CBSE Class 10 Mathematics Previous Year Question Paper 2018 All India Set - 3

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Q 1. What is the value of (cos267° - sin23°)?

Q 2. In an AP. if the common difference (d) = -4, and the seventh term (a7) is 4, then find the first term. 

Q 3. Given that ∆ABC ∼ ∆PQR begin mathsize 12px style AB over PQ equals 1 third comma space then space find space fraction numerator ar left parenthesis ΔABC right parenthesis over denominator ar left parenthesis ΔPQR right parenthesis end fraction end style.

Q 4. What is the HCF of smallest prime number and the smallest composite number?

Q 5. Find the distance of point P(x, y) from the origin.

Q 6. If x = 3 is one root of the quadratic equation x- 2kx - 6 = 0, then find the value of k.

Q 7. Two different dice are tossed together. Find the probability:

  1. Of getting a doublet
  2. Of getting a sum 10, of the numbers on the two dice.

Q 8.  Find the radio in which P(4, m ) divides the line segment joining the points A(2,3) And B(6,-3). Hence find m. 

Q 9. An integer is chosen at random between 1 and 100. Find the probability that it is: 

  1. Divisible by 8
  2. Not divisible by 8

Q 10. In the given figure □ABCD is a rectangle. Find the values of x and y.

Q 11. Find the sum of first 8 multiples of 3.

Q 12. Given that begin mathsize 12px style square root of 2 end style is irrational, prove that begin mathsize 12px style left parenthesis 5 plus 3 square root of 2 right parenthesis end style is an irrational number. 

Q 13. If A (-2, 1), B (a, 0), C (4, b) and D (1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.

OR


If A (-5, 7), B (-4, -5), C (-1, -6) and D (4, 5) are the vertices of a quadrilateral, find the area of quadrilateral ABCD.

Q 14. Find all zeros of the polynomial (2x- 9x+ 5x+ 3x - 1) if two of its zeros are begin mathsize 12px style left parenthesis 2 plus square root of 3 right parenthesis end style and begin mathsize 12px style left parenthesis 2 minus square root of 3 right parenthesis end style.

Q 15. Find HCF and LCM of 404 and 96 and verify HCF × LCM = Product of the two given numbers.

Q 16. Prove that the lengths of tangents drawn from an external point to circle are equal.

Q 17. Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.

OR

If the area of two similar triangle are equal, prove that they are congruent.

Q 18. A plane left 30 minute late than its scheduled time and in order to reach the destination 1500 km away in time it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

Q 19.  The table below shows the salaries of 280 persons:

Salary(In thousand)

No.of persons

5-10

49

10-15

133

15-20

63

20-25

15

25-30

6

30-35

7

35-40

4

40-45

2

45-50

1

 Calculate the median salary of the data.

Q 20.  A wooden article was made by scooping out of hemisphere from each end of a solid cylinder, as shown in the given figure. If the height of the cylinder is 10cm and its base is of radius 3.5 cm. find the total surface area of the article.

 

 

 

OR

 

A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

 

Q 21. Find the area of the shaded region in Fig.3, where areas drawn with centres  A,B,C and D intersect in pairs at mid-points P,Q, R and S of the sides AB, BC, CD and DA respectively of a square ABCD of side 12cm (Use 𝜋 =3.14)

Q 22. If 4 tan 𝛳=3, evaluate begin mathsize 12px style open parentheses fraction numerator 4 text   end text sin text   end text straight theta text   end text minus cos text   end text straight theta plus 1 over denominator 4 text   end text sin text   end text straight theta text   end text minus cos text   end text straight theta text   end text minus 1 end fraction close parentheses end style

OR

If tan 2A = cot (A-18°), where 2A is an acute angle, find the value of A.

Q 23.  As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. (Use begin mathsize 12px style square root of 3 end style=1.732)

Q 24. The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find: 

  1. The area of the metal sheet used to make the bucket.
  2. Why we should avoid the bucket made by ordinary plastic? (Use = 3.14)

Q 25.  Prove that: begin mathsize 12px style fraction numerator sin text   end text straight A minus 2 text   end text sin cubed straight A over denominator 2 cos cubed straight A minus cosA end fraction equals tanA end style

 

Q 26. The mean of the following distribution is 18. Find the frequency f of the class 19-21.
 

Class

11-13

13-15

15-17

17-19

19-21

21-23

23-25

Frequency

3

6

9

13

f

5

4

 

OR

 

The following distribution gives the daily income of 50 workers of a factory:

Daily income(In)

100-120

120-140

140-160

160-180

180-200

Number of workers

12

14

8

6

10

 Convert the distribution above to a less than type of cumulative frequency distribution and draw its ogive.

Q 27. A motor boat whose speed is 18km/hr in still water takes 1hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

OR

A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km /hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed?

Q 28. The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15. Find the numbers. 

Q 29. Draw a triangle ABC with BC = 6 cm AB = 5 cm and ∠ ABC = 60° Then construct a triangle whose sides are begin mathsize 12px style 3 over 4 end style of the corresponding sides of the ∆ABC.

Q 30.  In an equilateral ∆ABC, D is a point on side BC such that BD = begin mathsize 12px style 1 third BC end style . Prove that 9[AD]= 7[AB]2

OR

Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the sequence on the other two side

 
 
 

 

Q 1. What is the value of (cos267° - sin23°)?

Solution:

Cos267° - Sin223°

= Cos(90° - 23°) - sin23°    (∵ cos= sin2(90° - 𝛳))

= sin223° - sin223°

= 0


Q 2. In an AP. if the common difference (d) = -4, and the seventh term (a7) is 4, then find the first term.

Solution: 

Given that d = -4, a= 4 , n = 7

a= 4

Now, an= a + (n-1) d
∴ a+ (7-1) (-4) =4

∴ a - 6 × 4 = 4

∴ a - 24 = 4

∴ a = 28

∴ First term of the given A.P. is 28.


Q 3. Given that ∆ABC ∼ ∆PQR begin mathsize 12px style AB over PQ equals 1 third comma space then space find space fraction numerator ar left parenthesis ΔABC right parenthesis over denominator ar left parenthesis ΔPQR right parenthesis end fraction end style.

