.AB - 2BE.AB
⇒ AC² + BD² = BC² + AD² - 2AB [AB - AF - BE]
= BC² + AD² - 2AB [AB - (AE + EF) - (BF + EF)]
= BC² + AD² - 2AB [AB - (AE + EF + BF + EF)]
= BC² + AD² - 2AB [AB - (AB + CD)] ($∴$ EF = DC)
= BC² + AD² - 2AB [(-CD)]
= AD² + BC² + 2AB × CD

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MCQ 1401 Mark

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

A
Prallelogram.
B
Rhombus.
C
Rectangle.
D
Square.

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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MCQ 1411 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 =

A
5cm.
B
6cm.
C
8cm.
D
10cm.

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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MCQ 1421 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:

A
$80^\circ,\ 160^\circ,\ 80^\circ,\ \text{and}\ 40^\circ$
B
$20^\circ,\ 30^\circ,\ 160^\circ,\ \text{and}\ 160^\circ$
C
$100^\circ,\ 100^\circ,\ 80^\circ,\ \text{and}\ 80^\circ$
D
$90^\circ,\ 90^\circ,\ 90^\circ,\ \text{and}\ 90^\circ$

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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MCQ 1431 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:

A
A trapezium.
B
A parallelogram.
C
A rectangle.
D
A rhombus.

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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MCQ 1441 Mark

Which of the following is not true for a parallelogram?

A
Opposite sides are equal.
B
Opposite angles are equal.
C
Opposite angles are bisected by the diagonals.
D
Diagonals bisect each other.

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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MCQ 1451 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?

A
20cm
B
80cm
C
51cm
D
52cm

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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MCQ 1461 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:

A
Kite
B
Parallelogram
C
Trapezium
D
Rhombus

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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MCQ 1471 Mark

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

A
E is the mid-Point of AC.
B
AB = BC.
C
DE = BC.
D
DE and BC meet at some point if we extend both of them indefinitely.

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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MCQ 1481 Mark

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

A
10cm
B
12cm
C
8cm
D
9cm

Answer

10cm
Solution:
Let us assume a rhombus ABCD where,
AB = BC = CD = DA
Now, in triangle OBC by using Pythagoras theorem we get:
BC² = OB² + OC²BC² = 6² + 8²BC² = 36 + 64
BC² = 100
$\text{BC}=\sqrt{100}$
BC = 10cm
$∴$ AB = BC = CD = DA = 10cm

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MCQ 1491 Mark

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

A
Rhombus
B
Triangle
C
Parallelogram
D
Rectangle

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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MCQ 1501 Mark

The bisectors of the angles of a parallelogram enclose a:

A
Rhombus.
B
Square.
C
Rectangle.
D
Parallelogram.

Answer

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

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Maths M.C.Q