1.Sodium carbonate is a basic salt because it is a salt of
(a) weak acid and weak base
(b) strong acid and weak base
(c) weak acid and strong base

Answer: (c) weak acid and strong base.
Explation= Sodium carbonate is a basic salt which is formed by the combination of sodium hydroxide and carbonic acid.

Sodium hydroxide is a strong base, while cabonic acid is a weak acid resulting in the formation of sodium carbonate salt.

2NaOH+H2CO3→Na2CO3+2H2O

2. Which gas is evolved when acids react with metals?
   (a) O2
   (b) CO2
   (c) H2
   (d) N2
   Answer: (c) H2
   Metals readily react with acid and produce hydrogen gas which burns with a pop sound.
3. Dilute acid does not produce carbon dioxide on being treated with: a. Marble
   b. Lime
   c. Baking soda
   d. Limestone
   Answer. b. Lime
   In the given options, marble, limestone, and baking soda are all either metal carbonates or bicarbonates. All these when react with a dilute acid then they produce carbon dioxide. Whereas lime which is calcium oxide when react with a dilute acid then they do not produce carbon dioxide.
4. Which of the following salt will give acidic solution when dissolved in water?
   a. NH4Cl
   b. NaCl
   c. Na2CO3
   d. CH3COONa
   Answer. a. NH4Cl
   NaCl is a neutral salt hence, it will give a neutral solution. NH4Cl is an acidic salt hence, it will give an acidic solution. CH3COONa is a basic salt hence, it will give a basic solution.
5. Bleaching powder is used as a disinfectant for water to: a. Make water tastier
   b. Remove all the dirt from water
   c. Make water germ-free
   d. Make water clear
   Answer. Make water germ-free

Bleaching powder, CaOCl2 is used to kill germs and bacteria of water because it is a good disinfecting agent

5. Which one of the following salts will dissolve in water to form an alkaline solution?
   a. Potassium carbonate
   b. Sodium chloride
   c. Sodium carbonate
   d. Potassium sulphate
   Answer. a. Potassium carbonate
   Potassium carbonate will dissolve in water to give alkaline
   solution. This is because they are formed by strong base and weak acid and thus are basic in nature
6. Which among the following represents the chemical formula for ‘Plaster of Paris’?

Explanition= The chemical formula for the plaster of Paris is (CaSO4) H2O and is better known as calcium sulfate hemihydrate.

7. The nature of calcium phosphate is present in tooth enamel is
   (a) Basic
   (b) Amphoteric
   (c) Acidic
   (d) NeutralRead
   ANSWER .Basic
   Calcium phosphate is a family of substances and minerals which consist of calcium ions ( Ca2 + Ca2 + ) with inorganic phosphate anions. In some calcium phosphates, oxides and hydroxides are also present. They are white solids of nutritious value. Calcium phosphates are found in many living organisms.
   Calcium phosphate is basic salt since it is a source of weak phosphoric acid and a slightly stronger base of calcium hydroxide. Calcium phosphate is a mineral comprising calcium ion and phosphate ion which is inorganic in nature. It is present in the crown area of the tooth and also in bones.
   Although calcium phosphate is a very hard substance, it can be dissolved in acids. Hence, we can state that the nature of calcium phosphate is basic and hence, the correct option is option (A).
8. Which one of the following is acidic?
   (a) Lemon juice
   (b) Tomatoes
   (c) Milk
   (d) All
   Answer . All
9. Acids react with metals to liberate _________gas
   (a) Carbon dioxide
   (b) Carbon monoxide
   (c) Hydrogen
   (d) Water
   Answer . Hydrogen
   Explation= acid reacts with metal, hydrogen gas is released.
10. . Generally, when certain metals react with an acid they release _ gas.
    A. Nitrogen
    B. Oxygen
    C. Hydrogen
    D. Argon
    Answer. Hydrogen
11. Which one of the given is commonly known as blue vitriol and is used as a fungicide?
    A. Potassium nitrate

B. Copper sulphate

C. Sodium carbonate

D. Sodium chloride
Answer Copper sulphate
Copper(II) sulfate, also known as cupric sulfate, or copper sulphate, is the inorganic compound with the chemical formula CuSO4(H2O)x, where x can range from 0 to 5. The pentahydrate (x = 5) is the most common form. Older names for this compound include blue vitriol, bluestone, the vitriol of copper, and Roman vitriol.

12. Vinegar is used in pickling as it

A. Is an acid

B. Prevents the growth of microbes

C. Prevents drying of a pickle

D. Increases taste
Answer= Prevents the growth of microbes

13. PH scale of a neutral solution is
    A. 14
    B. 7
    C. 10
    D. 12
    Answer. 7
    The pH scale ranges from 0 to 14. A pH of 7 is neutral. A pH less than 7 is acidic.
14. Butyric acid is found in
    A. Rancid butter
    B. Rancid cake
    C. Stings of bees
    D. All of these
    Answer. Rancid cake
15. An indicator is one kind of the following compound
    A. Strong acid only
    B. Reducing agent
    C. Weak base or acid only
    D. Complex salt
    Answer= Weak base or acid only
    indicators are chemical based compounds used to identify the acidity/alkalinity of a given compound so they should be a very weak acid /base

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Quick Revision Notes :Chapter 2 Acids, Bases, and Salts

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1.In which mode of nutrition an organism de-rives its food from the body of another living organism without killing it?
(a) Saprotrophic nutrition
(b) Parasitic nutrition
(c) Holozoic nutrition
(d) Autotrophic nutrition

Answer: b

2. The mode of nutrition found in fungi is:
(a) Parasitic nutrition
(b) Holozoic nutrition
(c) Autotrophic nutrition
(d) Saprotrophic nutrition

Answer (c )

the mode of nutrition found in fungi is saprotrophic nutrition.

