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.

Frank Solutions for Class 9 Maths Chapter 19 Quadrilaterals – Download the PDF

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

∠A = 75⁰

Then ∠C = 75⁰ ….. (Opposite angles of a parallelogram are equal)

Now,

∠A + ∠D = 180⁰ …. (Interior angles)

75⁰ + ∠D = 180⁰

∠D = 180⁰ – 75⁰

We get,

∠D = 105⁰

Hence, ∠B = ∠D = 105⁰ ……. (Opposite angles of a parallelogram are equal)

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

(b) Given

PQRS is a parallelogram

∠Q = 60⁰

Then ∠S = 60⁰ ….(Opposite angles of a parallelogram are equal)

Now, in △PQR,

∠RPQ + ∠PQR + ∠PRQ = 180⁰ ……. (Angle sum property of a triangle)

50⁰ + 60⁰ + ∠PRQ = 180⁰

110⁰ + ∠PRQ = 180⁰

We get,

∠PRQ = 70⁰

And, ∠SPR = ∠PRQ = 70⁰ …… (Alternate angles)

∠SPQ = ∠SPR + ∠RPQ

∠SPQ = 70⁰ + 50⁰

We get,

∠SPQ = 120⁰

Then, ∠SRQ = 120⁰ ……(Opposite angles of a parallelogram are equal)

Hence,

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

(c) ∠PQR + 65⁰ = 180⁰ …..(Linear pair angles)

∠PQR = 180⁰ – 65⁰

We get,

∠PQR = 115⁰

PQRS is a parallelogram

∠S = ∠Q = 115⁰ …….(Opposite angles of a parallelogram are equal)

And, ∠P + ∠S = 180⁰ ……..(Interior angles)

∠P + 115⁰ = 180⁰

∠P = 180⁰ – 115⁰

We get,

∠P = 65⁰

∠R = ∠P = 65⁰ …….(Opposite angles of a parallelogram are equal)

Hence,

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

(d) PQRS is a parallelogram

∠R + ∠Q = 180⁰ …..(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

∠R = 144⁰

(x⁰ / 4) = (144⁰ / 4)

(x⁰ / 4) = 36⁰

∠Q = 36⁰

Hence,

∠P = ∠ R = 144⁰ and ∠S = ∠Q = 36⁰ …(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 = 180⁰

x + 70⁰ + 90⁰ = 180⁰

x = 180⁰ – 160⁰

We get,

x = 20⁰

In △QSP,

PS = PQ

∠QSP = ∠PQS = x = 20⁰

And,

∠QSP + ∠PQS + ∠SPQ = 180⁰

20⁰ + 20⁰ + ∠SPQ = 180⁰

∠SPQ = 180⁰ – 40⁰

We get,

∠SPQ = 140⁰

∠SRQ = 140⁰ …….(Opposite angles of a parallelogram are equal)

Now,

∠SPQ + ∠PSR = 180⁰

140⁰ + ∠PSR = 180⁰

∠PSR = 40⁰

∠PQR = 40⁰ …….(Opposite angles of a parallelogram are equal)

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

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

Solution:

Given

ABCD is a parallelogram

∠C = 98⁰

Hence,

∠A = ∠C = 98⁰ …(Opposite angles of a parallelogram are equal)

Now,

∠A + ∠B + ∠C + ∠D = 360⁰ …..(Sum of all the angles of a quadrilateral = 360⁰)

98⁰ + ∠B + 98⁰ + ∠D = 360⁰

∠B + 196⁰ + ∠D = 360⁰

∠B + ∠D = 360⁰ – 196⁰

∠B + ∠D = 164⁰

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

2∠B = 164⁰

We get,

∠B = ∠D = 82⁰

Hence, ∠B = 82⁰ and ∠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

∠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 = 180⁰ (Co-interior angles are supplementary)

3x + 6x = 180⁰

9x = 180⁰

We get,

x = 20⁰

Therefore, ∠A = 3 x 20⁰ = 60⁰ and

∠B = 6 x 20⁰ = 120⁰

We know that,

Opposite angles of a parallelogram are equal

Hence,

∠C = ∠A = 60⁰ and ∠D = ∠B = 120⁰

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

Solution:

