Class 11
A card Is drawn at random from a well-shuffled deck o f playing cards. Find the probability that the card was drawn Is
One card Is drawn from a well-shuffled deck of 52 cards. Find the probability o f getting(v) a Jack o f hearts (vl) a spade.
Find the value of sec (-1470 )
Answer: To find: Value of sec (-1470°) We have, sec (-1470°) = sec (1470°) [∵ sec(-θ) = sec θ] = sec [90° × 16 + 30°] Clearly, 1470° is in I Quadrant and the multiple of 90° is even = sec 30° $\sec...
One card Is drawn from a well-shuffled deck of 52 cards. Find the probability o f getting
(III)a red face card
(Iv) a queen o f black suit
Find the value of sin 405°
Answer: To find: Value of sin 405° We have, sin 405° = sin [90° × 4 + 45°] = sin 45° [Clearly, 405° is in I Quadrant and the multiple of 90° is even] $\sin {{405}^{\circ...
Find the value of 
Answer: We have: $ co\sec \left( -\frac{41\pi }{4} \right) $ We know that: $ co\sec \left( -\frac{41\pi }{4} \right)=-co\sec \left( \frac{41\pi }{4} \right) $ [∵ cosec(-θ) = -cosec θ] Putting π =...
Find the value of 
.
Answer: We have $\sec \left( -\frac{25\pi }{3} \right)$ We know that: $\sec \left( -\frac{25\pi }{3} \right)=\sec \left( \frac{25\pi }{3} \right)$ [∵ sec(-θ) = sec θ] Putting π = 180° = sec[25 ×...
Find the value of 
Answer: We have: = cot (13 × 45°) = cot (585°) = cot [90° × 6 + 45°] = cot 45° [Clearly, 585° is in III Quadrant and the multiple of 90° is even] = 1 [∵ cot 45° = 1]
Find the general solution of each of the following equations: cot x + tan x = 2 cosec x
Answer: Given, cot x + tan x = 2 cosec x cos2x + sin2x = 2 sinx cosx cosec x 1 = sin 2x cosec x
Find the general solution of each of the following equations: sin x tan x – 1 = tan x – sin x
Answer: Given, sin x tan x – 1 = tan x – sin x sin x(tan x + 1) = tan x + 1
Find the general solution of each of the following equations: 2 tan x – cot x + 1 = 0
Answer: Given, 2 tan x – cot x + 1 = 0 ⇒ 2tan2x – 1 + tan x = 0 ⇒ 2tan2x – 1 + 2tan x – tan x = 0 ⇒ 2tanx(tan x +1) – (1+ tan x) = 0
Find the general solution of each of the following equations: √???? cos x + sin x = 1
Answer: Given, √???? cos x + sin x = 1
Find the general solution of each of the following equations: cos x – sin x = -1
Answer: Given, cos x – sin x = -1 $\begin{array}{l} = > cox\left( {x + \frac{\pi }{4}} \right) = \frac{{ - 1}}{{\sqrt 2 }}\\ = > \cos \frac{{3x}}{4} \end{array}$
One card Is drawn from a well-shuffled deck of 52 cards. Find the probability o f getting
(I)a king o f red suit
(II) a face card
A card Is drawn at random from a well-shuffled deck o f 52 cards. Find the probability of getting
(I)A queen
(II)A diamond
(III) A king or an ace
(Iv) A red ace.
One card Is drawn from a well-shuffled deck of 52 cards. Find the probability o f drawing
(I) An ace
(II)A 4 of spades
(III) A ‘9’ o f a black suit
(Iv) a red king.
Find the general solution of each of the following equations: 
Answers: (i) Given, $\begin{array}{l} sin 2x = \frac{1}{2} \end{array}$ (ii) Given, $\begin{array}{l} \cos 3x = \frac{1}{{\sqrt 2 }}\\ \end{array}$...
Find the general solution of the following equation:
Answer: (i) Given, $\begin{array}{l} \tan \frac{{2x}}{3} = \sqrt 3 \\ \end{array}$
Find the general solution of each of the following equations: 
Answers: (i) Given, $\begin{array}{l} \sec 3x = - 2 \end{array}$ (ii) Given, $\begin{array}{l} \cot 4x = - 1\\...
Find the general solution of the following equation: 
Answer: Given, $\begin{array}{l} \cos ec3x = \frac{{ - 2}}{{\sqrt 3 }}\\ \end{array}$
Find the general solution of the following equation: 
Answer: Given, $\begin{array}{l} {\tan ^2}x = 1\\ \end{array}$
Find the general solution of each of the following equations: 
Answers: (i) Given, $\begin{array}{l} 4{\cos ^2}x = 1 \end{array}$ (ii) Given, $\begin{array}{l} 4{\sin ^2}x - 3 = 0\\ \end{array}$ ...
Find the general solution of each of the following equations: 
Answers: (i) Given, $\begin{array}{l} \cos 3x = \cos 2x \end{array}$ (ii) Given, $\begin{array}{l} \cos 5x = \sin 3x\\ \end{array}$ ...
There are 40 students In a class o f whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. He writes the name o f each student on a separate, the card being Identical. Then she puts cards In a bag and stirs them thoroughly. She then draws one card from the bag. What Is the probability that the name written on the card Is the name of
(I) A girl?
(II)A boy?
Find the general solution of the following equation: 
Answer: Given, $\begin{array}{l} \cos mx = \sin nx \end{array}$ Formula used: cos ???? = cos ???? ⇒ ???? = 2k ???? ± ???? , k ∈ I
Find the general solution of each of the following equations: sin x = tan x
Answer: Given, sin x = tan x ⇒ sin x = sin x ÷ cos x sin x = 0 or cos x = 1 = cos(0)
Find the general solution of each of the following equations: 4sin x cos x + 2sin x + 2cos x + 1 = 0
Answer: Given, 4sin x cos x + 2sin x + 2cos x + 1 = 0 ⇒ 2sin x (2cos x + 1) + 2cos x + 1 = 0 (2cos x + 1) (2sin x + 1) = 0
Find the general solution of each of the following equations: 
Answer: Given, ${\sec ^2}2x = 1 - \tan 2x$ ⇒ 1 + tan22x+ tan 2x = 1 ⇒ tan 2x (1+tan 2x) = 0
Find the general solution of each of the following equations: 
Answer: Given, ${\tan ^3}x - 3\tan x = 0$ => tan x(tan2 x-3)= 0 => tan x = 0 (or) tan x = $ \pm \sqrt 3 $
Find the general solution of each of the following equations: sin x + sin 3x + sin 5x = 0
Answer: Given, sin x + sin 3x + sin 5x = 0 ⇒ sin 3x + 2sin 3x cos 2x= 0 ⇒ sin 3x (1 + 2cos 2x) = 0
Find the general solution of each of the following equations: sin x tan x – 1 = tan x – sin x
Answer; Given, sin x tan x – 1 = tan x – sin x ⇒ sin x(tan x + 1) = tan x + 1
Find the general solution of each of the following equations: cos x + sin x = 1
Answer: Given, cos x + sin x = 1
A bag contains lemon-flavored candles only. Hema takes out 1 candy without looking Into the bag. What Is the probability that she takes out
(I)An orange-flavored candy
(II)A lemon-flavored candy
(I) A lo t o f 20 bulbs contain 4 defective ones. 1 bulb Is drawn at random from the lot. What Is the probability that this bulb Is defective? (II) Suppose the ball drawn In
(I) Is not defective and not replaced. Now, ball Is drawn at random from the rest. What Is the probability that this bulb Is not defective?
