## Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals

**Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals**

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### Special Types of Quadrilaterals Exercise 17 – Selina Concise Mathematics Class 8 ICSE Solutions

**Question 1.**

In parallelogram ABCD, ∠A = 3 times ∠B. Find all the angles of the parallelogram. In the same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD.

**Solution:**

Let ∠B = x

∠A = 3 ∠B = 3x

AD||BC

∠A + ∠B = 180°

3x + x = 180°

⇒ 4x = 180°

⇒ x = 45°

∠B = 45°

∠A = 3x = 3 x 45 = 135°

and ∠B = ∠D = 45°

opposite angles of || gm are equal.

∠A = ∠C = 135°

opposite sides of //gm are equal.

AB = CD

5x – 7 = 3x + 1

⇒ 5x – 3x = 1+7

⇒ 2x = 8

⇒ x = 4

CD = 3 x 4+1 = 13

Hence 135°, 45°, 135° and 45° ; 13

**Question 2.**

In parallelogram PQRS, ∠Q = (4x – 5)° and ∠S = (3x + 10)°. Calculate : ∠Q and ∠R.

**Solution:**

In parallelogram PQRS,

∠Q = (4x – 5)° and ∠S = (3x + 10)°

opposite ∠s of //gm are equal.

∠Q = ∠S

4x – 5 = 3x + 10

4x – 3x = 10+5

x = 15

∠Q = 4x – 5 =4 x 15 – 5 = 55°

Also ∠Q + ∠R = 180°

55° + ∠R = 180°

∠R = 180°-55° = 125°

∠Q = 55° ; ∠R = 125°

**Question 3.**

In rhombus ABCD ;

(i) if ∠A = 74° ; find ∠B and ∠C.

(ii) if AD = 7.5 cm ; find BC and CD.

**Solution:**

AD || BC

∠A + ∠B = 180°

74° + ∠B = 180°

∠B =180° – 74°= 106°

opposite angles of Rhombus are equal.

∠A = ∠C = 74°

Sides of Rhombus are equal.

BC = CD = AD = 7.5 cm

(i) ∠B = 106° ; ∠C = 74°

(ii) BC = 7.5 cm and CD = 7.5 cm Ans.

**Question 4.**

In square PQRS :

(i) if PQ = 3x – 7 and QR = x + 3 ; find PS

(ii) if PR = 5x and QR = 9x – 8. Find QS

**Solution:**

(i) sides of square are equal.

PQ = QR

=> 3x – 7 = x + 3

=> 3x – x = 3 + 7

=> 2x = 10

x = 5

PS=PQ = 3x – 7 = 3 x 5 – 7 =8

(ii) PR = 5x and QS = 9x – 8

As diagonals of square are equal.

PR = QS

5x = 9x – 8

=> 5x – 9x = -8

=> -4x = -8

=> x = 2

QS = 9x – 8 = 9 x 2 – 8 =10

**Question 5.**

ABCD is a rectangle, if ∠BPC = 124°

Calculate : (i) ∠BAP (ii) ∠ADP

**Solution:**

Diagonals of rectangle are equal and bisect each other.

∠PBC = ∠PCB = x (say)

But ∠BPC + ∠PBC + ∠PCB = 180°

124° + x + x = 180°

2x = 180° – 124°

2x = 56°

=> x = 28°

∠PBC = 28°

But ∠PBC = ∠ADP [Alternate ∠s]

∠ADP = 28°

Again ∠APB = 180° – 124° = 56°

Also PA = PB

∠BAP = \(\frac { 1 }{ 2 }\) (180° – ∠APB)

= \(\frac { 1 }{ 2 }\) x (180°- 56°) = \(\frac { 1 }{ 2 }\) x 124° = 62°

Hence (i) ∠BAP = 62° (ii) ∠ADP =28°

**Question 6.**

ABCD is a rhombus. If ∠BAC = 38°, find :

(i) ∠ACB

(ii) ∠DAC

(iii) ∠ADC.

**Solution:**

ABCD is Rhombus (Given)

AB = BC

∠BAC = ∠ACB (∠s opp. to equal sides)

But ∠BAC = 38° (Given)

∠ACB = 38°

In ∆ABC,

∠ABC + ∠BAC + ∠ACB = 180°

∠ABC + 38°+ 38° = 180°

∠ABC = 180° – 76° = 104°

But ∠ABC = ∠ADC (opp. ∠s of rhombus)

∠ADC = 104°

∠DAC = ∠DCA ( AD = CD)

∠DAC = \(\frac { 1 }{ 2 }\) [180° – 104°]

∠DAC = \(\frac { 1 }{ 2 }\) x 76° = 38°

Hence (i) ∠ACB = 38° (ii) ∠DAC = 38° (iii) ∠ADC = 104° Ans.