Solution:

Given that ∆ABC ∼ ∆PQR

begin mathsize 12px style AB over PQ equals 1 third end style                                        

Using theorem "If two triangles are similar then the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

begin mathsize 12px style table attributes columnalign left end attributes row cell fraction numerator ar left parenthesis ΔABC right parenthesis over denominator ar left parenthesis ΔPQR right parenthesis end fraction equals open parentheses AB over PQ close parentheses squared end cell row cell therefore fraction numerator ar left parenthesis ΔABC right parenthesis over denominator ar left parenthesis ΔPQR right parenthesis end fraction equals open parentheses 1 third close parentheses squared end cell row cell therefore fraction numerator ar left parenthesis ΔABC right parenthesis over denominator ar left parenthesis ΔPQR right parenthesis end fraction equals 1 over 9 end cell end table end style

 

 

Q 4. What is the HCF of smallest prime number and the smallest composite number?

Solution:

Smallest prime number is 2 and smallest composite number is 4.

2 = 2 × 1

4 = 2 × 2

∴HCF of 2 and 4 is 2.

 


Q 5. Find the distance of point P(x, y) from the origin.

Solution:

According to the question,

Origin O (0,0) and point P (x, y)

Distance of a point P(x, y) from O (0, 0) =  begin mathsize 12px style OP equals square root of left parenthesis straight x minus 0 right parenthesis squared left parenthesis straight y minus 0 right parenthesis squared end root equals square root of straight x squared plus straight y squared end root units end style

 

 

Q 6. If x = 3 is one root of the quadratic equation x- 2kx - 6 = 0, then find the value of k.

Solution:

Let p(x) = x2 – 2kx – 6

x = 3 is one root of given quadratic equation.

∴ p (3) = 0

∴ 3- 2k  × 3 - 6 = 0

∴ 9 - 6k - 6 = 0

∴ 3 = 6k

begin mathsize 12px style therefore straight k equals 1 half end style

 

 

Q 7. Two different dice are tossed together. Find the probability:

  1. Of getting a doublet
  2. Of getting a sum 10, of the numbers on the two dice. 

Solution:

When two dice are tossed together, the sample space S is given by

begin mathsize 12px style table attributes columnalign left end attributes row cell straight s equals open square brackets table row cell left parenthesis 1 comma 1 right parenthesis end cell cell left parenthesis 1 comma 2 right parenthesis end cell cell left parenthesis 1 comma 3 right parenthesis end cell cell left parenthesis 1 comma 4 right parenthesis end cell cell left parenthesis 1 comma 5 right parenthesis end cell cell left parenthesis 1 comma 6 right parenthesis end cell row cell left parenthesis 2 comma 1 right parenthesis end cell cell left parenthesis 2 comma 2 right parenthesis end cell cell left parenthesis 2 comma 3 right parenthesis end cell cell left parenthesis 2 comma 4 right parenthesis end cell cell left parenthesis 2 comma 5 right parenthesis end cell cell left parenthesis 2 comma 6 right parenthesis end cell row cell left parenthesis 3 comma 1 right parenthesis end cell cell left parenthesis 3 comma 2 right parenthesis end cell cell left parenthesis 3 comma 3 right parenthesis end cell cell left parenthesis 3 comma 4 right parenthesis end cell cell left parenthesis 3 comma 5 right parenthesis end cell cell left parenthesis 3 comma 6 right parenthesis end cell row cell left parenthesis 4 comma 1 right parenthesis end cell cell left parenthesis 4 comma 2 right parenthesis end cell cell left parenthesis 4 comma 3 right parenthesis end cell cell left parenthesis 4 comma 4 right parenthesis end cell cell left parenthesis 4 comma 5 right parenthesis end cell cell left parenthesis 4 comma 6 right parenthesis end cell row cell left parenthesis 5 comma 1 right parenthesis end cell cell left parenthesis 5 comma 2 right parenthesis end cell cell left parenthesis 5 comma 3 right parenthesis end cell cell left parenthesis 5 comma 4 right parenthesis end cell cell left parenthesis 5 comma 5 right parenthesis end cell cell left parenthesis 5 comma 6 right parenthesis end cell row cell left parenthesis 6 comma 1 right parenthesis end cell cell left parenthesis 6 comma 2 right parenthesis end cell cell left parenthesis 6 comma 3 right parenthesis end cell cell left parenthesis 6 comma 4 right parenthesis end cell cell left parenthesis 6 comma 5 right parenthesis end cell cell left parenthesis 6 comma 6 right parenthesis end cell end table close square brackets end cell row cell therefore straight n left parenthesis straight s right parenthesis equals 36 end cell end table end style

  1.  Let A be the event of getting a doublet.
    ∴ A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
    ∴ n(A) = 6

    begin mathsize 12px style therefore straight P left parenthesis straight A right parenthesis equals fraction numerator straight n left parenthesis straight A right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 6 over 36 equals 1 over 6 end style
  2. Let B be the event of getting a sum of 10 of the numbers on two dice.
    ∴ B = {(5,5), (6,4), (4,6)}
    ∴ n(B) = 3
    begin mathsize 12px style therefore straight P left parenthesis straight B right parenthesis equals fraction numerator straight n left parenthesis straight B right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 3 over 36 equals 1 over 12 end style
 
 
 
Q 8.  Find the ratio in which P(4, m ) divides the line segment joining the points A(2,3) And B(6,-3). Hence find m. 

Solution:

Let P (4, m) divides the line segment joining the points A (2, 3) and B (6, –3) in the ratio t : 1.

