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#### Question 1:

Diagonals of a parallelogram WXYZ intersect each other at point O. If $\angle$XYZ = ${135}^{°}$ then what is the measure of  $\angle$XWZ and  $\angle$YZW ?
If l(OY)= 5 cm then l(WY)= ?

In a parallelogram, opposite angles are congruent.
$\angle$XYZ = $\angle$XWZ = ${135}^{°}$
Also, WX || ZY
So, $\angle$YZW + $\angle$XWZ = 180$°$                   (Since, interior angles on the same side of the transversal are supplementary)
$⇒\angle \mathrm{YZW}+135°=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{YZW}=45°$
Now l(OY)= 5
we know that diagonals of a parallelogram bisect each other.
So, l(WY) =

#### Question 2:

In a parallelogram ABCD, If $\angle$A = , $\angle$B =  then find the value of x and then find the measures of  $\angle$C and  $\angle$D.

Interior angles on the same side of the transversal are supplementary.
$\angle$A +$\angle$B = 180$°$
$⇒$ +  = 180$°$
$⇒5x-20=180°\phantom{\rule{0ex}{0ex}}⇒5x=200\phantom{\rule{0ex}{0ex}}⇒x=40°$
Thus, $\angle$A = $3×40°+12=132°$
$\angle \mathrm{B}=2x-32=2×40°-32=48°$
Also, opposite angles of a parallelogram are congruent.
$\angle \mathrm{A}=\angle \mathrm{C}=132°\phantom{\rule{0ex}{0ex}}\angle \mathrm{B}=\angle \mathrm{D}=48°$

#### Question 3:

Perimeter of a parallelogram is 150 cm. One of its sides is greater than the other side by 25 cm. Find the lengths of all sides.

Perimeter = 150 cm
Let one of the sides be x.
Other side = $x+25$
Perimeter = Sum of all sides
$x+x+\left(x+25\right)+\left(x+25\right)=150\phantom{\rule{0ex}{0ex}}⇒4x+50=150\phantom{\rule{0ex}{0ex}}⇒4x=100\phantom{\rule{0ex}{0ex}}⇒x=25$
Thus, the sides are 25 cm, 25 cm, 50 cm and 50 cm

#### Question 4:

If the ratio of measures of two adjacent angles of a parallelogram is 1 : 2, find the measures of all angles of the parallelogram.

Let the two adjacent angles be .
Interior angles on the same side of the transversal are supplementary.
So,
$x+2x=180°\phantom{\rule{0ex}{0ex}}⇒3x=180°\phantom{\rule{0ex}{0ex}}⇒x=60°$
Thus, the adjacent angles are .
The opposite angles of a parallelogram are congruent.
So, the angles of the parallelogram will be $60°,120°$.

#### Question 5:

Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO = 12 and AB = 13 then show that $\square$ABCD is a rhombus.

In $△$AOB,
AO = 5 cm
OB = 12 cm
AB = 13 cm
5, 12 and 13 form a Pythagorean triplet.
Thus, $△$AOB is a right angle triangle, right angled at O.
ABCD is a parallelogram.
So, diagonals bisect each other.
$⇒$AO = OC = 5 cm
Also, OB = OD = 12 cm.
Thus, in parallelogram ABCD, diagonals bisect at right angles.
Hence, ABCD is a rhombus.

#### Question 6:

In the given figure, $\square$PQRS and $\square$ABCR are two parallelograms. If  $\angle$P = ${110}^{°}$ then find the measures of  all angles of $\square$ABCR.

In $\square$PQRS,
$\angle$P = ${110}^{°}$
We know that opposite angles of a parallelogram are congruent.
So, $\angle$P = $\angle$R = ${110}^{°}$
In $\square$ABCR,
$\angle$R =$\angle$B = ${110}^{°}$        (Opposite angles of a parallelogram are congruent)
Interior angles on the same side of the transversal are supplementary.
So,
$\angle \mathrm{R}+\angle \mathrm{A}=180°\phantom{\rule{0ex}{0ex}}⇒110°+\angle \mathrm{A}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{A}=180°-110°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{A}=70°$
And $\angle$A =$\angle$C = 110$°$

#### Question 7:

In the given figure, $\square$ABCD is a parallelogram. Point E is on the ray AB such that BE = AB then prove that line ED bisects seg BC at point F.

