Raju

RS Aggarwal Class 7 Solutions Chapter 3 Decimals Ex 3E

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

OBJECTIVE QUESTIONS
Mark (✓) against the correct answer in each of the following:
Question 1.
Solution:
(b)

Question 2.
Solution:
(c)

Question 3.
Solution:
(b)

= \(\frac { 208 }{ 100 }\) = 2.08

Question 4.
Solution:

Question 5.
Solution:
(b) 70g = \(\frac { 70 }{ 1000 }\) = 0.07 kg

Question 6.
Solution:
(c) 5 kg 6 g = 5\(\frac { 6 }{ 1000 }\) kg = 5.006 kg

Question 7.
Solution:
(c) 2 km 5 m = 2\(\frac { 5 }{ 1000 }\) km = 2.005 km

Question 8.
Solution:
(c)
1.007 – 0.7 = 1.007 – 0.700 = 0.307

Question 9.
Solution:
(b)
0.1 – 0.03 = 0.10 – 0.03 = 0.07

Question 10.
Solution:
(c)
3.5 – 3.07 = 3.50 – 3.07 = 0.43

Question 11.
Solution:
(c)
0.23 x 0.3 = 0.069

Question 12.
Solution:
(b)
0.02 x 30 = .60 = .6

Question 13.
Solution:
(b)
0.25 x 0.8 = 0.200 = 0.2

Question 14.
Solution:
(c)
0.4 x 0.4 x 0.4 = 0.064

Question 15.
Solution:
(b)
1.1 x .1 x .01 = .0011

Question 16.
Solution:
(a)

Question 17.
Solution:
(b)
1.02 ÷ 6 = \(\frac { 1.02 }{ 6 }\) = 0.17

Question 18.
Solution:

Question 19.
Solution:
(b)

Question 20.
Solution:
(a)

Question 21.
Solution:
(c)

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4C
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4F
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4G
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper

Question 1.
Solution:
(i) Additive inverse of 5 = -5
(ii) Additive inverse of -9 = – (-9) = 9

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
The required number = \(\frac { -1 }{ 1 }\) – \(\frac { 2 }{ 9 }\)

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 2 Fractions CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper

Question 1.
Solution:
(i) A number of the form \(\frac { a }{ b }\), where a and b are rational numbers, is called a natural number.
Here, a is the numerator and b is the denominator.
(a) \(\frac { 2 }{ 3 }\) is a fraction with 2 as the numerator and 3 as the denominator.
(b) \(\frac { 12 }{ 5 }\) is a fraction with 12 as the numerator and 5 as the denominator.
(ii) A fraction whose denominator is a whole number other than 10, 100, 1000, etc., is called a vulgar faction.
Examples : \(\frac { 2 }{ 5 }\) and \(\frac { 4 }{ 15 }\).
(iii) A fraction whose numerator is greater than or equal to its denominator is called an improper fraction.
Example : \(\frac { 11 }{ 3 }\) and \(\frac { 41 }{ 35 }\)

Question 2.
Solution:
Required number to be added

Hence, the required number is 8\(\frac { 2 }{ 5 }\)

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Mark (√) against the correct answer in each of the following:

Question 10.

Solution:
(d) \(\frac { 5 }{ 8 }\)
\(\frac { 5 }{ 8 }\) is a vulgar fraction, because its denominator is other than 10,100, 1000, etc.

Question 11.
Solution:
(c) \(\frac { 46 }{ 63 }\)
A fraction \(\frac { a }{ b }\) is said to be irreducible or in its lowest terms if the HCF of a and b is 1
46 = 2 x 23 x 1
63 = 3 x 3 x 21 x 1
Clearly, the HCF of 46 and 63 is 1.
Hence, \(\frac { 46 }{ 63 }\) is an irreducible fraction.

Question 12.
Solution:
(d) None of these
Reciprocal of 1\(\frac { 3 }{ 5 }\) = Reciprocal of \(\frac { 8 }{ 5 }\) = \(\frac { 5 }{ 8 }\)

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
(b) 33 km
Distance covered by the car on

Question 17.
Solution:

Question 18.
Solution:
(i) False.
By cross multiplication, we have:
9 x 24 = 216 and 13 x 16 = 208
However, 216 > 208
\(\frac { 9 }{ 16 }\) > \(\frac { 13 }{ 24 }\)

Hope given RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 1 Integers Ex 1D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1A
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1B
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1C
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D
• RS Aggarwal Solutions Class 7 Chapter 1 Integers CCE Test Paper

OBJECTIVE QUESTIONS
Mark (√) against the correct answer in each of the following.
Question 1.
Solution:
(c)
6 – (-8) = 6 + 8 = 14

Question 2.
Solution:
(b)
-9 – (-6) = -9 + 6 = -3

Question 3.
Solution:
(d)
-3 + 5 = 2

Question 4.
Solution:
(a)
-1 – (+5) = -1 – 5 = -6

Question 5.
Solution:
(a)
-2 – (4) = -2 – 4 = -6

Question 6.
Solution:
(b)
-4 – (+4) = -4 – 4 = -8

Question 7.
Solution:
(b)
-3 – (-5) = -3 + 5 = 2

Question 8.
Solution:
(c)
-3 – (-9) = -3 + 9 = 6

Question 9.
Solution:
(c)
-5 – (6) = -5 – 6 = -11

Question 10.
Solution:
(c)
-8 – (-13) = -8 + 13 = 5

Question 11.
Solution:
(a)
(-36) ÷ (-9) = 4

Question 12.
Solution:
(b)
0 ÷ (-5) = 0
(Zero divided by any integer other than zero, is zero)