3. The site of photosynthesis in the cells of a leaf is
(a) chloroplast
(b) mitochondria
(c) cytoplasm
(d) protoplasm

Answer. (a) chloroplastThe process of photosynthesis takes place in the chloroplasts, using

chlorophyll, the green pigment involved in photosynthesis.

 4.In amoeba, food is digested in the:
(a) food vacuole
(b) mitochondria
(c) pseudopodia
(d) chloroplast

Answer. (a) food vacuole

 Food vacuole is formed when food is engulfed through phagocytosis.

5.The contraction and expansion movement of the walls of the food pipe is called:
(a) translocation
(b) transpiration
(c) peristaltic movement
(d) digestion

Answer. (c)

Peristalsis Is the Contraction of Muscle Tissue That Helps Move and Break Down Foodstuffs

6. . When a few drops of iodine solution are added to rice water, the solution turns blue- black in colour. This indicates that rice water contains:
(a) fats
(b) complex proteins
(c) starch
(d) simple proteins

Amylose in starch is the main factor for colour formation as iodine comes in contact with beta coils structure of amylose and gives blue – black colour with iodine. Thus, the reaction of iodine with rice water indicates that rice water contains starch.

7. The exit of unabsorbed food material is regu-lated by
(a) liver
(b) anus
(c) small intestine
(d) anal sphincter

(d) anal sphincter

Anal sphincter regulates the unabsorbed food , which releases it in small amounts into the small intestine.

8. The breakdown of pyruvate to give carbon di-oxide, water and energy takes place in
(a) cytoplasm
(b) mitochondria
(c) chloroplast
(d) nucleus

Answer.

(b) mitochondriaThe break down of pyruvate to give carbon dioxide, energy and water takes place in the presence of oxygen and is termed as aerobic respiration. This process takes place in the mitochondria of a cell

9.The kidneys in human beings are a part of the system for

 (a)  nutrition.

(b)  respiration.

(c)  excretion.

(d)  transportation.

Ans. (c) excretion.

Excretion refers to removal of metabolic waste from body. This makes kidneys part of excretory system.

10.  The xylem in plants are responsible for

     (a)  transport of water.

     (b)  transport of food.

     (c)  transport of amino acids.

     (d) transport of oxygen.

    Ans. (a) transport of water.

The role of xylem tissue in plants is to transport water and minerals.

11.  The autotrophic mode of nutrition requires

         (a)  carbon dioxide and water.

          (b)  chlorophyll.

          (c)  sunlight.

          (d)  all of the above.

Ans. (d) All of the above.

12.  The breakdown of pyruvate to give carbon dioxide, water and energy takes place in

          (a)  cytoplasm.

          (b)  chloroplast.

          (c) mitochondria.

(d)  nucleus.

Ans. (c) mitochondria.

The break down of pyruvate to give carbon dioxide, energy and water takes place in the presence of oxygen and is termed as aerobic respiration. This process takes place in the mitochondria of a cell.

 13.Name the pores in a leaf through which respi-ratory exchange of gases takes place.

(a) Lenticels
(b) Vacuoles
(c) Xylem
(d) Stomata

Answer.

(d) Stomata

The pore through which the exchange of respiratory gases takes place is known as stomata. These are the tiny pores that are mainly found under the surfaces of the leaves, epidermis, stems etc.

14.Which plant tissue transports water and min-erals from the roots to the leaf?
(a) Xylem
(b) Phloem
(c) Parenchyma
(d) Collenchyma

Answer. XylemXylem tissue transports water and nutrients from the roots to different parts of the plant, and also plays a

role in structural support in the stem.

15. The movement of food in phloem is called:
(a) transpiration
(b) translocation
(c) respiration
(d) evaporation

Answer. (b) translocation

The transport of food in plants is called translocation. It takes place with the help of a conducting tissue called phloem. Phloem transports glucose, amino acids and other substances from leaves to root, shoot, fruits and seeds.

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1. For some integer m, every odd integer is of the form

(A) m                                      

(B) m + 1

(C) 2m                                    

(D) 2m + 1

Answer:  D

Explanation: As the number 2m will always be even, so if we add 1 to it then, the number will always be odd.

2. If two positive integers a and b are written as a = p³q² and b = pq³; p, q are prime numbers, then HCF (a, b) is:

(A) pq                                                 

(B) pq²

(C) p³q³                                               

(D) p²q²

Answer:  B

Explanation:  Since a = p × p × p × q × q,

                                           b = p × q × q × q

Therefore H.C.F of a and b = pq²

3. The product of a non-zero number and an irrational number is:

(A) always irrational                          

(B) always rational

(C) rational or irrational                     

(D) one

Answer:  A

Explanation: Product of a non-zero rational and an irrational number is always irrational i.e.,

4. If the HCF of 65 and 117 is expressible in the form 65 m – 117, then the value of m is

(A) 4                                                   

(B) 2

(C) 1                                                   

(D) 3

Answer: B

Explanation: By Euclid’s division algorithm,

5 The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is

(A) 13                                                             

(B) 65

(C) 875                                                           

(D) 1750

Answer: A

Explanation: Since 5 and 8 are the remainders of 70 and 125, respectively. Thus after subtracting these remainders from the numbers, we have the numbers

65 = (70 − 5), 117 = (125 − 8) which is divisible by the required number.