**Question 7.**

ABCD is a rhombus. If ∠BCA = 35°. find ∠ADC.

**Solution:**

**Given :** Rhombus ABCD in which ∠BCA = 35°

**To find :** ∠ADC

**Proof :** AD || BC

∠DAC = ∠BCA (Alternate ∠s)

But ∠BCA = 35° (Given)

∠DAC = 35°

But ∠DAC = ∠ACD ( AD = CD) & ∠DAC +∠ACD + ∠ADC = 180°

35°+ 35° + ∠ADC = 180°

∠ADC = 180° – 70° = 110°

Hence ∠ADC = 110°

**Question 8.**

PQRS is a parallelogram whose diagonals intersect at M.

If ∠PMS = 54°, ∠QSR = 25° and ∠SQR = 30° ; find :

(i) ∠RPS

(ii) ∠PRS

(iii) ∠PSR.

**Solution:**

**Given :** ||gm PQRS in which diagonals PR & QS intersect at M.

∠PMS = 54° ; ∠QSR = 25° and ∠SQR=30°

**To find :** (i) ∠RPS (ii) ∠PRS (iii) ∠PSR

**Proof :** QR || PS

=> ∠PSQ = ∠SQR (Alternate ∠s)

But ∠SQR = 30° (Given)

∠PSQ = 30°

In ∆SMP,

∠PMS + ∠ PSM +∠MPS = 180° or 54° + 30° + ∠RPS = 180°

∠RPS = 180°- 84° = 96°

Now ∠PRS + ∠RSQ = ∠PMS

∠PRS + 25° =54°

∠PRS = 54° – 25° = 29°

∠PSR = ∠PSQ + ∠RSQ = 30°+25° = 55°

Hence (i) ∠RPS = 96° (ii) ∠PRS = 29° (iii) ∠PSR = 55°

**Question 9.**

**Given :** Parallelogram ABCD in which diagonals AC and BD intersect at M.

**Prove :** M is mid-point of LN.

**Solution:**

**Proof :** Diagonals of //gm bisect each other.

MD = MB

Also ∠ADB = ∠DBN (Alternate ∠s)

& ∠DML = ∠BMN (Vert. opp. ∠s)

∆DML = ∆BMN

LM = MN

M is mid-point of LN.

Hence proved.

**Question 10.**

In an Isosceles-trapezium, show that the opposite angles are supplementary.

**Solution:**

**Given :** ABCD is isosceles trapezium in which AD = BC

**To Prove :** (i) ∠A + ∠C = 180°

(ii) ∠B + ∠D = 180°

**Proof :** AB || CD.

=> ∠A + ∠D = 180°

But ∠A = ∠B [Trapezium is isosceles)]

∠B + ∠D = 180°

Similarly ∠A + ∠C = 180°

Hence the result.

**Question 11.**

ABCD is a parallelogram. What kind of quadrilateral is it if :

(i) AC = BD and AC is perpendicular to BD?

(ii) AC is perpendicular to BD but is not equal to it ?

(iii) AC = BD but AC is not perpendicular to BD ?

**Solution:**

**Question 12.**

Prove that the diagonals of a parallelogram bisect each other.

**Solution:**

**Given :** ||gm ABCD in which diagonals AC and BD bisect each other.

**To Prove :** OA = OC and OB = OD

**Proof :** AB || CD (Given)

∠1 = ∠2 (alternate ∠s)

∠3 = ∠4 = (alternate ∠s)

and AB = CD (opposite sides of //gm)

∆COD = ∆AOB (A.S.A. rule)

OA = OC and OB = OD

Hence the result.

**Question 13.**

If the diagonals of a parallelogram are of equal lengths, the parallelogram is a rectangle. Prove it.

**Solution:**

**Given :** //gm ABCD in which AC = BD

**To Prove :** ABCD is rectangle.

**Proof :** In ∆ABC and ∆ABD

AB = AB (Common)

AC = BD (Given)

BC = AD (opposite sides of ||gm)

∆ABC = ∆ABD (S.S.S. Rule)

∠A = ∠B

But AD // BC (opp. sides of ||gm are ||)

∠A + ∠B = 180°

∠A = ∠B = 90°

Similarly ∠D = ∠C = 90°

Hence ABCD is a rectangle.