Question 13.
Solution:
(c)
Division by zero is not defined

Question 14.
Solution:
(b)

Question 15.
Solution:
(b)
-3 + 9 = 6

Question 16.
Solution:
(a)
-4 – (-10) = -4 + 10 = 6

Question 17.
Solution:
(a)
Sum = 14
One integer = -8
Second = 14 – (-8) = 14 + 8 = 22

Question 18.
Solution:
(c)

Question 19.
Solution:
(b)
(-15) x 8 + (-15) x 2
= (-15) {8 + 2}
= -15 x 10 = -150

Question 20.
Solution:
(b)
(-12) x 6 – (-12) x 4 = (-12) (6 – 4) = -12 x 2 = -24

Question 21.
Solution:
(b)
(-27) x (-16)+ (-27) x (-14)
= (-27) {-16 – 14}
= (-27) x (-30)
= 810

Question 22.
Solution:
(a)
30 x (-23) + 30 x 14
= 30 x (-23 + 14)
= 30 x (-9)
= -270

Question 23.
Solution:
(c)
Sum of two integers = 93
One integer = -59
Second = 93 – (-59) = 93 + 59 = 152

Question 24.
Solution:
(b)
(?) ÷ (-18) = -5
Let x ÷ (-18) = -5
⇒ \(\frac { x }{ -18 }\) = -5
⇒ x = (-5) x (-18) = 90

Hope given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 3 Decimals Ex 3B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

Add:
Question 1.
Solution:
Converting them into like decimals 16.00, 8.70, 0.94, 6.80 and 7.77
Now, adding them,
16.0 + 8.70 + 0.94 + 6.80 + 7.77 = 40.21

Question 2.
Solution:
Converting them into like decimals 18.600, 206.370, 8.008, 26.400, 6.900
Adding we get
18.600 + 206.370 + 8.008 + 26.400 + 6.900 = 266.278

Question 3.
Solution:
Converting them into like decimals, 63.50, 9.70, 0.80, 26.66, 12.17
Adding we get:
63.50 + 9.70 + 0.80 + 26.66 + 12.17 = 112.83

Question 4.
Solution:
Converting them into like decimals 17.400, 86.390, 9.435, 8.800, 0.060
Adding we get:
17.400 + 86.390 + 9.435 + 8.800 + 0.060 = 122.085

Question 5.
Solution:
Converting them into like decimals 26.900, 19.740, 231.769, 0.048
Now adding we get:
26.900 + 19.740 + 231.769 + 0.048 = 278.457

Question 6.
Solution:
Converting them into like decimals 23.800, 8.940, 0.078 and 214.600
Now adding we get:
23.800 + 8.940 + 0.078 + 214.600 = 247.418

Question 7.
Solution:
Converting them into like decimals.
6.606, 66.600, 666.000,0.066, 0.660
Now adding we get:
6.606 + 66.600 + 666.000 + 0.066 + 0,660 = 739.932

Question 8.
Solution:
9.090, 0.909, 99.900, 9.990, 0.099
Now adding we get:
9.090 + 0.909 + 99.900 + 9.990 + 0.099 = 119.988

Subtract:
Question 9.
Solution:
14.79 from 72.43
72.43 – 14.79 = 57.64

Question 10.
Solution:
Converting them into like decimals, We get
36.74 and 52.60
Now 52.60 – 36.74 = 15.86

Question 11.
Solution:
Converting them into like decimals, We get
13.876 and 22.000
22.000 – 13.876 = 8.124

Question 12.
Solution:
Converting them into like decimals, We get
15.079 and 24.160
24.160 – 15.079 = 9.081

Question 13.
Solution:
Converting them into like decimals We get
0.680 and 1.007
1.007 – 0.680 = 0.327

Question 14.
Solution:
Converting them into like decimals,
We get 0.4678 and 5.0500
5.0500 – 0.4678 = 4.5822

Question 15.
Solution:
Converting them into like decimals,
We get 2.5307 and 8.0000
8.0 – 2.5307 = 5.4693

Question 16.
Solution:
There are like decimals
9.1 – 6.732 = 2.269

Question 17.
Solution:
Converting them into like decimals,
We get 5.746 and 9.100
9.100 – 5.746 = 3.354

Question 18.
Solution:
Converting into like decimals, we get,
63.59 and 92.00
Required number = 92.00 – 63.58 = 28.42

Question 19.
Solution:
Converting into like decimals, we get:
8.100 and 0.813
Required number = 8.100 – 0.813 = 7.287

Question 20.
Solution:
Converting them into like decimals, we get: 32.67 and 60.10
Required number = 60.10 – 32.67 = 27.43

Question 21.
Solution:
Converting into like decimals, we get 74.3 and 26.87
Required number = 74.30 – 26.87 = 47.43

Question 22.
Solution:
Cost of notebook = Rs. 23.75
Cost ofpencil = Rs. 2.85
Costofpen =Rs. 15.90
Total cost = Rs. 42.50
Amount gave to the shop keeper = 50 rupees
Balance amount got = Rs 50.00 – Rs 42.50 = 7.50

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 3 Decimals Ex 3A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:
(i) 6.5, 16.03, 0.274, 119.4
In these decimals, the greatest places of decimal is 3
6.5 = 6.500
16.03 = 16.030
0. 274 = 0.274
119.4 = 119.400 are like decimals.
(ii) 3.5, 0.67, 15.6, 4
In these decimal, the greatest place of decimal is 2
3.5 = 3.50
0.67 = 0.67
15.6 = 15.60
4 = 4.00 are the like decimals

