# 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)

(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.

Given,

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.

Let ABCD is a parallelogram in which AD || BC

∠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: 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

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.

ABCD is a parallelogram

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)

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.

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°

(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. 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

In △ABD,

∠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.

Let ABCD be a parallelogram

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.

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,

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,

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,

ND || OM (AD || QS)

DM || ON (DC || PR)

Therefore,

OMDN is a parallelogram

∠MDN = ∠NOM

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

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.

In the given figure, join AC and BD

In △ABC,

P and Q are midpoints of AB and BC respectively

Hence,

PQ || AC and PQ = (1/2) AC …(i)

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.

Join PR to intersect QS at point O

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.

Join AC and BD

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.

(i)

(ii)

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

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

Prove that:

(i) SNR QMP

(ii) SN = QM

(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.

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.

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.

Join AC

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

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

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

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

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.

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.

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.

(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).

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.

(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.

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.

(a)

(b)

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