**Question 14.**

In parallelogram ABCD, E is the mid-point of AD and F is the mid-point of BC. Prove that BFDE is a parallelogram.

**Solution:**

**Given :** //gm ABCD in which E and F are mid-points of AD and BC respectively.

**To Prove :** BFDE is a ||gm.

**Proof :** E is mid-point of AD. (Given)

DE = \(\frac { 1 }{ 2 }\) AD

Also F is mid-point of BC (Given)

BF = \(\frac { 1 }{ 2 }\) BC

But AD = BC (opp. sides of ||gm)

BF = DE

Again AD || BC

=> DE || BF

Now DE || BF and DE = BF

Hence BFDE is a ||gm.

**Question 15.**

In parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD. Prove that :

(i) AE = AD,

(ii) DE bisects and ∠ADC and

(iii) Angle DEC is a right angle.

**Solution:**

**Given :** ||gm ABCD in which E is mid-point of AB and CE bisects ZBCD.

**To Prove :** (i) AE = AD

(ii) DE bisects ∠ADC

(iii) ∠DEC = 90°

**Const.** Join DE

**Proof :** (i) AB || CD (Given)

and CE bisects it.

∠1 = ∠3 (alternate ∠s) ……… (i)

But ∠1 = ∠2 (Given) …………. (ii)

From (i) & (ii)

∠2 = ∠3

BC = BE (sides opp. to equal angles)

But BC = AD (opp. sides of ||gm)

and BE = AE (Given)

AD = AE

∠4 = ∠5 (∠s opp. to equal sides)

But ∠5 = ∠6 (alternate ∠s)

=> ∠4 = ∠6

DE bisects ∠ADC.

Now AD // BC

=> ∠D + ∠C = 180°

2∠6+2∠1 = 180°

DE and CE are bisectors.

∠6 + ∠1 = \(\frac { { 180 }^{ 0 } }{ 2 }\)

∠6 + ∠1 = 90°

But ∠DEC + ∠6 + ∠1 = 180°

∠DEC + 90° = 180°

∠DEC = 180° – 90°

∠DEC = 90°

Hence the result.

**Question 16.**

In the following diagram, the bisectors of interior angles of the parallelogram PQRS enclose a quadrilateral ABCD.

Show that:

(i) ∠PSB + ∠SPB = 90°

(ii) ∠PBS = 90°

(iii) ∠ABC = 90°

(iv) ∠ADC = 90°

(v) ∠A = 90°

(vi) ABCD is a rectangle

Thus, the bisectors of the angles of a parallelogram enclose a rectangle.

**Solution:**

**Given :** In parallelogram ABCD bisector of angles P and Q, meet at A, bisectors of ∠R and ∠S meet at C. Forming a quadrilateral ABCD as shown in the figure.

**To prove :**

(i) ∠PSB + ∠SPB = 90°

(ii) ∠PBS = 90°

(iii) ∠ABC = 90°

(iv) ∠ADC = 90°

(v) ∠A = 9°

(vi) ABCD is a rectangle

**Proof :** In parallelogram PQRS,

PS || QR (opposite sides)

∠P +∠Q = 180°

and AP and AQ are the bisectors of consecutive angles ∠P and ∠Q of the parallelogram

∠APQ + ∠AQP = \(\frac { 1 }{ 2 }\) x 180° = 90°

But in ∆APQ,

∠A + ∠APQ + ∠AQP = 180° (Angles of a triangle)

∠A + 90° = 180°

∠A = 180° – 90°

(v) ∠A = 90°

Similarly PQ || SR

∠PSB + SPB = 90°

(ii) and ∠PBS = 90°

But, ∠ABC = ∠PBS (Vertically opposite angles)

(iii) ∠ABC = 90°

Similarly we can prove that

(iv) ∠ADC = 90° and ∠C = 90°

(vi) ABCD is a rectangle (Each angle of a quadrilateral is 90°)

Hence proved.

**Question 17.**

In parallelogram ABCD, X and Y are midpoints of opposite sides AB and DC respectively. Prove that:

(i) AX = YC

(ii) AX is parallel to YC

(iii) AXCY is a parallelogram.