CO-ordinates P are begin mathsize 12px style open parentheses fraction numerator 6 straight t plus 2 over denominator straight t plus 1 end fraction comma fraction numerator negative 3 straight t plus 3 over denominator straight t plus 1 end fraction close parentheses end style

Given co-ordinates of point P are (4, m).

begin mathsize 12px style table attributes columnalign left end attributes row cell therefore open parentheses fraction numerator 6 straight t plus 2 over denominator straight t plus 1 end fraction comma fraction numerator negative 3 straight t plus 3 over denominator straight t plus 1 end fraction close parentheses equals left parenthesis 4 comma straight m right parenthesis end cell row cell therefore fraction numerator 6 straight t plus 2 over denominator straight t plus 1 end fraction equals 4 text   end text and text   end text fraction numerator negative 3 straight t plus 3 over denominator straight t plus 1 end fraction equals straight m end cell row cell Consider text   end text fraction numerator 6 straight t plus 2 over denominator straight t plus 1 end fraction equals 4 end cell row cell therefore 6 straight t plus 2 equals 4 straight t plus 4 end cell row cell therefore 2 straight t equals 2 end cell row cell therefore straight t equals 1 end cell row cell Put text   end text straight t equals 1 text   end text in text   end text fraction numerator negative 3 straight t plus 3 over denominator straight t plus 1 end fraction equals straight m end cell row cell therefore fraction numerator negative 3 cross times 1 plus 3 over denominator 1 plus 1 end fraction equals straight m end cell row cell therefore straight m equals 0 end cell end table end style

∴ The ratio in which point P divides the line segment joining the points A and B is 1:1.

 

Q 9. An integer is chosen at random between 1 and 100. Find the probability that it is: 

  1. Divisible by 8
  2. Not divisible by 8

Solution:

According to the question sample space S is as follows: 
s = {2, 3, 4, 5… 99}

∴ n(s) = 98

  1.  Let A be the event that chosen integer is divisible by 8.
    ∴ A = {8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96}
    ∴ n(A) = 12

    begin mathsize 12px style therefore straight P left parenthesis straight A right parenthesis equals fraction numerator straight n left parenthesis straight A right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 12 over 98 equals 6 over 49 end style
    Thus, probability that a chosen integer is divisible by 8 is begin mathsize 12px style 6 over 49 end style .
  2. Probability that a chosen integer is not divisible by 8 = P(A’) = 1 - P(A) = begin mathsize 12px style 43 over 49 end style . 
 
 
 

Q 10. In the given figure □ABCD is a rectangle. Find the values of x and y.


Solution:

Since □ABCD is a rectangle,

⇒ AB = CD and AD = BC

∴ x + y = 30 … (i)

and x – y = 14  …(ii)

Adding (i) and (ii),

∴ 2x = 44
∴ x = 22

Put x = 22 in (i)

∴ 22 + y = 30

∴ y = 30 - 22

∴ y = 8

∴ The value of x and y are 22 and 8 respectively.

 

 

Q 11. Find the sum of first 8 multiples of 3.

Solution:

First 8 multiples of 3 are {3, 6, 9, 12, 15, 18, 21, 24}.

This form an A. P. with

first term, a = 3,

Common difference, d = 6 - 3 = 3,

Last term, l = 24

And n = 8

begin mathsize 12px style table attributes columnalign left end attributes row cell straight S subscript straight n equals straight n over 2 left parenthesis straight a plus 1 right parenthesis end cell row cell therefore straight S subscript 8 equals 8 over 2 left parenthesis 3 plus 24 right parenthesis end cell row cell therefore straight S subscript 8 equals 4 cross times 27 end cell row cell therefore straight S subscript 8 equals 108 end cell end table end style

∴ Sum of first 8 multiples of 3 is 108.

 


Q 12. Given that begin mathsize 12px style square root of 2 end style is irrational, prove that begin mathsize 12px style left parenthesis 5 plus 3 square root of 2 right parenthesis end style is an irrational number. 

Solution:

Let us assume that begin mathsize 12px style left parenthesis 5 plus 3 square root of 2 right parenthesis end style is a rational number.

Hence, we can find co-prime p and q (q ≠ 0) such that

 begin mathsize 12px style table attributes columnalign left end attributes row cell 5 plus 3 square root of 2 equals straight p over straight q end cell row cell therefore straight p over straight q minus 5 equals 3 square root of 2 end cell row cell therefore fraction numerator straight p over denominator 3 straight q end fraction minus 5 over 3 equals square root of 2 end cell row cell therefore fraction numerator straight p minus 5 straight q over denominator 3 straight q end fraction equals square root of 2 end cell end table end style

But begin mathsize 12px style square root of 2 end style is irrational.

Thus, our assumption is wrong.

Hence, begin mathsize 12px style left parenthesis 5 plus 3 square root of 2 right parenthesis end style is an irrational.

 

Q 13. If A (-2, 1), B (a, 0), C (4, b) and D (1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.

OR


If A (-5, 7), B (-4, -5), C (-1, -6) and D (4, 5) are the vertices of a quadrilateral, find the area of quadrilateral ABCD.

Solution:

Diagonals of parallelogram bisect each other

So midpoint of AC= midpoint of BD

begin mathsize 12px style table attributes columnalign left end attributes row cell therefore open parentheses fraction numerator 4 minus 2 over denominator 2 end fraction comma fraction numerator straight b plus 1 over denominator 2 end fraction close parentheses equals open parentheses fraction numerator straight a plus 1 over denominator 2 end fraction comma fraction numerator 0 plus 2 over denominator 2 end fraction close parentheses end cell row cell therefore open parentheses 2 over 2 comma fraction numerator straight b plus 1 over denominator 2 end fraction close parentheses equals open parentheses fraction numerator straight a plus 1 over denominator 2 end fraction comma 2 over 2 close parentheses end cell row cell therefore open parentheses 1 comma fraction numerator straight b plus 1 over denominator 2 end fraction close parentheses equals open parentheses fraction numerator straight a plus 1 over denominator 2 end fraction comma 1 close parentheses end cell end table end style

After comparing the coordinates, we get

 begin mathsize 12px style table attributes columnalign left end attributes row cell 1 equals fraction numerator straight a plus 1 over denominator 2 end fraction space space space space text           end text and space space text       end text fraction numerator straight b plus 1 over denominator 2 end fraction equals 1 end cell row cell therefore straight a plus 1 equals 2 space space text        end text and space space text        end text straight b plus 1 equals 2 end cell row cell therefore straight a equals 1 space space space space space space space space text              end text and space space text       end text straight b equals 1 end cell end table end style

