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#### Question 1:

Write the correct answer in each of the following:
Three angles of a quadrilateral are 75º, 90º and 75º. The fourth angle is
(A) 90º
(B) 95º
(C) 105º
(D) 120º

Given that, .

The sum of all the angles of a quadrilateral is 360°.
$\therefore \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360°\phantom{\rule{0ex}{0ex}}⇒75°+90°+75°+\angle \mathrm{D}=360°$
$\begin{array}{rcl}& ⇒& \angle \mathrm{D}=360°-\left(75°+90°+75°\right)\\ & =& 360°-240°\\ & =& 120°\end{array}$
Thus, the fourth angle of the quadrilaterals is 120°.
Hence, the correct answer is option D.

#### Question 2:

A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonals is
(A) 55°
(B) 50°
(C) 40°
(D) 25°

The diagonals of rectangle are equal in length.

Thus, the acute angle between diagonals is 50°.
Hence, the correct answer is option B.

#### Question 3:

ABCD is a rhombus such that ∠ACB = 40º. Then ∠ADB is
(A) 40º
(B) 45º
(C) 50º
(D) 60º

Given, ABCD is a rhombus,
$\angle \mathrm{ACB}=40°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{OCB}=40°$

As, $\mathrm{AD}\parallel \mathrm{BC}$

Also,

By angle sum property of triangles, the sum of all angles of $△$ADO is 180­°.
$\therefore \angle \mathrm{ADO}+\angle \mathrm{DOA}+\angle \mathrm{OAD}=180°$
$\begin{array}{rcl}& ⇒& \angle \mathrm{ADO}=180°-\left(40°+90°\right)\\ & =& 180°-130°\\ & =& 50°\end{array}$
$⇒\angle \mathrm{ADB}=\angle \mathrm{ADO}=50°$
Hence, the correct answer is option C.

#### Question 4:

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if
(A) PQRS is a rectangle
(B) PQRS is a parallelogram
(C) diagonals of PQRS are perpendicular
(D) diagonals of PQRS are equal

Rectangle ABCD is formed by joining the mid points of quadrilateral PQRS.

Applying midpoint theorem in $△$PQR and $△$PSR,
AB = CD = $\frac{1}{2}\mathrm{PR}$
and
AB ∥ PR ∥ CD

Similarly, applying midpoint theorem in $△$PQS and $△$QRS,
AD = BC = $\frac{1}{2}\mathrm{QS}$
and

Since, QS ∥ BC therefore,
∠OXB + ∠XBY = 180°            (∵ sum of co-interior angles is 180°)        .....(1)

Now, ABCD is a rectangle.
∴ ∠XBY = 90°             .....(2)

Putting (2) in (1),
∠OXB = 90°              .....(3)

Again, AB ∥ PR.
⇒ BX ∥ OY
Therefore,
∠OXB + ∠XOY = 180°            (∵ sum of co-interior angles is 180°)        .....(4)

Using (3),
∴ ∠XOY = 90°
Thus, PR is perpendicular to QS.

Therefore, the diagonals of PQRS are perpendicular to each other.
Hence, the correct answer is option C.

#### Question 5:

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if
(A) PQRS is a rhombus
(B) PQRS is a parallelogram
(C) diagonals of PQRS are perpendicular
(D) diagonals of PQRS are equal

Given, quadrilateral ABCD is a rhombus.
So, AB = BC = CD = AD.

In $△$PQS,
D and A are the mid points of PS and PQ respectively.
So, by mid point theorem,
AD = BC = $\frac{1}{2}$QS           .....(1)
and

Similarly in $△$PSR,
D and C are the mid points of PS nad RS respectively.
So, by mid point theorem,
DC = AB = $\frac{1}{2}$PR                 .....(2)
and
DC ∥ PR ∥ AB

Since, ABCD is a rhombus.
Therefore, all its sides are equal.
⇒ AB = BC

Using (1) and (2),
$\frac{1}{2}$QS = $\frac{1}{2}$PR
$⇒\mathrm{QS}=\mathrm{PR}$
Thus, diagonals of PQRS are equal.
Hence, the correct answer is option D.

#### Question 6:

If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3 : 7 : 6 : 4, then ABCD is a
(A) rhombus
(B) parallelogram
(C) trapezium
(D) kite

Given that, the angles of quadrilaterals in ratio 3 : 7 : 6 : 4.

