RS Aggarwal Solutions for Class 9 Maths Chapter 10 … · 2020. 11. 2. · RS Aggarwal Solutions...

RS Aggarwal Solutions for Class 9 Maths Chapter 10

Quadrilaterals

Exercise 10(B) 1.

Solution:

We know that in a parallelogram opposite angles are equal

So we get

∠ A = ∠ C = 72o

We that that in a parallelogram the sum of all the angles is 360o

It can be written as

∠ A + ∠ B + ∠ C + ∠ D = 360o

By substituting the values in the above equation

72o + ∠ B + 72o + ∠ D = 360o

We know that ∠ A = ∠ C and ∠ B = ∠ D

So we can write it as

2 ∠ B + 144o = 360o

On further calculation

2 ∠ B = 360o – 144o

By subtraction

2 ∠ B = 216o

By division

∠ B = 108o

Therefore, ∠ B = 108o, ∠ C = 72o and ∠ D = 108o.

2.

Solution:

It is given that ABCD is a parallelogram in which ∠ DAB = 80o and ∠ DBC = 60o

We know that opposite angles are equal in parallelogram

So we get

∠ C = ∠ A = 80o

From the figure we know that AD || BC and BD is a transversal

We know that ∠ ADB and ∠ DBC are alternate angles

So we get

∠ ADB = ∠ DBC = 60o

Consider △ ABD

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Quadrilaterals

Using the sum property of triangle

∠ A + ∠ ADB + ∠ ABD = 180o

By substituting values in the above equation

80o + 60o + ∠ ABD = 180o


∠ ABD = 180o – 80o – 60o

By subtraction

∠ ABD = 180o – 140o

So we get

∠ ABD = 40o


∠ ABC = ∠ ABD + ∠ DBC

By substituting values we get

∠ ABC = 40o + 60o

By addition

∠ ABC = 100o

We know that the opposite angles are equal in a parallelogram

∠ ADC = ∠ ABC = 100o

We get

∠ ADC = ∠ CDB + ∠ ADB


∠ CDB = ∠ ADC – ∠ ADB

By substituting values

∠ CDB = 100o – 60o

By subtraction

∠ CDB = 40o

Therefore, ∠ ADB = 60o and ∠ CDB = 40o.

3.

Solution:

It is given that ABCD is a parallelogram

So we know that AD || BC

From the figure we know that ∠ DAM and ∠ AMB are alternate angles



Quadrilaterals

So we get

∠ DAM = ∠ AMB

We know that ∠ BAM = ∠ DAM


∠ BAM = ∠ AMB

From the figure we know that the sides opposite to equal angles are equal

So we get

BM = AB

We know that the opposite sides of a parallelogram are equal

AB = CD

So we can write it as

BM = AB = CD ……. (1)

We know that M is the midpoint of the line BC

So we get

BM = ½ BC

We know that BC = AD

We get

BM = ½ AD

Based on equation (1)

CD = ½ AD

By cross multiplication

AD = 2CD.

Therefore, it is proved that AD = 2CD.

4.

Solution:

(i) We know that opposite angles are equal in a parallelogram.

So we get

∠ C = ∠ A = 60o

We know that the sum of all the angles in a parallelogram is 360o


∠ A + ∠ B + ∠ C + ∠ D = 360o

So we get

∠ B + ∠ D = 360o – (∠ A + ∠ C)



Quadrilaterals

By substituting values in the above equation we get

∠ B + ∠ D = 360o – (60o + 60o)

On further calculation we get

∠ B + ∠ D = 360o – 120o

By subtraction

∠ B + ∠ D = 240o

We know that ∠ B = ∠ B

So the above equation becomes

∠ B + ∠ B = 240o

2 ∠ B = 240o

By division

∠ B = ∠ D = 120o

We know that AB || DP and AP is a transversal

From the figure we know that ∠ APD and ∠ PAD are alternate angles

∠ APD = ∠ PAD = 60o/2

∠ APD = ∠ PAD = 30o ……. (1)

We know that AB || PC and BP is a transversal

So we get

∠ ABP = ∠ CPB = ∠ B/2

i.e. ∠ ABP = ∠ CPB = 120o/2

We get ∠ ABP = ∠ CPB = 60o …….. (2)

