Maths M.C.Q

3. 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.

2. 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.

2. 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.

2. 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.

1. 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.

2. 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.

1. 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)

$\angle\text{OCD}=180^\circ-90^\circ-50^\circ=40^\circ$

$\Rightarrow\angle\text{OAB}=\angle\text{OCD}=40^\circ$

3. 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)

$\Rightarrow\angle\text{DPC}=80^\circ$

3. 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)

4. 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)]

$\Rightarrow\angle\text{AOB}=180^\circ-(\angle\text{OAB}+\angle\text{OBA})=180^\circ-90^\circ$

$\Rightarrow\angle\text{AOB}=90^\circ$

1. $\angle\text{P}=100^\circ,\angle\text{Q}=80^\circ,\angle\text{R}=100^\circ$

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

$\angle\text{Q}+\angle\text{R}=80^\circ+100^\circ=180^\circ$

1. 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. $\text{AB}=\frac{10}{2}=5\text{cm}$