**Solution:**

**Given :** In parallelogram ABCD, X and Y are the mid-points of sides AB and DC respectively AY and CX are joined

**To prove :**

(i) AX = YC

(ii) AX is parallel to YC

(iii) AXCY is a parallelogram

**Proof :** AB || DC and X and Y are the mid-points of the sides AB and DC respectively

(i) AX = YC ( \(\frac { 1 }{ 2 }\) of opposite sides of a parallelogram)

(ii) and AX || YC

(iii) AXCY is a parallelogram (A pair of opposite sides are equal and parallel)

Hence proved.

**Question 18.**

The given figure shows parallelogram ABCD. Points M and N lie in diagonal BD such that DM = BN.

Prove that:

(i) ∆DMC = ∆BNA and so CM = AN

(ii) ∆AMD = ∆CNB and so AM CN

(iii) ANCM is a parallelogram.

**Solution:**

**Given :** In parallelogram ABCD, points M and N lie on the diagonal BD such that DM = BN

AN, NC, CM and MA are joined

**To prove :**

(i) ∆DMC = ∆BNA and so CM = AN

(ii) ∆AMD = ∆CNB and so AM = CN

(iii) ANCM is a parallelogram

**Proof :**

(i) In ∆DMC and ∆BNA.

CD = AB (opposite sides of ||gm ABCD)

DM = BN (given)

∠CDM = ∠ABN (alternate angles)

∆DMC = ∆BNA (SAS axiom)

CM =AN (c.p.c.t.)

Similarly, in ∆AMD and ∆CNB

AD = BC (opposite sides of ||gm)

DM = BN (given)

∠ADM = ∠CBN – (alternate angles)

∆AMD = ∆CNB (SAS axiom)

AM = CN (c.p.c.t.)

(iii) CM = AN and AM = CN (proved)

ANCM is a parallelogram (opposite sides are equal)

Hence proved.

**Question 19.**

The given figure shows a rhombus ABCD in which angle BCD = 80°. Find angles x and y.

**Solution:**

In rhombus ABCD, diagonals AC and BD bisect each other at 90°

∠BCD = 80°

Diagonals bisect the opposite angles also ∠BCD = ∠BAD (Opposite angles of rhombus)

∠BAD = 80° and ∠ABC = ∠ADC = 180° – 80° = 100°

Diagonals bisect opposite angles

∠OCB or ∠PCB = \(\frac { { 80 }^{ 0 } }{ 2 }\) = 40°

In ∆PCM,

Ext. CPD = ∠OCB + ∠PMC

110° = 40° + x

=> x = 110° – 40° = 70°

and ∠ADO = \(\frac { 1 }{ 2 }\) ∠ADC = \(\frac { 1 }{ 2 }\) x 100° = 50°

Hence x = 70° and y = 50°

**Question 20.**

Use the information given in the alongside diagram to find the value of x, y and z.

**Solution:**

ABCD is a parallelogram and AC is its diagonal which bisects the opposite angle

Opposite sides of a parallelogram are equal

3x + 14 = 2x + 25

=> 3x – 2x = 25 – 14

=> x = 11

∴ x = 11 cm

∠DCA = ∠CAB (Alternate angles)

y + 9° = 24

y = 24° – 9° = 15°

∠DAB = 3y° + 5° + 24° = 3 x 15 + 5 + 24° = 50° + 24° = 74°

∠ABC =180°- ∠DAB = 180° – 74° = 106°

z = 106°

Hence x = 11 cm, y = 15°, z = 106°

**Question 21.**

The following figure is a rectangle in which x : y = 3 : 7; find the values of x and y.

**Solution:**

ABCD is a rectangle,

x : y = 3 : 1

In ∆BCE, ∠B = 90°

x + y = 90°

But x : y = 3 : 7

Sum of ratios = 3 + 7=10

Hence x = 27°, y = 63°

**Question 22.**

In the given figure, AB // EC, AB = AC and AE bisects ∠DAC. Prove that:

(i) ∠EAC = ∠ACB

(ii) ABCE is a parallelogram.

**Solution:**

ABCE is a quadrilateral in which AC is its diagonal and AB || EC, AB = AC

BA is produced to D

AE bisects ∠DAC

**To prove:**

(i) ∠EAC = ∠ACB

(ii) ABCE is a parallelogram

**Proof:**

(i) In ∆ABC and ∆ZAEC

AC=AC (common)

AB = CE (given)

∠BAC = ∠ACE (Alternate angle)

∆ABC = ∆AEC (SAS Axiom)

(ii) ∠BCA = ∠CAE (c.p.c.t.)

But these are alternate angles

AE || BC

But AB || EC (given)

∴ ABCD is a parallelogram