∴ The coordinates are A (-2, 1), B (1, 0), C (4, 1) and D (1, 2) Now using distance formula,

As we know, opposite sides of a Parallelogram are equal in length.

begin mathsize 12px style table attributes columnalign left end attributes row cell therefore AB equals DC equals square root of left parenthesis 1 plus 2 right parenthesis squared plus left parenthesis 0 plus 1 right parenthesis squared end root equals square root of 9 plus 1 end root equals square root of 10 text       end text space space space space space left parenthesis By text   end text distamce text    end text formula right parenthesis end cell row cell therefore AD equals BC equals square root of left parenthesis 1 plus 2 right parenthesis squared plus left parenthesis 2 minus 1 right parenthesis squared end root equals square root of 9 plus 1 end root equals square root of 10 end cell end table end style

 

 

 

OR


We will join B and D so that we will be getting two triangle ABD and BCD.

Then we will find their individual areas and add them together

To find the Area of a Quadrilateral ABCD.

 begin mathsize 12px style table attributes columnalign left end attributes row cell Area text   end text of text   end text ΔABD equals 1 half open square brackets negative 5 left parenthesis negative 5 minus 5 right parenthesis minus 4 left parenthesis 5 minus 7 right parenthesis plus 4 left parenthesis 7 plus 5 right parenthesis close square brackets....... by text    end text formula text   end text left parenthesis Coordinate text   end text Geometry right parenthesis end cell row cell text                                       end text equals 1 half left parenthesis 50 plus 8 plus 48 right parenthesis end cell row cell text                                             end text equals 106 over 2 end cell row cell text                                             end text equals 53 text   end text sq. units end cell row cell Area text   end text of text   end text ΔBCD equals 1 half open square brackets negative 4 left parenthesis negative 6 minus 5 right parenthesis minus 1 left parenthesis 5 plus 5 right parenthesis plus 4 left parenthesis negative 5 plus 6 right parenthesis close square brackets........ by space text   end text formula text   end text left parenthesis Coordinate text   end text Geometry right parenthesis end cell row cell text                                              end text equals 1 half left parenthesis 44 minus 10 plus 4 right parenthesis end cell row cell text                                             end text equals 38 over 2 end cell row cell text                                             end text equals 19 text   end text sq. units end cell row cell therefore Area text   end text of text   end text square ABCD equals Area text    end text of text   end text ΔABD text   end text plus Area text   end text of text   end text ΔBCD end cell row cell text                                                          end text equals 53 plus 19 end cell row cell text                                                          end text equals 72 text   end text sq. units end cell end table
end style

 

Q 14. Find all zeros of the polynomial (2x- 9x+ 5x+ 3x - 1) if two of its zeros are begin mathsize 12px style left parenthesis 2 plus square root of 3 right parenthesis end style and begin mathsize 12px style left parenthesis 2 minus square root of 3 right parenthesis end style

Solution:

Now begin mathsize 12px style left parenthesis 2 plus square root of 3 right parenthesis end style  and begin mathsize 12px style left parenthesis 2 minus square root of 3 right parenthesis end style are the two zeroes of the given polynomial

So the product begin mathsize 12px style open square brackets straight x minus open parentheses 2 plus square root of 3 close parentheses close square brackets open square brackets straight x minus open parentheses 2 minus square root of 3 close parentheses close square brackets end style will be a factor of the given polynomial

 begin mathsize 12px style table attributes columnalign left end attributes row cell therefore open square brackets straight x minus open parentheses 2 plus square root of 3 close parentheses close square brackets open square brackets straight x minus open parentheses 2 minus square root of 3 close parentheses close square brackets equals left parenthesis straight x minus 2 right parenthesis squared minus left parenthesis square root of 3 right parenthesis squared space end cell row cell text                                                                                             end text equals straight x squared minus 4 straight x plus 4 minus 3 end cell row cell text                                                                                        end text equals straight x squared minus 4 straight x plus 1 end cell row cell let text   end text straight f left parenthesis straight x right parenthesis equals 2 straight x to the power of 4 minus 9 straight x cubed plus 5 straight x squared plus 3 straight x minus 1 end cell row cell and text   end text straight g left parenthesis straight x right parenthesis equals straight x squared minus 4 straight x plus 1 end cell row cell Find fraction numerator straight f left parenthesis straight x right parenthesis over denominator straight g left parenthesis straight x right parenthesis end fraction. end cell end table

end style

 

 ∴ f(x) = (x- 4x + 1) (2x- x + 1)

∴ 2x- 9x+ 5x+ 3x - 1 = (x- 4x + 1) (2x- x - 1)

Hence, the other zeroes of f(x) are the zeroes of the polynomial 2x- x - 1.

∴ 2x-x - 1 = 2x- 2x + x - 1 = (2x + 1) (x - 1)

So, 2x- 9x+ 5x+ 3x - 1 = (x- 4x + 1) (2x- x - 1)

                                         = begin mathsize 12px style open square brackets straight x minus open parentheses 2 plus square root of 3 close parentheses close square brackets open square brackets straight x minus open parentheses 2 minus square root of 3 close parentheses close square brackets open parentheses 2 x plus 1 close parentheses open parentheses x minus 1 close parentheses end style     

Hence the roots of the Polynomial f(x) are begin mathsize 12px style left parenthesis 2 plus square root of 3 right parenthesis comma left parenthesis 2 minus square root of 3 right parenthesis comma fraction numerator negative 1 over denominator 2 end fraction end style and 1.

 


Q 15. Find HCF and LCM of 404 and 96 and verify HCF x LCM = Product of the two given numbers.

Solution:

We will find the Prime factors of 404 and 96.

∴ 404 = 2 × 2 × 101

∴ 96 = 2 × 2 × 2 × 2 × 2 × 3

 So,

LCM (404, 96) = 2 × 2 × 2 × 2 × 2 × 3 × 101 = 9696

HCF = 2 × 2 = 4

Here,

LCM × HCF = 9696 × 4 = 38784...............i)

Product of numbers = 404 × 96 = 38784.......ii)

∴ LCM × HCF = Product of numbers (from i and ii)

Hence verified.