We know that DPC is a straight line


∠ APD + ∠ APB + ∠ CPB = 180o

By substituting the values we get

30o + ∠ APB + 60o = 180o


∠ APB = 180o – 30o – 60o

By subtraction

∠ APB = 180o – 90o

∠ APB = 90o

Therefore, it is proved that ∠ APB = 90o



Quadrilaterals

(ii) From equation (1) we know that

∠ APD = 30o

We know that

∠ DAP = 60o/2

By division

∠ DAP = 30o

So we get

∠ APD = ∠ DAP ……… (3)

From the figure we know that the sides of an isosceles triangle are equal

So we get

DP = AD

We know that

∠ CPB = 60o and ∠ C = 60o

By sum property of triangle

We get

∠ C + ∠ CPB + ∠ PBC = 180o


60o + 60o + ∠ PBC = 180o

On further calculation we get

∠ PBC = 180o – 60o – 60o

By subtraction

∠ PBC = 180o – 120o

∠ PBC = 60o

We know that all the sides of an equilateral triangle are equal

So we get

PB = PC = BC ………. (4)

Therefore, it is proved that AD = AP and PB = PC = BC.

(iii) We know that ∠ DPA = ∠ PAD based on equation (3)

We also know that all the sides are equal in an isosceles triangle

DP = AD

From the figure we know that the opposite sides are equal

So we get



Quadrilaterals

DP = BC

Considering equation (4)

DP = PC

From the figure we know that DP = PC and P is the midpoint of the line DC

So we get

DP = ½ DC

By cross multiplication we get

DC = 2AD

Therefore, it is proved that DC = 2AD.

5.

Solution:

(i) From the figure we know that ∠ AOB and ∠ COD are vertically opposite angles

So we get

∠ AOB = ∠ COD = 105o

Consider △ AOB

By sum property of a triangle

∠ OAB + ∠ AOB + ∠ ABO = 180o

By substituting the values in above equation

35o + 105o + ∠ ABO = 180o


∠ ABO = 180o – 35o – 105o

By subtraction

∠ ABO = 180o – 140o

∠ ABO = 40o

(ii) We know that AB || DC and BD is a transversal

From the figure we know that ∠ ABD and ∠ CDB are alternate angles


∠ CDO = ∠ CDB = ∠ ABD = ∠ ABO = 40o

So we get

∠ ODC = 40o

(iii) We know that AB || CD and AC is a transversal

From the figure we know that ∠ ACB and ∠ DAC are alternate opposite angles



Quadrilaterals

So we get

∠ ACB = ∠ DAC = 40o

(iv) We know that ∠ B can be written as

∠ B = ∠ CBD + ∠ ABO

So we get

∠ CBD = ∠ B – ∠ ABO

In a parallelogram we know that the sum of all the angles is 360o

So we get

∠ A + ∠ B + ∠ C + ∠ D = 360o


2 ∠ A + 2 ∠ B = 360o


2 (40o + 35o) + 2 ∠ B = 360o


2 (75o) + 2 ∠ B = 360o

So we get

150o + 2 ∠ B = 360o

2 ∠ B = 360o – 150o

By subtraction

2 ∠ B = 210o

By division

∠ B = 105o

So we get

∠ CBD = ∠ B – ∠ ABO


∠ CBD = 105o – 40o

By subtraction

∠ CBD = 65o

6.

Solution:


Consider parallelogram ABCD



Quadrilaterals

So we get

∠ A = ∠ C = (2x + 25) o

∠ B = ∠ D = (3x – 5) o

We know that the sum of all the angles of a parallelogram is 360o

So it can be written as

∠ A + ∠ B + ∠ C + ∠ D = 360o


(2x + 25) + (3x – 5) + (2x + 25) + (3x – 5) = 360o

By addition we get

10x + 40o = 360o

By subtraction

10x = 360o – 40o

So we get

10x = 320o

By division we get

x = 32o

Now substituting the value of x

∠ A = ∠ C = (2x + 25) o = (2(32) + 25) o

∠ A = ∠ C = (64 + 25) o

By addition

∠ A = ∠ C = 89o

∠ B = ∠ D = (3x – 5) o = (3(32) – 5) o

∠ B = ∠ D = (96 – 5) o

By subtraction

∠ B = ∠ D = 91o

Therefore, x = 32o, ∠ A = ∠ C = 89o and ∠ B = ∠ D = 91o.