Now required number = H.C.F of (65,117)

6. If two positive integers p and q can be expressed as p = ab² and q = a³b; a, b being prime numbers, then LCM (p, q) is

(A) ab                                     

(B) a²b²

(C) a³b²                                   

(D) a³b³

Answer: C

Explanation:

p = a × b × b

q = a × a × a × b

Since L.C.M is the product of the greatest power of each prime factor involved in the numbers

Therefore, L.C.M of p and q = a³b²

7. The values of the remainder r, when a positive integer a is divided by 3 are:

(A) 0, 1, 2, 3                           

(B) 0, 1

(C) 0, 1, 2                               

(D) 2, 3, 4

Answer: C

Explanation:

According to Euclid’s division lemma,

a = 3q + r, where 0  r < 3

As the number is divided by 3.So the remainder cannot be greater than divisor 3 also r is an integer. Therefore, the values of r can be 0, 1 or 2.

(A) one decimal place       

(A) Terminating decimal expansion   

(B) Non–Terminating Non repeating decimal expansion       

(C) Non–Terminating repeating decimal expansion   

(D) None of these

Answer: A

Explanation: After simplification,

As the denominator has factor 5³ × 2² and which is of the type 5^m × 2ⁿ, So this is a terminating decimal expansion.

10. A rational number in its decimal expansion is 327.7081. What would be the prime factors of q when the number is expressed in the p/q form?

(A) 2 and 3                                         

(B) 3 and 5

(C) 2, 3 and 5                                     

(D) 2 and 5

Answer: D

Explanation: This can be explained as,

11. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

(A) 10                                                             

(B) 100

(C) 2060                                                         

(D) 2520

Answer: D

Explanation: Factors of 1 to 10 numbers

L.C.M of numbers from 1 to 10 is =

12. n² – 1 is divisible by 8, if n is

(A) an integer                               

(B) a natural number

(C) an odd integer greater than 1

(D) an even integer

Answer: C

Explanation: n can be even or odd

Case 1: If n is even

Case 2: If n is odd

Which is divisible by 8.

Similarly we can check for any integer.

13. If n is a rational number, then 5²ⁿ − 2²ⁿ is divisible by

(A) 3                                                                           

(B) 7

(C) Both 3 and 7                                                        

(D) None of these

Answer: C

Explanation:

5²ⁿ −2²ⁿ is of the form a²ⁿ − b²ⁿ which is divisible by both (a + b) and (a – b).

So, 5²ⁿ − 2²ⁿ is divisible by both 7, 3.

14. The H.C.F of 441, 567 and 693 is

(A) 1                                                                           

(B) 441

(C) 126                                                                       

(D) 63

Answer: D

Explanation:

693 = 3×3×7×7

567 = 3×3×3×3×7

441 = 3×3×7×11

Therefore H.C.F of 693, 567 and 441 is 63.

15. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?

(A) 2520cm                                                                

(B) 2525cm

(C) 2555cm                                                                

(D) 2528cm

Answer: A

Explanation: We need to find the L.C.M of 40, 42 and 45 cm to get the required minimum distance.

40 = 2×2×2×5

42 = 2×3×7

45 = 3×3×5

L.C.M. = 2×3×5×2×2×3×7 = 2520

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Quick Revision Notes : Real Numbers

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MCQs: Real Numbers

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Question 1.
In the given figure, AD = BC and BD = AC, prove that ∠DAB = ∠CBA.
Solution:

In ∆DAB and ∆CBA, we have
AD = BC [given]
BD = AC [given]
AB = AB [common]
∴ ∆DAB ≅ ∆CBA [by SSS congruence axiom]
Thus, ∠DAB =∠CBA [c.p.c.t.]

Question 2.
In the given figure, ∆ABD and ABCD are isosceles triangles on the same base BD. Prove that ∠ABC = ∠ADC.
Solution:

In ∆ABD, we have
AB = AD (given)
∠ABD = ∠ADB [angles opposite to equal sides are equal] …(i)
In ∆BCD, we have
CB = CD
⇒ ∠CBD = ∠CDB [angles opposite to equal sides are equal] … (ii)
Adding (i) and (ii), we have
∠ABD + ∠CBD = ∠ADB + ∠CDB
⇒ ∠ABC = ∠ADC

Question 3.
In the given figure, if ∠1 = ∠2 and ∠3 = ∠4, then prove that BC = CD.
Solution:

In ∆ABC and ACDA, we have
∠1 = ∠2 (given)
AC = AC [common]
∠3 = ∠4 [given]
So, by using ASA congruence axiom
∆ABC ≅ ∆CDA
Since corresponding parts of congruent triangles are equal
∴ BC = CD

Question 4.
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:
Here, ∠B < ∠A
⇒ AO < BO …..(i)
and ∠C < ∠D
⇒ OD < CO …..(ii)
[∴ side opposite to greater angle is longer]
Adding (i) and (ii), we obtain
AO + OD < BO + CO
AD < BC

Question 5.
In the given figure, AC > AB and D is a point on AC such that AB = AD. Show that BC > CD.
Solution:

Here, in ∆ABD, AB = AD
∠ABD = ∠ADB
[∠s opp. to equal sides of a ∆]
In ∆BAD
ext. ∠BDC = ∠BAD + ∠ABD
⇒ ∠BDC > ∠ABD ….(ii)
Also, in ∆BDC .
ext. ∠ADB > ∠CBD …(iii)
From (ii) and (iii), we have
∠BDC > CD [∵ sides opp. to greater angle is larger]