Let the angles be
∴ Sum of all angles of quadrilaterals = 360°
$⇒3x+7x+6x+4x=360°\phantom{\rule{0ex}{0ex}}⇒20x=360°\phantom{\rule{0ex}{0ex}}⇒x=18°$

Now, the angles of quadrilaterals are
$\angle \mathrm{A}=3×18=54°\phantom{\rule{0ex}{0ex}}\angle \mathrm{B}=7×18=126°\phantom{\rule{0ex}{0ex}}\angle \mathrm{C}=6×18=108°\phantom{\rule{0ex}{0ex}}\angle \mathrm{D}=4×18=72°$

From the figure,

Here, $\angle \mathrm{BCE}=\angle \mathrm{ADC}=70°$.

Since, corresponding angle are equal.
$\therefore \mathrm{BC}\parallel \mathrm{AD}$

Then, sum of co-interior angles,
$\angle \mathrm{A}+\angle \mathrm{B}=126°+54°=180°\phantom{\rule{0ex}{0ex}}\angle \mathrm{C}+\angle \mathrm{D}=180°+72°=180°$
Thus, ABCD is a trapezium.

Hence, the correct answer is option C.

#### Question 7:

If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠C and ∠D at R and of ∠D and ∠A at S, then PQRS is a
(A) rectangle
(B) rhombus
(C) parallelogram
(D) quadrilateral whose opposite angles are supplementary

Given that, ABCD is a quadrilateral and all angles bisectors from a quadrilateral PQRS.

The sum of all angles in a quadrilateral is 360°.

In $△$APB, by angle sum property,

Similarly, in $△$RDC,

Substituting (2) and (3) in (1),
$180°-\angle \mathrm{BPA}+180°-\angle \mathrm{DRC}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{BPA}+\angle \mathrm{DRC}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{SPQ}+\angle \mathrm{SRQ}=180°$
[∵$\angle \mathrm{BPA}=\angle \mathrm{SPQ}$ and $\angle \mathrm{DRC}=\angle \mathrm{SRQ}$ are vertically opposite angles]
Thus, PQRS is a quadrilateral whose opposite angles are supplementary.
Hence, the correct answer is option D.

#### Question 8:

If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form
(A) a square
(B) a rhombus
(C) a rectangle
(D) any other parallelogram

Given that, APB and CQD are two parallel lines.
The bisector of the angles APQ and CQP meet at point M.
Similarly, the bisectors of angles BPQ and PQD meet at point N.
Join PM, MQ, QN and NP.

Since, $\mathrm{AB}\parallel \mathrm{CD}$.
Then,
$⇒2\angle \mathrm{MPQ}=2\angle \mathrm{NQP}\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{MPQ}=\angle \mathrm{NQP}$
$\therefore \mathrm{PM}\parallel \mathrm{QN}$

Similarly,
$\therefore \mathrm{PN}\parallel \mathrm{QM}$

So, quadrilateral PMQN is a parallelogram.

Thus, PMQN is a rectangle.
Hence, the correct answer is option C.

#### Question 9:

The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is
(A) a rhombus
(B) a rectangle
(C) a square
(D) any parallelogram

Let ABCD be a rhombus in which P, Q, R, and S are the mid-points of side AB, BC, CD and DA respectively.

Join AC, PR and SQ.
In $△$ABC,
P is the mid-point of AB and Q the mid-point of BC.

Similarly, in $△$DAC,

From (1) and (2),

So, PQRS is a parallelogram.

So, the diagonals of a parallelogram are equal.

Thus, PQRS is a rectangle.
Hence, the correct answer is option B.

#### Question 10:

D and E are the mid-points of the sides AB and AC of $△$ABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is
(A) a square
(B) a rectangle
(C) a rhombus
(D) a parallelogram

In $△$ABC, D and E are the mid-point of sides AB and AC, respectively.

By mid-point theorem,

Now, P and Q are the mid-points of OB and OC.
$\mathrm{DE}=\frac{1}{2}\left[\mathrm{BP}+\mathrm{PO}+\mathrm{OQ}+\mathrm{QC}\right]\phantom{\rule{0ex}{0ex}}⇒\mathrm{DE}=\frac{1}{2}\left[2\mathrm{PO}+2\mathrm{OQ}\right]$

Again, in $△$AOC, Q and E are the mid-points of OC and AC.
[mid-point theorem]            .....(3)
And, in $△$ABO,
[mid-point theorem]

From (1) and (2),
DE$\parallel$BC (or DE$\parallel$PQ) and DE = PQ

From (3) and (4),
EQ$\parallel$PD and EQ = PD

Thus, DEQP is a parallelogram.
Hence, the correct answer is option D.