7.

Solution:

Consider ABCD as a parallelogram

If ∠ A = xo

We know that ∠ B is adjacent to A which can be written as 4/5 xo

Opposite angles are equal in a parallelogram



Quadrilaterals

So we get

∠ A = ∠ C = xo and ∠ B = ∠ D = 4/5 xo



∠ A + ∠ B + ∠ C + ∠ D = 360o


x + (4/5) x + x + (4/5) x = 360o

By addition we get

2x + (8/5) x = 360o

By taking the LCM as 5

(18/5) x = 360o

By cross multiplication

x = (360 × 5)/18


x = 100o

By substituting the value of x

So we get

∠ A = ∠ C = x = 100o

∠ B = ∠ D = 4/5 xo = (4/5) (100o) = 80o

Therefore, ∠ A = ∠ C = x = 100o and ∠ B = ∠ D = 80o.

8.

Solution:

Consider ABCD as a parallelogram

Let us take ∠ A as the smallest angle

So we get

∠ B = 2 ∠ A – 30o


∠ A = ∠ C and ∠ B = ∠ D = 2 ∠ A – 30o



∠ A + ∠ B + ∠ C + ∠ D = 360o




Quadrilaterals

∠ A + (2 ∠ A – 30o) + ∠ A + (2 ∠ A – 30o) = 360o


∠ A + 2 ∠ A – 30o + ∠ A + 2 ∠ A – 30o = 360o

So we get

6 ∠ A – 60o = 360o

By addition

6 ∠ A = 360o + 60o

6 ∠ A = 420o

By division

∠ A = 70o

By substituting the value of ∠ A

∠ A = ∠ C = 70o

∠ B = ∠ D = 2 ∠ A – 30o = 2 (70o) – 30o

∠ B = ∠ D = 110o

Therefore, ∠ A = ∠ C = 70o and ∠ B = ∠ D = 110o.

9.

Solution:

We know that the perimeter of parallelogram ABCD can be written as

Perimeter = AB + BC + CD + DA

We know that opposite sides of parallelogram are equal

AB = CD and BC = DA

By substituting the values

Perimeter = 9.5 + BC + 9.5 + BC

It is given that perimeter = 30 cm

So we get

30 = 19 + 2BC


2BC = 30 – 19

By subtraction

2BC = 11

By division we get

BC = 5.5 cm



Quadrilaterals

Therefore, AB = 9.5 cm, BC = 5.5 cm, CD = 9.5 cm and DA = 5.5 cm.

10.

Solution:

(i) We know that all the sides are equal in a rhombus

Consider △ ABC

We know that AB = BC


∠ CAB = ∠ ACB = xo

By sum property of a triangle

We get

∠ CAB + ∠ ABC + ∠ ACB = 180o


x + 110o + x = 180o

By addition we get

2x + 110o = 180o


2x = 180o – 110o

By subtraction

2x = 70o

By division

x = 35o

Therefore, x = 35o and y = 35o.

(ii) We know that all the sides are equal in a rhombus

Consider △ ABD

We get

AB = AD and

∠ ABD = ∠ ADB


x = y ……. (1)

Consider △ ABC

We get

AB = BC and



Quadrilaterals

∠ CAB = ∠ ACB

We know that

∠ ACB = 40o

By using the sum property of a triangle

∠ B + ∠ CAB + ∠ ACB = 180o


∠ B + 40o + 40o = 180o


∠ B = 180o – 40o – 40o

By subtraction

∠ B = 180o – 80o

So we get

∠ B = 100o

∠ DBC can be written as

∠ DBC = ∠ B – xo


∠ DBC = 100o – xo

From the figure we know that ∠ DBC and ∠ ADB are alternate angles

∠ DBC = ∠ ADB = yo

By substituting the value of ∠ DBC

100o – xo = yo

Consider the equation (1) we know that x = y

100o – xo = xo


2xo = 100o

By division

xo = 50o

Therefore, x = y = 50o.