Question 5.
Solution:
(i) Among 78.23 and 69.85,
78.23 is greater than 69.85 (78 > 69)
78.23 > 69.85
(ii) Among 3.406 and 3.46,
3.406 is less than 3.46 (40 < 46)
3.406 < 3.46
(iii) Among 5.68 and 5.86,
5.68 is less than 5.86 (68 < 86)
5.68 < 5.86
(iv) Among 14.05 and 14.005
14.5 is greater than 14.005 (05 > 00)
14.5 >14.005
(v) Among 1.85 and 1.805,
1.85 is greater than 1.805 (85 > 80)
1.85 > 1.805
(vi) Among 0.98 and 1.07,
0.98 is less than 1.07 (0 < 1)
0.98 < 1.07

Question 6.
Solution:
(i) 4.6, 7.4, 4.58, 7.32, 4.06
Converting the given decimals into like decimals, we get:
4.60, 7.40, 4.58, 7.32, 4.06.
We see that 4.06 < 4.58 < 4.60 < 7.32 < 7.40.
Writing in ascending order, 4.06, 4.58, 4.6, 7.32, 7.4
(ii) 0.5, 5.5, 5.05, 0.05, 5.55
Converting the given decimals into like decimals, we get:
0. 50, 5.50, 5.05, 0.05, 5.55
We see that 0.05 < 0.50 < 5.05 < 5.50 < 5.55.
Writing in ascending order, 0.05, 0.50, 5.05, 5.5, 5.55
(iii) 6.84, 6.84, 6.8, 6.4, 6.08
Converting the given decimals into like decimals
6.84, 6.48, 6.80, 6.40, 6.08
We see that 6.08 < 6.40 < 6.48 < 6.80 < 6.84
Writing in ascending order,
6.08, 6.4, 6.48, 6.8, 6.84
(iv) 2.2, 2.202, 2.02, 22.2, 2.002
Converting them into like decimals
2.200, 2.202, 2.020, 22.200, 2.002 we see that
2.002 < 2.020 < 2.200 < 2.202 < 22.200
Now writing in ascending order,
2.002, 2.020, 2.2, 2.202, 22.2

Question 7.
Solution:
(i) 7.4, 8.34, 74.4, 7.44, 0.74
Converting them into like decimals,
7.40, 8.34, 74.40, 7.44, 0.74
we see that
74.40 > 8.34 > 7.44 > 7.40 > 0.74
Writing in descending order,
74.4, 8.34, 7.44, 7.4, 0.74
(ii) 2.6, 2.26, 2.06, 2.007, 2.3
Converting them into like decimals,
2.600, 2.260, 2.060, 2.007, 2.300
We see that
2.600 > 2.300 > 2.260 > 2.060 > 2.007
Writing in descending order,
2.6, 2.3, 2.26, 2.06, 2.007

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12A
• RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12B
• RS Aggarwal Solutions Class 10 Chapter 12 Circles MCQS
• RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself

MCQ
Question 1.
Solution:
In the given figure,
PT is the tangent and PQ is the chord of the circle with centre O.
∠OPT = 50°

OP is radius and PT is the tangent.
OP ⊥ PT ⇒ ∠OPT = 90°
∠OPQ + ∠QPT = 90°
⇒ ∠OPQ + 50° = 90°
⇒ ∠OPQ = 90° – 50° = 40°
In ∆OPQ,
OP = OQ (radii of the same circle)
∠OQP = ∠OPQ = 40°
In ∆OPQ,
∠POQ = 180° – (∠OPQ + ∠OQP)
= 180° – (40° + 40°) = 180° – 80°
= 100° (b)

Question 2.
Solution:
Angle between two radii of a circle = 130°

Then ∠APB = 180° – ∠AOB
= 180°- 130° = 50° (c)

Question 3.
Solution:
In the given figure,
PA and PB are the tangents drawn from P to the circle with centre O
∠APB = 80°

OA is radius of the circle and AP is the tangent
OA ⊥ AP ⇒ ∠OAP = 90°
OP bisects ∠APB,
∠APO = \(\frac { 1 }{ 2 }\) x 80 = 40°
∠POA = 180° – (90° + 40°)
= 180° – 130° = 50° (b)

Question 4.
Solution:
In the given figure, AD and AE are tangents to the circle with centre O.
BC is the tangent at F which meets AD at C and AE at B
AE = 5 cm

AE and AD are the tangents to the circle
AE = AD = 5 cm
Tangents from an external point drawn to the circle are equal
CD = CF and BE = BF
Now, perimeter of ∆ABC = AB + AC + BC
= AB + AC + BF + CF (BE = BF and CF = CD)
= AB + AC + BE + CD
= AB + BE + AC + CD
= AE + AD
= 5 + 5 = 10 cm (b)

Short-Answer Questions
Question 5.
Solution:
In the given figure, a quadrilateral ABCD is circumscribed a circle touching its sides at P, Q, R and S respectively.
AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm

A circle touches the sides of a quadrilateral ABCD.
AB + CD = BC + AD …(i)
Now, AP and AS are tangents to the circle
AP = AS = 5 cm …(ii)
Similarly, CQ = CR = 3 cm
BP = BQ = x – 5 = 4
BQ = BC – CQ = 7 – 3 = 4 cm
x – 5 = 4
⇒ x = 4 + 5 = 9cm

