# Frank Solutions for Chapter 19 Quadrilaterals Class 9 Mathematics ICSE

**Exercise 19.1**

**1. In the following figures, find the remaining angles of the parallelogram.**

**(a)**

**(b)**

**(c)**

**(d)**

**(e)**

**Answer**

**(a)** Given

ABCD is a parallelogram

∠A = 75°

Then ∠C = 75° **…(Opposite angles of a parallelogram are equal)**

Now,

∠A + ∠D = 180° …(Interior angles)

⇒ 75° + ∠D = 180°

⇒ ∠D = 180° – 75°

We get,

∠D = 105°

Hence, ∠B = ∠D = 105° **…(Opposite angles of a parallelogram are equal)**

Therefore, the remaining angles of the parallelogram are ∠B = 105°, ∠C = 75° and ∠D = 105°

**(b)** Given

PQRS is a parallelogram

∠Q = 60°

Then ∠S = 60° **…(Opposite angles of a parallelogram are equal)**

Now, in △PQR,

∠RPQ + ∠PQR + ∠PRQ = 180° **…(Angle sum property of a triangle)**

⇒ 50° + 60° + ∠PRQ = 180°

⇒ 110° + ∠PRQ = 180°

We get,

∠PRQ = 70°

And, ∠SPR = ∠PRQ = 70° **...(Alternate angles)**

⇒ ∠SPQ = ∠SPR + ∠RPQ

⇒ ∠SPQ = 70° + 50°

We get,

∠SPQ = 120°

Then, ∠SRQ = 120° **...(Opposite angles of a parallelogram are equal)**

Hence,

The remaining angles of a parallelogram are ∠P = 120°, ∠S = 60° and ∠R = 120°

**(c)** ∠PQR + 65° = 180° **…(Linear pair angles)**

⇒ ∠PQR = 180° – 65°

We get,

∠PQR = 115°

PQRS is a parallelogram

∠S = ∠Q = 115° **…(Opposite angles of a parallelogram are equal)**

And, ∠P + ∠S = 180° **…(Interior angles)**

⇒ ∠P + 115° = 180°

⇒ ∠P = 180° – 115°

We get,

∠P = 65°

∠R = ∠P = 65° **…(Opposite angles of a parallelogram are equal)**

Hence,

The remaining angles of a parallelogram are ∠P = 65°, ∠Q = 115°, ∠R = 65° and ∠S = 115°

**(d)** PQRS is a parallelogram

∠R + ∠Q = 180° **…(Interior angles)**

x° + (x°/4) = 180°

⇒ 4x° + x° = 180°×4

⇒ 5x° = 180°×4

On further calculation, we get,

x° = (180°×4)/5

⇒ x° = 36×4

We get,

x° = 144°

⇒ ∠R = 144°

⇒ (x°/4) = (144°/4)

⇒ (x°/4) = 36°

⇒ ∠Q = 36°

Hence,

∠P = ∠ R = 144° and ∠S = ∠Q = 36° **…(Opposite angles of a parallelogram are equal)**