(iii) From the figure we know that

∠ A = ∠ C = 62o

Consider △ BCD

We get



Quadrilaterals

BC = DC


∠ CDB = ∠ DBC = yo


∠ BDC + ∠ DBC + ∠ BCD = 180o


y + y + 62o = 180o


2y = 180o – 62o

By subtraction

2y = 118o

By division

y = 59o

We know that the diagonals of a rhombus are perpendicular to each other

Consider △ COD as a right angle triangle

∠ DOC = 90o

∠ ODC = y = 59o


∠ DCO + ∠ ODC = 90o

To find ∠ DCO

∠ DCO = 90o – ∠ ODC


∠ DCO = 90o – 59o

∠ DCO = x = 31o


11.

Solution:

We know that ABCD is a rhombus

It is given that AC = 24cm and BD = 18cm

In a rhombus we know that the diagonals bisect each other at right angles

Consider △ AOB

We know that



Quadrilaterals

∠ AOB = 90o

To find AO and BO

We know that

AO = ½ AC

By substituting AC

AO = ½ (24)

AO = 12 cm

BO = ½ BD

By substituting BD

BO = ½ (18)

BO = 9 cm

Based on the Pythagoras Theorem

AB2 = AO2 + OB2


AB2 = 122 + 92

So we get

AB2 = 144 + 81

By addition



Quadrilaterals

AB2 = 225

AB = √ 225

So we get

AB = 15 cm

Therefore, the length of each side of the rhombus is 15 cm.

12.

Solution:

We know that the diagonals of a rhombus bisect at right angles

We get

AO = OC = ½ AC

We know that AC = 16 cm

AO = OC = ½ (16) = 8 cm

Consider △ AOB

Using the Pythagoras Theorem

AB2 = AO2 + OB2


102 = 82 + OB2


OB2 = 100 – 64

By subtraction

OB2 = 36

So we get

OB = √ 36

OB = 6 cm

We know that the length of other diagonal

BD = 2 × OB

By substituting the value

BD = 2 × 6

BD = 12 cm



Quadrilaterals

From the figure the area of

△ ABC = ½ × AC × OB


△ ABC = ½ × 16 × 6

By multiplication

△ ABC = 48 cm2

△ ACD = ½ × AC × OD


△ ACD = ½ × 16 × 6

By multiplication

△ ACD = 48 cm2

So we can calculate the area of rhombus as

Area of rhombus = area of △ ABC + area of △ ACD

We get

Area of rhombus = 48 + 48 = 96 cm2

Therefore, the length of other diagonal is 12 cm and the area of rhombus is 96 cm2.

13.

Solution:



Quadrilaterals

(i) From the figure we know that the diagonals are equal and bisect at point O.

Consider △ AOB

We get AO = OB

We know that the base angles are equal

∠ OAB = ∠ OBA = 35o

By using the sum property of triangle

∠ AOB + ∠ OAB + ∠ OBA = 180o

By substituting the values we get

∠ AOB + 35o + 35o = 180o


∠ AOB = 180o – 35o – 35o

By subtraction

∠ AOB = 180o – 70o

∠ AOB = 110o

From the figure we know that the vertically opposite angles are equal

∠ DOC = ∠ AOB = y = 110o

In △ ABC

We know that ∠ ABC = 90o

Consider △ OBC

We know that

∠ OBC = xo = ∠ ABC – ∠ OBA


∠ OBC = 90o – 35o

By subtraction

∠ OBC = 55o


(ii) From the figure we know that the diagonals of a rectangle are equal and bisect each other.