Question 6.
In a triangle ABC, D is the mid-point of side AC such that BD = 12 AC. Show that ∠ABC is a right angle.
Solution:

Here, in ∆ABC, D is the mid-point of AC.
⇒ AD = CD = 12AC …(i)
Also, BD = 12AC… (ii) [given]
From (i) and (ii), we obtain
AD = BD and CD = BD
⇒ ∠2 = ∠4 and ∠1 = ∠3 …..(iii)
In ∆ABC, we have
∠ABC + ∠ACB + ∠CAB = 180°
⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠1 + ∠2 + ∠1 + ∠2 = 180° [using (iii)]
⇒ 2(∠1 + ∠2) = 180°
⇒ ∠1 + ∠2 = 90°
Hence, ∠ABC = 90°

Question 7.
ABC is an isosceles triangle with AB = AC. P and Q are points on AB and AC respectively such that AP = AQ. Prove that CP = BQ.
Solution:

In ∆ABQ and ∆ACP, we have
AB = AC (given)
∠BAQ = ∠CAP [common]
AQ = AP (given)
∴ By SAS congruence criteria, we have
∆ABQ ≅ ∆ACP
CP = BQ

Question 8.
In the given figure, ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC, AD is extended to intersect BC at P. Show that : (i) ∆ABD ≅ ∆ACD (ii) ∆ABP ≅ ∆ACP

Solution:
(i) In ∆ABD and ∆ACD
AB = AC [given]
BD = CD [given]
AD = AD [common)]
∴ By SSS congruence axiom, we have
∆ABD ≅ ∆ACD
(ii) In ∆ABP and ∆ACP
AB = AC [given]
∠BAP = ∠CAP [c.p.cit. as ∆ABD ≅ ∆ACD]
AP = AP [common]
∴ By SAS congruence axiom, we have
∆ABP ≅ ∆ACP

Question 9.
In the given figure, it is given that AE = AD and BD = CE. Prove that ∆AEB ≅ ∆ADC.

Solution:
We have AE = AD … (i)
and CE = BD … (ii)
On adding (i) and (ii),
we have AE + CE = AD + BD
⇒ AC = AB
Now, in ∆AEB and ∆ADC,
we have AE = AD [given]
AB = AC [proved above]
∠A = ∠A [common]
∴ By SAS congruence axiom, we have
∆AEB = ∆ADC

Question 10
In the given figure, in ∆ABC, ∠B = 30°, ∠C = 65° and the bisector of ∠A meets BC in X. Arrange AX, BX and CX in ascending order of magnitude.

Solution:
Here, AX bisects ∠BAC.
∴ ∠BAX = ∠CAX = x (say)
Now, ∠A + ∠B + C = 180° [angle sum property of a triangle]
⇒ 2x + 30° + 65° = 180°
⇒ 2x + 95 = 180°
⇒ 2x = 180° – 95°
⇒ 2x = 85°
⇒ x = 85∘2 = 42.59
In ∆ABX, we have x > 30°
BAX > ∠ABX
⇒ BX > AX (side opp. to larger angle is greater)
⇒ AX < BX
Also, in ∆ACX, we have 65° > x
⇒ ∠ACX > ∠CAX
⇒ AX > CX [side opp. to larger angle is greater]
⇒ CX > AX … (ii)
Hence, from (i) and (ii), we have
CX < AX < BX

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Quick Revision Notes : Triangles

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MCQs: Triangles

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Question 1.
Define :
(a) a square (b) perpendicular lines.
Solution:
(a) A square : A square is a rectangle having same length and breadth. Here, undefined terms are length, breadth and rectangle.
(b) Perpendicular lines : Two coplanar (in a plane) lines are perpendicular, if the angle between them at the point of intersection is one right angle. Here, the term one right angle is undefined.

Question 2.
In the given figure, name the following :
(i) Four collinear points
(ii) Five rays
(iii) Five line segments
(iv) Two-pairs of non-intersecting line segments.

Solution:
(i) Four collinear points are D, E, F, G and H, I, J, K
(ii) Five rays are DG, EG, FG, HK, IK.
(iii) Five line segments are DH, EI, FJ; DG, HK.
(iv) Two-pairs of non-intersecting line segments are (DH, EI) and (DG, HK).

Question 3.
In the given figure, AC = DC and CB = CE. Show that AB = DE. Write the Euclid’s axiom to support this.

Solution:
We have
AC = DC
CB = CE
By using Euclid’s axiom 2, if equals are added to equals, then wholes are equal.
⇒ AC + CB = DC + CE
⇒ AB = DE.

Question 4.
In figure, it is given that AD=BC. By which Euclid’s axiom it can be proved that AC = BD?

Solution:
We can prove it by Euclid’s axiom 3. “If equals are subtracted from equals, the remainders are equal.”
We have AD = BC
⇒ AD – CD = BC – CD
⇒ AC = BD

Question 5.
In the given figure, AB = BC, BX = BY, show that
AX = CY.

Solution:
Given that AB = BC
and BX = BY
By using Euclid’s axiom 3, equals subtracted from equals, then the remainders are equal, we have
AB – BX = BC – BY
AX = CY

Question 6.

