M.C.Q - MATHS STD 9 Questions - Vidyadip

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Question 11 Mark

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

$\frac{3}{2}\text{AB}$
$2\text{AB}$
$3\text{AB}$
$\frac{5}{4}\text{AB}$

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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Question 21 Mark

In a parallelogram ABCD, if $\angle\text{DAB}=75^\circ$ and $\angle\text{DBC}=60^\circ,$ then $\angle\text{BDC}=$

75°
60°
45°
55°

Answer

45°

Solution:

In parallelogram ABCD,
$\angle\text{A}+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{D}=180^\circ-75^\circ=105^\circ$
$\angle\text{ADB}=\angle\text{DBC}$ (Alternate angles)
$\Rightarrow\angle\text{ADB}=60^\circ$
$\angle\text{BDC}=\angle\text{ADC}-\angle\text{ADB}=105^\circ-60^\circ=45^\circ$

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Question 31 Mark

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

Prallelogram.
Rhombus.
Rectangle.
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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Question 41 Mark

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

70°
45°
50°
60°

Answer

50°

Solution:

Consider $\triangle\text{AOD} \ \&\ \triangle\text{COB}$ $$
$\angle\text{AOD}=\angle\text{COB}=90^\circ$
AD = BC (Sides of Rhombus)
AO = CO (Diagonals bisects each other)
So by RHS property, $\triangle\text{AOD}\cong\triangle\text{COB}$
$\Rightarrow\angle\text{OAD}=\angle\text{OCB}=40^\circ$
$\angle\text{ADB}=\angle\text{ADO}=180^\circ-90^\circ-40^\circ=50^\circ$

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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:

108°
54°
72°
81°

Answer

72°

Solution:

Let ABCD be a parallelogram and $\angle\text{A}=\frac{2}{3}\angle\text{B}$
Also, $\angle\text{A}+\angle\text{B}=180^\circ$ (Adjacent angles in a parallelogram are supplementry)
$\Rightarrow\frac{2}{3}\angle\text{B}+\angle\text{B}=180^\circ$
$\Rightarrow\angle\text{B}=108^\circ$ and $\angle\text{A}=72^\circ$
⇒ Smallest angle is 72°.

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Question 61 Mark

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

Parallelogram.
Rhombus.
Rectangle.
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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Question 71 Mark

The bisectors of any two adjacent angles of a parallelogram intersect at:

30°
45°
60°
90°

Answer

90°

Solution:

In a parallelogram, sum of adjacent angles = 180°
$\Rightarrow \angle \text{A}+\angle\text{B}=180^\circ$
$\Rightarrow\frac{\angle\text{A}}{2}+\frac{\angle\text{B}}{2}=90^\circ\ ...(1)$
$\Rightarrow\angle\text{OAB}=\frac{\angle\text{A}}{2}$ and $\angle\text{OBA}=\frac{\angle\text{B}}{2}$
Thus, $\angle\text{OAB}+\angle\text{OBA}=90^\circ$ [From eq (1)]
$\Rightarrow\angle\text{AOB}=180^\circ-(\angle\text{OAB}+\angle\text{OBA})=180^\circ-90^\circ$
$\Rightarrow\angle\text{AOB}=90^\circ$

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Question 81 Mark

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

Square.
Rhombus.
Trapezium.
None of these.

Answer

Rhombus.

Solution:

PQ || AC (since in $\triangle\text{ABC}$ mid-points of AB & BC are meeting by PQ)
Similarly, SR || AC
⇒ PQ || SR
Now in $\triangle\text{ABD}$ and $\triangle\text{CBD},$
PS || BD and QR || BD
⇒ PS || QR
Hence, PQRS is a parallelogram.
But $\text{PR }\bot \text{ QS}$
⇒ Diagonals cut at 90°
⇒ PQRS is a Rhomus.

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Question 91 Mark

The consecutive sides of a quadrilateral have:

No common point.
One common point.
Two common points.
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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Question 101 Mark

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

Rectangle.
Parallelogram.
Rhombus.
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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Question 111 Mark

Digonals necessarily bisect opposite angles in a:

Ractangle.
Parallelogram.
Isosceles trapezium.
Square.

Answer

Square.

Solution:
Diagonals necessarily bisect opposite angles in a square.

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Question 121 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 =

5cm.
6cm.
8cm.
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 131 Mark

If one angle of a parallelogram is $24^\circ$ less than twice the smallest angle, then the measure of the largest angle of the largest angle of the parallelogram is:

A
$176^\circ$
B
$68^\circ$
✓
$112^\circ$
D
$102^\circ$

Answer

Correct option: C.