A box contains 90 discs which are numbered from 1 to 90 If one disc Is drawn at random from the box, find the probability that It bears
(I) A two-digit number
(II) A perfect square number
(III) A number divisible by 5.
A lot consists o f 144 ballpoint pens o f which 20 are defective and others good. Tanvl will buy a pen If It Is a good but will not buy If It Is defective. The shopkeeper draws 1 pen at random and gives It to her. What Is the probability that
(I) She will buy It,
(II) She will not buy It?
12 defective pens are accidentally mixed with 132 good ones, It Is not possible to Just look at pen and tell whether or not It Is defective. 1 pen Is taken out at random from this lot. Find the probability that the pen taken out Is good one.
A game of chance consists of spinning and arrow which Is equally likely to come to the rest pointing to one of the numbers 1 , 2 ,3 ,4 12 as shown In the figure. What Is the probability that It will point to
(III)A prime number
(Iv) A number which Is a multiple o f 5
A game of chance consists of spinning and arrow which Is equally likely to come to the rest pointing to one of the numbers 1 , 2 ,3 ,4 12 as shown In the figure. What Is the probability that It will point to
(I)6
(II)An even number
Cards marked with numbers 5 to 50 are placed In a box and mixed thoroughly. A card Is drawn from the box at random. Find the probability that the number on the taken out card Is
Two dice are rolled together. Find the probability o f getting such numbers on the two dice whose product Is 12.
Find the general solution of each of the following equations: 
Answer: Given, $\begin{array}{l} \tan x = - 1\\ \end{array}$
Find the general solution of each of the following equations: 
Answers: (i) Given, $\begin{array}{l} \cos x = \frac{{ - 1}}{2} \end{array}$ (ii) Given, $\begin{array}{l} \cos ecx = - \sqrt 2 \\ \end{array}$ ...
Find the general solution of the following equation: 
Answer: Given, $\begin{array}{l} \sec x = \sqrt 2 \\ \end{array}$
Find the general solution of each of the following equations: 
Answers: (i) Given, $\begin{array}{l} \sin x = \frac{{\sqrt 3 }}{2} \end{array}$ (ii) Given, $\begin{array}{l} \cos x = 1\\ \end{array}$...
Find the general solution of the following equation: 
Answer: Given, $\begin{array}{l} \tan \left( {2x - \frac{\pi }{4}} \right) = 0\\ \end{array}$
Find the general solution of each of the following equations: 
Answers: (i) Given, $\begin{array}{l} \tan 2x = 0 \end{array}$ (ii) Given, $\begin{array}{l} \tan \left( {3x + \frac{\pi }{6}} \right) = 0\\ \end{array}$...
Find the general solution of each of the following equations: 
Answers: (i) Given, $\begin{array}{l} \cos \frac{{5x}}{2} = 0 \end{array}$ (ii) Given, $\begin{array}{l} \cos \left( {x + \frac{\pi }{{10}}} \right) = 0\\...
Find the general solution of each of the following equations: 
Answers: (i) Given, $\begin{array}{l} \sin \left( {x + \frac{\pi }{5}} \right) = 0 \end{array}$ (ii) Given, $\begin{array}{l} \cos 2x = 0\\ \end{array}$ ...
Find the general solution of each of the following equations: 
Answers: (i) Given, $\begin{array}{l} \sin 3x = 0 \end{array}$ (ii) Given, $\begin{array}{l} \sin \frac{{3x}}{2} = 0\\ \end{array}$
Find the principal solutions of each of the following equations : 
Answers: (i) Given, $\begin{array}{l} \tan x = - \sqrt 3 \end{array}$ (ii) Given, $\begin{array}{l} \sqrt 3 \sec x + 2 = 0\\ \end{array}$...
Find the principal solutions of each of the following equations : 
Answers: (i) Given, $\begin{array}{l} \tan x = - 1 \end{array}$ (ii) Given, $\begin{array}{l} \sqrt 3 \cos ecx + 2 = 0\\ \end{array}$ ...
Find the principal solutions of each of the following equations : 
Answers: (i) Given, $\begin{array}{l} \sin x = - \frac{1}{2} \end{array}$ (ii) Given, $\begin{array}{l} \sqrt 2 \cos x + 1 = 0\\ \end{array}$ ...
Find the principal solutions of each of the following equations: 
Answers: (i) Given, $\begin{array}{l}\cos ecx = 2\\ \end{array}$ (ii) Given, $\begin{array}{l}\sec x = \frac{2}{{\sqrt 3 }}\\ \end{array}$...
Find the principal solution of the following equation: 
Answer: Given, $\begin{array}{l}\tan x = \sqrt 3\\ \end{array}$ ...
Find the principal solutions of each of the following equations: 
Answers: (i) Given, $\begin{array}{l}\sin x = \frac{{\sqrt 3 }}{2}\\ \end{array}$ (ii) Given, $\begin{array}{l}\cos x = \frac{1}{2}\\ \end{array}$...
Find the value of 
.
Answer: We know that, tan(-θ) = - tan θ = - tan 60° = -√3 [∵ tan 60° = √3]
Find the value of 
Answer:
Find the value of 
Answer:
If 
, find the values of all the other five trigonometric functions.
Answer: Given: $\sec \,x=-2$ Given that: So, x lies in III Quadrant. So, sin and cos will be negative but tan will be positive. Now, we know that We know that, cos x + sin x = 1 Putting the values,...
If 
, find the value of sin x.
Answer: Given: $\cos \,x=-\frac{\sqrt{15}}{4}$ To find: value of sinx Given that: $\frac{\pi }{2}<x<\pi $ So, x lies in II quadrant, and sin will be positive. We know that, cos θ + sin θ = 1...
If 
and x lies in Quadrant III, find the values of cos x and cot x.
Answer: Given: $\sin \,x=-\frac{2\sqrt{6}}{5}$ Since, x is in III Quadrant. So, sin and cos will be negative but tan will be positive. We know that, sin x + cos x = 1 Putting the values, we get...