**(e)** PQRS is a parallelogram with all sides equal and opposite sides parallel

Hence,

PQRS is a Rhombus

Diagonals of a Rhombus bisect each other

In △POS,

∠OSP + ∠SPO + ∠POS = 180°

⇒ x + 70° + 90° = 180°

⇒ x = 180° – 160°

We get,

x = 20°

In △QSP,

PS = PQ

∠QSP = ∠PQS = x = 20°

And,

∠QSP + ∠PQS + ∠SPQ = 180°

⇒ 20° + 20° + ∠SPQ = 180°

∠SPQ = 180° – 40°

We get,

∠SPQ = 140°

∠SRQ = 140° **…(Opposite angles of a parallelogram are equal)**

Now,

∠SPQ + ∠PSR = 180°

⇒ 140° + ∠PSR = 180°

⇒ ∠PSR = 40°

∠PQR = 40° **…(Opposite angles of a parallelogram are equal)**

Therefore, ∠P = ∠R = 140° and ∠S = ∠Q = 40°

**2. In a parallelogram ABCD ∠C = 98°. Find ∠A and ∠B.**

AnswerGiven,

Answer

ABCD is a parallelogram

∠C = 98°

Hence,

∠A = ∠C = 98° **…(Opposite angles of a parallelogram are equal)**

Now,

∠A + ∠B + ∠C + ∠D = 360° **…(Sum of all the angles of a quadrilateral = 360°)**

⇒ 98° + ∠B + 98° + ∠D = 360°

⇒ ∠B + 196° + ∠D = 360°

⇒ ∠B + ∠D = 360° – 196°

⇒ ∠B + ∠D = 164°

Here, ∠B = ∠D **…(Opposite angles of a parallelogram are equal)**

2∠B = 164°

We get,

∠B = ∠D = 82°

Hence, ∠B = 82° and ∠A = 98°

**3. The consecutive angles of a parallelogram are in the ratio 3: 6. Calculate the measures of all the angles of the parallelogram.**

AnswerLet ABCD is a parallelogram in which AD || BC

Answer

∠A and ∠B are consecutive angles

∠A: ∠B = 3: 6

Hence,

∠A = 3x and ∠B = 6x

AD || BC and AB is the transversal

∠A + ∠B = 180° **...(Co-interior angles are supplementary)**

⇒ 3x + 6x = 180°

⇒ 9x = 180°

We get,

x = 20°

Therefore, ∠A = 3×20° = 60° and

∠B = 6×20° = 120°

We know that,

Opposite angles of a parallelogram are equal

Hence,

∠C = ∠A = 60° and ∠D = ∠B = 120°

4. Find the measures of all the angles of the parallelogram shown in the figure:

4. Find the measures of all the angles of the parallelogram shown in the figure:

**Answer**

In △BDC,

∠BDC + ∠DCB + ∠CBD = 180°

2a + 5a + 3a = 180°

⇒ 10a = 180°

We get,

a = 18°

∠BDC = 2a = 2×18° = 36°

∠DCB = 5a = 5×18° = 90°

∠CBD = 3a = 3×18° = 54°

∠DAB = ∠DCB = 90° **…(Opposite angles of a parallelogram are equal)**

∠DBA = ∠BDC = 36° **…(alternate angles since AB || CD)**

∠BDA = ∠CBD = 54° **…(alternate angles since AB || CD)**

Hence, ∠DAB = ∠DCB = 90°, ∠DBA + ∠CBD = 90°, ∠BDA + ∠BDC = 90°

**5. In the given figure, ABCD is a parallelogram, find the values of x and y**

**Answer**

ABCD is a parallelogram

Opposite angles of a parallelogram are equal

Hence,

∠A = ∠C

⇒ 4x + 3y – 6 = 9y + 2

⇒ 4x – 6y = 8

⇒ 2x – 3y = 4 **…(1)**

AB || CD and AD is the transversal

∴ ∠A + ∠D = 180°^{ }**…(Co–interior angles are supplementary)**

(4x + 3y – 6) + (6x + 22) = 180°

⇒ 10x + 3y + 16 = 180°

We get,

10x + 3y = 164 **…(2)**

Adding equations (1) and (2), we get,

12x = 168

⇒ x = 14

Substituting the value of x in equation (1), we get,

2(14) – 3y = 4

⇒ 28 – 3y = 4

⇒ 3y = 24

We get,

y = 8

Therefore, x = 14 and y = 8

**6.** **The angles of a triangle formed by 2 adjacent sides and a diagonal of a parallelogram are in the ratio 1: 5: 3. Calculate the measures of all the angles of the parallelogram.**

**Answer**

Let ∠CAB = x°

Then,

∠ABC = 5x° and ∠BCA = 3x°

In △ABC,

∠CAB + ∠ABC + ∠BCA = 180° **…(Sum of angles of a triangle)**

⇒ x ° + 5x° + 3x° = 180°

⇒ 9x° = 180°

We get.