Question 6.
Solution:
In the given figure, PA and PB are the tangents drawn from P to the circle.
OA and OB are the radii of the circle and AP and BP are the tangents.
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
In quad. AOBP
∠A + ∠B = 90° + 90° = 180°
But these are opposite angles of a quadrilateral
AOBP is a cyclic quadrilateral
A, O, B, P are concyclic

Question 7.
Solution:
In the given figure, PA and PB are two tangents to the circle with centre O from an external point P.
∠PBA = 65°,
To find : ∠OAB and ∠APB
In ∆APB
AP = BP (Tangents from P to the circle)
∠PAB = ∠PBA = 65°
∠APB = 180° – (∠PAB + ∠PBA)
= 180° – (65° + 65°) = 180° – 130° = 50°
OA is radius and AP is tangent
OA ⊥ AP
∠OAP = 90°
∠OAB = ∠OAP – ∠PAB = 90° – 65° = 25°
Hence, ∠OAB = 25° and ∠APB = 50°

Question 8.
Solution:
Given : In the figure,
BC and BD are the tangents drawn from B
to the circle with centre O.
∠CBD = 120°
To prove : OB = 2BC
Construction : Join OB.
Proof: OB bisects ∠CBD (OC = OD and BC = BD)

Question 9.
Solution:
(i) A line intersecting a circle in two distinct points is called a secant.
(ii) A circle can have two parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the point of contact.
(iv) A circle can have infinitely many tangents.

Question 10.
Solution:
Given : In a circle, from an external point P, PA and PB are the tangents drawn to the circle with centre O.
To prove : PA = PB
Construction : Join OA, OB and OP.
Proof : OA and OB are the radii of the circle and AP and BP are tangents.

OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = ∠OBP = 90°
Now, in right ∆OAP and ∆OBP,
Hyp. OP = OP (common)
Side OA = OB (radii of the same circle)
∆OAP = ∆OBP (RHS axiom)
PA = PB (c.p.c.t.)
Hence proved.

Short-Answer Questions
Question 11.
Solution:
Given : In a circle with centre O and AB is its diameter.
From A and B, PQ and RS are the tangents drawn to the circle

To prove : PQ || RS
Proof : OA is radius and PAQ is the tangent
OA ⊥ PQ
∠PAO = 90° …(i)
Similarly, OB is the radius and RBS is tangent
∠OBS = 90° …(ii)
From (i) and (ii)
∠PAO = ∠OBS
But there are alternate angles
PQ || RS

Question 12.
Solution:
Given : In the given figure,
In ∆ABC,
AB = AC.
A circle is inscribed the triangle which touches it at D, E and F
To prove : BE = CE
Proof: AD and AF are the tangents drawn from A to the circle
AD = AF
But, AB = AC
AB – AD = AC -AF
⇒ BD = CF
But BD = BE and CF = CE (tangent drawn to the circle)
But BD = CF
BE = CE
Hence proved.

Question 13.
Solution:
Given : In a circle from an external point P, PA and PB are the tangents to the circle
OP, OA and OB are joined.

To prove: ∠POA = ∠POB
Proof: OA and OB are the radii of the circle and PA and PB are the tangents to the circle
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
Now, in right ∆OAP and ∆OBP,
Hyp. OP = OP (common)
Side OA = OB (radii of the same circle)
∆OAP = ∆OBP (RHS axiom)
∠POA = ∠POB (c.p.c.t.)
Hence proved.

Question 14.
Solution:
Given : A circle with centre O, PA and PB are the tangents drawn from A and B which meets at P.
AB is chord of the circle
To prove : ∠PAB = ∠PBA
Construction : Join OA, OB and OP

Proof: OA is radius and AP is tangent
OA ⊥ AP ⇒ ∠OAP = 90°
Similarly, OB ⊥ BP ⇒ ∠OBP = 90°
In ∆OAB, OA = OB (radii of the circle)
∠OAB = ∠OBA
⇒ ∠OAP – ∠OAB = ∠OBP – ∠OBA
⇒ ∠PAB = ∠PBA
Hence proved.

Question 15.
Solution:
Given : A parallelogram ABCD is circumscribed a circle.
To prove : ABCD is a rhombus.
Proof: In a parallelogram ABCD.
Opposite sides are equal and parallel.

AB = CD and AD = BC
Tangents drawn from an external point of a circle to the circle are equal.
AP = AS BP = BQ
CQ = CR and DR = DS
Adding, we get
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC [AB = CD and AD = BC]
⇒ AB + AB = BC + BC
⇒ 2AB = 2BC
⇒ AB = BC
But AB = CD and BC = AD
AB = BC = CD = AD
Hence || gm ABCD is a rhombus.

Question 16.
Solution:
Given : O is the centre of two concentric circles with radii 5 cm and 3 cm respectively.
AB is the chord of the larger circle which touches the smaller circle at P.