If 
and θ lies in Quadrant IV, find the values of all the other five trigonometric functions.
Given : $\sec \theta =\sqrt{2}$ Since, θ is in IV Quadrant. So, sin and tan will be negative but cos will be positive. Now, we know that We know that, cos θ + sin θ = 1 Putting the values, we get...
If 
and θ lies in Quadrant II, find the values of all the other five trigonometric functions.
Answer: Given: $\cos ec\,\theta =\frac{5}{3}$ Since, θ is in II Quadrant. So, cos and tan will be negative but sin will be positive. Now, we know that We know that, sin θ + cos θ = 1 Putting the...
If 
and θ lies in Quadrant IV, find the values of all the other five trigonometric functions.
Answer: Since, θ is in IV Quadrant. So, sin and tan will be negative but cos will be positive. We know that, sin θ + cos θ = 1 Putting the values, we get Since, θ in IV quadrant and cosθ is positive...
If 
and θ lies in Quadrant III, find the value of all the other five trigonometric functions.
Answer: We are given : $\cos \theta =-\frac{\sqrt{3}}{2}$ Since, θ is in III Quadrant. So, sin and cos will be negative but tan will be positive. We know that, cos θ + sin θ = 1 Putting the values,...
Two dice are rolled together. Find the probability o f getting such numbers on two dice whose product Is perfect square.
Solution: When two different dice are thrown, then total number of outcomes = 36. Let E be the event of getting the product of numbers, as a perfect square. These numbers are (1,1), (1,4), (2,2),...
When two dice are tossed together, find the probability that the sum o f the numbers on their tops Is less than 7.
Solution: When two different dice are thrown, the total number of outcomes = 36. Let E be the event of getting the sum of the numbers less than 7. These numbers are (1,1), (1,2), (1,3), (1,4),...
Two different dice are rolled simultaneously. Find the probability that the sum o f the numbers on the two dice Is 10.
Solution: When two different dice are thrown, the total number of outcomes = 36. Let E1 be the event of getting the sum of the numbers on the two dice is 10. These numbers are (4 ,6), (5,5)...
Solve |x| > 4, when x ϵ R.
Answer : |x| > 4 Square ⇒ x2 > 16 ⇒ x2 – 16 > 0 ⇒ x2 – 42 > 0 ⇒ (x + 4)(x – 4) > 0 Observe that when x is greater than 4, (x + 4)(x – 4) is positive And for each root the sign changes...
Solve |x| < 4, when x ϵ R.
Answer : |x| < 4 Square ⇒ x2 < 16 ⇒ x2 – 16 < 0 ⇒ x2 – 42 < 0 ⇒ (x + 4)(x – 4) < 0 Observe that when x is greater than 4, (x + 4)(x – 4) is positive And for each root the sign changes...
Solve x/x-5>1/2, when x ϵ R.
Answer :Observe that is zero at x = -5 and not defined at x = 5 Hence plotting these two points on number line Now for x > 5 is positive For every root and not defined value of the sign...
Solve 3/x-2< 1, when x ϵ R.
Answer :Observe that zero at x = 5 and not defined at x = 2 Hence plotting these two points on number line Now for x > 5, is negative For every root and not defined value of the sign will change...
Solve x + 5 > 4x – 10, when x ϵ R.
Answer : x + 5 > 4x – 10 ⇒ 5 + 10 > 4x – x ⇒ 15 > 3x Divide by 3 ⇒ 5 > x ⇒ x < 5 x < 5 means x is from -∞ to 5 that is x ∈ (-∞, 5) Hence solution of x + 5 > 4x – 10 is x ∈ (-∞,...
Solve –4x > 16, when x ϵ Z.
Answer : We have to find integer values of x for which -4x > 16 (why only integer values because it is given that x ∈ Z that is set of integers) -4x > 16 ⇒ -x > 4 ⇒ x < -4 The integers...
Solve the system of in equation x – 2 ≥ 0, 2x – 5 ≤ 3.
Answer : We have to find values of x for which both the equations hold true x – 2 ≥ 0 and 2x – 5 ≤ 3 We will solve both the equations separately and then their intersection set will be solution of...
Find the solution set of the in equation
Find the solution set of the in equation
Answer : means we have to find values of x for which is negative Observe that the numerator |x – 2| is always positive because of mod, hence for negative quantity the...
Find the solution set of the in equation |2x – 3| < 1.
Answer : |2x – 3| < 1 Square both sides ⇒ (2x – 3)2 < 12 ⇒ 4x2 – 12x + 9 < 1 ⇒ 4x2 – 12x + 8 < 0 Divide throughout by 4 ⇒ x2 – 3x + 2 < 0 ⇒ x2 – 2x – x + 2 < 0 ⇒ x(x – 2) – 1(x –...
Find the solution set of the in equation |x – 1| < 2.
Answer : |x – 1| < 2 Square both sides ⇒ (x – 1)2 < 4 ⇒ x2 – 2x + 1 < 4 ⇒ x2 – 2x – 3 < 0 ⇒ x2 – 3x + x – 3 < 0 ⇒ x(x – 3) + 1(x – 3) < 0 ⇒ (x + 1)(x – 3) < 0 Observe that when...
Find the solution set of the in equation
Answer : We have to find values of x for which is less than zero that is negative Now for negative x – 2 should be negative that is x – 2 < 0 ⇒ x – 2 < 0 ⇒ x < 2 Hence x should be less than...
To receive grade A in a course one must obtain an average of 90 marks or more in five papers, each of 100 marks. If Tanvy scored 89, 93, 95 and 91 marks in first four papers, find the minimum marks that she must score in the last paper to get grade A in the course.
Answer : Let x marks be scored by Tanvy in her last paper. It is given that Tanvy scored 89, 93, 95 and 91 marks in first 4 papers. To receive grade A, she must obtain an average of 90 marks or...
How many litres of water will have to be added to 600 litres of the 45% solution of acid so that the resulting mixture will contain more than 25%, but less than 30% acid content?
Answer : Let x litres of water be added. Then total mixture = x + 600 Amount of acid contained in the resulting mixture is 45% of 600 litres. It is given that the resulting mixture contains more...
A manufacturer has 640 litres of an 8% solution of boric acid. How many litres of 2% boric and acid solution be added to it so that the boric acid content in the resulting mixture will be more than 4% but less than 6%.
Answer : Let x litres of 2% boric and acid solution be added to 640 litres of 8% solution of boric acidl
The watering acidity in a pool is considered normal when the average pH reading of three daily measurements is between 8.2 and 8.5. If the first two pH reading 8.48 and 8.35, find the range of the pH values for the third reading that will result in the acidity level is normal.