 


Q 16. Prove that the lengths of tangents drawn from an external point to circle are equal.

Solution:

Here

PA and PB are tangents to the circle with centre O,

And AO and OB are the radii of the Circle.

begin mathsize 12px style therefore open table attributes columnalign right end attributes row cell PA perpendicular AO end cell row cell PB perpendicular BO end cell end table close curly brackets.............. tangent text   end text perpendicular text   end text to text   end text radius
end style

In ∆OPA and ∆OPB

∠OAP = ∠OBP…...............each 90° (radius and tangent are⊥ at their point of contact)

OA = OB           ...............(radii of the same circle)

OC = OC           ...............(Common)

∆OPA ≅ ∆OPB.................. (by RHS Theorem)

∴ PA = PB........................ (CPCT)

Hence Proved

 

Q 17. Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.

OR

If the area of two similar triangle are equal, prove that they are congruent.

Solution:

Let sides of a square be x units

∴ Diagonal length = begin mathsize 12px style square root of 2 straight x end root end style units

Since, Area of an Equilateral triangle = begin mathsize 12px style fraction numerator square root of 3 over denominator 4 end fraction open parentheses side close parentheses squared end style

∴ Area of an Equilateral triangle described on side =  begin mathsize 12px style fraction numerator square root of 3 over denominator 4 end fraction open parentheses straight x close parentheses squared end style ........i)

∴ Area of equilateral triangle described on diagonal =  begin mathsize 12px style fraction numerator square root of 3 over denominator 4 end fraction open parentheses square root of 2 straight x close parentheses squared end style .........ii)

begin mathsize 12px style table attributes columnalign left end attributes row cell therefore text Ratio of their areas end text equals fraction numerator text Area of equilateral triangle described on side end text over denominator text Area of equilateral triangle described on diagonal end text end fraction end cell row cell text                                            end text space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator fraction numerator square root of 3 over denominator 4 end fraction left parenthesis straight x right parenthesis squared over denominator fraction numerator square root of 3 over denominator 4 end fraction left parenthesis square root of 2 straight x end root right parenthesis squared end fraction...................... open parentheses from space straight i space and space ii close parentheses end cell row cell text                                         end text space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half end cell row cell because fraction numerator text Area of equilateral triangle described on side end text over denominator text Area of equilateral triangle described on diagonal end text end fraction equals 1 half end cell row cell therefore text Area of equilateral triangle on side end text equals 1 half cross times text Area of equilateral triangle on diagonal end text end cell end table end style

 

 OR

Suppose two similar triangles are ∆ABC and ∆PQR.

Given,

Area of ∆ABC = Area of ∆PQR
Now using properties of similar triangles, we get

 begin mathsize 12px style therefore AB over PQ equals AC over PR equals BC over QR.......... left parenthesis CPCT right parenthesis.................... straight i right parenthesis end style

Also, we know that ratio of areas of two Similar Triangles is equal to the ratio of the square of their corresponding sides.

 begin mathsize 12px style table attributes columnalign left end attributes row cell therefore fraction numerator Area text    end text of text   end text ΔABC over denominator Area text    end text of text   end text ΔPQR end fraction equals AB squared over PQ squared end cell row cell therefore 1 equals AB squared over PQ squared end cell row cell therefore 1 equals AB over PQ end cell end table end style

∴ AB = PQ

So from i)

 begin mathsize 12px style AB over PQ equals AC over PR space equals BC over QR equals space 1 end style

∴ AB = PQ

∴ AC = PR

∴ BC = QR

∆ABC ≅ ∆PQR… (SSS test)

 


Q 18. A plane left 30 minute late than its scheduled time and in order to reach the destination 1500 km away in time it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

Solution:

Here,
Distance =1500km

As we know,

 begin mathsize 12px style Time equals Distance over Speed end style

Let the usual speed of plane be x km/hr.

 begin mathsize 12px style therefore straight T subscript 1 equals 1500 over straight x hr end style

After increasing the speed by 100 km/hr it's speed becomes (x+100) km/hr.

 begin mathsize 12px style therefore straight T subscript 2 equals fraction numerator 1500 over denominator straight x plus 100 end fraction hr end style

 Given that difference in speed is 30 mins which is begin mathsize 12px style 1 half end style hours

∴ T- T=  begin mathsize 12px style 1 half end style

begin mathsize 12px style table attributes columnalign left end attributes row cell therefore 1500 over straight x minus fraction numerator 1500 over denominator straight x plus 100 end fraction equals 1 half end cell row cell therefore fraction numerator 1500 left parenthesis straight x plus 100 right parenthesis minus 1500 straight x over denominator straight x left parenthesis straight x plus 100 right parenthesis end fraction equals 1 half end cell end table end style

 

∴ x+ 100x - 300000 = 0

∴ x+ 600x - 500x - 300000 = 0

∴ (x + 600) (x - 500) = 0

∴ x = -600 or x = 500

Speed can’t be negative

∴ x = 500 km/hr

Therefore, the usual speed of plane is 500km/hr.



Q 19.  The table below shows the salaries of 280 persons:

Salary(In thousand)

No.of persons

5-10

49

10-15

133

15-20

63

20-25

15

25-30

6

30-35

7

35-40

4

40-45

2

45-50

1

Calculate the median salary of the data.

Solution: 

Salary

Frequency

CF

5-10

49

49

10-15

133

182

15-20

63

245

20-25

15

260

25-30

6

266

30-35

7

273

35-40

4

277

40-45

2

279

45-50

1

280

begin mathsize 12px style therefore straight N over 2 equals 280 over 2 equals 140 end style

The cumulative frequency which is greater and nearest to 140 is 182.

So,

Medium Class is 10-15

Thus,

l = 10

h = 5

N = 280

cf = 49

f = 133

Median for the grouped date is given by,

Error converting from MathML to accessible text.

 


Q 20.  A wooden article was made by scooping out of hemisphere from each end of a solid cylinder, as shown in the given figure. If the height of the cylinder is 10cm and its base is of radius 3.5 cm. find the total surface area of the article.