$112^\circ$

Let the smallest angle $=\angle\text{ADC}=\text{x}^\circ$
Other angle $\angle\text{BCD}$
$\Rightarrow\angle\text{BCD}=2\text{x}^\circ-24^\circ$
Also, $\angle\text{ACD}+\angle\text{BCD}=180^\circ ($Sum of adjacent angles in $\|^{gram} = 180^\circ )$
$\Rightarrow\text{x}^\circ+2\text{x}^\circ-24^\circ=180^\circ$
$\Rightarrow3\text{x}^\circ=204^\circ$
$\Rightarrow\text{x}=68^\circ$
$\Rightarrow$ Largest angle $=\angle\text{BCD}=2\times68^\circ-24^\circ=112^\circ$

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Question 141 Mark

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

3cm.
3.5cm.
2.5cm.
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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Question 151 Mark

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

70°
90°
80°
100°

Answer

80°

Solution:

In $\triangle\text{ABD},$
$\angle\text{BDA}+\angle\text{ABD}+\angle\text{DAB}=180^\circ$
$\angle\text{ABD}=50^\circ$ and $\angle\text{DAB}=90^\circ$
$\Rightarrow\angle\text{BDA}=180^\circ-90^\circ-50^\circ=40^\circ$
Consider $\triangle\text{ABD}\ \&\ \triangle\text{BAC}$
$\text{AD}=\text{BC},\ \angle\text{DAB}=\angle\text{ABC}=90^\circ,\text{BD}=\text{AC}$
Hence, by RHS property $\triangle\text{ABD}\cong\triangle\text{BAC}$
$\Rightarrow\angle\text{ABD}=\angle\text{BAC}=50^\circ$
Now, consider $\triangle\text{ABP}$
$\angle\text{PAB}+\angle\text{PBA}+\angle\text{APB}=180^\circ$
$\angle\text{PAB}=\angle\text{BAC}=50^\circ$
$\angle\text{PAB}=\angle\text{ABD}=50^\circ$
$\Rightarrow\angle\text{APB}=180^\circ-50^\circ-50^\circ=80^\circ$
Now, $\angle\text{APB}=\angle\text{DPC}$ (Opposite angles)
$\Rightarrow\angle\text{DPC}=80^\circ$

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Question 161 Mark

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

Rhombus.
Square.
Rectangle.
parallelogram.

Answer

Square.

Solution:

PS || QR, PQ || SR ...(1)
{Because lines joining the mid-points of any two sides of a triangle are parallel to the third side}
$\text{AC } \bot \text{ BD}$ & $\text{BR } \bot \text{ QS}$ (From Figure)
SR || AC and QR || BD
$\text{AC } \bot \text{ BD}$
$\Rightarrow \text{SR }\bot \text{ QR}$
Hence $\angle\text{SRQ}=90^\circ\ ...(2)$
Also $\triangle\text{APS}\cong\triangle\text{DSR}$
$\Rightarrow\text{PS} = \text{SR}\dots(3)$
From equations (1), (2), (3)
PQRS is a square.

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Question 171 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 $\angle\text{B}=$

60°
80°
120°
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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Question 181 Mark

The bisectors of the angle of a parallelogram enclose a:

Parallelogram.
Rhombus.
Rectangle.
Square.

Answer

Rectangle.

Solution:

AR, BR, CP, DP are the bisectors of angles of parallelogram.
Because two bisectors of adjacent angles make 90° between them So PQRS is a Rectangle
Because DP and BR are acute angle bisectors so the distance between them PQ < PS (The distance between other two bisectors) So $\text{PQ}\neq\text{PS}$ (So PQRS is not a square, but only a rectangle)

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Question 191 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:

70°, 70°, 40°
60°, 40°, 80°
30°, 40°, 110°
60°, 70°, 50°

Answer

30°, 40°, 110°

Solution:

If in any triangle, all the mid-points (of each sides) are joined to form a triangle, then that triangle is similian to a parent triangle.
i.e.$\triangle\text{QPR}\sim\triangle\text{ABC}$
So angles of $\triangle\text{PQR}$ will be same as angles of $\triangle\text{ABC}.$
Thus, angles are 30°, 40°, 110°

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Question 201 Mark

Which of the following quadrilateral is not a rhombus?

All four sides sre equal.
Diagonals bisect each other.
Diagonals bisect opposite angles.
One angle between the diagonals is 60°.

Answer

One angle between the diagonals is 60°

Solution:
For a rhombus, the angle between the diagonals is 90° and not 60°.

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Question 211 Mark

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

AE.
BE.
CE.
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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Question 221 Mark

The opposite sides of a quadrilateral have:

No common point.
One common point.
Two common points.
Infiniely many common points.

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 231 Mark

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

16cm.
15cm.
20cm.
17cm.