#### Question 11:

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if,
(A) ABCD is a rhombus
(B) diagonals of ABCD are equal
(C) diagonals of ABCD are equal and perpendicular
(D) diagonals of ABCD are perpendicular.

Given that, ABCD is a quadrilateral and P, Q, R and S are the mid-points of sides of AB, BC, CD and DA.

Also, PQRS is a square.
∴ PQ = QR = RS = PS        .....(1)
Now, the diagonals of a square are equal. Thus, PR = SQ.
But PR = BC and SQ = AB
∴ AB = BC
Therefore, all the sides of quadrilateral ABCD are equal.

Now, quadrilateral ABCD is either a square or a rhombus.

In $△$ADB, by mid-point theorem

Now, in $△$ABC, (by mid-point theorem)

From (1),

Thus, diagonals of ABCD are equal and therefore quadrilateral ABCD is a square not rhombus, So, diagonals of quadrilateral are also perpendicular.
Hence, the correct answer is option C.

#### Question 12:

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC = 32° and ∠AOB = 70°, then ∠DBC is equal to
(A) 24°
(B) 86°
(C) 38°
(D) 32°

Given that, .

Now, $\angle \mathrm{AOB}+\angle \mathrm{BOC}=180°$           [linear pair axiom]
$\begin{array}{rcl}& ⇒& \angle \mathrm{BOC}=180°-\angle \mathrm{AOB}\\ & =& 180°-70°\\ & =& 110°\end{array}$

Now, in $△$BOC,
$\angle \mathrm{BOC}+\angle \mathrm{BCO}+\angle \mathrm{OBC}=180°$      [angle sum property of a triangle]
$⇒110°+32°+\angle \mathrm{OBC}=180°$           []
$\begin{array}{rcl}& ⇒& \angle \mathrm{OBC}=180°-\left(110°+32°\right)\\ & =& 38°\\ \therefore \angle \mathrm{DBC}& =& \angle \mathrm{OBC}=38°\end{array}$

Hence, the correct answer is option C.

#### Question 13:

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.

In a parallelogram opposite sides are equal, opposite angles are equal, diagonal bisect each other, but opposite angles are not bisected by the diagonal.
Hence, the correct answer is option C.

#### Question 14:

D and E are the mid-points of the sides AB and AC respectively of $△$ABC. DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is
(A) ∠DAE = ∠EFC
(B) AE = EF
(C) DE = EF

In $△$ADE and $△$CFE, assume DE = EF as E is the mid-point of DF.
Now,  AE = CE.

Let DE = EF
and   [vertically opposite angles]
$\therefore △\mathrm{ADE}\cong △\mathrm{CFE}$     [SAS congruence rule]
⇒ AD = CF                   [CPCT rule]

Also, $\angle \mathrm{ADE}=\angle \mathrm{CFE}$       [CPCT]
Hence, $\mathrm{AD}\parallel \mathrm{CF}$                [alternate interior angles are equal]

Therefore, we need an additional information which is DE = EF.
Hence, the correct answer is option C.

#### Question 1:

Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, determine the lengths of AC and BD.

Given that, ABCD is a parallelogram. Also, OA = 3 cm and OD = 2 cm.

The diagonals of a parallelogram bisects each other.
∴ Diagonal AC = 2 × OA = 6 cm
and Diagonal BD = 2 × OD = 4 cm
Hence, the length of the diagonal AC and BD are 6 cm and 4 cm respectively.

#### Question 2:

Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reason for your answer.

No, diagonals of a parallelogram are not perpendicular to each other, but they bisect each other.

#### Question 3:

Can the angles 110°, 80°, 70° and 95° be the angles of a quadrilateral? Why or why not?

As, the sum of the angles of the quadrilateral is 360°.
Here,
110° + 80° + 70° + 95° = 355°$\ne$ 360°

Hence, the angles cannot be the angles of a quadrilateral.

#### Question 4:

In quadrilateral ABCD, ∠A + ∠D = 180º. What special name can be given to this quadrilateral?

As, the sum of adjacent opposite angle is 180°. Thus, the given quadrilateral is a trapezium.

#### Question 5:

All the angles of a quadrilateral are equal. What special name is given to this quadrilateral?

The sum of all the angles of a quadrilateral is 360°.

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360°$

If all angles are equal, i.e.,
$\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=\angle \mathrm{D}$
$⇒4\angle \mathrm{A}=360°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{A}=90°$
$\therefore \angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=\angle \mathrm{D}=90°$

As, all the angles are 90°.
Hence, the given quadrilateral is either a rectangle or a square.

