Maths M.C.Q

2. SSA

Solution:

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).

1. SAS

Solution:

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).

2. ASA

Solution:

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).

2. 60º

Solution:

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).

3. 7cm

Solution:

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)

4. $\triangle\text{FEC}$

Solution:

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).

2. 110º

Solution:

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).

3. 4cm

Solution:

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).

1. 72º

Solution:

$\angle\text{ABE}=\angle\text{EBC}$ $($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).

4. 104º

Solution:

$\triangle\text{ABC}$ 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).

2. 120º

Solution:

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).

1. 20°

Solution:

$\text{AB}=\text{AC}$

$\Rightarrow\angle\text{ABC}=\angle\text{ACB}$ (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).

1. $\angle\text{P}$

Solution:

$\triangle\text{PQR}\cong\triangle\text{EFD},$

$\Rightarrow\angle\text{E}=\angle\text{P}$ (congruent angles of congruent triangles)

Hence, correct option is (a).

3. PR

Solution:

$\triangle\text{PQR}\cong\triangle\text{EFD},$

$\Rightarrow\text{ED}=\text{PR}$ (congruent sides of congruent triangles)

Hence, correct option is (c)

1. $\text{BC}=\text{PQ}$

Solution:

$\triangle\text{ABC}\cong\triangle\text{PQR}$

$\Rightarrow\text{AB}=\text{PR},\text{AC}=\text{PR},\text{BC}=\text{QR}$

$\triangle\text{ABC}\not\cong\triangle\text{RQP}$

$\Rightarrow\text{AB}\not=\text{QR},\text{AC}\not=\text{RP},\text{BC}\not=\text{PQ}$

Hence, correct option (a).

3. LM

Solution:

If $\triangle\text{ABC}\cong\triangle\text{LKM},$ then from figure $\text{AC}=\text{LM}.$

Hence, correct option is (c).

1. $\text{AB}=\text{AC}$

Solution:

$\triangle\text{ABC}\cong\triangle\text{ACB}$

$\Rightarrow\text{AB}=\text{AC}$

or

$\text{AC}=\text{AB}$

So, in $\triangle\text{ABC}$ is isosceles with $\text{AB}=\text{AC}.$

Hence, correct option (a).

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

Solution:

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).

4. AFE, BFD, CDE

Solution:

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.

1. 55°

Solution:

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).