M.C.Q

MCQ 11 Mark

If $\triangle\text{ABC}\cong\triangle\text{LKM},$ then side of $\triangle\text{LKM}$ equal to side AC of $\triangle\text{ABC}$ is:

A
LK
B
KM
✓
LM
D
None of these

Answer

Correct option: C.

If $\triangle\text{ABC}\cong\triangle\text{LKM},$ then from figure $\text{AC}=\text{LM}.$
Hence, correct option is (c).

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MCQ 21 Mark

In Fig. The measure of $\angle\text{B}'\text{A}'\text{C}'$ is:

A
50º
✓
60º
C
70º
D
80º

Answer

Correct option: B.

60º

In $\triangle\text{ABC}$ and $\triangle\text{A}'\text{B}'\text{C},$
$\text{AB}=\text{A}'\text{B}'$
$\text{BC}=\text{B}'\text{C}'$
$\angle\text{ABC}=\angle\text{A}'\text{B}'\text{C}'$
So $\triangle\text{ABC}\cong\triangle\text{A}'\text{B}'\text{C}'$ by SAS creterion
$\Rightarrow\angle\text{BAC}=\angle\text{B}'\text{A}'\text{C}'$
$\Rightarrow3\text{x}=2\text{x}+20$
$\text{x}=20^\circ$
$2\text{x}+20=2\times20+60^\circ=\angle\text{B}'\text{A}'\text{C}'$
Hence, correct option is (b).

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MCQ 31 Mark

In Fig. ABC is a triangle in which $\angle\text{B}=2\angle\text{C}.$ D is a point on side BC such that AD bisects $\angle\text{BAC}$ and $\text{AB}=\text{CD}.$ Be is the bisector of $\angle\text{B}.$ The measure of $\angle\text{BAC}$ is:

✓
72º
B
95º
C
73º
D
74º

Answer

Correct option: A.

72º

$\angle\text{ABE}=\angle\text{EBC}$ $($EBC is bisector of $\angle\text{B})$
and $\angle\text{C}=\frac{\angle\text{B}}{2}$
$\Rightarrow\angle\text{EBC}=\angle\text{ECB}$
So $\triangle\text{EBC}$ is isosceles triangle.
$\Rightarrow\text{EB}=\text{EC}\ ....(1)$
Now Consider $\triangle\text{ABE}$ and $\triangle\text{DCE}$
$\text{AB}=\text{DC}$ (Given)
$\text{BE}=\text{CE}$ [From (1)]
$\angle\text{ABE}=\angle\text{DCE}$ (From above data)
So $\triangle\text{ABE}\cong\triangle\text{DCE}$ by SAS property
$\Rightarrow\text{AE}=\text{DE}$
$\angle\text{BAE}=\angle\text{CDE}=\angle\text{A}$
Now consider $\triangle\text{AED},$
$\text{AE}=\text{DE}$ (above proved)
$\Rightarrow\triangle\text{AED}$ is isosceles triangle
$\Rightarrow\angle\text{EAD}=\angle\text{EDA}=\frac{\angle\text{A}}{2}$ $($AD is Bisector of $\angle\text{A})\ ....(2)$
Now, consider $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+2\angle\text{C}+\angle\text{C}=180^\circ(\angle\text{B}=2\angle\text{C)}$
$\Rightarrow\angle\text{A}+3\angle\text{C}=180^\circ\ .....(3)$
Consider $\triangle\text{ADE},$
$\Rightarrow\frac{\angle\text{A}}{2}+\angle\text{ADC}+\angle\text{}\text{C}=180^\circ$
$\Rightarrow\frac{\angle\text{A}}{2}+(\angle\text{EDA}+\angle\text{CDE})+\angle\text{C}=180^\circ$
$\Rightarrow\frac{\angle\text{A}}{2}+\frac{\angle\text{A}}{2}+\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{}A+\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow2\angle\text{A}+\angle\text{C}=180^\circ\ .....(4)$
Right hand side of equations (3) and (4) are equal, hence Left hand side.
$\Rightarrow\angle\text{A}+3\angle\text{C}=2\angle\text{A}+\angle\text{C}$
$\Rightarrow\angle\text{A}=2\angle\text{C}$
Substituting in equation (3),
$2\angle\text{C}+3\angle\text{C}=180^\circ$
$\Rightarrow5\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=36^\circ$
$\Rightarrow\angle\text{A}=2\times36^\circ=72^\circ$
Hence, correct option is (a).

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MCQ 41 Mark

In Fig. X is a point in the interior of square ABCD. AXYZ is also a square. If DY = 3cm and AZ = 2cm, then BY =

A
5cm
B
6cm
✓
7cm
D
8cm

Answer

Correct option: C.