Answer : Let x be the third pH value. Now, it is given that the average pH reading of three daily measurements is between 8.2 and 8.5 Also, the first two pH readings are 8.48 and 8.35 Therefore, 8.2...
A company manufactures cassettes. Its cost and revenue function are C(x) = 25000 + 30x and R(x) = 43x respectively, where x is the number of cassettes produced and sold in a week. How many cassettes must be sold by the company to realize some profit?
Answer : Given: Cost function C(x) = 25000 + 30x Revenue function R(x) = 43x To Find: Number of cassettes to be sold to realize some profit In order, to gain profit: R(x) > C(x) Therefore, 43x...
Find all pairs of consecutive even positive integers both of which are larger than 8 such that their sum is less than 25.
Answer : Let the pair of consecutive even positive integers be x and x + 2. So, it is given that both the integers are greater than 8 Therefore, x > 8 and x + 2 > 8 When, x + 2 > 8...
Find all pairs of consecutive even positive integers both of which are larger than 8 such that their sum is less than 25.
Answer : Let the pair of consecutive even positive integers be x and x + 2. So, it is given that both the integers are greater than 8 Therefore, x > 8 and x + 2 > 8 When, x + 2 > 8...
Draw the graph of each of the following functions: Sin 3x
Table: Graph: The frequency of the function sin(x) is increased by 3 times.
Draw the graph of each of the following functions: 3sin x
Table: Graph: The amplitude of the function sin(x) is increased by 3 times.
Draw the graph of each of the following functions: 2sin 3x
Table: Graph:
Draw the graph of each of the following functions: 2cos 3x
Table: Graph: The amplitude and frequency of the function cos(x) is increased by 2 and 3 times.
Draw the graph of each of the following functions: 
Table: Graph: The frequency of the function sin(x) is decreased by 0.5 times.
Draw the graphs of y = sin x and ![\begin{array}{l} y = \cos xin[0,2\pi ] \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-523111e3d29d4af4c4181c8fa4667dd9_l3.png "Rendered by QuickLaTeX.com")
on the same axes.
Table: (i) For sinx (ii) For cosx Graph: The green line represents the curve for sin(x) and blue for cos(x) for...
Draw the graphs of y = cos x and ![\begin{array}{l} y = \cos 2xin[0,2\pi ] \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-40846a1c50086ba5eac3bfb33637ca3a_l3.png "Rendered by QuickLaTeX.com")
on the same axes.
Table: (i) For cosx (ii) For cos(2x) Graph: The blue line depicts curve cos(2x) and the purple lines depict...
Solve the following systems of linear in equations: 1 ≤ |x – 2| ≤ 3
Answer : 1 ≤ |x - 2| and |x - 2| ≤ 3 When, |x - 2| ≥ 1 Then, x – 2 ≤ -1 and x -2 ≥ 1 Now when, x – 2 ≤ - 1 Adding 2 to both the sides in above equation x – 2 + 2 ≤ -1 + 2 x ≤ 1 Now when, x – 2 ≥ 1...
Solve the following systems of linear in equations:
Solve the following systems of linear in equations:
Solve the following systems of linear in equations: 3x – x > x + > 3
Answer : 3x – x > 2 and x > 3 When, 3x – x > x Subtracting x from both the sides in above equation 2x – x > x + - x Multiplying both the sides by 3 in the above equation 3x > 3...
Solve the following systems of linear in equations: 5x – 7 < (x + 3), 1 – x – 4
Answer : When, 5x – 7 < x + 3 Adding 7 to both the sides in the above equation 5x – 7 + 7 < x + 3 + 7 5x < x + 10 Now, subtracting x from both the sides 5x – x < x + 10 -x 4x < 10...
Solve the following systems of linear in equations: –11 ≤ 4x – 3 ≤ 13
Answer : -11 ≤ 4x – 3 and 4x – 3 ≤ 13 When, -11 ≤ 4x - 3 4x – 3 ≥ -11 Adding 3 to both the sides 4x – 3 + 3 ≥ -11 + 3 4x ≥ - 8 Divide both the sides by 4 in above equation x ≥ -2 Now when, 4x – 3 ≤...
Solve the following systems of linear in equations:
Solve each of the following in equations and represent the solution set on the number line. x – 4 > 1, x ≠ 4.
Answer : Given: > 1, x ≠ 4. Adding 4 to both the sides in above equation x – 4 + 4 > 1 + 4 x > 5 Therefore, x є (5, ∞)
Solve each of the following in equations and represent the solution set on the number line. |x + a| + |x| > 3, x ϵ R.
Answer : Given: |x + a| + |x| > 3, x ϵ R. |x + a| = -(x + a) or (x + a) |x| = -x or x When |x + a| = -(x + a) and |x| = -x Then, |x + a| + |x| > 3 → -(x + a) + (-x) > 3 -x -a – x > 3 -2x...
Solve each of the following in equations and represent the solution set on the number line.
Solve each of the following in equations and represent the solution set on the number line.
Show that the vectors 
and 
form a right – angled triangle.
Solution: $\begin{array}{l} \vec{a}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \\ \vec{b}=\hat{\imath}-3 \hat{\jmath}+5 \hat{k} \\ \vec{c}=2 \hat{\imath}+\hat{\jmath}-4 \hat{k} \\ \mathrm{I} \vec{a}...
Solve each of the following in equations and represent the solution set on the number line.
each of the following in equations and represent the solution set on the number line.
If 
and 
, find 
and 
Solution: $\mathrm{I} \vec{a} \mathrm{I}=8 \mathrm{I} \vec{b} \mathrm{I}$ $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$ $\mathrm{I} \vec{a} \mathrm{I}_{2}-\mathrm{I} \vec{b} \mathrm{I}_{2}=8$ $(64...
Solve each of the following in equations and represent the solution set on the number line.
Solve each of the following in equations and represent the solution set on the number line.
Solve each of the following in equations and represent the solution set on the number line.
If 
and 
, find the angle between 
and 
.
Solution: $\begin{array}{l} \vec{a}+\vec{b}+\vec{c}=0 \\ \vec{a}+\vec{b}=-\vec{c}-\cdots \\ \mid \vec{a} \mathrm{I}=3 \\ \mathrm{I} \vec{b} \mathrm{I}=5 \\ \mid \vec{c} \mathrm{I}=7 \end{array}$...
Solve each of the following in equations and represent the solution set on the number line.
Solve each of the following in equations and represent the solution set on the number line.
Solve each of the following in equations and represent the solution set on the number line.
Solve each of the following in equations and represent the solution set on the number line.
A beam is supported at its ends by supports which are 12 m apart. Since the load is concentrated at its centre, there is a deflection of 3 cm at the centre, and the deflected beam is in the shape of a parabola. How far from the centre is the deflection 1 cm?