In the above figure, if AB = PQ, PQ = XY, then AB = XY. State True or False. Justify your answer.
Solution:
True. ∵ By Euclid’s first axiom “Things which are equal to the same thing are equal to one another”.
∴ AB = PQ and XY = PQ ⇒ AB = XY

Question 7.
In the given figure, if ∠1 = ∠3, ∠2 = ∠4 and ∠3 = ∠4, write the relation between ∠1 and ∠2, using an Euclid’s axiom.

Solution:
Here, ∠3 = ∠4, ∠1 = ∠3 and ∠2 = ∠4. Euclid’s first axiom says, the things which are equal to equal thing are equal to one another. So ∠1 = ∠2.,

Question 8.
In the given figure, we have ∠1 = ∠2, ∠3 = ∠4. Show that ∠ABC = ∠DBC. State the Euclid’s Axiom used.
Solution:
Here, we have 1 = ∠2 and ∠3 = ∠4. By using Euclid’s Axiom 2. If equals are added to
equals, then the wholes are equal..
∠1 + ∠3 = ∠2 + ∠4
∠ABC = ∠DBC.

Question 9.
In the figure, we have BX and 12 AB =12 BC. Show that BX = BY.

Solution:
Here, BX = 12 AB and BY = 12 BC …(i) [given]
Also, AB = BC [given]
⇒ 12AB = 12BC …(ii)
[∵ Euclid’s seventh axiom says, things which are halves of the same thing are equal to one another]
From (i) and (ii), we have BX = BY

Question 10.
In the given figure, AC = XD, C is mid-point of AB and D is mid-point of XY. Using an Euclid’s axiom, show that AB = XY.

Solution:
∵ C is the mid-point of AB
AB = 2AC
Also, D is the mid-point of XY
XY = 2XD
By Euclid’s sixth axiom “Things which are double of same things are equal to one another.”
∴ AC = XD = 2AC = 2XD ⇒ AB = XY

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MCQs: Introduction to Euclid’s Geometry

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Question 1.
Represent √3 on the number line.
Solution:


On the number line, take OA = 1 unit. Draw AB = 1 unit perpendicular to OA. Join OB.
Again, on OB, draw BC = 1 unit perpendicular to OB. Join OC.
By Pythagoras Theorem, we obtain OC = √3. Using
compasses, with centre O and radius OC, draw an arc, which intersects the number line at point
D. Thus, OD = √3 and D corresponds to √3.

Question 2.
Represent √3.2 on the number line.
Solution:
First of all draw a line of length 3.2 units such that AB = 3.2 units. Now, from point B, mark a distance of 1 unit. Let this point be ‘C’. Let ‘O’ be the mid-point of the distance AC. Now, draw a semicircle with centre ‘O’ and radius OC. Let us draw a line perpendicular to AC passing through the point ‘B’ and intersecting the semicircle at point ‘D’.
∴ The distance BD = √3.2

Now, to represent √3.2 on the number line. Let us take the line BC as number line and point ‘B’ as zero, point ‘C’ as ‘1’ and so on. Draw an arc with centre B and radius BD, which intersects the number line at point ‘E’.
Then, the point ‘E’ represents √3.2.

Question 3.
Express 1.32 + 0.35 as a fraction in the simplest form.
Solution:
Let . x = 1.32 = 1.3222…..(i)

Multiplying eq. (i) by 10, we have
10x = 13.222…
Again, multiplying eq. (i) by 100, we have
100x = 132.222… …(iii)
Subtracting eq. (ii) from (iii), we have
100x – 10x = (132.222…) – (13.222…)
90x = 119
⇒ x = 11990
Again, y = 0.35 = 0.353535……
Multiply (iv) by 100, we have …(iv)
100y = 35.353535… (v)
Subtracting (iv) from (u), we have
100y – y = (35.353535…) – (0.353535…)
99y = 35
y = 3599

Question 4.
Find the square root of 10 + √24 + √60 + √40.
Solution:

Question 5.
If x = 9 + 4√5, find the value of √x – 1x√.
Solution:
Here,
x = 9 + 4√5
x = 5 + 4 + 2 x 2√5
x = (√5² + (2² + 2 x 2x √5).
x = (√5 + 2)²
√x = √5 + 2
Now, 1x√ = 15√+2

Question 6.
If x = 15√−2 , find the value of x³ – 3² – 5x + 3
Solution:

∴ x – 2 = √5
Squaring both sides, we have
x² – 4x + 4 = 5
x² – 4x – 1 = 0 …(i)
Now, x³ – 3² – 5x + 3 = (x² – 4x – 1) (x + 1) + 4
= 0 (x + 1) + 4 = 4 [using (i)]

Question 7.
Find ‘x’, if 2^x-7 × 5^x-4 = 1250.
Solution:
We have 2^x-7 × 5^x-4 = 1250
⇒ 2^x-7 × 5^x-4 = 2 5 × 5 × 5 × 5
⇒ 2^x-7 × 5^x-4 = 21 × 54
Equating the powers of 2 and 5 from both sides, we have
⇒ x – 7 = 1 and x – 4 = 4
⇒ x = 8 and x = 8
Hence, x = 8 is the required value.

Question 8.
Evaluate:

Solution:

`Question 9.
If x = p+q√+p−q√p+q√−p−q√, then prove that q² – 2px + 9 = 0.
Solution:


Squaring both sides, we have
⇒ q²x² + p² – 2pqx = p² – q²
⇒ q²x² – 2pqx + q² = 0
⇒ q(q² – 2px + q) = 0
⇒ qx² – 2px + q = 0 (∵ q ≠ 0)

Question 10.
If a = 13−11√ and b = 1a, then find a² – b²
Solution:

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Question 1.
In the given figure, AB || CD, ∠FAE = 90°, ∠AFE = 40°, find ∠ECD.