7cm

Consider $\triangle\text{ AZD}$ and $\triangle\text{AXB}$
$\text{AZ}=\text{AX}=2\text{cm}$ (AXYZ is a square)
$\angle\text{AZD}=\angle\text{AXB}=90^\circ$
$\text{AD}=\text{AB}$ (ABCD is a square)
So by RHS creterion, $\triangle\text{AZD}\cong\triangle\text{AXB}$
$\Rightarrow\text{ZD}=\text{XB}$
Now, $\text{ZD}=\text{ZY}+\text{DY}$
=2cm + 3cm (ZY = AZ = 2cm)
=5cm
⇒ XB = 5cm
⇒ BY = YX + XB = 2cm + 5cm = 7cm
Hence, correct option is (c)

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MCQ 51 Mark

In Fig. ABC ids an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is:

A
52º
B
76º
C
156º
✓
104º

Answer

Correct option: D.

104º

$\triangle\text{ABC}$ is isosceles
$\angle\text{ABC}=\angle\text{ACB}=52^\circ$
then $\angle\text{BAC}=180^\circ-52^\circ-52^\circ=76^\circ$
If $\text{AB}\parallel\text{CD},$ AC is transversal
then $\angle\text{BAC}=\angle\text{ACD}$ (alternate angles)
$\Rightarrow\angle\text{ACD}=76^\circ$
Now from figure,
$\angle\text{ACD}+\text{x}^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-76^\circ$
$\Rightarrow\text{x}^\circ=104^\circ$
Hence, correct option is (d).

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MCQ 61 Mark

D, E, F are the mid-point of the sides BC, CA and AB respectively of $\triangle\text{ABC}.$ Then $\triangle\text{DEF}$ is congruent to triangle.

A
ABC
B
AEF
C
BFD, CDE
✓
AFE, BFD, CDE

Answer

Correct option: D.

AFE, BFD, CDE

In anytriangle, a line joining the mid-points of any two sides is parallel to the third side.
⇒ EF || BC EF || DC and BD
Similiarly DF || EC
⇒DF || AE and EC
Also DE || AB.
⇒ DE || AF and BF
From this information it is clear that EFDC, EFBD, EAFD
are the parallelogram by property.
Now consider one parallelogram EFDC
Consider $\triangle\text{DEF}$ and $\triangle\text{EDC}$
$\text{DE}=\text{ED}$ (common)
$\angle\text{DEF}=\angle\text{EDC}$
$\angle\text{EDF}=\angle\text{DEC}$ (ASA property)
$\Rightarrow\triangle\text{DEF}\cong\triangle\text{EDC}$
Similiarly in parallelogram EAFD,
$\triangle\text{DEF}\cong\triangle\text{AFC}$
And in parallelogram EFBD
$\triangle\text{DEF}\cong\triangle\text{FBD}$
Hence, correct option is (d).
Note: Option (d) modified.

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MCQ 71 Mark

If ABC and DEF are two triangles such that $\triangle\text{ABC}\cong\triangle\text{FDE}$ and $\text{AB}=5\text{m},\angle\text{B}=40^\circ$ and $\angle\text{A}=80^\circ.$ Then, which of the following is true?

A
$\text{DF}=5\text{cm},\angle\text{F}=60^\circ$
B
$\text{DE}=5\text{cm},\angle\text{E}=60^\circ$
✓
$\text{DF}=5\text{cm},\angle\text{E}=60^\circ$
D
$\text{DE}=5\text{cm},\angle\text{D}=40^\circ$

Answer

Correct option: C.

$\text{DF}=5\text{cm},\angle\text{E}=60^\circ$

In $\triangle\text{ABC},$
$\angle\text{C}=180^\circ-\angle\text{A}+\angle\text{B}=180^\circ-80^\circ-40^\circ=60^\circ$
$\triangle\text{ABC}\cong\triangle\text{FDE}$
$\Rightarrow\text{AB}=\text{FD}=5\text{cm}$
$\Rightarrow\angle\text{B}=\angle\text{D}=40^\circ$
$\Rightarrow\angle\text{A}=\angle\text{F}=80^\circ$
$\Rightarrow\angle\text{C}=\angle\text{E}=60^\circ$
$\Rightarrow\text{DF}=\text{FD}=5\text{cm}$ and $\angle\text{E}=60^\circ$
Hence, correct option is (c).

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MCQ 81 Mark

In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of vertex angle of the triangle is:

A
100º
✓
120º
C
110º
D
130º

Answer

Correct option: B.