Given, A beam is supported at its ends by supports which are 12 m apart. There is a deflection of 3 cm at the centre, and the deflected beam is in the shape of a parabola. EF are the...
A rod of length 15 cm moves with its ends always touching the coordinate axes. Find the equation of the locus of a point P on the rod, which is at a distance of 3 cm from the end in contact with the x-axis.
Answer: Given, A rod of length 15 cm moves with its ends always touching the coordinate axes. A point P on the rod, which is at a distance of 3 cm from the end in contact with...
The towers of bridge, hung in the form of a parabola, have their tops 30 m above the roadway, and are 200 m apart. If the cable is 5 m above the roadway at the centre of the bridge, find the length of the vertical supporting cable, 30 m from the centre.
Answer: Given, Top of the towers are 30 m above the roadway and are 200 m apart. Cable is 5 m above the roadway at centre. A and B are the top of the towers. AE and BF are the...
A parabolic reflector is 5 cm deep and its diameter is 20 cm. How far is its focus from the vertex?
Answer: Given, Parabolic reflector = 5 cm deep Diameter = 20 cm Reflector is 5 cm deep, OD = 5 cm Diameter of the mirror is 20 com, BC = 20 cm The equation...
The focus of a parabolic mirror is at a distance of 6 cm from its vertex. If the mirror is 20 cm deep, find its diameter.
Answer: Given, The focus of a parabolic mirror is at a distance of 6 cm from its vertex. And the mirror is 20 cm deep. O is the vertex A is the Focus OA = a = 6...
Solve each of the following in equations and represent the solution set on the number line.
Solve each of the following in equations and represent the solution set on the number line.
Solve each of the following in equations and represent the solution set on the number line.
i. Find the projection of on if 
and 
.
ii. Write the projection of the vector 
on the vector 
.
Solution: i) $\begin{array}{l} \vec{b}=2 \hat{\imath}+6 \hat{\jmath}+3 \hat{k} \\ \vec{a} \cdot \vec{b}=8 \end{array}$ To find the projection of $\vec{a}$ on $\vec{b}$, we are required to find...
Solve each of the following in equations and represent the solution set on the number line.
Find the projection of 
in the direction of 
.
Solution: $\begin{array}{l} \vec{a}=8 \hat{\imath}+\hat{\jmath} \\ \vec{b}=\hat{\imath}+2 \hat{\jmath}-2 \hat{k} \end{array}$ Direction of $\vec{b}$ can be given by $\begin{array}{l}...
Solve each of the following in equations and represent the solution set on the number line.
Find a vector 
which is perpendicular to both 
and 
, and is such that 
.
Solution: $\begin{array}{l} \vec{a}=4 \hat{\imath}+5 \hat{\jmath}-\hat{k} \\ \vec{b}=\hat{\imath}-4 \hat{\jmath}+5 \hat{k} \end{array}$ $\vec{c}=3 \hat{\imath}+\hat{\jmath}-\hat{k}$ Let...
Find the value of 
for which and are perpendicular, wl
i. 
and 
ii. 
and 
iii. 
and 
iv. 
and 
Solution: If $\vec{a}$ and $\vec{b}$ are perpendicular to each other, then their scalar product or dot product $\vec{a} \cdot \vec{b}=0$ $\begin{array}{r} \text { i) } \quad \vec{a}=2...
Solve each of the following in equations and represent the solution set on the number line.
∆ABC ~ ∆DEF and the perimeters of ∆ABC and ~ ∆DEF are 32cm and 24cm respectively. If AB = 10cm, then DE = ? (a) 8cm (b) 7.5cm (c) 15cm (d) 5√3cm
Correct Answer: (b) 7.5 cm Explanation: ∆ABC ~ ∆DEF $\begin{array}{l} \frac{{Perimeter(\Delta ABC)}}{{Perimeter(\Delta DEF)}} = \frac{{AB}}{{DE}}\\ \frac{{32}}{{24}} = \frac{{10}}{{DE}}\\ DE =...
In ∆ABC, DE║BC. If DE = 5cm, BC = 8cm and AD = 3.5cm, then AB = ? (a) 5.6cm (b) 4.8cm (c) 5.2cm (d) 6.4cm
Correct Answer: (a) 5.6 cm Explanation: DE ‖ BC $\begin{array}{l} \frac{{AD}}{{AB}} = \frac{{AE}}{{AC}} = \frac{{DE}}{{BC}}\\ \frac{{3.5}}{{AB}} = \frac{5}{8}\\ AB = \frac{{3.5 \times 8}}{5}\\ AB =...
Two poles of heights 6m and 11m stand vertically on a plane ground. If the distance between their feet is 12m, find the distance between their tops. (a) 12m (b) 13m (c) 14m (d) 15m
Correct Answer: (b)13 m Explanation: Let the poles be and CD AB = 6 m CD = 11 m Let AC be 12 m Draw a perpendicular from CD, meeting CD at E. BE = 12 m Applying...
The areas of two similar triangles are 25
and 36 respectively. If the altitude of the first triangle is 3.5cm, then the corresponding altitude of the other triangle. (a) 5.6cm (b) 6.3cm (c) 4.2cm (d) 7cm
Correct Answer: (c) 4.2cm Explanation: The ratio of areas of similar triangles is equal to the ratio of squares of their corresponding altitudes. Let ɦ be the altitude of the other triangle....
If ∆ABC~∆DEF such that 2AB = DE and BC = 6cm, find EF.
Answer: ∆ABC ~ ∆ DEF $\begin{array}{l} \frac{{AB}}{{DE}} = \frac{{BC}}{{EF}}\\ \frac{1}{2} = \frac{6}{{EF}}\\ EF = 12cm \end{array}$
In the given figure, DE║BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.
Answer: DE || BC $\begin{array}{l} \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\\ \frac{x}{3x+4} = \frac{x+3}{{3x+19}}\\ \end{array}$ ???? (3???? + 19) = (???? + 3)(3???? + 4) 3????2 +...
Solve each of the following in equations and represent the solution set on the number line.
A ladder 10m long reaches the window of a house 8m above the ground. Find the distance of the foot of the ladder from the base of the wall.
Answer: Let the ladder be AB and BC be the height of the window from the ground. Given, AB 10 m BC = 8 m Applying theorem in right-angled triangle ACB, ????????2 = ????????2 +...
Find the length of the altitude of an equilateral triangle of side 2a cm.
Answer: Let the triangle be ABC with AD as its altitude. D is the midpoint of BC. In right-angled triangle ABD, ????????2 = ????????2 + ????????2 ????????2 = ????????2 − ????????2...
∆ABC~∆DEF such that ar(∆ABC) = 64 and ar(∆DEF) = 169. If BC = 4cm, find EF.