x° = 20°

∠CAB = x° = 20°

∠ABC = 5x° = 5×20° = 100°

∠BCA = 3x° = 3×20° = 60°

∠ADC = ∠ABC = 100° **(opposite angles of a parallelogram are equal)**

∠ACD = ∠CAB = 20° **(alternate angles since BC || AD)**

∠CAD = ∠BCA = 60° **(alternate angles since BC || AD)**

Hence,

∠ADC = ∠ABC = 100°, ∠ACD + ∠BCA = 80°, ∠CAD + ∠CAB = 80°

**7. PQR is a triangle formed by the adjacent sides PQ and QR and diagonal PR of a parallelogram PQRS. If in △PQR, ∠P: ∠Q: ∠R = 3: 8: 4, calculate the measures of all the angles of parallelogram PQRS.**

**Answer**

PQRS is a parallelogram

Let ∠RPQ = 3x°, ∠PQR = 8x° and ∠QRP = 4x°

In △PQR,

∠RPQ + ∠PQR + ∠QRP = 180° **…(Sum of angles of a triangle = 180°)**

⇒ 3x° + 8x° + 4x° = 180°

⇒ 15x° = 180°

We get,

x° = 12°

∠RPQ = 3x° = 3×12° = 36°

∠PQR = 8x° = 8×12° = 96°

∠QRP = 4x° = 4×12° = 48°

∠PSR = ∠PQR = 96° **(Opposite angles of a parallelogram are equal)**

∠RPS = ∠QRP = 48° **(Alternate angles since QR || PS)**

∠PRS = ∠RPQ = 36° **(Alternate angles since QR || PS)**

Hence, ∠PSR = ∠PQR = 96°, ∠RPS + ∠RPQ = 84°, ∠PRS + ∠QRP = 84°

**8. PQRS is a parallelogram. T is the mid-point of PQ and ST bisects ∠PSR.**

**Prove that:**

**(i) QR = QT**

**(ii) RT bisects angle R**

**(iii) ****∠****RTS = 90****°**

**Answer**

**(i)**∠PST = ∠TSR

**…(i)**

∠PTS = ∠TSR **…(ii) (alternate angles since SR || PQ)**

From (i) and (ii)

∠PST = ∠PTS

Hence,

PT = PS **(sides opposite to equal angles are equal)**

But PT = QT **(T is the midpoint of PQ)**

And PS = QR **(PS and QR are opposite and equal sides of a parallelogram)**

Therefore, QT = QR

**(ii)** Since,

QT = QR

∠QTR = ∠QRT **(angles opposite to equal sides are equal)**

But ∠QTR = ∠TRS **(alternate angles since SR || PQ)**

∠QRT = ∠TRS

Hence, RT bisects ∠R

**(iii)** ∠PST = ∠TSR

∠QRT = ∠TRS

∠QRS + ∠PSR = 180° **(adjacent angles of a parallelogram are supplementary)**

Now, multiplying by (1/2)

(1/2) ∠QRS + (1/2) ∠PSR = (1/2) x 180°

∠TRS + ∠TSR = 90°

In △STR,

∠TSR + ∠RTS + ∠TRS = 180°

⇒ ∠TRS + ∠TSR + ∠RTS = 180°

⇒ 90° + ∠RTS = 180°

We get,

∠RTS = 180° – 90°

⇒ ∠RTS = 90°

**9. PQRS is a square whose diagonals PR and QS intersect at O. M is a point on QR such that OQ = MQ. Find the measures of ∠MOR and ∠QSR.**