 

OR

A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

Solution: 


Radius of cylindrical part = Radius of Hemispherical Part = 3.5 cm

Height of cylindrical part = h =10 cm

Here,

Surface Area of the article = Curved surface Area of Cylinder + 2 x Curved surface Area of Hemisphere

                                      = 2𝜋rh + 2 x 2𝜋r2

                                      = 2𝜋 x 3.5 x 10 + 2 x 2𝜋 x 3.5 x 3.5

                                      = 70𝜋 + 49𝜋

                                      =119 x 𝜋

                                     = 374cm2

 

 

OR

Volume of rice (heap) = volume of cone = begin mathsize 12px style 1 third πr squared straight h end style

r = 12m, h = 3.5m

 Error converting from MathML to accessible text.

Area of canvas cloth = curved surface area of the cone = 𝜋rl

 Error converting from MathML to accessible text.

Area of canvas cloth = 𝜋rl = Error converting from MathML to accessible text.

Therefore Area of canvas required to cover the heap of the rice is 471.42m2



Q 21. Find the area of the shaded region in fig.3, where areas drawn with centres  A,B,C and D intersect in pairs at mid-points P,Q, R and S of the sides AB, BC, CD and DA respectively of a square ABCD of side 12cm (Use 𝜋 =3.14)

Solution:

According to the question,

Here side of the square = 12 cm and and begin mathsize 12px style straight r equals side over 2 equals 6 cm end style 𝛳=90° (angle of the square)

∴ Area of shaded region = Area of the square ABCD - 4(Area of quadrant)

                                 Error converting from MathML to accessible text.

                                   = 144 - 113.04

                                   = 30.96 cm2

∴ Area of shaded region is 30.96 cm2.




Q 22. If 4 tan 𝛳=3, evaluate begin mathsize 12px style open parentheses fraction numerator 4 text   end text sin text   end text straight theta text   end text minus cos text   end text straight theta plus 1 over denominator 4 text   end text sin text   end text straight theta text   end text minus cos text   end text straight theta text   end text minus 1 end fraction close parentheses end style

OR

If tan 2A = cot (A-18°), where 2A is an acute angle, find the value of A.

Solution:

Let ∆XYZ is a right angle triangle at Y and ∠Z = 𝛳

Given, 4tan 𝛳 = 3

       begin mathsize 12px style therefore tanθ equals 3 over 4 end style

Let XY=3k and YZ = 4k

By Pythagoras Theorem, we get

 ∴ XZ2  = XY+ YZ2

∴  XZ= (3k) 2 + (4k) 2

∴ XZ= 9k+ 16k2

∴ XZ= 25k2

∴ XZ = 5k

begin mathsize 12px style therefore sinθ equals XY over XZ equals fraction numerator 3 straight k over denominator 5 straight k end fraction equals 3 over 5 text   end text and text   end text cosθ text   end text equals YZ over XZ equals fraction numerator 4 straight k over denominator 5 straight k end fraction equals 4 over 5 end style

 Substituting the values in , begin mathsize 12px style fraction numerator 4 sinθ minus cosθ plus 1 over denominator 4 sinθ plus cosθ minus 1 end fraction end style , we get

begin mathsize 12px style table attributes columnalign left end attributes row cell fraction numerator 4 cross times 3 over 5 minus 4 over 5 plus 1 over denominator 4 cross times 3 over 5 plus 4 over 5 minus 1 end fraction equals fraction numerator fraction numerator 12 minus 5 plus 5 over denominator 5 end fraction over denominator fraction numerator 12 plus 4 minus 5 over denominator 5 end fraction end fraction end cell row cell equals 13 over 11 end cell row cell therefore fraction numerator 4 sinθ minus cosθ plus 1 over denominator 4 sinθ plus cosθ minus 1 end fraction equals 13 over 11 end cell end table end style

 

 

 OR

 

According to the question,

tan2A = cot (A-18°)

∴ tan2A = tan [90° - (A - 18°)]   [∵cot A = tan (90°-A)]

∴ tan2A = tan (90° - A + 18°)

∴ 2A = (90° - A + 18°)

∴ 2A + A = 108°

∴3A = 108°

∴ A = 36° 


Q 23.  As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. (Use begin mathsize 12px style square root of 3 end style=1.732)

Solution: 
Let A and B be the two ships.

Let d be the distance between the two ships.
i.e. AB = d metres.
Suppose that the observer is at the point P.
It is given that PC = 100 m.
Let h be the distance (in metres) from B to C.

In right ∆PCA,

 

 begin mathsize 12px style table attributes columnalign left end attributes row cell cot 30 degree equals Ac over PC end cell row cell rightwards double arrow square root of 3 equals fraction numerator straight d plus straight h over denominator 100 end fraction end cell row cell rightwards double arrow straight d plus straight h equals 100 square root of 3 text   end text end root text    end text........................ left parenthesis 1 right parenthesis end cell end table end style  

In right ∆PCB,

 begin mathsize 12px style table attributes columnalign left end attributes row cell cot 45 degree equals BC over PC end cell row cell rightwards double arrow 1 equals straight h over 100 end cell row cell rightwards double arrow straight h equals 100 straight m text   end text................ left parenthesis 2 right parenthesis end cell end table end style

Substituting the value of h in (1), we get

d + 100 = begin mathsize 12px style square root of 3 end style

⟹ d = 100begin mathsize 12px style square root of 3 end style - 100 

⟹ d = begin mathsize 12px style 100 left parenthesis square root of 3 minus 1 end root right parenthesis end style = 100 x 0.732 = 73.2 m

Thus, the distance between two ships is 73.2 m.