$\square$ABCD is a parallelogram
$⇒$AD || BF
Given: AB = BE so, B is the midpoint of AE.
By converse of mid-point theorem,
F is the midpoint of DE. So, DF = FE
In $△$EBF and $△$DCF,
DF = FE                         (Proved above)
DC = BE                        (Since DC = AB and AB = BE)
$\angle$FDC = $\angle$FEB             (Alternate interior angles of the parallel lines AE and CD)
Thus, $△$EBF $\cong$ $△$DCF     (SAS congruency)
Therefore, FB = FC             (CPCT)
Hence, ED bisects seg BC at F.

#### Question 1:

In the given figure, $\square$ABCD is a parallelogram, P and Q are midpoints of side AB and DC respectively, then prove $\square$APCQ is a parallelogram.

$\square$ABCD is a parallelogram,
AB $\cong$ CD and AB || CD
P and Q are the mid points of AB and CD respectively.
So, AP $\cong$ CQ and AP || CQ
Thus, $\square$APCQ is also a parallelogram.

#### Question 2:

Using opposite angles test for parallelogram, prove that every rectangle is a parallelogram.

Let ABCD be a rectangle.
$\angle$A = $\angle$B = $\angle$C = $\angle$D = $90°$.
For any quadrilateral to be a parallelogram, pair of opposite angles should be congruent.
In rectangle ABCD,
$\angle$A = $\angle$C =  $90°$
$\angle$B = $\angle$D = $90°$
Thus, rectangle ABCD is a paralleogram.

#### Question 3:

In the given figure, G is the point of concurrence of medians of $∆$DEF. Take point H on ray DG such that D-G-H and DG = GH, then prove that
$\square$GEHF is a parallelogram.

G is the point of concurrence of the medians of $∆$DEF.
Let the point where the median divides EF into two equal parts be A.
Thus, EA = AF.                     .....(1)
we know that the point of concurrence of the medians, divides each median in the ratio 2 : 1.
So, let DG = 2x  and GA = x
Given that DG = GH
So, GA = AH = x
Thus, point A dividess EF and GH into two equal parts.
Hence, $\square$GEHF is a parallelogram as the diagonals EF and GH bisect each other.

#### Question 4:

Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.(shown in the given figure)

ABCD is a parallelogram.

Similarly, $\angle \mathrm{PQR}=90°$

Thus, PQRS is a rectangle.

#### Question 5:

In the given figure, if points P, Q, R, S are on the sides of parallelogram such that AP = BQ = CR = DS then prove that $\square$PQRS is a parallelogram.

ABCD is a parallelogram
So, opposite pair of sides will be congruent and parallel.

Given, AP = BQ = CR = DS

In $△$APS and $△$RCQ
AS = CQ                         (From (1))
AP = CR                          (Given)

Similarly, PQ = SR
Thus, opposite pair of sides are congruent.
Hence, PQRS is a parallelogram.

#### Question 1:

Diagonals of a rectangle ABCD intersect at point O. If AC = 8 cm then find BO and if $\angle$CAD = ${35}^{°}$ then find $\angle$ACB

Diagonals of a rectangle are congruent.
So, AC = BD = 8 cm
Also, rectangle is a parallelogram so, the diagonals bisect each other.
Thus, BO = OD = 4 cm
So, $\angle$CAD = $\angle$ACB = ${35}^{°}$     (Since, alternate interior angles are equal)

#### Question 2:

In a rhombus PQRS if PQ = 7.5 then find QR. If $\angle$QPS = ${75}^{°}$ then find the measure of $\angle$PQR and $\angle$SRQ.

All the sides of a rhombus are equal.
So, PQ = QR = 7.5 cm
A rhombus is also a parallelogram so,
$\angle$QPS = $\angle$SRQ = ${75}^{°}$                      (opposite angles of a parallelogram are congruent)
Also, $\angle$QPS + $\angle$PQR = $180°$           (interior angles on the same side of the transversal are supplementary)
$⇒75°+\angle \mathrm{PQR}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{PQR}=105°$

#### Question 3:

Diagonals of a square IJKL intersects at point M, Find the measures of $\angle$IMJ, $\angle$JIK and $\angle$LJK .

Diagonals of a square are perpendicular bisectors of each other.
So, $\angle$IMJ = 90$°$.
A square has all the 4 angles as right angles.
Also, diagonals of a square bisect the opposite angles.
So, $\angle$JIK = $\frac{1}{2}×90°=45°$
Similarly, $\angle$LJK = $\frac{1}{2}×90°=45°$

#### Question 4:

Diagonals of a rhombus are 20 cm and 21 cm respectively, then find the side of rhombus and its perimeter.

Let ABCD be the rhombus.
AC = 20 cm and BD = 21 cm
We know that the diagonals of a rhombus bisect at right angles.
So, AO = OC = 10 cm
And BO = OD = 10.5 cm
In $△$AOB,

Thus, each side of rhombus = 14.5 cm
Perimeter = AB + BC + CD + DA = 14.5 + 14.5 + 14.5 + 14.5 = 58 cm.