120º

Let $\triangle\text{ABC}$ be an isosceles triangle with
vertex angle = $\angle\text{A}$ and base angles = $\angle\text{B}$ and $\angle\text{C}$
Now, $\angle\text{A}=2(\angle\text{B}+\angle\text{C})$
$\Rightarrow\frac{\angle\text{A}}{2}=\angle\text{B}+\angle\text{C}\ ....(1)$
Also in $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+(\angle\text{B}+\angle\text{C})=180^\circ$
$\Rightarrow\angle\text{A}+\frac{\angle\text{A}}{2}=180^\circ$ .....[From (1)]
$\Rightarrow\frac{3}{2}\angle\text{A}=180^\circ$
$\Rightarrow\angle\text{A}=\frac{180^\circ\times2}{3}$
$\Rightarrow\angle\text{A}=120^\circ$
Hence, correct option is (b).

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MCQ 91 Mark

In triangles ABC and PQR three equality relation between some parts are as follows: $\text{AB}=\text{QP},\angle\text{B}=\angle\text{P}$ and $\text {BC}=\text{PR}$ State which of the congruence conditions applies:

✓
SAS
B
ASA
C
SSS
D
RHS

Answer

Correct option: A.

SAS

From given conditions, we have
$\text{AB}=\text{PQ}$
$\text{BC}=\text{PR}$
And the angle between these sides are also equal
i.e. $\angle\text{B}=\angle\text{P}$
So SAS property.
Hence, correct option is (a).

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MCQ 101 Mark

In a $\triangle\text{ABC},$ if $\text{AB}=\text{AC}$ and BC is produced to D such that $\angle\text{ACD}=100^\circ$ then $\angle\text{A}=$

✓
20°
B
40°
C
60°
D
80°

Answer

Correct option: A.

20°

$\text{AB}=\text{AC}$
$\Rightarrow\angle\text{ABC}=\angle\text{ACB}$ (Isoscles $\triangle$ Property)
$\angle\text{ACB}=180^\circ-100^\circ=80^\circ$
$\Rightarrow\angle\text{ABC}-=\angle\text{ACB}=80^\circ$
$\angle\text{A}=180^\circ-\angle\text{ACB}-\angle\text{ABC}=180^\circ-80^\circ-80^\circ=20^\circ$
Hence, correct option is (a).

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MCQ 111 Mark

If $\triangle\text{ABC}\cong\triangle\text{PQR}$ and $\triangle\text{ABC}$ is not congruent to $\triangle\text{RPQ},$ then which of the following is not true:

✓
BC = PQ
B
AC = PR
C
AB = PQ
D
RQ = BC

Answer

Correct option: A.

BC = PQ

$\triangle\text{ABC}\cong\triangle\text{PQR}$
$\Rightarrow\text{AB}=\text{PR},\text{AC}=\text{PR},\text{BC}=\text{QR}$
$\triangle\text{ABC}\not\cong\triangle\text{RQP}$
$\Rightarrow\text{AB}\not=\text{QR},\text{AC}\not=\text{RP},\text{BC}\not=\text{PQ}$
Hence, correct option (a).

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MCQ 121 Mark

In Fig, if AC is bisector of $\angle\text{BAD}$ such that AB = 3cm and AC = 5cm, then CD =

A
2cm
B
3cm
✓
4cm
D
5cm

Answer

Correct option: C.

4cm

Consider $\triangle\text{ABC}$ and $\triangle\text{ADC}$
$\angle\text{ABC}=\angle\text{ADC}=90^\circ$
$\angle\text{BAC}=\angle\text{CAD}$ $($ AC is bisector of $\angle\text{A})$
Also if two angles are equal, then the third angle will also be equal.
$\Rightarrow\angle\text{BCA}=\angle\text{DCA}$
Now, $\text{AC}=\text{AC}$ (common)
So by ASA property, $\triangle\text{ABC}\cong\triangle\text{ADC}$
$\Rightarrow\text{BC}=\text{CD}$
And, BC $=\sqrt{\text{AC}^2-\text{AB}^2}=\sqrt{25-9}=4\text{cm}$
$\Rightarrow\text{CD}=4\text{cm}$
Hence, correct option is (c).

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MCQ 131 Mark

If $\triangle\text{PQR}\cong\triangle\text{EFD},$ then ED =

A
PQ
B
QR
✓
PR
D
None of these

Answer

Correct option: C.

$\triangle\text{PQR}\cong\triangle\text{EFD},$
$\Rightarrow\text{ED}=\text{PR}$ (congruent sides of congruent triangles)
Hence, correct option is (c)

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MCQ 141 Mark

In Fig. if $\text {DC}\parallel\text{DC}$ and $\text{AB}=\text{AC},$ the value of $\angle\text{ABD}$ is:

A
70º
✓
110º
C
120º
D
130º

Answer

Correct option: B.