 

Q 24. The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find:

  1.  The area of the metal sheet used to make the bucket.
  2. Why we should avoid the bucket made by ordinary plastic? (Use ∏= 3.14) 
Solution:
 
Height of the cone, h = 24 cm

Upper radius of the cone, R = 15 cm

Lower radius of the cone, r = 5cm

Slant height of the cone,

begin mathsize 12px style straight l equals square root of straight h squared plus left parenthesis straight R minus straight r right parenthesis squared end root equals square root of 24 squared plus left parenthesis 15 minus 5 right parenthesis squared end root equals square root of 576 plus 100 end root equals square root of 676 equals 26 cm end style

Now, area of metal sheet used to make the bucket

= Total surface area of the bucket

𝜋(R+r)l + 𝜋r2

= 𝜋 (15 + 5) x 26 + 𝜋 x (5)2

= 𝜋 [520 + 25]

= 𝜋 x 545

= 3.14 x 545
= 1711.3 cm3
 
Plastics are not biodegradable. That is, plastic material mostly end as harmful waste that pollutes the environment and causes health problems, we should avoid using plastic.
 


Q 25.  Prove that: begin mathsize 12px style fraction numerator sin text   end text straight A minus 2 text   end text sin cubed straight A over denominator 2 cos cubed straight A minus cosA end fraction equals tanA end style
 
Solution:
begin mathsize 12px style table attributes columnalign left end attributes row cell straight L. straight H. straight S equals fraction numerator sin space straight A minus 2 text   end text sin cubed text end text straight A over denominator 2 cos cubed straight A minus cos space straight A end fraction end cell row cell text           end text equals fraction numerator sin space straight A left parenthesis 1 minus 2 text   end text sin squared text end text straight A right parenthesis over denominator cos space straight A space left parenthesis 2 cos squared space straight A minus 1 right parenthesis end fraction end cell row cell text           end text equals fraction numerator sin space text   end text straight A space left parenthesis 1 minus sin squared space straight A minus sin squared space straight A right parenthesis over denominator cos space straight A space left square bracket 2 cos squared space straight A minus left parenthesis sin squared space straight A plus cos to the power of 2 space end exponent straight A right parenthesis right square bracket end fraction end cell row cell text           end text equals fraction numerator sin space straight A left parenthesis cos squared space straight A minus sin squared space straight A right parenthesis over denominator cos space straight A left parenthesis cos squared space straight A minus sin squared space straight A right parenthesis end fraction end cell row cell text           end text equals fraction numerator sin space straight A over denominator cos space straight A end fraction end cell row cell text           end text equals tan space straight A end cell row cell text           end text equals straight R. straight H. straight S. end cell end table end style

 
 
Q 26. The mean of the following distribution is 18. Find the frequency f of the class 19-21.
 

Class

11-13

13-15

15-17

17-19

19-21

21-23

23-25

Frequency

3

6

9

13

f

5

4

 

OR

 

The following distribution gives the daily income of 50 workers of a factory:

Daily income(In)

100-120

120-140

140-160

160-180

180-200

Number of workers

12

14

8

6

10

Convert the distribution above to a less than type of cumulative frequency distribution and draw its ogive.

Solution: 

We have,

Class interval

Frequency

fi

Mid-value

Xi

fx xi

11-13

3

12

36

13-15

6

14

84

15-17

9

16

144

17-19

13

18

234

19-21

F

20

20f

21-23

5

22

110

23-25

4

24

96

 

∑f= 40 + f

 

∑fix= 704 + 20f

 Now, Mean = begin mathsize 12px style fraction numerator sum straight f subscript straight i straight x subscript straight i over denominator sum straight f subscript straight i end fraction end style

begin mathsize 12px style rightwards double arrow 18 equals fraction numerator 704 plus 20 straight f over denominator 40 plus straight f end fraction end style

 ⟹ 18 (40 + f) = 704 + 20f

⟹ 720 + 18f = 704 + 20f

⟹ 2f = 16

⟹ f = 8

Thus, the missing frequency is 8.

 

 

OR


Less Than Series:

Class interval

Cumulative frequency

Less than 120

12

Less than 140

26

Less than 160

34

Less than 180

40

Less than 200

50

 Now, we mark the upper class limits along X-axis and the cumulative frequencies along Y-axis.

Thus, we plot the points (120, 12), (140, 26), (160, 34), (180, 40), (200, 50) to get 'less than type’ ogive.

 

 

Q 27. A motor boat whose speed is 18km/hr in still water takes 1hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

OR

A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km /hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed?

Solution:

Let x be the speed of the stream.

∴ Speed of the boat while going upstream= (18 - x) km/ hr

 

And, speed of the boat while going downstream= (18 + x) km/ hr

Time taken by the boat to go upstream = begin mathsize 12px style fraction numerator 24 over denominator 18 minus straight x end fraction hours end style

Time taken by the boat to go downstream =  begin mathsize 12px style fraction numerator 24 over denominator 18 plus straight x end fraction hours end style

It is given that time taken for going upstream is 1 hour more than the time taken for going downstream.

Thus, we have

 begin mathsize 12px style table attributes columnalign left end attributes row cell fraction numerator 24 over denominator 18 minus straight x end fraction minus fraction numerator 24 over denominator 18 plus straight x end fraction equals 1 end cell row cell rightwards double arrow 24 open square brackets fraction numerator 1 over denominator 18 minus straight x end fraction minus fraction numerator 1 over denominator 18 plus straight x end fraction close square brackets equals 1 end cell row cell rightwards double arrow fraction numerator 18 plus straight x minus 18 plus straight x over denominator left parenthesis 18 minus straight x right parenthesis left parenthesis 18 plus straight x right parenthesis end fraction equals 1 over 24 end cell row cell rightwards double arrow fraction numerator 2 straight x over denominator 324 minus straight x squared end fraction equals 1 over 24 end cell end table end style

⟹ 48x = 324 - x2

⟹ x+ 48x - 324 = 0

⟹ x+ 54x - 6x - 324 = 0

⟹ x(x + 54) - 6(x + 54) =0

⟹ (x - 6) (x + 54) = 0

⟹ x - 6 = o or x + 54 = 0

⟹ x = 6 or x = -54

As speed cannot be negative, the speed of the stream is 6km/hr.



OR

Let the original average speed of the train be x km/hr.