#### Question 6:

Diagonals of a rectangle are equal and perpendicular. Is this statement true? Give reason for your answer.

No, diagonals of a rectangle are equal but need not be perpendicular to each other.

#### Question 7:

No, all the 4 angles of a quadrilateral cannot be obtuse. This is because the sum of the angles of a quadrilateral is 360°.
At max 3 obtuse angles can exist.

#### Question 8:

In $△$ABC, AB = 5 cm, BC = 8 cm and CA = 7 cm. If D and E are respectively the mid-points of AB and BC, determine the length of DE.

In $△$ABC,
AB = 5 cm, BC = 8 cm and CA = 7 cm

As, D and E are mid-points of AB and BC.

By mid-point theorem,
$\mathrm{DE}\parallel \mathrm{AC}$
and

Hence, the length of DE = 3.5 cm.

#### Question 9:

In the figure, it is given that BDEF and FDCE are parallelograms. Can you say that BD = CD? Why or why not?

Yes, in the given figure, BDEF is a parallelogram.

Also, FDCE is a parallelogram.

From (1) and (2),
BD = CD = EF
Hence, we can say BD = CD.

#### Question 10:

In the figure, ABCD and AEFG are two parallelograms. If ∠C = 55º, determine ∠F.

Given that, ABCD and AEFG are two parallelograms.
Also, $\angle \mathrm{C}=55°$.

As ABCD is a parallelogram, then its opposite angles are equal.
$⇒\angle \mathrm{A}=\angle \mathrm{C}=55°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{A}=55°$

Also, AEFG is a parallelogram.
$\therefore \angle \mathrm{A}=\angle \mathrm{F}=55°$

Hence, $\angle \mathrm{F}=55°$.

#### Question 11:

No.

The sum of the angles of quadrilateral is 360°.

all the angles of a quadrilateral cannot be acute angles. In this case, the sum of the angles will be less than 360°. Thus, a maximum 3 acute angles are possible.

#### Question 12:

Yes, all the angles of a quadrilateral can be right angles. In this case, the quadrilateral is rectangle or a square.

#### Question 13:

Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 35º, determine ∠B.

As, the diagonals of a quadrilateral bisect each other, thus it can be a parallelogram.

Therefore, the sum of co-interior angles between two parallel lines is 180° i.e.,

#### Question 14:

Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD.

Given that, the opposite angles of a quadrilateral ABCD are equal. Thus, ABCD is a parallelogram. As per the property of parallelogram, its opposites sides are equal.

Hence, CD = 4 cm.

#### Question 1:

One angle of a quadrilateral is of 108° and the remaining three angles are equal. Find each of the three equal angles.

The sum of angle of quadrilateral is 360°.
$⇒\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360°$

From (1),
$⇒108°+x+x+x=360°\phantom{\rule{0ex}{0ex}}⇒3x=360°-108°=252°\phantom{\rule{0ex}{0ex}}⇒x=\frac{252°}{3}=84°$

Hence, each of the three angles is 84°.

#### Question 2:

ABCD is a trapezium in which AB || DC and ∠A = ∠B = 45º. Find angles C and D of the trapezium.

Given that, ABCD is a trapezium.
Also, $\mathrm{AB}\parallel \mathrm{DC}$ and $\angle \mathrm{A}=\angle \mathrm{B}=45°$.

As $\mathrm{AB}\parallel \mathrm{DC}$, then BC is transversal.
The sum of two co-interior angle is 180°.

Similarly, $\angle \mathrm{A}+\angle \mathrm{D}=180°$

Hence,  are 135° each.

#### Question 3:

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60º. Find the angles of the parallelogram.

Consider a parallelogram ABCD, in which  are obtuse angles.
Let DE and DF be the two altitudes of parallelogram and angle between them is 60°.

$\angle \mathrm{BED}=\angle \mathrm{BFD}=90°$.
$\begin{array}{rcl}\therefore \angle \mathrm{FBE}& =& 360°-\left(\angle \mathrm{FDE}+\angle \mathrm{BED}+\angle \mathrm{BFD}\right)\\ & =& 360°-\left(60°+90°+90°\right)\\ & =& 360°-240°\\ & =& 120°\end{array}$

In ABCD parallelogram, opposite angles are equal.
$\angle \mathrm{ADC}=120°$
Now,

Also,
$\angle \mathrm{C}=\angle \mathrm{A}=60°$

Hence, angle of the parallelogram are 60°, 120°, 60° and 120°.

#### Question 4:

ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus.

Let the sides of a rhombus be AB = BC = CD = DA = $x$.
Then, join DB.

