**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

âˆ A = 75^{0}

Then âˆ C = 75^{0} â€¦.. (Opposite angles of a parallelogram are equal)

Now,

âˆ A + âˆ D = 180^{0} â€¦. (Interior angles)

75^{0} + âˆ D = 180^{0}

âˆ D = 180^{0} – 75^{0}

We get,

âˆ D = 105^{0}

Hence, âˆ B = âˆ D = 105^{0} â€¦â€¦. (Opposite angles of a parallelogram are equal)

Therefore, the remaining angles of the parallelogram are âˆ B = 105^{0}, âˆ C = 75^{0} and âˆ D = 105^{0}

(b) Given

PQRS is a parallelogram

âˆ Q = 60^{0}

Then âˆ S = 60^{0} â€¦.(Opposite angles of a parallelogram are equal)

Now, in â–³PQR,

âˆ RPQ + âˆ PQR + âˆ PRQ = 180^{0} â€¦â€¦. (Angle sum property of a triangle)

50^{0} + 60^{0} + âˆ PRQ = 180^{0}

110^{0} + âˆ PRQ = 180^{0}

We get,

âˆ PRQ = 70^{0}

And, âˆ SPR = âˆ PRQ = 70^{0} â€¦â€¦ (Alternate angles)

âˆ SPQ = âˆ SPR + âˆ RPQ

âˆ SPQ = 70^{0} + 50^{0}

We get,

âˆ SPQ = 120^{0}

Then, âˆ SRQ = 120^{0} â€¦â€¦(Opposite angles of a parallelogram are equal)

Hence,

The remaining angles of a parallelogram are âˆ P = 120^{0}, âˆ S = 60^{0} and âˆ R = 120^{0}

(c) âˆ PQR + 65^{0} = 180^{0} â€¦..(Linear pair angles)

âˆ PQR = 180^{0} â€“ 65^{0}

We get,

âˆ PQR = 115^{0}

PQRS is a parallelogram

âˆ S = âˆ Q = 115^{0} â€¦â€¦.(Opposite angles of a parallelogram are equal)

And, âˆ P + âˆ S = 180^{0} â€¦â€¦..(Interior angles)

âˆ P + 115^{0} = 180^{0}

âˆ P = 180^{0} – 115^{0}

We get,

âˆ P = 65^{0}

âˆ R = âˆ P = 65^{0} â€¦â€¦.(Opposite angles of a parallelogram are equal)

Hence,

The remaining angles of a parallelogram are âˆ P = 65^{0}, âˆ Q = 115^{0}, âˆ R = 65^{0} and âˆ S = 115^{0}

(d) PQRS is a parallelogram

âˆ R + âˆ Q = 180^{0} â€¦..(Interior angles)

x^{0} + (x^{0} / 4) = 180^{0}

4x^{0} + x^{0} = 180^{0} x 4

5x^{0} = 180^{0} x 4

On further calculation, we get,

x^{0} = (180^{0} x 4) / 5

x^{0} = 36 x 4

We get,

x^{0} = 144^{0}

âˆ R = 144^{0}

(x^{0} / 4) = (144^{0} / 4)

(x^{0 }/ 4) = 36^{0}

âˆ Q = 36^{0}

Hence,

âˆ P = âˆ R = 144^{0} and âˆ S = âˆ Q = 36^{0} 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 = 180^{0}

x + 70^{0} + 90^{0} = 180^{0}

x = 180^{0} – 160^{0}

We get,

x = 20^{0}

In â–³QSP,

PS = PQ

âˆ QSP = âˆ PQS = x = 20^{0}

And,

âˆ QSP + âˆ PQS + âˆ SPQ = 180^{0}

20^{0} + 20^{0} + âˆ SPQ = 180^{0}

âˆ SPQ = 180^{0} – 40^{0}

We get,

âˆ SPQ = 140^{0}

âˆ SRQ = 140^{0} â€¦â€¦.(Opposite angles of a parallelogram are equal)

Now,

âˆ SPQ + âˆ PSR = 180^{0}

140^{0} + âˆ PSR = 180^{0}

âˆ PSR = 40^{0}

âˆ PQR = 40^{0} â€¦â€¦.(Opposite angles of a parallelogram are equal)

Therefore, âˆ P = âˆ R = 140^{0} and âˆ S = âˆ Q = 40^{0}

**2. In a parallelogram ABCD âˆ C = 98 ^{0}. Find âˆ A and âˆ B.**

**Solution:**

Given

ABCD is a parallelogram

âˆ C = 98^{0}

Hence,

âˆ A = âˆ C = 98^{0} â€¦(Opposite angles of a parallelogram are equal)

Now,

âˆ A + âˆ B + âˆ C + âˆ D = 360^{0} â€¦..(Sum of all the angles of a quadrilateral = 360^{0})

98^{0} + âˆ B + 98^{0} + âˆ D = 360^{0}

âˆ B + 196^{0} + âˆ D = 360^{0}

âˆ B + âˆ D = 360^{0} – 196^{0}

âˆ B + âˆ D = 164^{0}

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

2âˆ B = 164^{0}

We get,

âˆ B = âˆ D = 82^{0}

Hence, âˆ B = 82^{0} and âˆ A = 98^{0}

**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 = 180^{0} (Co-interior angles are supplementary)

3x + 6x = 180^{0}

9x = 180^{0}

We get,

x = 20^{0}

Therefore, âˆ A = 3 x 20^{0} = 60^{0} and

âˆ B = 6 x 20^{0} = 120^{0}

We know that,

Opposite angles of a parallelogram are equal

Hence,

âˆ C = âˆ A = 60^{0} and âˆ D = âˆ B = 120^{0}

**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 = 180^{0}

2a + 5a + 3a = 180^{0}

10a = 180^{0}

We get,

a = 18^{0}

âˆ BDC = 2a = 2 x 18^{0} = 36^{0}

âˆ DCB = 5a = 5 x 18^{0} = 90^{0}

âˆ CBD = 3a = 3 x 18^{0} = 54^{0}

âˆ DAB = âˆ DCB = 90^{0} â€¦(Opposite angles of a parallelogram are equal)