OP and OA are joined.
To find : Length of AB
Proof: OP is the radius of the smaller circle and touches the smaller circle at P
OP ⊥ AB and also bisects AB at P
AP = PB = \(\frac { 1 }{ 2 }\) AB
Now, in right ∆OAP,
OA² = OP² + AP² (Pythagoras Theorem)
⇒ (5)² = (3)² + AP²
⇒ 25 = 9 + AP²
⇒ AP² = 25 – 9 = 16 = (4)²
AP = 4 cm
Hence AB = 2 x AP = 2 x 4 = 8 cm

Long-Answer Questions
Question 17.
Solution:
In the figure, quad. ABCD is circumscribed about a circle which touches its sides at P, Q, R and S respectively

To prove : AB + CD = AD + BC
Proof: Tangents drawn from an external point to a circle are equal
AP = AS
BP = BQ
CR = CQ
DR = DS
Adding, we get,
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
Hence AB + CD = AD + BC

Question 18.
Solution:
Given : A quad. ABCD circumscribe a circle with centre O and touches at P, Q, R and S respectively
OA, OB, OC and OD are joined forming angles AOB, BOC, COD and DOA
To prove : ∠AOB + ∠COD = 180°
and ∠BOC + ∠AOD = 180°
Construction : Join OP, OQ, OR and OS

Proof: In right ∆AOP and ∆AOS,
Side OP = OS (radii of the same circle)
Hyp. OA = OA (common)
∆AOP = ∆AOS (RHS axiom)
∠1 = ∠2 (c.p.c.t.)
Similarly, we can prove that
∠4 = ∠3
∠5 = ∠6
∠8 = ∠7
Adding, we get
∠1 + ∠4 + ∠5 + ∠8 = ∠2 + ∠3 + ∠6 + ∠7
⇒ (∠1 + ∠8) + (∠4 + ∠5) = (∠2 + ∠3) + (∠6 + ∠7)
⇒ ∠AOB + ∠COD = ∠AOD + ∠BOC
But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (angles at a point)
∠AOB + ∠COD = ∠AOD + ∠BOC = 180°
Hence proved

Question 19.
Solution:
Given : From an external point P, PA and PB are the tangents drawn to the circle,
OA and OB are joined.
To prove : ∠APB + ∠AOB = 180°
Construction : Join OP.

Proof : Now, in ∆POA and ∆PBO,
OP = OP (common)
PA = PB (Tangents from P to the circle)
OA = OB (Radii of the same circle)
∆POA = ∆PBO (SSS axiom)
∠APO = ∠BPO (c.p.c.t.)
and ∠AOP = ∠BOP (c.p.c.t.)
OA and OB are the radii and PA and PB are the tangents
OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = 90° and ∠OBP = 90°
In ∆POA,
∠OAP = 90°
∠APO + ∠AOP = 90°
Similarly, ∠BPO + ∠BOP = 90°
Adding, we get
(∠APO + ∠BPO) + (∠AOP + ∠BOP) = 90° + 90°
⇒ ∠APB + ∠AOB = 180°.
Hence proved.

Question 20.
Solution:
Given : PQ is chord of a circle with centre O.
TP and TQ are tangents to the circle
Radius of the circle = 10 cm
i.e. OP = OQ = 10 cm and PQ = 16 cm

To find : The length of TP.
OT bisects the chord PQ at M at right angle.
PM = MQ = \(\frac { 16 }{ 2 }\) = 8 cm
In right ∆PMO,
OP² = PM² + MO² (Pythagoras Theorem)
⇒ (10)² = (8)² + MO²
⇒ 100 = 64 + MO²
⇒ MO² = 100 – 64 = 36 = (6)²
MO = 6 cm
Let TP = x and TM = y
In right ∆TPM,
TP² = TM² + PM²
⇒ x² = y² + 8²
⇒ x² = y² + 64 …(i)
and in right ∆TPM
OT² = TP² + OP²
⇒ (y + 6)² = x² + 10²
⇒ y² + 12y + 36 = x² + 100
⇒ y² + 12y + 36 = y2 + 64 + 100 {From (i)}
⇒ 12y = 64 + 100 – 36 = 128

Hope given RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Choose the correct answer in each of the following questions.
Question 1.
Solution:
Distance between P (-6, 8) and origin O (0, 0)

Question 2.
Solution:
Distance of the point (-3, 4) from x-axis = 4 (c)

Question 3.
Solution:
Let point P (x, 0) is on x-axis which is equidistant from points A (-1, 0) and B (5, 0)

Question 4.
Solution:
R (5, 6) is the midpoint of the line segment joining the points A (6, 5) and B (4, y)

Question 5.
Solution:
Point C (k, 4) divides the join of points A (2, 6) and B (5, 1) in the ratio 2 : 3

Question 6.
Solution:
Vertices of ∆ABC are A (0, 4), B (0, 0) and C (3, 0)

Question 7.
Solution:
A (1, 3), B (-1, 2), C (2, 5) and D (x, 4) are the vertices of a ||gm ABCD.
AB = CD

Question 8.
Solution:
Points A (x, 2), B (-3, -4) and C (7, -5) are collinear

Question 9.
Solution:
Area of ∆ABC whose vertices are A (5, 0), B (8, 0) and C (8, 4)

Question 10.
Solution:
Area of ∆ABC with vertices A (a, 0), O (0, 0) and B (0, b)

Question 11.
Solution:
P( \(\frac { a }{ 2 }\), 4) is midpoint of the line segment joining the points A (-6, 5) and B (-2, 3)

Question 12.
Solution:
ABCD is a rectangle whose three vertices are B (4, 0), C (4, 3) and D (0, 3)

Question 13.
Solution:
Let coordinates of P be (x, 7)
P divides the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1

Question 14.
Solution:
Let coordinates of one end of diameter of a circle are A (2, 3) coordinates of centre are (-2, 5)
Let coordinates of other end B of the diameter be (x, y)

Question 15.
Solution:
In the given figure, P (5, -3) and Q (3, y) are the points of trisection of the line segment joining A (7, -2) and B (1, -5)

Question 16.
Solution:
Midpoint of AB is P (0, 4) coordinates of B are (-2, 3)

Question 17.
Solution:
Let point P (x, y) divides AB with vertices A(2, -5) and B (5, 2) in the ratio 2 : 3

Question 18.
Solution:
A (-6, 7) and B (-1, -5) two points, then

Question 19.
Solution:
Let point P (x, 0) on x-axis is equidistant from A (7, 6) and B (-3, 4)

Question 20.
Solution:
Distance of P (3, 4) from the x-axis = 4 units. (b)

Question 21.
Solution:
Let a point P (x, 0) on x-axis divides the line segment.
AB joining the points A (2, -3) and B (5, 6) in the ratio m : n.