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 241 Mark

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

115°
120°
125°
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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Question 251 Mark

ABCD is a parallelogram, M is the mid-point of BD and BM bisects $\angle\text{B}.$ Then, $\angle\text{AMB}=$

45°
60°
90°
75°

Answer

90°

Solution:

$\angle\text{ABM}=\angle\text{CBM}\ ...(1)$ $($BM bisects $\angle\text{B})$
$\angle\text{ABM}=\angle\text{MDC}\ ...(2)$ (Alternate angles)
$\angle\text{CBM}=\angle\text{ADM}\ ...(3)$ (Alternate angles)
From equations (1), (2) & (3)
$\angle\text{MDC}=\angle\text{ADM}...(4)$
Now, consider $\triangle\text{ABM}$ & $\triangle\text{CBD}$
$\angle\text{CBD}=\angle\text{ABD}$ {from eq (1)}
DB = DB (Common)
$\angle\text{ADB}=\angle\text{CDB}$ {from eq (4)}
Hence, by ASA property,
$\triangle\text{ADB}\cong\triangle\text{CBD}$
⇒ AB = CB, AD = CD
Hence, it becomes a Rhombus.
So now diagonals of a Rhombus bisect each other at 90°.
$\Rightarrow\angle\text{AMB}=90^\circ$

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Question 261 Mark

If the degree measures of the angles of quadrilateral are 4x, 7x, 9x and 10, what is the sum of the measures of the smallest angle and largest angle?

140°
150°
168°
180°

Answer

168°

Solution:
Sum of all angles of a Quadrilateral = 360°
4x + 7x + 9x + 10x = 360°
30x = 360°
x = 12°
So, sum of smallest and largest angle,
i.e. 4x + 10x = 14x = 14 × 12 = 168°

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Question 271 Mark

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

Square.
Rectangle.
Trapezium.
None of these.

Answer

Rectangle.

Solution:

In $\triangle\text{ABD}$ and $\triangle\text{CBD}$
PS || BD and QR || BD
{A line joining mid-points of two sides of triangle is parallel to third side}
⇒ PS || QR
Similiarly PQ || SR
Because SR || AC and QR || BD,
And angle between the diagonals of a Rhombus AC and BD =90°,
Angle between SR and QR = 90°
⇒ PQRS is a rectangle.

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Question 281 Mark

ABCD is a trpezium in which AB || DC. M and N are then mid-points of AD and BC respectively. If AB = 12cm, MN = 14cm, then CD =

10cm.
12cm.
14cm.
16cm.

Answer

16cm.

Solution:

Let a line BP is drawn || to AD to meet DC at P.
ABPD is a parallelogram.
AB || PD, AD || BP
So AB = DP
Let BP cuts MN at Q.
MQ is also || to AB || PD
So AB = MQ = PD = 12cm ...(1)
QN = MN - MQ = 14 - 12 = 2cm
Consider $\triangle\text{BPC}.$
Q and N are the mid-points of BP & BC, and the line joining them QN || PC.
Then by property, $\frac{\text{QN}}{\text{PC}}=\frac{1}{2}$
⇒ PC = 2QN = 2 × 2 = 4cm
Now, DC = DP + PC
DP = 12cm [From (1)]
⇒ DC = 12 + 4 = 16cm

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Question 291 Mark

ABCD is a parallelogram in which diagonal AC bisects $\angle\text{BAD}.$ if $\angle\text{BAC}=35^\circ,$ then $\angle\text{ABC}=$

70°
110°
90°
120°

Answer

110°

Solution:

AC bisects $\angle\text{DAB}.$
$\Rightarrow\angle\text{DAC}=\angle\text{BAC}=35^\circ$
$\Rightarrow\angle\text{BAD}=2\times35^\circ=70^\circ$
$\angle\text{A}+\angle\text{B}=180^\circ$ (Sum of any two adjacent angles in parallelogram =180°)
$\Rightarrow\angle\text{B}=\angle\text{ABC}=180^\circ-\angle\text{BAD}$
$=180^\circ-70^\circ=110^\circ$

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Question 301 Mark

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

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

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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Question 311 Mark

The diagonals of a parallelogram ABCD intersect at O. if $\angle\text{BOC}=90^\circ$ and $\angle\text{BDC}=50^\circ,$ then $\angle\text{AOB}=$

40°
50°
10°
90°

Answer

40°

Solution:

In a parallelogram ABCD,
$\angle\text{OAB}=\angle\text{OCB}$
In $\triangle\text{OCB}$
$\angle\text{OCD}+\angle\text{COD}+\angle\text{ODC}=180^\circ$
$\angle\text{COD}=90^\circ$
$\angle\text{ODC}=50^\circ$ (given)
$\angle\text{OCD}=180^\circ-90^\circ-50^\circ=40^\circ$
$\Rightarrow\angle\text{OAB}=\angle\text{OCD}=40^\circ$

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Question 321 Mark

The two digonals are equal in a:

parallelogram.
Rhombus.
Rectangle.
Trapezium.

Answer

Rectangle.

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

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M.C.Q - MATHS STD 9 Questions - Vidyadip