Answer: ∆ABC ~ ∆DEF $\begin{array}{l} \frac{{ar(\Delta ABC)}}{{ar(\Delta DEF)}} = \frac{{B{C^2}}}{{E{F^2}}}\\ \frac{{64}}{{169}} = \frac{{{4^2}}}{{E{F^2}}}\\ E{F^2} = \frac{{16 \times 169}}{{64}}\\...
In a trapezium ABCD, it is given that AB║CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar(∆AOB) = 84. Find ar(∆COD).
Answer: In ∆AOB and ∆COD, ∠???????????? = ∠???????????? (???????????????????????????????????????? ???????????????????????????????? ????????????????????????) ∠????????????...
The corresponding sides of two similar triangles are in the ratio 2: 3. If the area of the smaller triangle is 48, find the area of the larger triangle
Answer: Given, The triangles are similar and the ratio of areas of similar triangles will be equal to the ratio of squares of their corresponding sides. $\begin{array}{l} \frac{{48}}{{Area\ of\...
In the given figure, LM║CB and LN║CD. Prove that 
Answer: Given, LM || CB and LN || CD Applying Thales’ theorem, $\begin{array}{l} \frac{{AB}}{{AM}} = \frac{{AC}}{{AL}}\\ \end{array}$ And $\begin{array}{l} \frac{{AD}}{{AN}} =...
Prove that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
Answer: Let the triangle be ABC with AD as the bisector of ∠???? which meets BC at D. Draw CE || DA, meeting BA produced at E. CE || DA ∠2 = ∠3...
In an equilateral triangle with side a, prove that area = 
Answer: Let ABC be the equilateral triangle with each side equal to a. Let AD be the altitude from A, meeting BC at D. D is the midpoint of BC. Let AD be h. Applying...
Find the length of each side of a rhombus whose diagonals are 24cm and 10cm long.
Answer: Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O. The diagonals of a rhombus bisect each other at right angles. If AC – 24 cm and BD = 10...
Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.
Answer: Let the two triangles be ABC and PQR. BC = a AC = b AB = c PQ = r PR = q QR = p ∆ABC ~ ∆PQR; therefore, their corresponding sides will be proportional. $\begin{array}{l} \frac{a}{p} =...
In the given figure, ∆ABC and ∆DBC have the same base BC. If AD and BC intersect at O, prove that 
.
Answer: ???????????????? ???????? ⊥ ???????? ???????????? ???????? ⊥ ????O $\begin{array}{l} \frac{{ar(\Delta ABC)}}{{ar(\Delta DBC)}} = \frac{{\frac{1}{2} \times AX \times BC}}{{\frac{1}{2} \times...
In the given figure, XY║AC and XY divides ∆ABC into two regions, equal in area. Show that 
Answer: In ∆ ABC and ∆BXY, ∠???? = ∠???? ∠???????????? = ∠???????????? (???????????????????????????????????????????????????? ????????????????????????) ????ℎ????????,...
In the given figure, ∆ABC is an obtuse triangle, obtuse-angled at B. If AD⊥CB, prove that 
Answer: Applying Pythagoras theorem in right-angled triangle ADC, ????????2 = ????????2 + ????????2 ????????2 − ????????2 = ????????2 ????????2 = ????????2 − ????????2...
In the given figure, each one of PA, QB and RC is perpendicular to AC. If AP = x, QB = z, RC = y, AB = a and BC = b, show that 
Answer: In ∆ PAC and ∆QBC, ∠???? = ∠???? (????????????ℎ ???????????????????????? ???????????? 900) ∠???? = ∠???? (????????????????????????????????????????????????????...
Solve each of the following in equations and represent the solution set on the number line.
Solve each of the following in equations and represent the solution set on the number line.
. Solve each of the following in equations and represent the solution set on the number line.
Solve each of the following in equations and represent the solution set on the number line. 3 – 2x ≥ 4x – 9, where x ϵ R.
Solve each of the following in equations and represent the solution set on the number line. 3x – 4 > x + 6, where x ϵ R.
Match the following columns:
Column I Column II (a) A man goes 10m due east and then 20m due north. His distance from the starting point is ……m. (p) 25√3 (b) In an equilateral triangle with each side 10cm, the altitude is...
Match the following columns:
Column I Column II (a) In a given ∆ABC, DE║BC and $\frac{{AD}}{{DB}} = \frac{{3}}{{5}}$. If AC = 5.6cm, then AE = ……..cm. (p) 6 (b) If ∆ABC~∆DEF such that 2AB = 3DE and BC = 6cm, then EF = …….cm....
Which of the following is a false statement? (a) If the areas of two similar triangles are equal, then the triangles are congruent. (b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding sides. (c) The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding. (d) The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.
Correct Answer: (b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding sides. Explanation: The ratio of the areas of two similar triangles is equal to the...
Which of the following is a true statement? (a) Two similar triangles are always congruent (b) Two figures are similar if they have the same shape and size. (c)Two triangles are similar if their corresponding sides are proportional. (d) Two polygons are similar if their corresponding sides are proportional.
Correct Answer: (c)Two triangles are similar if their corresponding sides are proportional. Explanation: Given, ∆ABC~ ∆DEF $\frac{{AB}}{{DE}} = \frac{{AC}}{{DF}} = \frac{{BC}}{{EF}}$
In ∆ABC, if AB = 16cm, BC = 12cm and AC = 20cm, then ∆ABC is (a) acute-angled (b) right-angled (c) obtuse-angled
Correct Answer: (b) right-angled Explanation: Given, ????????2 + ????????2 = 162 + 122 => 256 + 144 => 400 ????????2 = 202 => 400 ∴ ????????2 + ????????2 = ????????2 ∆ABC is a right-angled...
In an isosceles ∆ABC, if AC = BC and 
, then ∠C = ? (a) 
(b) 
(c) 
(d) 
Correct Answer: (d) ${90^0}$ Explanation: Given, AC = BC ????????2 = 2????????2 = ????????2 + ????????2 = ????????2 + ????????2 Applying Pythagoras theorem, ∆ABC is right angled at C ∠???? =...
In the given figure, O is the point of intersection of two chords AB and CD such that OB = OD and ∠AOC = 
. Then, ∆OAC and ∆ODB are (a) equilateral and similar (b) equilateral but not similar (c) isosceles and similar (d) isosceles but not similar
Correct Answer: (c) isosceles and similar Explanation: In ∆AOC and ∆ODB, ∠???????????? = ∠???????????? (???????????????????????????????????????? ????????????????????????????????...
If ∆ABC~∆QRP, 
, AB = 18cm and BC = 15cm, then PR = ? (a) 18cm (b) 10cm (c) 12 cm (d) 
cm
Correct Answer: (b) 10 cm Explanation: ∆ABC ~ ∆QRP $\begin{array}{l} \frac{{AB}}{{QR}} = \frac{BC}{PR}\\ \end{array}$ $\frac{{ar(\Delta ABC)}}{{ar(\Delta QRP)}} = \frac{9}{4}$ $\begin{array}{l}...