Consider △ AOB

We get

OA = OB

We know that the base angles are equal

∠ OAB = ∠ OBA



Quadrilaterals

By using the sum property of triangle

∠ AOB + ∠ OAB + ∠ OBA = 180o


110o + ∠ OAB + ∠ OBA = 180o

We know that ∠ OAB = ∠ OBA

So we get

2 ∠ OAB = 180o – 110o

By subtraction

2 ∠ OAB = 70o

By division

∠ OAB = 35o

We know that AB || CD and AC is a transversal

From the figure we know that ∠ DCA and ∠ CAB are alternate angles

∠ DCA = ∠ CAB = yo = 35o

Consider △ ABC

We know that

∠ ACB + ∠ CAB = 90o

So we get

∠ ACB = 90o – ∠ CAB


∠ ACB = 90o – 35o

By subtraction

∠ ACB = x = 55o


14.

Solution:



Quadrilaterals

From the figure consider DP bisects the side AB at point P

Construct a line at BD

Consider △ AMD and △ BMD

We know that M is the mid-point of AB

AM = BM

From the figure we know that

∠ AMD = ∠ BMD = 90o

MD is common i.e. MD = MD

By SAS congruence criterion

△ AMD ≅ △ BMD

AD = BD (c. p. c. t)

We know that the sides of a rhombus are equal

So we get

AD = AB




Quadrilaterals

AD = AB = BD

We know that △ ADB is an equilateral triangle

So we get

∠ A = 60o

From the figure we know that the opposite angles are equal

∠ C = ∠ A = 60o

We know that

∠ B + ∠ A = 180o


∠ B = 180o – ∠ A

∠ B = 180o – 60o

By subtraction

∠ B = 120o

We know that ∠ D = ∠ B = 120o

Therefore, the angles are ∠ C = ∠ A = 60o and ∠ D = ∠ B = 120o.

15.

Solution:

In △ ABD

We know that AB = AD

From the figure we know that the base angles are equal

∠ ADB = ∠ ABD

We know that ∠ A = 90o


∠ ADB + ∠ ABD = 90o

We know that ∠ ADB = ∠ ABD

So we get

2 ∠ ADB = 90o

By division

∠ ADB = 45o

Consider △ OXB

From the figure we know that ∠ XOB and ∠ DOC are vertically opposite angles

∠ XOB = ∠ DOC = 80o



Quadrilaterals

We also know that

∠ ABD = ∠ XBD = 45o

We can write it as

Exterior ∠ AXO = ∠ XOB + ∠ XBD


xo = 80o + 45o

By addition

xo = 125o

Therefore, the value of x is 125o.

16.

Solution:

Consider △ ABC and △ ADC

We know that the sides of rhombus are equal

AB = AD and BC = CD

AC is common i.e. AC = AC

By SSS congruence criterion

△ ABC ≅ △ ADC

We know that

∠ BAC = ∠ DAC and ∠ BCA = ∠ DCA (c. p. c. t)

Therefore, AC bisects ∠ A as well as ∠ C.

Consider △ BAD and △ BCD

We know that the sides of rhombus are equal

AB = BC and AD = CD

BD is common i.e. BD = BD


△ BAD ≅ △ BCD



Quadrilaterals

We know that

∠ ABD = ∠ CBD and ∠ ADB = ∠ CDB (c. p. c. t)

Therefore, BD bisects ∠ B as well as ∠ D.

17.

Solution:

Consider △ AMO and △ CNO

We know that AB || CD

From the figure we know that ∠ MAO and ∠ NCO are alternate angles

It is given that AM = CN

We know that ∠ AOM and ∠ CON are vertically opposite angles

∠ AOM = ∠ CON

By ASA congruence criterion

△ AMO ≅ △ CNO

So we get

AO = CO and MO = NO (c. p. c. t)

Therefore, it is proved that AC and MN bisect each other.

18.

Solution:

Consider △ ABQ and △ CDP

We know that the opposite sides of a parallelogram are equal

AB = CD

So we get ∠ B = ∠ D

We know that

DP = AD – PA

i.e. DP = 2/3 AD

BQ = BC – CQ

i.e. BQ = BC – 1/3 BC

BQ = (3-1)/3 BC

We know that AD = BC

So we get

BQ = 2/3 BC = 2/3 AD

We get BQ = DP



Quadrilaterals


△ ABQ ≅ △ CDP

AQ = CP (c. p. c. t)

We know that

PA = 1/3 AD

We know that AD = BC

CQ = 1/3 BC = 1/3 AD

So we get

PA = CQ

∠ QAB = ∠ PCD (c. p. c. t)… (1)

We know that

∠ QAP = ∠ A – ∠ QAB

Consider equation (1)

∠ A = ∠ C

∠ QAP = ∠ C – ∠ PCD

From the figure we know that the alternate interior angles are equal

∠ QAP = ∠ PCQ

So we know that AQ and CP are two parallel lines.