Solution:
In AFAE,
ext. ∠FEB = ∠A + F
= 90° + 40° = 130°
Since AB || CD
∴ ∠ECD = FEB = 130°
Hence, ∠ECD = 130°.

Question 2.
In the fig., AD and CE are the angle bisectors of ∠A and ∠C respectively. If ∠ABC = 90°, then find ∠AOC.

Solution:
∵ AD and CE are the bisector of ∠A and ∠C

In ∆AOC,
∠AOC + ∠OAC + ∠OCA = 180°
⇒ ∠AOC + 45o = 180°
⇒ ∠AOC = 180° – 45° = 135°.

Question 3.
In the given figure, prove that m || n.

Solution:
In ∆BCD,
ext. ∠BDM = ∠C + ∠B
= 38° + 25° = 63°
Now, ∠LAD = ∠MDB = 63°
But, these are corresponding angles. Hence,
m || n

Question 4.
In the given figure, two straight lines PQ and RS intersect each other at O. If ∠POT = 75°, find the values of a, b, c.

Solution:
Here, 4b + 75° + b = 180° [a straight angle]
5b = 180° – 75° = 105°
b – 105∘5 = 21°
∴ a = 4b = 4 × 21° = 84° (vertically opp. ∠s]
Again, 2c + a = 180° [a linear pair]
⇒ 2c + 84° = 180°
⇒ 2c = 96°
⇒ c = 96∘2 = 48°
Hence, the values of a, b and c are a = 84°, b = 21° and c = 48°.

Question 5.
In figure, if AB || CD. If ∠ABR = 45° and ∠ROD = 105°, then find ∠ODC.

Solution:

Through O, draw a line ‘l’ parallel to AB.
⇒ line I will also parallel to CD, then
∠1 = 45°[alternate int. angles]
∠1 + ∠2 + 105° = 180° [straight angle]
∠2 = 180° – 105° – 45°
⇒ ∠2 = 30°
Now, ∠ODC = ∠2 [alternate int. angles]
= ∠ODC = 30°

Question 6.
In the figure, ∠X = 72°, ∠XZY = 46°. If YO and ZO are bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OYZ and ∠YOZ.

Solution:
In ∆XYZ, we have
∠X + XY + ∠Z = 180°
⇒ ∠Y + ∠Z = 180° – ∠X
⇒ ∠Y + ∠Z = 180° – 72°
⇒ Y + ∠Z = 108°
⇒ 12 ∠Y + 12∠Z = 12 × 108°
∠OYZ + ∠OZY = 54°
[∵ YO and ZO are the bisector of ∠XYZ and ∠XZY]
⇒ ∠OYZ + 12 × 46° = 54°
∠OYZ + 23° = 54°
⇒ ∠OYZ = 549 – 23° = 31°
In ∆YOZ, we have
∠YOZ = 180° – (∠OYZ + ∠OZY)
= 180° – (31° + 23°) 180° – 54° = 126°

Question 7.
In the given figure, m and n are two plane mirrors perpendicular to each other. Show that incident rays CA is parallel to reflected ray BD.

Solution:
Let normals at A and B meet at P.
As mirrors are perpendicular to each other, therefore, BP || OA and AP || OB.
So, BP ⊥ PA i.e., ∠BPA = 90°
Therefore, ∠3 + ∠2 = 90° [angle sum property] …(i)
Also, ∠1 = ∠2 and ∠4 = ∠3 [Angle of incidence = Angle of reflection]
Therefore, ∠1 + ∠4 = 90° [from (i)) …(ii]
Adding (i) and (ii), we have
∠1 + ∠2 + ∠3 + ∠4 = 180°
i.e., ∠CAB + ∠DBA = 180°
Hence, CA || BD

Question 8
If in ∆ABC, the bisectors of ∠B and ∠C intersect each other at O. Prove that ∠BOC = 90° + 12∠ A.

Solution:
Let ∠B = 2x and ∠C = 2y
∵OB and OC bisect ∠B and ∠C respectively.
∠OBC = 12∠B = 12 × 2x = x
and ∠OCB = 12∠C = 12 × 2y = y
Now, in ∆BOC, we have
∠BOC + ∠OBC + ∠OCB = 180°
⇒ ∠BOC + x + y = 180°
⇒ ∠BOC = 180° – (x + y)
Now, in ∆ABC, we have
∠A + 2B + C = 180°
⇒ ∠A + 2x + 2y = 180°
⇒ 2(x + y) = 12(180° – ∠A)
⇒ x + y = 90° – 12∠A …..(ii)
From (i) and (ii), we have
∠BOC = 180° – (90° – 12∠A) = 90° + 12 ∠A

Question 9.
In figure, if I || m and ∠1 = (2x + y)°, ∠4 = (x + 2y)° and ∠6 = (3y + 20)°. Find ∠7 and ∠8.