In ,

.
Then, $△\mathrm{ADB}$ is an equilateral triangle
$\therefore \angle \mathrm{A}=\angle \mathrm{ADB}=\angle \mathrm{ABD}=60°$

Similarly, $△$DBC is an equilateral triangle.
$\therefore \angle \mathrm{C}=\angle \mathrm{BDC}=\angle \mathrm{DBC}=60°$
Also, $\angle \mathrm{A}=\angle \mathrm{C}$.

Hence, .

#### Question 5:

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. Show that BFDE is a parallelogram.

Given that, ABCD is a parallelogram and AE = CF.
To show: OE = OF

Join BD, meet AC at point O.
Since, the diagonals of a parallelogram bisect each other.

Now, OA = OC and AE = CF
$⇒\mathrm{OA}-\mathrm{AE}=\mathrm{OC}-\mathrm{CF}\phantom{\rule{0ex}{0ex}}⇒\mathrm{OE}=\mathrm{OF}$
Thus, BEDE is a quadrilateral whose diagonals bisect each other.
Therefore, BFDE is a parallelogram.
Hence, proved.

#### Question 6:

E is the mid-point of the side AD of the trapezium ABCD with AB ∥ DC. A line through E drawn parallel to AB intersect BC at F. Show that F is the mid-point of BC.

Given that, ABCD is a trapezium
such that .
Join, the diagonal AC which intersects EF at O.

In $△$ADC,
E is mid-point of AD and $\mathrm{OE}\parallel \mathrm{CD}$.

By mid-point theorem, O is the mid point of AC.

In $△$CBA,
O is mid point of AC and $\mathrm{OF}\parallel \mathrm{AB}.$

So, by mid-point theorem, E is the mid-point of BC.

Hence, proved.

#### Question 7:

Through A, B and C, lines RQ, PR and QP have been drawn, respectively parallel to sides BC, CA and AB of a ∆ABC as shown in the figure. Show that $\mathrm{BC}=\frac{1}{2}\mathrm{QR}.$

In $△$ABC,
.

In quadrilateral BCAR, BR ∥ CA and BC ∥ RA.
Then, BCAR is a parallelogram.

∴ BC = AR           .....(1)

So, in quadrilateral BCQA, BC ∥ AQ and AB ∥ QC.

Thus, BCQA is a parallelogram.
∴ BC = AQ          .....(2)

2BC = AR + AQ
$⇒2\mathrm{BC}=\mathrm{RQ}\phantom{\rule{0ex}{0ex}}⇒\mathrm{BC}=\frac{1}{2}\mathrm{QR}$
Hence, proved

#### Question 8:

D, E and F are the mid-points of the sides BC, CA and AB, respectively of an equilateral triangle ABC. Show that $△$DEF is also an equilateral triangle.

In equilateral $△$ABC, D, E and F are the mid-point of sides BC, CA and AB respectively.

In $△$ABC, E and F are the mid-point of AC and AB.
Then, using mid point theorem,
EF ∥ BC and EF =

Similarly,
DF$\parallel$AC, DE$\parallel$AB
.....(2)

As, $△$ABC is an equilateral triangle.
∴ AB = BC = CA
$⇒\frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{BC}=\frac{1}{2}\mathrm{CA}$      [dividing by 2]
$⇒\mathrm{DE}=\mathrm{EF}=\mathrm{FD}$            [from (1) and (2)]
Thus, all sides of AEDF are equal.
Therefore, $△$DEF is an equilateral triangle.
Hence, proved.

#### Question 9:

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.

Given that, ABCD is a parallelogram and AP = CQ.

In $△$AMP and $△$CMQ,
$\angle \mathrm{MAP}=\angle \mathrm{MCQ}$ [alternate interior angles]
AP = CQ            (given)
and $\angle \mathrm{AMP}=\angle \mathrm{CMQ}$  [vertically opposite angles]
$\therefore △\mathrm{AMP}\cong △\mathrm{CMQ}$  [AAS congruence rule]

$⇒$AM = CM                    [CPCT rule]
and PM = MQ                 [CPCT rule]
Thus, AC and PQ bisect each other.
Hence, proved.

#### Question 10:

In the figure, P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. Prove that AD = 2CD.

Given that ABCD is a parallelogram, P is a mid-point of BC such that $\angle \mathrm{BAP}=\angle \mathrm{DAP}$.
Since, ABCD is a parallelogram.
$\therefore \mathrm{AD}\parallel \mathrm{BC}$ and AB is transversal, then

In $△$ABP,

On multiplying both sides by 2, we get
2AB = 2BP

[since, ABCD is a parallelogram, then AB = CD and BC = AD]
Hence, proved.