Therefore, it is proved that PAQC is a parallelogram.

19.

Solution:

We know that ABCD is a parallelogram whose diagonals intersect each other at O

Consider △ AOE and △ COF

We know that ∠ CAE and ∠ DCA are alternate angles

From the figure we know that the diagonals are equal and bisect each other

AO = CO

We know that ∠ AOE and ∠ COF are vertically opposite angles

∠ AOE =∠ COF

By ASA congruence criterion

△ AOE ≅ △ COF

OE = OF (c. p. c. t)

Therefore, it is proved that OE = OF.



Quadrilaterals

20.

Solution:

From the figure

∠ DCM = ∠ DCN + ∠ MCN


90o = ∠ DCN + 60o


∠ DCN = 90o – 60o

By subtraction

∠ DCN = 30o

Consider △ DCN

Using the sum property

∠ DNC + ∠ DCN + ∠ D = 180o


90o + 30o + ∠ D = 180o


∠ D = 180o – 90o – 30o

By subtraction



Quadrilaterals

∠ D = 180o – 120o

∠ D = 60o

We know that the opposite angles of parallelogram are equal

∠ B = ∠ D = 60o


∠ A + ∠ B = 180o


∠ A + 60o = 180o

By subtraction

∠ A = 180o – 60o

∠ A =120o

So we get ∠ A = ∠ C =120o

Therefore, ∠ B = ∠ D = 60o and ∠ A = ∠ C =120o.

21.

Solution:

(i) We know that ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C.

So we get

∠ BAC = ∠ DAC …….. (1)



Quadrilaterals

∠ BCA = ∠ DCA …….. (2)

We know that every rectangle is a parallelogram

So we get AB || DC and AC is a transversal

From the figure we know that ∠ BAC and ∠ DCA are alternate angles

∠ BAC = ∠ DCA

By considering equation (1)

We get

∠ DAC = ∠ DCA

Consider △ ADC

We know that the opposite sides of equal angles are equal

AD = CD

Since ABCD is a rectangle

We get AB = BC and CD = AD

So we get AB = BC = CD = AD

Therefore, it is proved that ABCD is a square.

(ii) Consider △ BAD and △ BCD

We know that AB = CD and AD = BC

BD is common i.e. BD = BD


△ BAD ≅ △ BCD

We know that

∠ ABD = ∠ CBD and ∠ ADB = ∠ CDB (c. p. c. t)

Therefore, diagonal BD bisects ∠ B as well as ∠ D.



Quadrilaterals

22.

Solution:

Consider △ OCD and △ OBE

From the figure we know that ∠ DOC and ∠ EOB are vertically opposite angles

∠ DOC = ∠ EOB

We know that AB || CD and BC is a transversal

∠ OCD and ∠ OBE are alternate angles

∠ OCD = ∠ OBE

From the figure we know that AB = CD and BE = AB

So we can write DC = BE

By AAS congruence criterion

△ OCD ≅ △ OBE

OC = OB (c. p. c. t)

Therefore, it is proved that ED bisects BC.

23.

Solution:

Consider △ DEC and △ FEB

From the figure we know that ∠ DEC and ∠ FEB are vertically opposite angles



Quadrilaterals

∠ DEC = ∠ FEB

∠ DCE and ∠ FBE are alternate angles

∠ DCE = ∠ FBE

It is given that CE = EB

By AAS congruence criterion

△ DEC ≅ △ FEB

DC = FB (c. p. c. t)

From the figure

AF = AB + BF

We know that BF = DC and AB = DC

So we get

AF = AB + DC

AF = AB + AB

By addition

AF = 2AB

Therefore, it is proved that AF = 2AB.

24.

