# Frank Solutions Class 9 Maths Chapter 19 Quadrilaterals

Frank Solutions Class 9 Maths Chapter 19 Quadrilaterals have solutions in a step-wise manner for students to get a better hold on the concepts. The solutions help them to self analyse their areas of weakness, which might require more practice. Practising Frank Solutions helps students to improve their skills in solving complex problems with ease. For better academic performance, learners can refer to the Frank Solutions Class 9 Maths Chapter 19 Quadrilateral PDF for free from the below-provided link.

Chapter 19 has problems based on quadrilaterals. A polygon with four edges and four vertices is known as a quadrilateral. It is essential to practise the solutions designed by experienced teachers for more conceptual knowledge of concepts covered in the chapter. It also helps them to face exams with confidence.

## Access Frank Solutions for Class 9 Maths Chapter 19 Quadrilaterals

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

(a)

(b)

(c)

(d)

(e)

Solution:

(a) Given

ABCD is a parallelogram

âˆ A = 750

Then âˆ C = 750 â€¦.. (Opposite angles of a parallelogram are equal)

Now,

âˆ A + âˆ D = 1800 â€¦. (Interior angles)

750 + âˆ D = 1800

âˆ D = 1800 – 750

We get,

âˆ D = 1050

Hence, âˆ B = âˆ D = 1050 â€¦â€¦. (Opposite angles of a parallelogram are equal)

Therefore, the remaining angles of the parallelogram are âˆ B = 1050, âˆ C = 750 and âˆ D = 1050

(b) Given

PQRS is a parallelogram

âˆ Q = 600

Then âˆ S = 600 â€¦.(Opposite angles of a parallelogram are equal)

Now, in â–³PQR,

âˆ RPQ + âˆ PQR + âˆ PRQ = 1800 â€¦â€¦. (Angle sum property of a triangle)

500 + 600 + âˆ PRQ = 1800

1100 + âˆ PRQ = 1800

We get,

âˆ PRQ = 700

And, âˆ SPR = âˆ PRQ = 700 â€¦â€¦ (Alternate angles)

âˆ SPQ = âˆ SPR + âˆ RPQ

âˆ SPQ = 700 + 500

We get,

âˆ SPQ = 1200

Then, âˆ SRQ = 1200 â€¦â€¦(Opposite angles of a parallelogram are equal)

Hence,

The remaining angles of a parallelogram are âˆ P = 1200, âˆ S = 600 and âˆ R = 1200

(c) âˆ PQR + 650 = 1800 â€¦..(Linear pair angles)

âˆ PQR = 1800 â€“ 650

We get,

âˆ PQR = 1150

PQRS is a parallelogram

âˆ S = âˆ Q = 1150 â€¦â€¦.(Opposite angles of a parallelogram are equal)

And, âˆ P + âˆ S = 1800 â€¦â€¦..(Interior angles)

âˆ P + 1150 = 1800

âˆ P = 1800 – 1150

We get,

âˆ P = 650

âˆ R = âˆ P = 650 â€¦â€¦.(Opposite angles of a parallelogram are equal)

Hence,

The remaining angles of a parallelogram are âˆ P = 650, âˆ Q = 1150, âˆ R = 650 and âˆ S = 1150

(d) PQRS is a parallelogram

âˆ R + âˆ Q = 1800 â€¦..(Interior angles)

x0 + (x0 / 4) = 1800

4x0 + x0 = 1800 x 4

5x0 = 1800 x 4

On further calculation, we get,

x0 = (1800 x 4) / 5

x0 = 36 x 4

We get,

x0 = 1440

âˆ R = 1440

(x0 / 4) = (1440 / 4)

(x0 / 4) = 360

âˆ Q = 360

Hence,

âˆ P = âˆ  R = 1440 and âˆ S = âˆ Q = 360 â€¦(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 â–³POS,

âˆ OSP + âˆ SPO + âˆ POS = 1800

x + 700 + 900 = 1800

x = 1800 – 1600

We get,

x = 200

In â–³QSP,

PS = PQ

âˆ QSP = âˆ PQS = x = 200

And,

âˆ QSP + âˆ PQS + âˆ SPQ = 1800

200 + 200 + âˆ SPQ = 1800

âˆ SPQ = 1800 – 400

We get,

âˆ SPQ = 1400

âˆ SRQ = 1400 â€¦â€¦.(Opposite angles of a parallelogram are equal)

Now,

âˆ SPQ + âˆ PSR = 1800

1400 + âˆ PSR = 1800

âˆ PSR = 400

âˆ PQR = 400 â€¦â€¦.(Opposite angles of a parallelogram are equal)

Therefore, âˆ P = âˆ R = 1400 and âˆ S = âˆ Q = 400

2. In a parallelogram ABCD âˆ C = 980. Find âˆ A and âˆ B.

Solution:

Given

ABCD is a parallelogram

âˆ C = 980

Hence,

âˆ A = âˆ C = 980 â€¦(Opposite angles of a parallelogram are equal)

Now,

âˆ A + âˆ B + âˆ C + âˆ D = 3600 â€¦..(Sum of all the angles of a quadrilateral = 3600)

980 + âˆ B + 980 + âˆ D = 3600

âˆ B + 1960 + âˆ D = 3600

âˆ B + âˆ D = 3600 – 1960

âˆ B + âˆ D = 1640

Here, âˆ B = âˆ D â€¦. (Opposite angles of a parallelogram are equal)

2âˆ B = 1640

We get,

âˆ B = âˆ D = 820

Hence, âˆ B = 820 and âˆ A = 980

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

Solution:

Let ABCD is a parallelogram in which AD || BC

âˆ A and âˆ B are consecutive angles

âˆ A: âˆ B = 3: 6

Hence,

âˆ A = 3x and âˆ B = 6x

AD || BC and AB is the transversal

âˆ A + âˆ B = 1800 (Co-interior angles are supplementary)

3x + 6x = 1800

9x = 1800

We get,

x = 200

Therefore, âˆ A = 3 x 200 = 600 and

âˆ B = 6 x 200 = 1200

We know that,

Opposite angles of a parallelogram are equal

Hence,

âˆ C = âˆ A = 600 and âˆ D = âˆ B = 1200

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

Solution:

In â–³BDC,

âˆ BDC + âˆ DCB + âˆ CBD = 1800

2a + 5a + 3a = 1800

10a = 1800

We get,

a = 180

âˆ BDC = 2a = 2 x 180 = 360

âˆ DCB = 5a = 5 x 180 = 900

âˆ CBD = 3a = 3 x 180 = 540

âˆ DAB = âˆ DCB = 900 â€¦(Opposite angles of a parallelogram are equal)

âˆ DBA = âˆ BDC = 360 â€¦ (alternate angles since AB || CD)

âˆ BDA = âˆ CBD = 540 â€¦.(alternate angles since AB || CD)

Hence, âˆ DAB = âˆ DCB = 900, âˆ DBA + âˆ CBD = 900, âˆ BDA + âˆ BDC = 900

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

Solution:

ABCD is a parallelogram

Opposite angles of a parallelogram are equal

Hence,

âˆ A = âˆ C

4x + 3y â€“ 6 = 9y + 2

4x â€“ 6y = 8

2x â€“ 3y = 4 â€¦â€¦..(1)

AB || CD and AD is the transversal

âˆ´ âˆ A + âˆ D = 1800 ………(Co â€“ interior angles are supplementary)

(4x + 3y â€“ 6) + (6x + 22) = 1800

10x + 3y + 16 = 1800

We get,

10x + 3y = 164 â€¦â€¦.(2)

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

12x = 168

x = 14

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

2(14) â€“ 3y = 4

28 â€“ 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.

Solution:

ABCD is a parallelogram

Let âˆ CAB = x0

Then,

âˆ ABC = 5x0 and âˆ BCA = 3x0

In â–³ABC,

âˆ CAB + âˆ ABC + âˆ BCA = 1800 â€¦.(Sum of angles of a triangle)

x 0 + 5x0 + 3x0 = 1800

9x0 = 1800

We get.

x0 = 200

âˆ CAB = x0 = 200

âˆ ABC = 5x0 = 5 x 200 = 1000

âˆ BCA = 3x0 = 3 x 200 = 600

âˆ ADC = âˆ ABC = 1000 (opposite angles of a parallelogram are equal)

âˆ ACD = âˆ CAB = 200 (alternate angles since BC || AD)

âˆ CAD = âˆ BCA = 600 (alternate angles since BC || AD)

Hence,

âˆ ADC = âˆ ABC = 1000, âˆ ACD + âˆ BCA = 800, âˆ CAD + âˆ CAB = 800

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

Solution:

PQRS is a parallelogram

Let âˆ RPQ = 3x0, âˆ PQR = 8x0 and âˆ QRP = 4x0

In â–³PQR,

âˆ RPQ + âˆ PQR + âˆ QRP = 1800 â€¦(Sum of angles of a triangle = 1800)

3x0 + 8x0 + 4x0 = 1800

15x0 = 1800

We get,

x0 = 120

âˆ RPQ = 3x0 = 3 x 120 = 360

âˆ PQR = 8x0 = 8 x 120 = 960

âˆ QRP = 4x0 = 4 x 120 = 480

âˆ PSR = âˆ PQR = 960 (Opposite angles of a parallelogram are equal)

âˆ RPS = âˆ QRP = 480 (Alternate angles since QR || PS)