#### Question 1:

A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

In an isosceles triangle $△$ABC, a square ADEF is inscribed.
To prove: CE = BE
Proof: In the isosceles $△$ABC, $\angle \mathrm{A}=90°$
and AB = AC     .....(1)

AD = AF          [all sides of squares are equal]     .....(2)

On subtracting (2) from (1),
AB − AD = AC − AE
⇒ BD = CF               .....(3)

Now, in ,

BD = CF         [From (3)]
DE = EF         [sides of square]
and $\angle \mathrm{CFE}=\angle \mathrm{EDB}=90°$

Hence, vertex E of the square bisects the hypotenuse BC.

#### Question 2:

In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of ∠A meets DC in E. AE and BC produced meet at F. Find the length of CF.

Given, a parallelogram ABCD in which AB = 10 cm and AD = 6 cm.
Now, draw a bisector $\angle \mathrm{A}$ meets DC in E and produces it to F and produce BC to meet at F.

Also, produce AD to H and join HF, so that ABFH, is a parallelogram.
Since, $\mathrm{HF}\parallel \mathrm{AB}$.

But

Since, CFHD is a parallelogram.
Thus, the opposite side are equal.
Hence, DH = CF = 4 cm

#### Question 3:

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.

Let ABCD be the quadrilateral, P, Q, R and S are the mid-point of sides AB, BC, CD and DA. Also, AC = BD.

Proof : In $△$ADC, S and R are the mid-points of AD and DC respectively.
Then by mid point theorem,

In $△$ABC, P and Q are mid-point of AB and BC respectively.
Then by mid-point theorem,

From (1) and (2),
SR = PQ =

Similarly, in $△$BCD,

and in $△$BAD,

From (4) and (5),

From (3) and (6),
$\mathrm{SR}=\mathrm{PQ}=\mathrm{SP}=\mathrm{RQ}$
Thus, all the sides of a quadrilateral PQRS are equal.
Hence, PQRS is a rhombus.

#### Question 4:

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD such that AC ⊥ BD. Prove that PQRS is a rectangle.

Given: A quadrilateral ABCD with P, Q, R and S as the mid-points of sides AB, BC, CD and DA respectively.
Also, $\mathrm{AC}\perp \mathrm{BD}$.
To prove: PQRS is a rectangle
Proof: As $\mathrm{AC}\perp \mathrm{BD}$
$\therefore \angle \mathrm{COD}=\angle \mathrm{AOD}=\angle \mathrm{AOB}=\angle \mathrm{COB}=90°$

In $△$ADC, S and R are mid-points of AD and CD respectively. Then by mid-point theorem,

In $△$ABC, P and Q are mid-points of AB and BC respectively. Then by mid-point theorem,

From (1) and (2),

Similarly,

Since, AC ∥ PQ.
$\therefore \angle \mathrm{P}=\angle \mathrm{Q}=90°$
So, PQRS is a rectangle.
Hence, proved.

#### Question 5:

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square.

Given: A quadrilateral ABCD with P, Q, R and S as the mid-points of sides AB, BC, CD and DA respectively.
Also, AC = BD and AC$\perp$BD.
To prove: PQRS is a square.
Proof: In $△$ADC, S and R are the mid-points of AD and CD respectively.
Then by mid-point theorem,

In $△$ABC, P and Q are the mid-points of AB and BC respectively.
Then by mid-point theorem,

From (1) and (2),

Similarly in $△$ABD, by mid point theorem,

In $△$BCD, by mid point theorem,

From (4) and (5),

From (3) and (6),
PQ = SR = SP = RQ
Thus, all 4 sides are equal.

Since, AC ∥ PQ.
∴ ∠P = ∠Q = 90°

So, PQRS is a square.
Hence, proved.

#### Question 6:

A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

Consider a parallelogram ABCD such that its diagonal AC bisects ∠A.
Thus,

To show: ABCD is a rhombus.
Proof: Since, ABCD is a parallelogram.
Therefore, AB ∥ CD and AC is the transversal.

Again, AD ∥ BC and AC is the transversal.

So,

Also,

As, AB = CD and AD = BC [opposite side of parallelogram are equal]
$\therefore$ AB = BC = CD = AD

Thus, all sides are equal.
So, ABCD is a rhombus.
Hence, proved.

#### Question 7:

P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP at S and BQ intersects CP at R. Show that PRQS is a parallelogram.