**Answer**

In △QOM,

∠OQM = 45° **(In square, diagonals make 45° with the sides)**

OQ = MQ

∠QOM = ∠QMO **…(i) (angles opposite to equal sides are equal)**

∠QOM + ∠QMO + ∠OQM = 180°

∠QOM + ∠QOM + 45° = 180°

On further calculation, we get,

2∠QOM = 180° – 45°

∠QOM = 67.5°

In △QOR,

∠QOR = 90° **(In square diagonals bisect at right angles)**

∠QOM + ∠MOR = 90°

67.5° + ∠MOR = 90°

∠MOR = 90° – 67.5°

We get,

∠MOR = 22.5°

In △ROS,

∠OSR = 45° **(In square diagonals make 45° with the sides)**

Therefore, ∠QSR = 45°

**10. ABCD is a rectangle with ∠ADB = 55°, calculate ∠ABD**

**Answer**

∠ADB = 55° **(given)**

∠DAB = 90° **(In rectangle angle between two sides is 90°)**

∠ADB + ∠DAB + ∠ABD = 180°

⇒ 55° + 90° + ∠ABD = 180°

On calculating further, we get,

∠ABD = 180° – 145°

∠ABD = 35°

**11. Prove that if the diagonals of a parallelogram are equal then it is a rectangle.**

AnswerLet ABCD be a parallelogram

Answer

In △ABC and △DCB,

AB = DC **(Opposite sides of a parallelogram are equal)**

BC = BC **(Common)**

AC = DB **(Given)**

Hence,

△ABC ≅ △DCB **(By SSS congruence rule)**

∠ABC = ∠DCB

We know that the sum of the measures of angles on the same side of transversal is 180°

∠ABC + ∠DCB = 180°

⇒ ∠ABC + ∠ABC = 180°

⇒ 2∠ABC = 180°

We get,

∠ABC = 90°

Since ABCD is a parallelogram and one of its interior angles is 90°

Therefore, ABCD is a rectangle

**12. ****The diagonals PR and QS of a quadrilateral PQRS are perpendicular to each other. A, B, C and D are mid-points of PQ, QR, RS and SP respectively. Prove that ABCD is a rectangle.**

**Answer**Given,

PQRS is a quadrilateral where A, B, C and D are mid-points of PQ, QR, RS and SP respectively.

In △PQS, A and D are mid-points of sides QP and PS respectively.

Hence,

AD || QS and AD = (1/2) QS **…(i)**

In △QRS

B and C are the mid-points of QR and RS respectively

Hence,

BC || QS and BC = (1/2) QS **…(ii)**

From equations (i) and (ii), we get,

AD || BC and AD = BC

Since in quadrilateral ABCD one pair of opposite sides are equal and parallel to each other.

Hence it is a parallelogram

Here, the diagonals of quadrilateral PQRS intersect each other at point O

Now,

In quadrilateral OMDN

ND || OM **(AD || QS)**

DM || ON **(DC || PR)**

Therefore,

OMDN is a parallelogram

∠MDN = ∠NOM

∠ADC = ∠NOM

But, ∠NOM = 90° **(diagonals are perpendicular to each other)**

∠ADC = 90°

Clearly ABCD is a parallelogram having one of its interior angle as 90°

Therefore,

ABCD is a rectangle

**13. ****ABCD is a quadrilateral P, Q, R and S are the midpoints of AB, BC, CD and AD. Prove that PQRS is a parallelogram.**

**Answer**

In △ABC,

P and Q are midpoints of AB and BC respectively

Hence,

PQ || AC and PQ = (1/2) AC **…(i)**

In △ADC,

S and R are midpoints of AD and DC respectively

Hence,

SR || AC and SR = (1/2) AC **…(ii)**

From equations (i) and (ii), we get,

PQ || SR and PQ = SR

Hence,

PQRS is a parallelogram

**14. ****PQRS is a parallelogram. T is the mid-point of RS and M is a point on the diagonal PR such that MR = (1/4) PR. TM is joined and extended to cut QR at N. Prove that QN = RN.**

**Answer**

Diagonals of a parallelogram bisect each other

Hence,

OP = OR

Given MR = (1/4) PR

Therefore,

MR = (1/4) (2×OR)

MR = (1/2) OR

Thus, M is the midpoint of OR

In △ROS,

T and M are the mid-points of RS and OR respectively

Hence,

TM || OS

TN || QS

Also,

In △RQS,

T is the midpoint of RS and TN || QS

Hence,

N is the mid-point of QR and TN = (1/2) QS

Therefore,

QN = RN

**15. ****Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to each of the parallel sides, and is equal to half the difference of these sides.**