âˆ DBA = âˆ BDC = 36^{0} â€¦ (alternate angles since AB || CD)

âˆ BDA = âˆ CBD = 54^{0} â€¦.(alternate angles since AB || CD)

Hence, âˆ DAB = âˆ DCB = 90^{0}, âˆ DBA + âˆ CBD = 90^{0}, âˆ BDA + âˆ BDC = 90^{0}

**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 = 180^{0 }………(Co â€“ interior angles are supplementary)

(4x + 3y â€“ 6) + (6x + 22) = 180^{0}

10x + 3y + 16 = 180^{0}

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 = x^{0}

Then,

âˆ ABC = 5x^{0} and âˆ BCA = 3x^{0}

In â–³ABC,

âˆ CAB + âˆ ABC + âˆ BCA = 180^{0} â€¦.(Sum of angles of a triangle)

x ^{0} + 5x^{0} + 3x^{0} = 180^{0}

9x^{0} = 180^{0}

We get.

x^{0} = 20^{0}

âˆ CAB = x^{0} = 20^{0}

âˆ ABC = 5x^{0} = 5 x 20^{0} = 100^{0}

âˆ BCA = 3x^{0} = 3 x 20^{0} = 60^{0}

âˆ ADC = âˆ ABC = 100^{0} (opposite angles of a parallelogram are equal)

âˆ ACD = âˆ CAB = 20^{0} (alternate angles since BC || AD)

âˆ CAD = âˆ BCA = 60^{0} (alternate angles since BC || AD)

Hence,

âˆ ADC = âˆ ABC = 100^{0}, âˆ ACD + âˆ BCA = 80^{0}, âˆ CAD + âˆ CAB = 80^{0}

**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 = 3x^{0}, âˆ PQR = 8x^{0} and âˆ QRP = 4x^{0}

In â–³PQR,

âˆ RPQ + âˆ PQR + âˆ QRP = 180^{0} â€¦(Sum of angles of a triangle = 180^{0})

3x^{0} + 8x^{0} + 4x^{0} = 180^{0}

15x^{0} = 180^{0}

We get,

x^{0} = 12^{0}

âˆ RPQ = 3x^{0} = 3 x 12^{0} = 36^{0}

âˆ PQR = 8x^{0} = 8 x 12^{0} = 96^{0}

âˆ QRP = 4x^{0} = 4 x 12^{0} = 48^{0}

âˆ PSR = âˆ PQR = 96^{0} (Opposite angles of a parallelogram are equal)

âˆ RPS = âˆ QRP = 48^{0} (Alternate angles since QR || PS)

âˆ PRS = âˆ RPQ = 36^{0} (Alternate angles since QR || PS)

Hence, âˆ PSR = âˆ PQR = 96^{0}, âˆ RPS + âˆ RPQ = 84^{0}, âˆ PRS + âˆ QRP = 84^{0}

**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 = 90 ^{0}**

**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 = 180^{0} (adjacent angles of a parallelogram are supplementary)

Now, multiplying by (1/2)

(1/2) âˆ QRS + (1/2) âˆ PSR = (1/2) x 180^{0}

âˆ TRS + âˆ TSR = 90^{0}

In â–³STR,

âˆ TSR + âˆ RTS + âˆ TRS = 180^{0}

âˆ TRS + âˆ TSR + âˆ RTS = 180^{0}

90^{0} + âˆ RTS = 180^{0}

We get,

âˆ RTS = 180^{0} â€“ 90^{0}

âˆ RTS = 90^{0}

**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 = 45^{0} (In square, diagonals make 45^{0} with the sides)

OQ = MQ

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

âˆ QOM + âˆ QMO + âˆ OQM = 180^{0}

âˆ QOM + âˆ QOM + 45^{0} = 180^{0}

On further calculation, we get,

2âˆ QOM = 180^{0} – 45^{0}

âˆ QOM = 67.5^{0}

In â–³QOR,

âˆ QOR = 90^{0} (In square diagonals bisect at right angles)

âˆ QOM + âˆ MOR = 90^{0}

67.5^{0} + âˆ MOR = 90^{0}

âˆ MOR = 90^{0} â€“ 67.5^{0}

We get,

âˆ MOR = 22.5^{0}

In â–³ROS,

âˆ OSR = 45^{0} (In square diagonals make 45^{0} with the sides)

Therefore, âˆ QSR = 45^{0}

**10. ABCD is a rectangle with âˆ ADB = 55 ^{0}, calculate âˆ ABD**

**Solution:**

In â–³ABD,

âˆ ADB = 55^{0} (given)

âˆ DAB = 90^{0} (In rectangle angle between two sides is 90^{0})

âˆ ADB + âˆ DAB + âˆ ABD = 180^{0}

55^{0} + 90^{0} + âˆ ABD = 180^{0}

On calculating further, we get,

âˆ ABD = 180^{0} – 145^{0}

âˆ ABD = 35^{0}

**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 180^{0}

âˆ ABC + âˆ DCB = 180^{0}

âˆ ABC + âˆ ABC = 180^{0}

2âˆ ABC = 180^{0}

We get,

âˆ ABC = 90^{0}

Since ABCD is a parallelogram and one of its interior angles is 90^{0}

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 = 90^{0} (diagonals are perpendicular to each other)

âˆ ADC = 90^{0}

Clearly ABCD is a parallelogram having one of its interior angle as 90^{0}

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

## Comments