Given that, a parallelogram ABCD such that P and Q are the mid-points of AB and CD respectively.
To show: PQRS is a parallelogram.
Proof: As, ABCD is a parallelogram.
⇒ AB$\parallel$CD
⇒ AP$\parallel$QC
Also, AB = DC as the opposite sides of a parallelogram are equal.

Here, AP ∥ QC and AP = QC.
Thus, APCQ is a parallelogram.

∴ AQ ∥ PC or SQ ∥ PR          .....(1)

Again, AB ∥ DC or BP ∥ DQ.

Also, AB = DC.
$⇒\frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{DC}$          [dividing both sides by 2]
⇒ BP = QD               [since, P and Q are the mid-point of AB and DC]
Here, BP ∥ QD and BP = QD.
So, BPDQ is a parallelogram.

From (1) and (2), PRQS is a parallelogram.
Hence, proved.

#### Question 8:

ABCD is a quadrilateral in which AB ∥ DC and AD = BC. Prove that ∠A = ∠B and ∠C = ∠D.

Given that, ABCD is a quadrilateral with AB ∥ DC and AD = BC.
Construction: Extend AB to E and draw a line CE parallel to AD.

Proof: As, AD ∥ CE and transversal AE cuts them at A and E.
Thus,

Now, AB ∥ CD and AD ∥ CE.
So, quadrilateral AECD is a parallelogram.

In $△$BCE,
CE = BC                     [shown above]
$⇒\angle \mathrm{CBE}=\angle \mathrm{CEB}$       [opposite angles of equal sides are equal]

From (1) and (2),
$\angle \mathrm{A}=\angle \mathrm{B}$
Hence, proved.

#### Question 9:

In the figure AB || DE, AB = DE, AC || DF and AC = DF. Prove that BC || EF and BC = EF.

In the given figure,

To prove:

Thus, ABED is a parallelogram.

Thus, ACFD is a parallelogram.

From (1) and (2),

BE = CF and BE$\parallel$CF           [from (3)]
Thus, BCFE is a parallelogram.
Hence, proved.

#### Question 10:

E is the mid-point of a median AD of ∆ABC and BE is produced to meet AC at F. Show that $\mathrm{AF}=\frac{1}{3}\mathrm{AC}.$

In $△$ABC, AD is a median and E is the mid-point of AD.
Construction: Draw DP ∥ EF.

Proof: In $△$ADP,
E is the mid-point of AD and EF ∥ DP.
Thus, by converse of mid-point theorem, F is mid-point of AP.

In $△$FBC,
D is mid-point of BC and DP ∥ BF.
So, by converse of mid-point theorem, P is mid-point of FC.
Thus, AF = FP = PC.
$⇒\mathrm{AF}=\frac{1}{3}\mathrm{AC}$
Hence, proved.

#### Question 11:

Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.

In the square ABCD, P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.
To show: PQRS is a square.
Construction : Join AC and BD.

Proof : As, ABCD is a square.
$\therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}$

Also, P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.

Now, in $△$ADC, by mid-point theorem,
SR ∥ AC
and $\mathrm{SR}=\frac{1}{2}\mathrm{AC}$      .....(1)

In $△$ABC,
PQ ∥ AC
and

From (1) and (2),

Similarly,

and $\mathrm{RQ}=\frac{1}{2}\mathrm{BD}$
$\therefore \mathrm{SP}=\mathrm{RQ}=\frac{1}{2}\mathrm{BD}$

Thus, the diagonals of a square bisect each other at 90° and they are equal.

From (3) and (4),

$\therefore \angle \mathrm{EOF}=\angle \mathrm{ERF}=90°$
Thus, PQRS is a square.
Hence, proved.

#### Question 12:

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF || AB and $\mathrm{EF}=\frac{1}{2}\left(\mathrm{AB}+\mathrm{CD}\right).$

Given: ABCD is a trapezium in which AB ∥ CD.
Also, E and F are the mid-points of sides AD and BC respectively.
To prove: EF ∥ AB and EF = $\frac{1}{2}\left(\mathrm{AB}+\mathrm{CD}\right)$
Construction: Join BE and extend it to meet CD produced at G. Thus draw BD with intersects EF at O.