In △BDC,

∠BDC + ∠DCB + ∠CBD = 180⁰

2a + 5a + 3a = 180⁰

10a = 180⁰

We get,

a = 18⁰

∠BDC = 2a = 2 x 18⁰ = 36⁰

∠DCB = 5a = 5 x 18⁰ = 90⁰

∠CBD = 3a = 3 x 18⁰ = 54⁰

∠DAB = ∠DCB = 90⁰ …(Opposite angles of a parallelogram are equal)

∠DBA = ∠BDC = 36⁰ … (alternate angles since AB || CD)

∠BDA = ∠CBD = 54⁰ ….(alternate angles since AB || CD)

Hence, ∠DAB = ∠DCB = 90⁰, ∠DBA + ∠CBD = 90⁰, ∠BDA + ∠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,

∠A = ∠C

4x + 3y – 6 = 9y + 2

4x – 6y = 8

2x – 3y = 4 ……..(1)

AB || CD and AD is the transversal

∴ ∠A + ∠D = 180⁰ ………(Co – interior angles are supplementary)

(4x + 3y – 6) + (6x + 22) = 180⁰

10x + 3y + 16 = 180⁰

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.

Solution:

ABCD is a parallelogram

Let ∠CAB = x⁰

Then,

∠ABC = 5x⁰ and ∠BCA = 3x⁰

In △ABC,

∠CAB + ∠ABC + ∠BCA = 180⁰ ….(Sum of angles of a triangle)

x ⁰ + 5x⁰ + 3x⁰ = 180⁰

9x⁰ = 180⁰

We get.

x⁰ = 20⁰

∠CAB = x⁰ = 20⁰

∠ABC = 5x⁰ = 5 x 20⁰ = 100⁰

∠BCA = 3x⁰ = 3 x 20⁰ = 60⁰

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

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

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

Hence,

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

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.

Solution:

PQRS is a parallelogram

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

In △PQR,

∠RPQ + ∠PQR + ∠QRP = 180⁰ …(Sum of angles of a triangle = 180⁰)

3x⁰ + 8x⁰ + 4x⁰ = 180⁰

15x⁰ = 180⁰

We get,

x⁰ = 12⁰

∠RPQ = 3x⁰ = 3 x 12⁰ = 36⁰

∠PQR = 8x⁰ = 8 x 12⁰ = 96⁰

∠QRP = 4x⁰ = 4 x 12⁰ = 48⁰

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

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

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

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

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 = 90⁰

Solution:

(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 = 180⁰ (adjacent angles of a parallelogram are supplementary)

Now, multiplying by (1/2)

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

∠TRS + ∠TSR = 90⁰

In △STR,

∠TSR + ∠RTS + ∠TRS = 180⁰

∠TRS + ∠TSR + ∠RTS = 180⁰

90⁰ + ∠RTS = 180⁰

We get,

∠RTS = 180⁰ – 90⁰

∠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 ∠MOR and ∠QSR.

Solution:

In △QOM,

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

OQ = MQ

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

∠QOM + ∠QMO + ∠OQM = 180⁰

∠QOM + ∠QOM + 45⁰ = 180⁰

On further calculation, we get,

2∠QOM = 180⁰ – 45⁰

∠QOM = 67.5⁰

In △QOR,

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

∠QOM + ∠MOR = 90⁰

67.5⁰ + ∠MOR = 90⁰

∠MOR = 90⁰ – 67.5⁰

We get,

∠MOR = 22.5⁰

In △ROS,

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

Therefore, ∠QSR = 45⁰

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

Solution:

In △ABD,

∠ADB = 55⁰ (given)

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

∠ADB + ∠DAB + ∠ABD = 180⁰

55⁰ + 90⁰ + ∠ABD = 180⁰

On calculating further, we get,

∠ABD = 180⁰ – 145⁰

∠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 △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 180⁰

∠ABC + ∠DCB = 180⁰

∠ABC + ∠ABC = 180⁰

2∠ABC = 180⁰

We get,

∠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 △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 = 90⁰ (diagonals are perpendicular to each other)

∠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 △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.

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 △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.

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)}

∠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) x difference of parallel sides AB and CD

Hence proved