The line segments joining the midpoints of the sides of a triangle form four triangles, each of which is (a) congruent to the original triangle (b) similar to the original triangle (c) an isosceles triangle (d) an equilateral triangle
Correct Answer: (b) similar to the original triangle Explanation: The line segments joining the midpoint of the sides of a triangle form four triangles,...
Two isosceles triangles have their corresponding angles equal and their areas are in the ratio 25: 36. The ratio of their corresponding heights is (a) 25 : 36 (b) 36 : 25 (c) 5 : 6 (d) 6: 5
Correct Answer: (c) 5:6 Explanation: Let x and y be the corresponding heights of the two triangles. The corresponding angles of the triangles are equal. The triangles are similar. (By AA criterion)...
∆ABC~∆DEF such that ar(∆ABC) = 36 and ar(∆DEF) = 49 . Then, the ratio of their corresponding sides is (a) 36 : 49 (b) 6 : 7 (c) 7 : 6 (d) √6 : √7
Correct Answer: (b) 6:7 Explanation: In ∆ABC ~ ∆DEF, $\begin{array}{l} \frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{AC}}{{DF}}\\ \frac{{ar(\Delta ABC)}}{{ar(\Delta DEF)}} =...
In ∆ABC and ∆DEF, we have: 
, then ar(∆ABC) : ∆(DEF) = ? (a) 5 : 7 (b) 25 : 49 (c) 49 : 25 (d) 125 : 343
Correct Answer: (b) 25 :49 Explanation: In ∆ABC and ∆DEF, $\frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{AC}}{{DF}} = \frac{5}{7}$ By SSS criterion, ∆ABC ~ ∆DEF $\begin{array}{l} \frac{{ar(\Delta...
In an equilateral ∆ABC, D is the midpoint of AB and E is the midpoint of AC. Then, ar(∆ABC) : ar(∆ADE) = ? (a) 2 : 1 (b) 4 : 1 (c) 1 : 2 (d) 1 : 4
Correct Answer: (b) 4:1 Explanation: In ∆ABC, D is the midpoint of AB and E is the midpoint of AC. By midpoint theorem and Basic Proportionality Theorem,...
It is given that ∆ABC~∆PQR and 
, then 
Correct Answer: (d)\frac{9}{4}$ Explanation: Given, ∆ABC ~ ∆PQR $\begin{array}{l} \frac{{BC}}{{QR}} = \frac{2}{3}\\ \end{array}$ $\begin{array}{l} \frac{{ar(\Delta PQR)}}{{ar(\Delta ABC)}} =...
Corresponding sides of two similar triangles are in the ratio 4:9 Areas of these triangles are in the ration (a) 2:3 (b) 4:9 (c) 9:4 (d) 16:81
Correct Answer: (d) 16:81 Explanation: If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides. $\begin{array}{l} \frac{{area\...
In the given figure, two line segment AC and BD intersect each other at the point P such that PA = 6cm, PB = 3cm, PC = 2.5cm, PD = 5cm, ∠APB = 
and ∠CDP = , then ∠PBA = ? (a) (b) (c) (d) 
Correct Answer: (d) ${100^0}$ Explanation: In ∆ APB and ∆ DPC, In ∆ APB and ∆ DPC, ∠???????????? = ∠???????????? = 500 $\begin{array}{l} \frac{{AP}}{{BP}} = \frac{6}{3} = 2\\...
If in ∆ABC and ∆PQR, we have: 
, then (a) ∆PQR ~ ∆CAB (b) ∆PQR ~ ∆ABC (c) ∆CBA ~ ∆PQR (d) ∆BCA ~ ∆PQR
Correct Answer: (a) ∆PQR ~ ∆CAB Explanation: In ∆ABC and ∆PQR, $\begin{array}{l} \frac{{AB}}{{QR}} = \frac{{BC}}{{PR}} = \frac{{CA}}{{PQ}}\\ \end{array}$ ∆ABC ~ ∆QRP
In ∆ABC and ∆DEF, it is given that ∠B = ∠E, ∠F = ∠C and AB = 3DE, then the two triangles are (a) congruent but not similar (b) similar but not congruent (c) neither congruent nor similar (d) similar as well as congruent
Correct Answer: (b) similar but not congruent Explanation: In ∆ABC and ∆DEF, ∠???? = ∠???? ∠???? = ∠???? Applying AA similarity theorem, ∆ABC - ∆DEF. AB = 3DE AB ≠ DE ∆ABC and ∆DEF are similar but...
If ∆ABC~∆EDF and ∆ABC is not similar to ∆DEF, then which of the following is not true? (a) BC.EF = AC.FD (b) AB.EF = AC.DE (c) BC.DE = AB.EF (d) BC.DE = AB.FD
Correct Answer: (c) BC. DE = AB. EF Explanation: ∆ABC ~ ∆EDF $\begin{array}{l} \frac{{AB}}{{DE}} = \frac{{AC}}{{EF}} = \frac{{BC}}{{DF}}\\ BC.DE \ne AB.EF \end{array}$
Solve each of the following in equations and represent the solution set on the number line. 5x + 2 < 17, where (i) x ϵ Z, (ii) x ϵ R.
Solve each of the following in equations and represent the solution set on the number line. 3x + 8 > 2, where (i) x ϵ Z, (ii) x ϵ R.
Solve each of the following in equations and represent the solution set on the number line. –2x > 5, where (i) x ϵ Z, (ii) x ϵ R.
Solve each of the following in equations and represent the solution set on the number line. 6x ≤ 25, where (i) x ϵ N, (ii) x ϵ Z.
Fill in the blanks with correct inequality sign (>, <, ≥, ≤).
(i) 5x < 20 ⇒ x4
(ii) –3x > 9 ⇒ x 3
(iii) 4x > –16 ⇒ x 4
(iv) –6x ≤ –18 ⇒ 3
(v) x > –3 ⇒ –2x 6
(vi)a < b and c > 0 ⇒
(vii) p – q = –3 ⇒ pq
(viii)u – v = 2 ⇒ u V
If 
and 
is the given point, find the coordinates of 
.
Solution: $\begin{array}{l} \overrightarrow{A B}=2 \hat{\imath}+\hat{\jmath}-3 \hat{k} \\ \overrightarrow{O A}=\hat{\imath}+2 \hat{\jmath}-\hat{k} \\ \overrightarrow{O B}=? \\ \overrightarrow{A...
Show that the points A, B and C having position vectors 
and 
respectively, form the vertices of a right-angled triangle.