Solution:
Here, ∠1 and ∠4 are forming a linear pair
∠1 + ∠4 = 180°
(2x + y)° + (x + 2y)° = 180°
3(x + y)° = 180°
x + y = 60
Since I || m and n is a transversal
∠4 = ∠6
(x + 2y)° = (3y + 20)°
x – y = 20
Adding (i) and (ii), we have
2x = 80 = x = 40
From (i), we have
40 + y = 60 ⇒ y = 20
Now, ∠1 = (2 x 40 + 20)° = 100°
∠4 = (40 + 2 x 20)° = 80°
∠8 = ∠4 = 80° [corresponding ∠s]
∠1 = ∠3 = 100° [vertically opp. ∠s]
∠7 = ∠3 = 100° [corresponding ∠s]
Hence, ∠7 = 100° and ∠8 = 80°

Question 10.
In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28o and ∠QRT = 65°. Find the values of x, y and z.

Solution:
Here, PQ || SR .
⇒ ∠PQR = ∠QRT
⇒ x + 28° = 65°
⇒ x = 65° – 28° = 37°
Now, in it. ∆SPQ, ∠P = 90°
∴ ∠P + x + y = 180° [angle sum property]
∴ 90° + 37° + y = 180°
⇒ y = 180° – 90° – 37° = 53°
Now, ∠SRQ + ∠QRT = 180° [linear pair]
z + 65° = 180°
z = 180° – 65° = 115°

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Q.1  In a rectangle, one diagonal is inclined to one of its sides at 25°. Measure the acute angle between the two diagonals.

Solution:

Let ABCD be a rectangle where AC and BD are the two diagonals which are intersecting at point O.

Now, assume ∠BDC = 25° (given)

Now, ∠BDA = 90° – 25° = 65°

Also, ∠DAC = ∠BDA, (as diagonals of a rectangle divide the rectangle into two congruent right triangles)

So, ∠BOA = the acute angle between the two diagonals = 180° – 65° – 65° = 50°

Q.2. Is it possible to draw a quadrilateral whose all angles are obtuse angles?

Solution:

It is known that the sum of angles of a quadrilateral is always 360°. To have all angles as obtuse, the angles of the quadrilateral will be greater than 360°. So, it is not possible to draw a quadrilateral whose all angles are obtuse angles.

Q.2 Prove that the angle bisectors of a parallelogram form a rectangle. 

Solution:

LMNO is a parallelogram in which bisectors of the angles L, M, N, and O intersect at P, Q, R and S to form the quadrilateral PQRS.
LM || NO (opposite sides of parallelogram LMNO)
L + M = 180 (sum of consecutive interior angles is 180o)
MLS + LMS = 90
In LMS, MLS + LMS + LSM = 180
90 + LSM = 180
LSM = 90
RSP = 90 (vertically opposite angles)
SRQ = 90, RQP = 90 and SPQ = 90
Therefore, PQRS is a rectangle.

Q3. In a trapezium ABCD, AB∥CD. Calculate ∠C and ∠D if ∠A = 55° and ∠B = 70°

Solution:

In a trapezium ABCD, ∠A + ∠D = 180° and ∠B + ∠C = 180°

So, 55° + ∠D = 180°

Or, ∠D = 125°

Similarly,

70° + ∠C = 180°

Or, ∠C = 110°

Q4. Calculate all the angles of a parallelogram if one of its angles is twice its adjacent angle.

Solution:

Let the angle of the parallelogram given in the question statement be “x”.

Now, its adjacent angle will be 2x.

It is known that the opposite angles of a parallelogram are equal.

So, all the angles of a parallelogram will be x, 2x, x, and 2x

As the sum of interior angles of a parallelogram = 360°,

x + 2x + x + 2x = 360°

Or, x = 60°

Thus, all the angles will be 60°, 120°, 60°, and 120°.

Q5. Calculate all the angles of a quadrilateral if they are in the ratio 2:5:4:1.

Solution:

As the angles are in the ratio 2:5:4:1, they can be written as-

2x, 5x, 4x, and x

Now, as the sum of the angles of a quadrilateral is 360°,

2x + 5x + 4x + x = 360°

Or, x = 30°

Now, all the angles will be,

2x =2 × 30° = 60°

5x = 5 × 30° = 150°

4x = 4 × 30° = 120°, and

x = 30°

Q 6 Prove that a diagonal of a parallelogram divide it into two congruent triangles. [CBSE March 2012]
Solution. Given : A parallelogram ABCD and AC is its diagonal.

Question. 7 ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see fig.). Show that :
(i) AAPB ≅ ACQD (ii) AP = CQ            [CBSE March 2012]
Solution.

Q.8

Solution.

Q.9

Solution.

Q.10

Solution.

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Question.1 In figure, TR ⊥ PS, PQ // TR and PS // QR. If QR = 8 cm, PQ = 3 cm and SP = 12 cm, find arquad. PQRS).[CBSE-14-17DIG1U]
Solution.

Question.2
]
Solution.

Question.3

Solution.

Question.4

Solution.

Question.5

Solution.

Question.6

Solution.

Question. 7 Prove that parallelogram on equal bases and between the same parallels are equal in area. [CBSE March 2012]
Solution.

Question.8

Solution.

Question.9

Solution.

Question.10

Solution.

Question. 21

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Question 1.
In the figure, O is the centre of a circle passing through points A, B, C and D and ∠ADC = 120°. Find the value of x.

Solution:
Since ABCD is a cyclic quadrilateral
∠ADC + ∠ABC = 180°
[∴ opp. ∠s of a cyclic quad. are supplementary]
120° + ∠ABC = 180°
∠ABC = 180° – 120° = 60°
Now, ∠ACB = 90° [angle in a semicircle]
In rt. ∠ed ∆CB, ∠ACB = 90°
∠CAB + ∠ABC = 90°
x + 60° = 90°
x = 90° -60°
x = 30°

Question 2.
In the given figure, O is the centre of the circle, ∠AOB = 60° and CDB = 90°. Find ∠OBC.

