CIRCLES C7

In the figure below, find the value of <RUS.

Solution

<RSU = <TRU

∴ <RSU = 54⁰ [angles in alternate segments are equal.]

Consider ∆SRU

<SRU + <RUS+ <RSU = 180⁰ [total degrees of a triangle]

 (46⁰ + 54⁰) + <RUS + 54= 180⁰

100⁰ + <RUS + 54⁰ = 180⁰

154⁰ + <RUS= 180⁰

<RUS= 180⁰ - 154⁰

<RUS= 26⁰  

∴ <RUS = 26⁰  

PROBABILITY C13

A bag contains 6 red stones and 8 blue stones. Two balls are taken from the bag. What is the probability that they are both red?

Solution

(This is a problem with replacement)

n(R) = 6, n(B) = 8, n(S) = 14

P(R) = n(R)

            n(S)

1st pick = 6/14

2nd pick = 6/14 as well.

P(R) =  6      x    6

          14          14

P(R) =    36    =      4 

             196           49  

Therefore Probability of drawing a red stone is 4/49     

LOGARITHMS B35

If log_x27 + log₂256 = 11; find x

solution

log_x27 + log₂256 = 11

log_x27 + log₂2⁸ = 11

log_x27 + 8log₂2 = 11

log_x27 + 8 x 1 = 11

log_x27 + 8 = 11

log_x27 = 11- 8

log_x27 = 3

27 = x³

3³ = x³      Powers cancel out

x = 3

Hence x = 3.

CIRCLES C6

In the following figure, prove that the angles in the same segment of a circle are equal.

Solution

Given : A circle with centre O and the angles ∠USV and ∠UTV in the same segment formed by the chord UV (or arc UAV)

Required to prove: ∠USV = ∠UTV

Construction : Join OU and OV.

Proof :

Try to remember that angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle, hence we have,

∠UOV = 2 ∠USV ...............(i)

and,

∠UOV = 2∠UTV …………...(ii)

Since (i) = (ii), we get,

2∠USV = 2∠UTV

2∠USV = 2∠UTV                 (dividing by 2 both sides)

2             2

∠USV = ∠UTV

∴ ∠USV =∠UTV   proved!

PROBABILITY C12

A certain man takes 2 cards from a standard deck of 52 cards.

What will be the probability of getting a pair of hearts?

Solution

(This is a problem with replacement)

n(E) = 13, n(S) = 52

P(E) = n(E)

           n(S)

1st pick = 13/52

2nd pick = 13/52 as well.

P(2 hearts) =   13      x    13

                       52            52

P(2 hearts) =    169    =      1 

                        2704         16  

Therefore Probability of drawing 2 heart cards is 1/16     

TRY THIS............

A certain man takes 2 cards from a standard deck of 52 cards.

What will be the probability of getting a pair of spades?

LOGARITHMS B34

If log_x625 + log₃81 = 8; find x

solution

log_x625 + log₃81 = 8

log_x625 + log₃3⁴ = 8

log_x625 + 4log₃3 = 8

log_x625 + 4 x 1 = 8

log_x625 + 4 = 8

log_x625 = 8- 4

log_x625 = 4

625 = x⁴

5⁴ = x⁴      Powers cancel out

x = 5

Hence x = 5.

CIRCLES C5

In the figure below, find the value of <HLJ.

Solution

<HJL = <LHK

∴ <HJL = 54⁰ [angles in alternate segments are equal.]

Consider ∆HJL

<JHL + <HLJ+ <HJL = 180⁰ [total degrees of a triangle]

 (44⁰ + 54⁰) + <HLJ + 54= 180⁰

98⁰ + <HLJ + 54⁰ = 180⁰

152⁰ + <HLJ= 180⁰

<HLJ= 180⁰ - 152⁰

<HLJ= 28⁰  

∴ <HLJ= 28⁰  

PROBABILITY C11

A bag contains 6 red balls and 8 blue balls. Two balls are taken from the bag. What is the probability that they are both blue?

Solution

(This is a problem with replacement)

n(R) = 6, n(B) = 8, n(S) = 14

P(B) = n(B)

            n(S)

1st pick = 8/14

2nd pick = 8/14 as well.

P(B) =  6      x    6

          14          14

P(B) =    36    =      4 

             196           49  

Therefore Probability of drawing a blue ball is 4/49     

TRY THIS ..............

A bag contains 10 purple stones and 12 blue stones. Two stones are taken from the bag. What is the probability that they are both blue?

LOGARITHMS B33

If log_x625 + log₃27 = 7; find x

solution

log_x625 + log₃27 = 7

log_x625 + log₃3³ = 7

log_x625 + 3log₃3 = 7

log_x625 + 3 x 1 = 7

log_x625 + 3 = 7

log_x625 = 7- 3

log_x625 = 4

625 = x⁴

5⁴ = x⁴      Powers cancel out

x = 5

Hence x = 5.

CIRCLES C4

In the figure below, find the value of <MNQ.

Solution

<MQN = <PMN.

∴ <MQN = 52⁰ [angles in alternate segments are equal.]

Consider ∆MQN

<QMN + <MNQ+ <MQN = 180⁰ [total degrees of a triangle]

 (38⁰ + 52⁰) + <MNQ + 52= 180⁰

90⁰ + <MNQ + 52⁰ = 180⁰

142⁰ + <MNQ= 180⁰

<MNQ= 180⁰ -142⁰

<MNQ= 38⁰  

∴ <MNQ= 38⁰