**Answer**

M and N are midpoints of AC and BD respectively

Join MN

Draw a line CN cutting AB at E

Now in triangles DNC and BNE,

DN = NB **[N is the midpoint of BD (given)]**

∠CDN = ∠EBN **(Alternate angles, since DC || AB)**

∠DNC = ∠BNE **(Vertically opposite angles)**

Therefore,

△DNC ≅ △BNE **(By ASA test)**

DC = BE

In △ACE, M and N are midpoints

MN = (1/2) AE and MN || AE or MN || AB

Also,

AB || CD

Hence,

MN || CD

MN = (1/2) {AB – BE}

⇒ MN = (1/2) {AB – CD} **(Since BE = CD)**

⇒ MN = (1/2)×difference of parallel sides AB and CD

Hence proved

**Exercise 19.2 **

**1. ABCD is a parallelogram. The bisector of ****∠****BAD meets DC at P, and AD is half of AB. **

**Prove that :**

**(i) BP bisects ****∠****ABC. **

**(ii) ****∠****APB is a right angle. **

**Answer**

(i)

**2. ABCD is a parallelogram. P and Q are mid-points of AB and CD. Prove that APCQ is also a parallelogram. **

**Answer**

**3. SN and QM are perpendiculars to the diagonal PR of parallelogram PQRS. **

**Prove that: **

**(i)**

**△**

**SNR**

**≅**

**△**

**QMP**

**(ii) SN = QM **

**Answer**

(i)

**(ii)** Since △SNR ≅ △QMP

Hence, SN = QM

**4. Point M and N are taken on the diagonal AC of a parallelogram ABCD such that AM = CN. Prove that BMDN is a parallelogram. **

**Answer**

Join BD.

**5. In the given figure, MP is the bisector of ****∠****P and RN is the bisector of ****∠****R of parallelogram PQRS. Prove that PMRN is a parallelogram. **

**Answer**

Construction: Join PR.

**6. ABCD is a parallelogram. P and T are points on AB and DC respectively and AP = CT. Prove that PT and BD bisect each other. **

**Answer**

Join AC

**7. PQRS is a parallelogram. PQ is produced to T so that PQ = QT. prove that ST bisects QR.**

**Answer**

**8. Prove that the quadrilateral formed by joining the mid-points of a square is also a square. **

**Answer**

**9. Prove that the quadrilateral formed by joining the mid-points of consecutive sides of a rectangle is a rhombus. **

**Answer**

**10. Prove that the quadrilateral formed by joining the mid-points of consecutive sides of a rhombus is a rectangle. **

**Answer**

**11. P is a point on side KN of a parallelogram KLMN such that KP : PN is 1 : 2. Q is a point on side LM such that LQ : MQ is 2 : 1. Prove that KQMP is a parallelogram. **

**Answer**

**12. PQRS is a parallelogram. M and N are the mid-points of the adjacent sides QR and RS. O is the mid-point of the diagonal PR. Prove that MONR is a rectangle and MN is half of PR. **

**Answer**

**13. (a) In a parallelogram PQRS, M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that RN and RM trisect QS. **

**(b) In a parallelogram PQRS. M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that PMRN is a parallelogram. **

**(c) In a parallelogram PQRS, M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that MN QS. **

**Answer**

(a)

(b)

(c)

**14. ABCD is a trapezium in which side AB is parallel to side DC. P is the mid-point of side AD. If Q is a point on the side BC such that the segment PQ is parallel to DC, prove that PQ = 1/2(AB + DC).**

**Answer**

**15. (a) In the given figure, PQRS is a parallelogram in which PA = AB = Prove that SA ****|| QB and SA = QB. **

**(b) In the given figure, PQRS is a parallelogram in which PA = AB = Prove that SAQB is a parallelogram.**

**Answer**

(a)

(b)

**16. (a) In the given figure, PQRS is a trapezium in which PQ ****|| SR and PS = QR. Prove that ****∠****PSR = ****∠****QRS and ****∠****SPQ = ****∠****RQP. **

**Answer**

**17. In a parallelogram ABCD, E is a midpoint of AB and DE bisects angle D. Prove that: **

**(a) BC = BE. **

**(b) CE is the bisector of angle C and angle DEC is a right angle. **

**Answer**

(a)

(b)**18. Prove that if the diagonals of a quadrilateral bisect each other at right angles then it is a rhombus.**

**Answer**

**19. Prove that the diagonals of a kite intersect each other at right angles. **

**Answer**

**20. Prove that the diagonals of a square are equal and perpendicular to each other. **

**Answer**