Proof: In $△$GCB, E and F are the mid-points of BG and BC respectively.
Then by mid-point theorem,
$\mathrm{EF}\parallel \mathrm{GC}$
But
$\therefore \mathrm{EF}\parallel \mathrm{AB}$

In $△$ADB,
$\mathrm{AB}\parallel \mathrm{EO}$ and E is the mid-point of AD.
Hence, by converse of mid-point theorem, O is mid-point of BD.
Also,

In $△$BDC,
OF ∥ CD and O is the mid-point of BD.
$\therefore \mathrm{OF}=\frac{1}{2}\mathrm{CD}$ [by converse of mid-point theorem]      ......(2)

$\mathrm{EO}+\mathrm{OF}=\frac{1}{2}\mathrm{AB}+\frac{1}{2}\mathrm{CD}$
$⇒\mathrm{EF}=\frac{1}{2}\left(\mathrm{AB}+\mathrm{CD}\right)$
Hence, proved.

#### Question 13:

Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.

Let ABCD be a parallelogram and AP, BR, CR and DP be are the bisectors .
To prove: Quadrilateral PQRS is a rectangle.

Proof: As, ABCD is a  parallelogram. Therefore, DC$\parallel$AB and DA is the transversal.
$\therefore \angle \mathrm{A}+\angle \mathrm{D}=180°$            [sum of co-interior angles of a parallelogram is 180°]

Thus, PQRS is a quadrilateral with each angle is 90°, i.e., right angle.
Hence, PQRS is a rectangle.

#### Question 14:

P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.

Given: ABCD is a parallelogram whose diagonals bisect each other at O.
To show: PQ is bisected at O.

In ,

and OB = OD        [∵ diagonals of a parallelogarm bisect each other]

Hence, proved.

#### Question 15:

ABCD is a rectangle in which diagonal BD bisects ∠B. Show that ABCD is a square.

Given: Rectangle ABCD, diagonal BD bisects $\angle \mathrm{B}$.
To show: ABCD is a square.
Construction: Join AC.

Proof: In
$\angle \mathrm{A}=\angle \mathrm{C}=90°$
BD = BD          (common)
AB = CD          (opposite sides of a rectangle are equal)

Thus, AB = BC
and AD = CD            [by CPCT rule]

Therefore,
AB = BC = CD = DA.
So, ABCD is a square.
Hence, proved.

#### Question 16:

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. Prove that by joining these mid-points D, E and F, the triangles ABC is divided into four congruent triangles.

In $△$ABC, D, E, F are the mid-points of the sides AB, BC and CA respectively.
To prove: $△$ABC is divided into four congruent triangles.

Proof: As, $△$ABC is a triangle and D, E and F are the mid-point of sides AB, BC and CA respectively.
Then,
AD = BD = $\frac{1}{2}$AB,
BE = EC = $\frac{1}{2}$BC
and AF = CF = $\frac{1}{2}$AC

Then, using the mid-point theorem,

and

In $△$ADF and $△$EFD,
AF = DE
DF = DF  (common)

So, $△$ABC is divided into four congruent triangles.
Hence, proved.

#### Question 17:

Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.

Consider a trapezium ABCD in which AB ∥ DC and let M and N be the mid-points of the diagonals AC and BD respectively.
To prove: MN ∥ AB ∥ CD
Construction: Join CN and extend it to meet AB at E as shown:

Proof: In ,
$\mathrm{DN}=\mathrm{BN}$        [∵ N is the mid-point of BD]
$\angle \mathrm{DCN}=\angle \mathrm{BEN}$       [alternate interior angles]
and $\angle \mathrm{CDN}=\angle \mathrm{EBN}$        [alternate interior angles]
$\therefore △\mathrm{CDN}\cong △\mathrm{EBN}$        [AAS congruence rule]
[by CPCT rule]

Thus, in $△\mathrm{CAE}$, the points M and N are the mid-points of AC and CE respectively.
$\therefore \mathrm{MN}\parallel \mathrm{AE}$           [by mid-point theorem]
$⇒$MN$\parallel$AB$\parallel$CD
Hence, proved.

#### Question 18:

P is the mid-point of the side CD of a parallelogram ABCD. A line through C parallel to PA intersects AB at Q and DA produced at R. Prove that DA = AR and CQ = QR.

Given: A parallelogram ABCD where P is the mid-point of DC.
To prove: DA = AR and CQ = QR
Proof: ABCD is a parallelogram.

Also,

As, P is the mid-point of DC.
$\therefore \mathrm{DP}=\mathrm{PC}=\frac{1}{2}\mathrm{DC}$

Now, .
Thus, APCQ is a parallelogram.

Therefore,

Now, in ,

and

Then, AR = BC  [by CPCT rule]
But BC = DA
$\therefore$ AR = DA
and CQ = QR  [CPCT rule]
Hence, proved.

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