Solution:
Since angle subtended at the centre by an arc is double the angle
subtended at the remaining part of the circle.
∴ ∠ACB = 13 ∠AOB = 13 x 60° = 30°
Now, in ACBD, by using angle sum property, we have
∠CBD + ∠BDC + ∠DCB = 180°
∠CBO + 90° + ∠ACB = 180°
[∵ ∠CBO = ∠CBD and ∠ACB = ∠DCB are the same ∠s]
∠CBO + 90° + 30° = 180°
∠CBO = 180o – 90° – 30° = 60°
or ∠OBC = 60°

Question 3.
In the given figure, O is the centre of the circle with chords AP and BP being produced to R and Q respectively. If ∠QPR = 35°, find the measure of ∠AOB.

Solution:
∠APB = ∠RPQ = 35° [vert. opp. ∠s]
Now, ∠AOB and ∠APB are angles subtended by an arc AB at centre and at the remaining part of the circle.
∴ ∠AOB = 2∠APB = 2 × 35° = 70°

Question 4.
In the figure, PQRS is a cyclic quadrilateral. Find the value of x.

Solution:
In ∆PRS, by using angle sum property, we have
∠PSR + ∠SRP + ∠RPS = 180°
∠PSR + 50° + 35o = 180°
∠PSR = 180° – 85o = 95°
Since PQRS is a cyclic quadrilateral
∴ ∠PSR + ∠PQR = 180°
[∵ opp. ∠s of a cyclic quad. are supplementary]
95° + x = 180°
x = 180° – 95°
x = 85°

Question 5.
In the given figure, ∠ACP = 40° and BPD = 120°, then find ∠CBD.

Solution:
∠BDP = ∠ACP = 40° [angle in same segment]
Now, in ∆BPD, we have
∠PBD + ∠BPD + ∠BDP = 180°
⇒ ∠PBD + 120° + 40° = 180°
⇒ ∠PBD = 180° – 160o = 20°
or ∠CBD = 20°

Question 6.
In the given figure, if ∠BEC = 120°, ∠DCE = 25°, then find ∠BAC.

Solution:
∠BEC is exterior angle of ∆CDE.
∴ ∠CDE + ∠DCE = ∠BEC
⇒ ∠CDE + 25° = 120°
⇒ ∠CDE = 95°
Now, ∠BAC = ∠CDE [∵ angle in same segment are equal]
⇒ ∠BAC = 95°

Question 7.
In the given figure, PQR = 100°, where P, Q and R are points on a circle with centre O. Find LOPR.

Solution:
Take any point A on the circumcircle of the circle.
Join AP and AR.
∵ APQR is a cyclic quadrilateral.
∴ ∠PAR + ∠PQR = 180° [sum of opposite angles of a cyclic quad. is 180°]
∠PAR + 100° = 180°
⇒ Since ∠POR and ∠PAR are the angles subtended by an arc PR at the centre of the circle and circumcircle of the circle.
∠POR = 2∠PAR = 2 x 80° = 160°
∴ In APOR, we have OP = OR [radii of same circle]
∠OPR = ∠ORP [angles opposite to equal sides]
Now, ∠POR + ∠OPR + ∠ORP = 180°
⇒ 160° + ∠OPR + ∠OPR = 180°
⇒ 2∠OPR = 20°
⇒ ∠OPR = 10°

Question 8.
In figure, ABCD is a cyclic quadrilateral in which AB is extended to F and BE || DC. If ∠FBE = 20° and DAB = 95°, then find ∠ADC.

Solution:
Sum of opposite angles of a cyclic quadrilateral is 180°
∴ ∠DAB + ∠BCD = 180°
⇒ 95° + ∠BCD = 180°
⇒ ∠BCD = 180° – 95° = 85°
∵ BE || DC
∴ ∠CBE = ∠BCD = 85°[alternate interior angles]
∴ ∠CBF = CBE + ∠FBE = 85° + 20° = 105°
Now, ∠ABC + 2CBF = 180° [linear pair]
and ∠ABC + ∠ADC = 180° [opposite angles of cyclic quad.]
Thus, ∠ABC + ∠ADC = ∠ABC + 2CBF
⇒ ∠ADC = CBF
⇒ ∠ADC = 105° [∵ CBF = 105°]

Question 9
Equal chords of a circle subtends equal angles at the centre.

Solution:
Given : In a circle C(O, r), chord AB = chord CD
To Prove : ∠AOB = ∠COD.
Proof : In ∆AOB and ∆COD
AO = CO (radii of same circle]
BO = DO [radii of same circle]
Chord AB = Chord CD (given]
⇒ ∆AOB = ACOD [by SSS congruence axiom]
⇒ ∠AOB = COD (c.p.c.t.]

Question 10.
In the given figure, P is the centre of the circle. Prove that : ∠XPZ = 2(∠X∠Y + ∠YXZ).

Solution:
Arc XY subtends ∠XPY at the centre P and ∠XZY in the remaining part of the circle.
∴ ∠XPY = 2 (∠X∠Y)
Similarly, arc YZ subtends ∠YPZ at the centre P and ∠YXZ in the remaining part of the circle.
∴ ∠YPZ = 2(∠YXZ) ….(ii)
Adding (i) and (ii), we have
∠XPY + ∠YPZ = 2 (∠XZY + ∠YXZ)
∠XP2 = 2 (∠XZY + ∠YXZ)

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