#### Question 5:

State with reasons whether the following statements are ‘true’ or ‘false’.

(i) Every parallelogram is a rhombus.
(ii) Every rhombus is a rectangle.
(iii) Every rectangle is a parallelogram.
(iv) Every squre is a rectangle.
(v) Every square is a rhombus.
(vi) Every parallelogram is a rectangle.

(i) Every parallelogram is a rhombus.
False, every rhombus is a paralellogram but the vice versa is not true.

(ii) Every rhombus is a rectangle.
False, as in a rectangle all the angles are right angles but the same is not true in a rhombus.

(iii) Every rectangle is a parallelogram.
True, all rectangle have the opposite pair of sides congruent and parallel and the diagonals bisect each other.

(iv) Every square is a rectangle.
True, as all the angles are right angles and the diagonals are congruent to each other.

(v) Every square is a rhombus.
True, as the diagonals of a rhombus are perpendicular bisectors of each other and they also bisect the pair of opposite angles.

(vi) Every parallelogram is a rectangle.
False, as in a parallelogram the opposite angles are congruent but not right angles.

#### Question 1:

In $\square$IJKL,  side IJ || side KL $\angle$I = ${108}^{°}$  $\angle$K = ${53}^{°}$ then find the measure of $\angle$J and  $\angle$L.

Given: side IJ || side KL
So, the interior angles on the same side of the transversal will be supplementary.
$\angle$I + $\angle$L = 180º
$⇒$${108}^{°}$ + $\angle$L = 180º
$⇒$$\angle$L = 72º
Similarly, $\angle$K + $\angle$J = 180º
$⇒$${53}^{°}$ + $\angle$J = 180º
$⇒$$\angle$J = 127º

#### Question 2:

In $\square$ABCD , side  BC || side AD, side AB $\cong$ side DC  If $\angle$A = ${72}^{°}$  then find the measure of  $\angle$B, and  $\angle$D.

Interior angles on the same side of the transversal are supplementary.
$\angle$A + $\angle$B = 180º
$⇒$${72}^{°}$ + $\angle$B = 180º
$⇒$$\angle$B = 180º − ${72}^{°}$
$⇒$$\angle$B = 108º
Construction: Draw BP || CD
So, BC || AD and BP || CD
$⇒$PBCD is a parallelogram
$⇒$CD $\cong$ BP
Now CD = BP and CD = AB
$⇒$BP = AB
$⇒$$\angle \mathrm{BAP}=\angle \mathrm{BPA}=72°$
BP || CD so,

$\angle$B = 108º
$\angle$D = 72º

#### Question 3:

In $\square$ABCD , side BC < side AD  (Figure 5.32) side BC || side AD and if side BA $\cong$ side CD then prove that $\angle$ABC$\cong$ $\angle$DCB.

Construction: Draw a line CE || AB
CE || AB          (By construction)
AE || BC          (Give)
So, ABCE is a parallelogram.
Since, AB || CE
⇒ ∠BAE = ∠CED = ∠x              (Corresponding angles)         .....(1)
Now, BA $\cong$ CE                           (ABCE is a parallelogram)
and BA $\cong$ CD                              (Given)
So, CE $\cong$ CD
⇒ ∠CED = ∠CDE = ∠x               (Isosceles triangle)               .....(2)
From (1) and (2) we have
∠BAE = ∠CDE = ∠x                                                                .....(3)
∠CEA = 180º − ∠CED = 180º − ∠x
In a parallelogram, the opposite angles are equal.
So, ∠ABC = ∠CEA = 180º − ∠x                                              .....(4)
Similarly, ∠BAE = ∠BCE = ∠
In ∆CED,
$\angle \mathrm{ECD}+\angle \mathrm{CED}+\angle \mathrm{EDC}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{ECD}+x+x=180º\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{ECD}=180º-2x$
Now ∠DCB = ∠BCE + ∠ECD = + 180º − 2x = 180º −
Thus, $\angle$ABC$\cong$ $\angle$DCB

#### Question 1:

In the given figure, points X, Y, Z are the midpoints of side AB, side BC and side AC of $∆$ABC respectively. AB = 5 cm, AC = 9 cm and BC = 11 cm. Find the length of XY, YZ, XZ.