Solution:



Quadrilaterals

We know that l || m and t is a transversal

From the figure we know that ∠ APR and ∠ PRD are alternate angles

∠ APR = ∠ PRD

We can write it as

½ ∠ APR = ½ ∠ PRD

We know that PS and RQ are the bisectors of ∠ APR and ∠ PRD

So we get

∠ SPR = ∠ PRQ

Hence, PR intersects PS and RQ at points P and R respectively

We get

PS || RQ

In the same way SR || PQ

Therefore, PQRS is a parallelogram

We know that the interior angles are supplementary

∠ BPR + ∠ PRD = 180o

From the figure we know that PQ and RQ are the bisectors of ∠ BPR and ∠ PRD

We can write it as



Quadrilaterals

2 ∠ QPR + 2 ∠ QRP = 180o

Dividing the equation by 2

∠ QPR + ∠ QRP = 90o …… (1)

Consider △ PQR


∠ PQR + ∠ QPR + ∠ QRP = 180o

By substituting equation (1)

∠ PQR + 90o = 180o


∠ PQR = 180o – 90o

By subtraction

∠ PQR = 90o

We know that PQRS is a parallelogram


∠ PQR = ∠ PSR = 90o

We know that the adjacent angles in a parallelogram are supplementary

∠ SPQ + ∠ PQR = 180o


∠ SPQ + 90o = 180o


∠ SPQ = 180o – 90o

By subtraction

∠ SPQ = 90o

We know that all the interior angles of quadrilateral PQRS are right angles

Therefore, it is proved that the quadrilateral formed by the bisectors of interior angles is a rectangle.

25.

Solution:



Quadrilaterals

It is given that AK = BL = CM = DN

ABCD is a square

So we get

BK = CL = DM = AN …… (1)

Consider △ AKN and △ BLK

It is given AK = BL

From the figure we know that ∠ A = ∠ B = 90o

Using equation (1)

AN = BK


△ AKN ≅ △ BLK

We get

∠ AKN = ∠ BLK and ∠ ANK = ∠ BKL (c. p. c. t)

We know that

∠ AKN + ∠ ANK = 90o

∠ BLK + ∠ BKL = 90o



Quadrilaterals

By adding both the equations

∠ AKN + ∠ ANK + ∠ BLK + ∠ BKL = 90o + 90o


2 ∠ ANK + 2 ∠ BLK = 180o

Dividing the equation by 2

∠ ANK + ∠ BLK = 90o

So we get

∠ NKL = 90o

In the same way

∠ KLM = ∠ LMN = ∠ MNK = 90o

Therefore, it is proved that KLMN is a square.

26.

Solution:

We know that AR || BC and AB || RC

From the figure we know that ABCR is a parallelogram

So we get

AR = BC …… (1)

We know that AQ || BC and QB || AC

From the figure we know that AQBC is a parallelogram

So we get

QA = BC ……… (2)

By adding both the equations

AR + QA = BC + BC

We know that AR + QA = QR

So we get

QR = 2BC

Dividing by 2

BC = QR/2

BC = ½ QR

Therefore, it is proved that BC = ½ QR.

27.



Quadrilaterals

Solution:

We know that AR || BC and AB || RC

From the figure we know that ABCR is a parallelogram

So we get

AR = BC …….. (1)

We know that AQ || BC and QB || AC

From the figure we know that AQBC is a parallelogram

So we get

QA = BC ……… (2)

By adding both the equations we get

AR + QA = BC + BC

We know that AR + QA = QR

So we get

QR = 2BC


BC = QR/2

BC = ½ QR

In the same way

AB = ½ RP and AC = ½ PQ

Perimeter of △ PQR = PQ + QR + RP


Perimeter of △ PQR = 2AC + 2BC + 2AB

By taking 2 as common

Perimeter of △ PQR = 2 (AC + BC + 2AB)

Perimeter of △ PQR = 2 (Perimeter of △ ABC)

Therefore, it is proved that the perimeter of △ PQR is double the perimeter of △ ABC.

RS Aggarwal Solutions for Class 9 Maths Chapter 10 … · 2020. 11. 2. · RS Aggarwal Solutions...

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