Question 22.
Solution:
Let a point A (0, y) on y-axis divides the line segment PQ joining the points P (-4, 2) and Q (8, 3) in the ratio of m : n.

Question 23.
Solution:
P (-1, 1) is the midpoint of line segment joining the points A (-3, b) and B (1, b + 4)

Question 24.
Solution:
Let the point P (x, y) divides the line segment joining the points A (2, -2) and B (3, 7) in the ratio k : 1, then

Question 25.
Solution:
AD is the median of ∆ABC with vertices A (4, 2), B (6, 5) and C (1, 4)

Question 26.
Solution:
A (-1, 0), B (5, -2) and C (8, 2) are the vertices of a ∆ABC then centroid

Question 27.
Solution:
Two vertices A (-1, 4), B (5, 2) of a ∆ABC and its centroid G is (0, -3)
Let (x, y) be the co-ordinates of vertex C, then

Question 28.
Solution:
Points A (-4, 0), B (4, 0) and C (0, 3) are vertices of a ∆ABC.

Question 29.
Solution:
Points P (0, 6), Q (-5, 3) and R (3, 1) are the vertices of a ∆PQR

Question 30.
Solution:
Points A (2, 3), B (5, k) and C (6, 7) are collinear
Area of ∆ABC = 0

Question 31.
Solution:
Points A (1, 2), B (0, 0) and C (a, b) are collinear
Area of ∆ABC = 0

Question 32.
Solution:
A (3, 0), B (7, 0) and C (8, 4)
Area ∆ABC

Question 33.
Solution:
AOBC is a rectangle with vertices A (0, 3), O (0, 0) and B (5, 0) and each diagonal of rectangle are equal.

Question 34.
Solution:
Points are A (4, p) and B (1, 0)
Distance = 5 units

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Question 1.
Solution:
(i) Let P (x, y) be the required point which divides the line joining the points A (-1, 7) and B (4, -3) in the ratio 2 : 3.

Question 2.
Solution:
Let (7, -2) and B (1, -5) be the given points and P (x, y) and Q (x’, y’) are the points of trisection.

Question 3.
Solution:
Let coordinates of P be (x, y) which divides the line segment A (-2, -2) and B (2, -4) in

Question 4.
Solution:
Let the coordinates of A be (x, y) which lies on line joining P (6, -6) and Q (-4, -1)

Question 5.
Solution:
Points P, Q, R and S divides a line segment joining the points A (1, 2) and B (6, 7) in 5 equal parts.

Question 6.
Solution:
Points P, Q and R in order divide a line segment joining the points A (1, 6) and B (5, -2) in 4 equal parts.

P divides AB in the ratio of 1 : 3 Let coordinates of P be (x, y), then

Question 7.
Solution:
The line segment joining the point A (3, -4) and B (1, 2) is trisected by the points P (p, -2) and Q(\(\frac { 1 }{ 2 }\), q).

Question 8.
Solution:
Mid point of the line segment joining the points A (3, 0) and B (-5, 4)

Question 9.
Solution:
(2, p) is the mid point of the line segment joining the points A (6, -5), B (-2, 11)

Question 10.
Solution:
Mid point of the line segment joining the points A (2a, 4) and B (-2, 3b) is C (1, 2a + 1)

Question 11.
Solution:
The line segment joining the points A (-2, 9) and B (6, 3) is a diameter of a circle with centre C.

Question 12.
Solution:
AB is diameter of a circle with centre C.
Coordinates of C (2, -3) and of B (1, 4)

Question 13.
Solution:
Let P (2, 5) divides the line segment joining the points A (8,2) and B (-6, 9) in the ratio m : n

Question 14.
Solution:

Question 15.
Solution:
Let P (m, 6) divides the join of A (-4, 3) and B (2, 8) in the ratio k : 1
Then coordinates of P will be

Question 16.
Solution:
Let point P (-3, k) divides the join of A (-5, -4) and B (-2, 3) in the ratio m : n, then

Question 17.
Solution:
Let point P on the x-axis divides the line segment joining the points A (2, -3) and B (5, 6) the ratio m : n
Let P is the point on x-axis whose coordinates are (x, 0)

Question 18.
Solution:
Let a point P on y-axis divides the line segment joining the points A (-2, -3) and B (3, 7) in the ratio m : n
Let the coordinates of P be (0, y)

Question 19.
Solution:
Let a point P (x, y) on the given line x – y – 2 = 0 divides the line segment joining the points A (3, -1) and B (8, 9) in the ratio m : n, then

Question 20.
Solution:
Vertices of ∆ABC are A (0, -1), B (2, 1) and C (0, 3)
Let AD, BE and CF are the medians of sides BC, CA and AB respectively, then
Coordinates of D will be =

Question 21.
Solution:
Centroid of ∆ABC where coordinates of A are (-1, 0), of B are (5, -2) and of C are (8, 2)

Question 22.
Solution:
G (-2, 1) is the centroid of ∆ABC whose vertex A is (1, -6) and B is (-5, 2)
Let the vertex C be (x, y), then

Question 23.
Solution:
O (0, 0) is the centroid of ∆ABC in which B is (-3, 1), C is (0, -2)
Let A be (x, y), then

Question 24.
Solution:
Points are A (3, 1), B (0, -2), C (1, 1) and D (4, 4)

Question 25.
Solution:
Points P (a, -11), Q (5, b), R (2, 15) and S (1, 1) are the vertices of a parallelogram PQRS.
Diagonals of a parallelogram bisect each other.
O is mid point of PR and QS.