Given that X, Y and Z are the midpoints of the side AB, BC and CA respectively.
By Midpoint theorem,
XZ || BC and XZ = $\frac{1}{2}$BC
BC = 11 cm
$⇒$XZ =
Similarly,

#### Question 2:

In the given figure, $\square$ PQRS and $\square$MNRL are rectangles. If point M is the midpoint of side PR then prove that,

(i) SL = LR, (ii) LN = $\frac{1}{2}$SQ.

(i) Given that M is the midpoint of PR.                                                                                   .....(1)
$\angle$PSR = $\angle$MLR = 90$°$                             (Since $\square$ PQRS and $\square$MNRL are rectangles)
Thus, LM || SP                                           (Corresponding angles are equal)                         .....(2)
From (1) and (2) we have
L is the midpoint of SR                              (Converse of midpoint theorem)
Thus, SL = LR

(ii)
$\angle$RNM = $\angle$RQP = 90$°$                             (Since $\square$ PQRS and $\square$MNRL are rectangles)
Thus, MN || PQ                                           (Corresponding angles are equal)                         .....(4)
From (1) and (4) we have
N is the midpoint of RQ                              (Converse of midpoint theorem)
Join LN and SQ

By midpoint theorem,
LN || SQ and LN = $\frac{1}{2}$SQ

#### Question 3:

In the given figure, $∆$ ABC is an equilateral traingle. Points F,D and E are midpoints of side AB, side BC, side AC respectively. Show that $∆$FED is an equilateral traingle.

$∆$ABC is an equilateral triangle.
So, AB = BC = CA

Points F,D and E are midpoints of side AB, side BC, side AC respectively.
By midpoint theorem,
$\mathrm{FE}=\frac{1}{2}\mathrm{BC}=\mathrm{BD}\phantom{\rule{0ex}{0ex}}\mathrm{DE}=\frac{1}{2}\mathrm{AB}=\mathrm{FB}\phantom{\rule{0ex}{0ex}}\mathrm{FD}=\frac{1}{2}\mathrm{AC}=\mathrm{EC}$
From (1) we have FB = BD = EC
So, FE = DE = FD
Thus, $∆$FED is an equilateral traingle.

#### Question 4:

In the given figure, seg PD is a median of $∆$ PQR. Point T is the mid point of seg PD. Produced QT intersects PR at M. Show that $\frac{\mathrm{PM}}{\mathrm{PR}}$= $\frac{1}{3}$.
[Hint: DN || QM]

PD is the median of QR.
So, D is the midpoint of QR.
DN is drawn parallel to QM.
By converse of midpoint theorem, N is the midpoint of MR.            .....(1)
Similarly, T is the midpoint of PD
Also, DN || QM
So, By converse of  midpoint theorem,
M is the midpoint of PN.                                                                    .....(2)
From (1) and (2) we have
PM = MN = NR
$⇒\frac{\mathrm{PM}}{\mathrm{PR}}=\frac{\mathrm{PM}}{\mathrm{PM}+\mathrm{MN}+\mathrm{NR}}=\frac{\mathrm{PM}}{\mathrm{PM}+\mathrm{PM}+\mathrm{PM}}=\frac{1}{3\mathrm{PM}}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{PM}}{\mathrm{PR}}=\frac{1}{3}$
Hence Proved.

#### Question 1:

Choose the correct alternative answer and fill in the blanks.

(i) If all pairs of adjacent sides of a quadrilateral are congruent then it is called ....
(A) rectangle (B) parallelogram (C) trapezium, (D) rhombus

(ii) If the diagonal of a square is 12 $\sqrt{2}$ cm then the perimeter of square is ......
(A) 24 cm (B) 24 $\sqrt{2}$  cm (C) 48 cm (D) 48 $\sqrt{2}$ cm

(iii) If opposite angles of a rhombus are (2x)° and (3x $-$ 40)° then value of x is ...
(A) 100° (B) 80° (C) 160° (D) 40°

(i) When all the pairs of the adjacent sides of a quadrilateral are congruent then it is a rhombus.
Hence, the correct answer is option D.

(ii)

Let the diagonal be AC = 12$\sqrt{2}$ cm
since it is a squarem so, all the angles are 90$°$.
In $△$ABC, we apply the Pythagoras theorem,

So, AB = BC = CD = AD = 12 cm
Perimeter of the square ABCD = 4AB =
Hence, the correct answer is option C.

(iii) A rhombus is also a parallelogram so, the opposite angles will be congruent.
Thus, (2x)° = (3$-$ 40)°
$⇒3x-2x=40°\phantom{\rule{0ex}{0ex}}⇒x=40°$
Hence, the correct answer is option D.

#### Question 2:

Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal.

In a rectangle, all the angles are equal to 90º.
On applying Pythagoras theorem in $△$ABD,

Thus, the length of the diagonal is 25 cm.