Solution: $\begin{array}{l} \overrightarrow{O A}=3 \hat{\imath}-4 \hat{\jmath}-4 \hat{k} \\ \overrightarrow{O B}=2 \hat{\imath}-\hat{\jmath}+\hat{k} \\ \overrightarrow{O C}=\hat{\imath}-3...
The position vectors of the points 
and 
are 
and 
respectively. Show that the points A, B and C are collinear.
Solution: $\begin{array}{l} \overrightarrow{O A}=2 \hat{\imath}+\hat{\jmath}-\hat{k} \\ \overrightarrow{O B}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \\ \overrightarrow{O C}=\hat{\imath}+4...
Find the direction ratios and direction cosines of the vector 
.
Solution: $\vec{a}=5 \hat{\imath}-3 \hat{\jmath}+4 \hat{k}$ Direction ratios are ratios of $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ component of the vector while direction cosine are cosines of...
If 
and 
are two given points, find a unit vector in the direction of 
.
Solution: $\mathrm{A}(-2,1,2)$ $\mathrm{B}(2,-1,6)$ Points given in such ways represents position vectors, $\overrightarrow{O A}$ and $\overrightarrow{O B}$ $\begin{array}{l} \overrightarrow{O A}=-2...
In ∆DEF and ∆PQR, it is given that ∠D = ∠Q and ∠R = ∠E, then which of the following is not true? 
Correct Answer: (b) $\frac{{DE}}{{PQ}} = \frac{{EF}}{{RP}}$ Explanation: In ∆DEF and ∆PQR, ∠???? = ∠???? ∠???? = ∠???? Applying AA similarity theorem, ∆DEF ~ ∆QRP $\frac{{DE}}{{PQ}} =...
In ∆ABC and ∆DEF, it is given that 
, then (a) ∠B = ∠E (b) ∠A = ∠D (c) ∠B = ∠D (d) ∠A = ∠F
Correct Answer: (c)∠???? = ∠D Explanation: ∆ ABC − EDF The corresponding angles, ∠???? ???????????? ∠???? ???????????????? ???????? ????????????????????. ∠???? = ∠D
Find a vector of magnitude 21 units in the direction of the vector 
.
Solution: $\vec{A}=2 \hat{\imath}-3 \hat{\jmath}+6 \hat{k}$ Given magnitude of required vector is 21 units $\begin{array}{l} \mathrm{I} \overrightarrow{\mathrm{I}}...
In ∆ABC, AB = 6√3 , AC = 12 cm and BC = 6cm. Then ∠B is
Answer: Given, ???????? = 6√3???????? ????????2 = 108 ????????2 AC = 12 cm ????????2 = 144 ????????2 BC = 6 cm ????????2 = 36 ???????? ∴ ????????2 = ????????2 + ????????2 The square of the longest...
In the given figure, ∠BAC = and AD⊥BC. Then, (a) BC.CD = 
(b) AB.AC = (c) BD.CD = 
(d) AB.AC =
Correct Answer: (c) BD.CD = $A{D^2}$ Explanation: In ∆ BDA and ∆ADC, ∠???????????? = ∠???????????? = 900 ∠???????????? = 900 − ∠???????????? => 900 − (900 − ∠????????????)...
It is given that ∆ABC~∆DFE. If ∠A = , ∠C = , , AB = 5cm, AC = 8cm and DF = 7.5cm, then which of the following is true? (a) DE = 12cm, ∠F = , (b) DE = 12cm, ∠F = , (c) DE = 12cm, ∠D = , (d) EF = 12cm, ∠D = ,
Correct Answer: (b) DE = 12cm, ∠F = ${100^0}$ Explanation: Given, In triangle ABC, ∠???? + ∠???? + ∠???? = 1800 ∠???? = 180 − 30 − 50 => 1000 ∵ ∆ABC ~ ∆DFE ∠???? = ∠???? = 300 ∠???? = ∠???? =...
ABC and BDE are two equilateral triangles such that D is the midpoint of BC. Ratio of these area of triangles ABC and BDE is (a) 2 : 1 (b) 1 : 4 (c) 1 : 2 (d) 4 : 1
Correct Answer: (d) 4 : 1 Explanation: Given, ABC and BDE are two equilateral triangles D is the midpoint of BC and BDE is also an equilateral triangle. E is also the midpoint of AB. D and E are the...
In ∆ABC, it is given that AB = 9cm, BC = 6cm and CA = 7.5cm. Also, ∆DEF is given such that EF = 8cm and ∆DEF~∆ABC. Then, perimeter of ∆DEF is (a) 22.5cm (b) 25cm (c) 27cm (d) 30cm
Correct Answer: (d) 30 cm Explanation: Perimeter of ∆ABC = AB + BC + CA => 9 + 6 + 7.5 => 22.5 cm $\begin{array}{l} \frac{{Perimeter(\Delta ABC)}}{{Perimeter(\Delta DEF)}} =...
∆ABC~∆DEF such that AB = 9.1cm and DE = 6.5cm. If the perimeter of ∆DEF is 25cm, what is the perimeter of ∆ABC? (a) 35cm (b) 28cm (c) 42cm (d) 40cm
Correct Answer: (a) 35 cm Explanation: Given, ∆ABC ~ ∆DEF $\begin{array}{l} \frac{{Perimeter(\Delta ABC)}}{{Perimeter(\Delta DEF)}} = \frac{{AB}}{{DE}}\\ \frac{{Perimeter(\Delta ABC)}}{{25}} =...
∆ABC~∆DEF and the perimeters of ∆ABC and ∆DEF are 30cm and 18cm respectively. If BC = 9cm, then EF = ? (a) 6.3cm (b) 5.4cm (c) 7.2cm (d) 4.5cm
Correct Answer: (b) 5.4 cm Explanation: Given, ∆ABC ~ ∆DEF $\begin{array}{l} \frac{{Perimeter(\Delta ABC)}}{{Perimeter(\Delta DEF)}} = \frac{{BC}}{{EF}}\\ \frac{{30}}{{18}} = \frac{9}{{EF}}\\ EF =...
In ∆ABC, DE ║ BC such that 
. If AC = 5.6cm, then AE = ? (a) 4.2cm (b) 3.1cm (c) 2.8cm (d) 2.1cm
Correct Answer: (d) 2.1 cm Explanation: Given, DE || BC. Applying Thales’ theorem, $\begin{array}{l} \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\\ \end{array}$ AE be x cm. EC = (5.6 –...
In ∆ABC, DE ║ BC so that AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x cm. Then, we have: (a) x = 3 (b) x = 5 (c) x = 4 (d) x = 2.5
Correct Answer: (c) x = 4 Explanation: Given, DE || BC Applying Thales’ theorem, $\begin{array}{l} \frac{{AD}}{{BD}} = \frac{{AE}}{{EC}}\\ \frac{{7x-4}}{{3x+4}} =...