110º

If $\text{AE}\parallel\text{DC}$ and AC is transversal,
then $\angle\text{FAC}=70^\circ$ (opposite angles)
Also $\angle\text{FAC}-=\angle\text{ACB}=70^\circ$ (Alternate angles)
Since $\text{AB}=\text{AC},\triangle\text{ABC}$ is isosceles.
So $\angle\text{ABC}=\angle\text{ACB}$
$\Rightarrow\angle\text{ABC}=70^\circ$
Now $\angle\text{ABD}=180^\circ-\angle\text{ABC}=180^\circ-70^\circ=100^\circ$
Hence, correct option is (b).

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MCQ 151 Mark

Which of the following is not a criterion for congruence of triangles?

A
SAS
✓
SSA
C
ASA
D
SSS

Answer

Correct option: B.

SSA

If two triangles have two congruent sides and a congruent non-included angle, then $\triangle$ s are not necessarily corgruent. This is why there is no 'side side angle'
i.e. SSA postulate.
Hence, correct option is (b).

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MCQ 161 Mark

In triangles ABC and PQR, if $\angle\text{A}=\angle\text{R},\angle\text{B}=\angle\text{P}$ and $\text{AB}=\text{RP},$ then which one of the following congruence conditioins applies:

A
SAS
✓
ASA
C
SSS
D
RHS

Answer

Correct option: B.

ASA

From given conditions,
$\angle\text{B}=\angle\text{P}$
$\angle\text{A}=\angle\text{R}$
And the side containing then is also equal
i.e $\text{AB}=\text{PR}$
So ASA property.
Hence, correct option is (b).

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MCQ 171 Mark

In Fig. $\text{AB}\perp\text{BE}$ and $\text{FE}\perp\text{BE},$ If $\text{BC}=\text{DE}$ and $\text{AB}=\text{EF},$ then $\triangle\text{ABD}$ is congruent to:

A
$\triangle\text{EFC}$
B
$\triangle\text{ECF}$
C
$\triangle\text{CEF}$
✓
$\triangle\text{FEC}$

Answer

Correct option: D.

$\triangle\text{FEC}$

AB = EF
BC = DE
BC + CD = DE + CD (adding CD both sides)
BC + CD = BD, DE + CD = CE
So BD = CE
Now Consider $\triangle\text{ABD},$ & $\triangle\text{FEC}$
$\text{AB}=\text{FE}$
$\text{BD}=\text{EC}$
$\angle\text{ABD}=\angle\text{FEC}=90^\circ$
So $\triangle\text{ABD}\cong\triangle\text{FEC}$ by SAS creterion.
Hence, correct option is (d).

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MCQ 181 Mark

If $\triangle\text{PQR}\cong\triangle\text{EFD},$ then $\angle\text{E}=$

✓
$\angle\text{P}$
B
$\angle\text{Q}$
C
$\angle\text{R}$
D
None of these

Answer

Correct option: A.

$\angle\text{P}$

$\triangle\text{PQR}\cong\triangle\text{EFD},$
$\Rightarrow\angle\text{E}=\angle\text{P}$ (congruent angles of congruent triangles)
Hence, correct option is (a).

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MCQ 191 Mark

ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, $\angle\text{BAD}=$

✓
55°
B
70°
C
35°
D
110°

Answer

Correct option: A.

55°

If AD is the median, then D is the mid-point of BC.
$\text{BD}=\text{DC}$
So consider $\triangle\text{ADB}$ and $\triangle\text{ADC}$
$\text{AD}=\text{AD}$ (common)
$\text{DB}=\text{DC}$
$\text{BA}=\text{CA}$
So by SSS, $\triangle\text{ADB}\cong\triangle\text{ADC}$
Now $\angle\text{B}=\angle\text{C}=35^\circ$
$\Rightarrow\angle\text{BAD}=\angle\text{DAC}$
So in $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow2\angle\text{BAD}+35^\circ+35^\circ=180^\circ$
$\Rightarrow2\angle\text{BAD}=110^\circ$
$\Rightarrow\angle\text{BAD}=55^\circ$
Hence, correct option is (a).

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MCQ 201 Mark

If $\triangle\text{ABC}\cong\triangle\text{ACB},$ then $\triangle\text{ABC}$ is isosceles with.

✓
AB = AC
B
AB = BC
C
AC = BC
D
None of these

Answer

Correct option: A.

AB = AC

$\triangle\text{ABC}\cong\triangle\text{ACB}$
$\Rightarrow\text{AB}=\text{AC}$
or
$\text{AC}=\text{AB}$
So, in $\triangle\text{ABC}$ is isosceles with $\text{AB}=\text{AC}.$
Hence, correct option (a).

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