#### Question 3:

If diagonal of a square is 13 cm then find its side.

Let PQRS be the square.
Let the diagonal SQ = 13 cm
All the angles of a square are right angles.
In $△$SPQ, applying Pythagoras theorem,

Thus, each side = $6.5\sqrt{2}$ cm.

#### Question 4:

Ratio of two adjacent sides of a parallelogram is 3 : 4, and its perimeter is 112 cm. Find the length of its each side.

Let the two sides be 3x and 4x .
In a parallelogram the opposite sides are congruent.
So, the sides are 3x, 3x, 4x and 4x
Perimeter = 112 cm

So,

Thus, the sides are 24 cm, 32 cm, 24 cm and 32 cm.

#### Question 5:

Diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm respectively. Find the length of side PQ.

Diagonals of a rhombus bisect each other at right angles.
PR = 20 cm

Similarly, QS = 48 cm

In $△$POQ, applying Pythagoras theorem,

#### Question 6:

Diagonals of a rectangle PQRS are intersecting in point M. If  $\angle$QMR = ${50}^{°}$ find the measure of $\angle$MPS.

In the given rectangle PQRS,
$\angle$QMR = $\angle$PMS = ${50}^{°}$                        (vertically opposite angles are equal)
Also, SQ = PR                                      (Diagonals of a rectangle are equal)
So,
$\frac{1}{2}\mathrm{SQ}=\frac{1}{2}\mathrm{PR}\phantom{\rule{0ex}{0ex}}⇒\mathrm{SM}=\mathrm{PM}$
$⇒\angle \mathrm{MSP}=\angle \mathrm{MPS}$                                (Angle opposite to equal sides are equal)
In $△$PMS,
$\angle \mathrm{PMS}+\angle \mathrm{MSP}+\angle \mathrm{MPS}=180°\phantom{\rule{0ex}{0ex}}⇒50°+2\angle \mathrm{MPS}=180°\phantom{\rule{0ex}{0ex}}⇒2\angle \mathrm{MPS}=130°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{MPS}=65°$

#### Question 7:

if seg AB || seg PQ , seg AB $\cong$seg PQ, seg AC || seg PR, seg AC $\cong$ seg PR then prove that, seg BC || seg QR and seg BC$\cong$seg QR.

Given: seg AB || seg PQ , seg AB $\cong$seg PQ
So, ABQP is a parallelogram as the pair of opposite sides are congruent and parallel.
Thus, seg AP || seg BQ , seg AP $\cong$seg BQ                       .....(1)
Similarly, seg AC || seg PR, seg AC $\cong$ seg PR
So, APRC is a parallelogram.
Thus, seg AP || seg CR, seg AP $\cong$ seg CR                       .....(2)
From (1) and (2) we have
seg BQ || seg CR
Also, seg BQ $\cong$ seg CR
Thus, BQRC is a parallelogram as the pair of opposite sides are congruent and parallel.
Therefore, seg BC || seg QR and seg BC$\cong$seg QR as BQRC is a parallelogram.

#### Question 8:

In the given Figure,$\square$ABCD is a trapezium. AB || DC .Points P and Q are midpoints of seg AD and seg BC respectively.
Then prove that, PQ || AB and  PQ = $\frac{1}{2}$( AB + DC ).

Construction: Join PB and extend it to meet CD produced at R.
To prove: PQ || AB and  PQ = $\frac{1}{2}$( AB + DC)
Proof: In $△$ABP and $△$DRP,

Thus, by ASA congruency, $△$ABP $\cong$ $△$DRP.
By CPCT, PB = PR and AB = RD.
In $△$BRC,
Q is the mid point of BC                     (Given)
P is the mid point of BR                      (As PB = PR)
So, by midpoint theorem, PQ || RC
$⇒$PQ || DC
But AB || DC                                       (Given)
So, PQ || AB.
Also, PQ = $\frac{1}{2}\left(\mathrm{RC}\right)$

#### Question 9:

In the adjacent figure, $\square$ABCD is a trapezium AB || DC . Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB .

Construction: Join MN. Also, join DM and produce it to AB in P.
To prove: MN || AB
Proof: AB || DC and AC is the transversal.
So,
In $△$AMP and $△$CMD,

(Vertically opposite angles)
AM = MC             (M is midpoint of AC)
So, $△$AMP $\cong$ $△$CMD          (ASA congruency criteria)
By CPCT, DM = MP.
So, M is the midpoint of DP.
In $△$DPB,
M is the midpoint of DP and N is the midpoint of DB.
By, midpoint theorem MN || AB.

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