Question 26.
Solution:
Three vertices of a parallelogram ABCD are A (1, -2), B (3, 6), C (5, 10).
Let fourth vertices D be (x, y)

Question 27.
Solution:
Let a point P (0, y) on 7-axis, divides the line segment joining the points (-4, 7) and (3, -7) in the ratio m : n

Question 28.
Solution:

Question 29.
Solution:
Let a point P (x, 0) divides the line segment joining the points A (3, -3) and B (-2, 2) in the ratio m : n

Question 30.
Solution:
Base QR of an equilateral APQR lies on x- axis is O (0, 0) is mid point PQR and coordinate of Q are (-4, 0).
Coordinate of R will be (4, 0)

Question 31.
Solution:
Base BC of an equilateral triangle ABC lies on y-axis in such a way that origin O (0, 0) lies is the middle of BC and coordinates of C are (0, -3).
Coordinates of B will be (0, 3)

Question 32.
Solution:
Let the points P (-1, y) lying on the line segment joining points A (-3, 10) and B (6, -8) divides it in the ratio m : n.

Question 33.
Solution:
In rectangle ABCD, A (-1, -1), B (-1, 4), C (5, 4), D (5, -1)
P, Q, R and S are the mid points of AB, BC, CD and DA respectively.

Question 34.
Solution:
P is mid point of line segment joining the points A (-10, 4) and B (-2, 0)

Question 35.
Solution:
Let coordinates of P and Q be (0, y) and (x, 0) respectively.
Let M (2, -5) be the mid-point of PQ.
By midpoint formula

Question 36.
Solution:

Question 37.
Solution:

Question 38.
Solution:
Let the other two vertices be (h, k) and (m, n).
Hence, the vertices in order are (3, 2), (-1, 0), (h, k) and (m, n).
It is to be kept in mind that the diagonals of a parallelogram bisect each other.
Hence, the point of intersection (2, -5) is the midpoint of the diagonal whose ends are (3, 2) and (h, k). Then

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 1 Integers Ex 1A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1A
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1B
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1C
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D
• RS Aggarwal Solutions Class 7 Chapter 1 Integers CCE Test Paper

Question 1.
Solution:
(i) 15 + (-8) =15 – 8 = 7
(ii) (-16) +9 = -7
(iii) (-7) + (-23)= -7 – 23 = -30
(iv) (-32) + 47 = -32 + 47 = 15
(v) 53 + (-26) = 53 – 26 = 27
(vi) (-48) + (-36) = -48 – 36 = -84

Question 2.
Solution:
(i) 153 and -302 = 153 + (-302) = 153 – 302 = -149
(ii) 1005 and -277 = 1005 + (-277) = 1005 – 277 = 728
(iii) -2035 and 297 = -2035 + 297 = – 1738
(iv) -489 and -324 = -489 + (-324) = -489 – 324 = -813
(v) -1000 and438 = -1000 + 438 = -562
(vi) -238 and 500 = -238 + 500 = 262

Question 3.
Solution:
Additive inverse of
(i) -83 is – (-83) = 83
(ii) 256 is -256
(iii) 0 is 0
(iv) -2001 is – (-2001) = 2001

Question 4.
Solution:
(i) 28 from – 42 = -42 – (28) = -42 – 28 = -70
(ii) – 36 from 42 = 42 – (-36) = 42 + 36 = 78
(iii) -37 from -53 = -53 – (-37) = -53 + 37= -16
(iv) -66 from -34 = -34 – (-66) = -34 + 66 = 32
(v) 318 from 0 = 0 – (318) = -318
(vi) -153 from -240= -240 – (-153) = -240 + 153 = -87
(vii) -64 from 0 = 0 – (-64) = 0 + 64 = 64
(viii) – 56 from 144 = 144 – (-56) = 144 + 56 = 200

Question 5.
Solution:
– 34 – (-1032 + 878)
= -34 – (-154) = -34 + 154 = 120

Question 6.
Solution:
38 + (-87) – 134
= (38 – 87) – 134
= -49 – 134 = -183

Question 7.
Solution:
(i) {(-13) + 27} + (-41) = (-13) + {27 + (-41)} (By Associative Law of Addition)
(ii) (-26) + {(-49) + (-83)} = {(-26) + (-49)} +(-83) (By Associative Law of Addition)
(iii) 53 + (-37) = (-37) + (53) (By Commutative Law of Addition)
(iv) (-68) + (-76) = (-76) + (-68) (By Commutative Law of Addition)
(v) (-72) + (0) = -72 (Existence of Additive identity)
(vi) -(-83) = 83
(vii) (-60) – (………) = -59 => -60 – (-1) = -59
(viii) (-31) + (……….) = -40 => -31 + (-9) = -40

Question 8.
Solution:
{-13 – (-27)} + {-25 – (-40)}
= {-13 + 27} + {-25 + 40}
= 14 + 15 = 29

Question 9.
Solution:
36 – (- 64) = 36 + 64 = 100
(-64) – 36= -64 – 36 = -100
They are not equal