âˆ PRS = âˆ RPQ = 360 (Alternate angles since QR || PS)

Hence, âˆ PSR = âˆ PQR = 960, âˆ RPS + âˆ RPQ = 840, âˆ PRS + âˆ QRP = 840

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

Prove that:

(i) QR = QT

(ii) RT bisects angle R

(iii) âˆ RTS = 900

Solution:

(i) âˆ PST = âˆ TSR â€¦â€¦(i)

âˆ PTS = âˆ TSR â€¦â€¦ (ii) {alternate angles since SR || PQ}

From (i) and (ii)

âˆ PST = âˆ 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

âˆ QTR = âˆ QRT (angles opposite to equal sides are equal)

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

âˆ QRT = âˆ TRS

Hence, RT bisects âˆ R

(iii) âˆ PST = âˆ TSR

âˆ QRT = âˆ TRS

âˆ QRS + âˆ PSR = 1800 (adjacent angles of a parallelogram are supplementary)

Now, multiplying by (1/2)

(1/2) âˆ QRS + (1/2) âˆ PSR = (1/2) x 1800

âˆ TRS + âˆ TSR = 900

In â–³STR,

âˆ TSR + âˆ RTS + âˆ TRS = 1800

âˆ TRS + âˆ TSR + âˆ RTS = 1800

900 + âˆ RTS = 1800

We get,

âˆ RTS = 1800 â€“ 900

âˆ RTS = 900

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 âˆ MOR and âˆ QSR.

Solution:

In â–³QOM,

âˆ OQM = 450 (In square, diagonals make 450 with the sides)

OQ = MQ

âˆ QOM = âˆ QMO â€¦(i) (angles opposite to equal sides are equal)

âˆ QOM + âˆ QMO + âˆ OQM = 1800

âˆ QOM + âˆ QOM + 450 = 1800

On further calculation, we get,

2âˆ QOM = 1800 – 450

âˆ QOM = 67.50

In â–³QOR,

âˆ QOR = 900 (In square diagonals bisect at right angles)

âˆ QOM + âˆ MOR = 900

67.50 + âˆ MOR = 900

âˆ MOR = 900 â€“ 67.50

We get,

âˆ MOR = 22.50

In â–³ROS,

âˆ OSR = 450 (In square diagonals make 450 with the sides)

Therefore, âˆ QSR = 450

10. ABCD is a rectangle with âˆ ADB = 550, calculate âˆ ABD

Solution:

In â–³ABD,

âˆ ADB = 550 (given)

âˆ DAB = 900 (In rectangle angle between two sides is 900)

âˆ ADB + âˆ DAB + âˆ ABD = 1800

550 + 900 + âˆ ABD = 1800

On calculating further, we get,

âˆ ABD = 1800 – 1450

âˆ ABD = 350

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

Solution:

Let ABCD be a parallelogram

In â–³ABC and â–³DCB,

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

BC = BC (Common)

AC = DB (Given)

Hence,

â–³ABC â‰… â–³DCB (By SSS congruence rule)

âˆ ABC = âˆ DCB

We know that the sum of the measures of angles on the same side of transversal is 1800

âˆ ABC + âˆ DCB = 1800

âˆ ABC + âˆ ABC = 1800

2âˆ ABC = 1800

We get,

âˆ ABC = 900

Since ABCD is a parallelogram and one of its interior angles is 900

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.

Solution

Given

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

In â–³PQS, A and D are mid-points of sides QP and PS respectively.

Hence,

AD || QS and AD = (1/2) QS â€¦â€¦â€¦.(i)

In â–³QRS

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

Hence,

BC || QS and BC = (1/2) QS â€¦â€¦â€¦(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

âˆ MDN = âˆ NOM

âˆ ADC = âˆ NOM

But, âˆ NOM = 900 (diagonals are perpendicular to each other)

âˆ ADC = 900

Clearly ABCD is a parallelogram having one of its interior angle as 900

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.

Solution:

In the given figure, join AC and BD

In â–³ABC,

P and Q are midpoints of AB and BC respectively

Hence,

PQ || AC and PQ = (1/2) AC â€¦â€¦.(i)

In â–³ADC,

S and R are midpoints of AD and DC respectively

Hence,

SR || AC and SR = (1/2) AC â€¦â€¦.(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.

Solution:

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 x OR)

MR = (1/2) OR

Thus, M is the midpoint of OR

In â–³ROS,

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

Hence,

TM || OS

TN || QS

Also,

In â–³RQS,

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.

Solution:

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)}

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

âˆ DNC = âˆ BNE (Vertically opposite angles)

Therefore,

â–³DNC â‰… â–³BNE (By ASA test)

DC = BE

In â–³ACE, 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 â€“ BE}

MN = (1/2) {AB â€“ CD} (Since BE = CD)

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

Hence proved