Question 10.
Solution:
(a + b) + c = {-8 + (-7)} + 6 = (-8 – 7) + 6 = -15 + 6 = -9
and a + (b + c) = -8 + (-7 + 6) = -8 + (-1) = -8 – 1 = -9 Hence proved

Question 11.
Solution:
LHS = (a -b) = -9 – (-6) = -9 + 6 = -3
RHS = (b – a) = -6 – (-9) = -6 + 9 = 3
LHS ≠ RHS.
Hence (a – b) ≠ (b – a)

Question 12.
Solution:
Sum of two integers = -16
One integer = 53
Second integer = -16 – (53) = -16 – 53 = (-69)

Question 13.
Solution:
Sum of two integers = 65
One integer = -31
Second integer = 65 – (-31) = 65 + 31 = 96

Question 14.
Solution:
Difference of a and (-6) = 4
a – (-6) = 4
⇒ a + 6 = 4
⇒ a = 4 – 6
⇒ a = -2

Question 15.
Solution:
(i) We can write any two integers having opposite signs
e.g. 5, -5
Sum = 5 + (-5) = 5 – 5 = 0
(ii) The sum is a negative integer
The greater integer must be negative and smaller integer be positive
e.g. -9, 6
Sum = -9 + 6 = -3
(iii) The sum is smaller than the both integers
Both integer will be negative -4, -6
Sum = -4 + (-6) = -4 – 6 = -10
(iv) The sum is greater than the both integers
Both integers will be positive
e.g. 6, 4
(v) The sum oftwo integers is smaller than one of these integers
The greater number will be positive and smaller be negative
e.g. 6, -4
Sum = 6 + (-4) = 2

Question 16.
Solution:
(i) False: Because, all negative integers are less than zero.
(ii) False: -10 is less than -7.
(iii) Tme: Every negative integer is less than zero.
(iv) True : Sum of two negative integers is negative.
(v) False: It is not always true.

 

Hope given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1A are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Question 1.
Solution:
(i) In ∆ABC, vertices are A (1, 2), B (-2, 3) and C (-3, -A)

Question 2.
Solution:
Vertices of quadrilateral ABCD are A (3,-1), B (9, -5), C (14, 0) and D (9, 19)
Join diagonal AC.

Question 3.
Solution:
PQRS is a quadrilateral whose vertices are P (-5, -3), Q (-4, -6), R (2, -3) and S (1, 2)
Join PR which forms two triangles PQR and PSR.

Question 4.
Solution:
ABCD is a quadrilateral whose vertices are A (-3, -1), B (-2, -4), C (4, -1) and D (3, 4)
Join AC which terms two triangles ABC and ADC

Question 5.
Solution:

Question 6.
Solution:
D, E and F are the midpoints of the sides CB, CA and AB respectively of ∆ABC.
Vertices are A (2, 1), B (4, 3) and C (2, 5)

Question 7.
Solution:
Vertices of ∆ABC are A (7, -3), B (5, 3) and C (3, -1)
D is midpoint of BC


Similarly area of ∆ACD = 5 sq. units (10 – 5 = 5 sq. units)
Hence, median divides the triangle into two triangles of equal in area.

Question 8.
Solution:
In ∆ABC, coordinates of A are (1, -4) and let C and E are the midpoints of AB and AC respectively.
Coordinates of F are (2, -1) and of E are (0, -1)
Let coordinates of B be (x1, y1) and of C be (x2, y2)

Question 9.
Solution:
A (6, 1), B (8, 2) and C (9, 4) are the three vertices of a parallelogram ABCD.
E is the midpoint of DC.
Join AE, AC and BD which intersects at O.
O is midpoint of AC

Question 10.
Solution:
(i) Vertices of a ∆ABC are A (1, -3), B (4, p) and C (-9, 7) and area = 15 sq. units

Question 11.
Solution:
Vertices of ∆ABC are A (k + 1, 1), B (4, -3) and C (7, -k) and area = 6 sq. units.

Question 12.
Solution:
Let the vertices of a triangle ABC are A (-2, 5), B (k, -4) and C (2k + 1, 10) and area = 53 sq. units

Question 13.
Solution:
(j) Points are A (2, -2), B (-3, 8) and C (-1, 4)

Question 14.
Solution:
Points are A (x, 2), B (-3, -4) and C (7, -5)

Question 15.
Solution:
Points are given A (-3, 12), B (7, 6) and C (x, 9)

Question 16.
Solution:
Points are given P (1, 4), Q (3, y) and R (-3, 16)

Question 17.
Solution:
The given points are A (-3, 9), B (2, y) and C (4, -5)

Question 18.
Solution:
The points are given A (8, 1), B (3, -2k) and C (k, -5)

Question 19.
Solution:
The points are given A (2, 1), B (x, y) and C (7, 5)

Question 20.
Solution:
The points are given A (x, y), B (-5, 7) and C (-4, 5)

Question 21.
Solution:
Points are given A (a, 0), B (0, b) and C (1,1)
Points are collinear.
Area of ∆ABC = 0

Question 22.
Solution:
The points are given P (-3, 9), Q (a, ti) and R (4, -5)
Points are collinear.
Area of ∆PQR = 0

Question 23.
Solution:
Vertices of ∆ABC are A (0, -1), B (2, 1) and C (0, 3)

Question 24.
Solution:
Let A (a, a²), B (b, b²) and C (0, 0)
For the points A, B and C to collinear area of ∆ABC must be zero.
Now, area of ∆ABC

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C are helpful to complete your math homework.

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