Time taken by train to cover a distance of 63 km = begin mathsize 12px style 63 over straight x end style hours

Time taken by train to cover a distance of 72 km =  begin mathsize 12px style fraction numerator 72 over denominator straight x space plus space 6 end fraction end style hours

According to given condition,

 begin mathsize 12px style table attributes columnalign left end attributes row cell 63 over straight x plus fraction numerator 72 over denominator straight x plus 6 end fraction equals 3 end cell row cell rightwards double arrow fraction numerator 63 left parenthesis straight x plus 6 right parenthesis plus 72 straight x over denominator straight x left parenthesis straight x plus 6 right parenthesis end fraction equals 3 end cell end table end style

⟹ 63x + 378 + 72x = 3x+ 18x

⟹ 135x + 378 = 3x+ 18x

⟹ 3x- 117x - 378 = 0

⟹ x- 39x - 126 = 0

⟹ x- 42x + 3x - 126 = 0

⟹ x(x-42) + 3(x-42) = 0

⟹ (x - 42) (x + 3) = 0

⟹ x - 42 = 0 or x + 3 = 0

⟹ x = 42 or x = -3

Since, speed cannot be negative, we reject x = - 3.

Hence, x = 42

Thus, the original average speed of the train is 42 km/hr.

 


Q 28. The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15. Find the numbers. 

Solution:

Let the four consecutive numbers in an A.P. be (a – 3d), (a – d), (a + d) and (a + 3d).

Sum of the numbers = 32

⟹ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 32

⟹ 4a = 32

⟹ a = 8

It is given that,

begin mathsize 12px style table attributes columnalign left end attributes row cell fraction numerator left parenthesis straight a minus 3 straight d right parenthesis left parenthesis straight a plus 3 straight d right parenthesis over denominator left parenthesis straight a minus straight d right parenthesis left parenthesis straight a plus straight d right parenthesis end fraction equals 7 over 15 end cell row cell rightwards double arrow fraction numerator straight a squared minus 9 straight d squared over denominator straight a squared minus straight d squared end fraction equals 7 over 15 end cell row cell rightwards double arrow fraction numerator 64 minus 9 straight d squared over denominator 64 minus straight d squared end fraction equals 7 over 15 end cell end table end style 

⟹ 960 - 135d= 448 - 7d2

⟹ 128d= 512

⟹ d= 4

⟹ d = ∓2

⟹ a - 3d = 8 - 3(2) = 8 - 6 = 2

⟹ a - d = 8 - 2 = 6

⟹ a + d = 8 + 2 = 10

⟹ a + 3d = 8 + 3(2) = 8 + 6 = 14

Thus, the four numbers are 2, 6, 10, 14.


 

Q 29. Draw a triangle ABC with BC = 6 cm AB = 5 cm and ∠ ABC = 60° Then construct a triangle whose sides are begin mathsize 12px style 3 over 4 end style of the corresponding sides of the ∆ABC.

Solution:

Steps of construction:

  1. Draw a line segment BC = 6cm
  2. At B, construct ∠XBC = 60°
  3. With B as centre and radius 5 cm, draw an arc intersecting XB at A.
  4. Join AC to obtain ∆ABC.
  5. Below BC, make an acute ∠CBY.
  6. Along BY, mark off 4 points begin mathsize 12px style left parenthesis greater text   end text of text   end text 3 text   end text and text   end text 4 text   end text in 3 over 4 right parenthesis end style
    B1, B2, B3, B4 such that BB= B1B= B2B= B3B4
  7. Join B4C
  8. From point B3, draw a line parallel to B4C intersecting BC at C’
  9. From point C’, draw a line parallel to CA intersecting AB at A’
  10. Thus, ∆A’BC’ is the required triangle.

 


Q 30.  In an equilateral ∆ABC, D is a point on side BC such that BD = begin mathsize 12px style 1 third BC end style . Prove that 9[AD]= 7[AB]2

OR

Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the sequence on the other two sides.

Solution:

Let ABC be an equilateral triangle and let D be a point on BC such that BD = begin mathsize 12px style 1 third BC end style.

Construction: Draw AE ⊥BC. Join AD.

∆ABC is an equilateral triangle.

⟹ E is the mid-point of BC

∴ BE = CE = begin mathsize 12px style 1 half BC end style

In right-angle ∆AEB, by Pythagoras theorem,

AB= AE+ BE2   ............(i)

In right-angle ∆AEB, by Pythagoras theorem,

AD= AE+ DE2  .............(ii)

Subtracting (ii) from (i), we have

AB- Ad= (AE+ BE2) - (AE+ DE2)

⟹ AB- AD= BE- DE2

 begin mathsize 12px style table attributes columnalign left end attributes row cell rightwards double arrow AB squared minus AD squared equals open parentheses 1 half AB close parentheses squared minus open parentheses 1 over 6 AB close parentheses squared text     end text open square brackets DE equals space BE minus BD equals 1 half AB minus 1 third AB equals 1 over 6 AB close square brackets end cell row cell rightwards double arrow AB squared minus AD squared equals 1 fourth AB squared minus 1 over 36 AB squared end cell row cell rightwards double arrow AB squared minus AD squared equals 8 over 36 AB squared end cell row cell rightwards double arrow AB squared minus AD squared equals 2 over 9 AB squared end cell end table end style

⟹ 9AB- 9AD= 2AB2

⟹ 7AB= 9AD2

⟹ 9(AD)= 7(AB)2

 

OR

Consider the following figure:

 

Given: In ∆ABC, ∠ABC = 90°

To prove: AC= AB+ BC2

Construction: Draw seg BD ⊥ hypotenuse AC and A-D-C

Proof:

In ABC, seg BD ⊥ hypotenuse AC.........(Construction)

∴ ∆ABC ∼ ∆ADB................................(Similarity in right angled triangles)

begin mathsize 12px style therefore AB over AD equals AC over AB end style..................................(Corresponding sides of similar triangles)

∴ AB= AC x DC...............................(i)

Similarly, ∆ABC ∼ ∆BDC....................(Similarity in right angled triangles)

begin mathsize 12px style therefore BC over DC equals AC over BC end style..................................(Corresponding sides of similar triangles) 

∴ BC= AC x DC   ............................(ii)

AB+ BC= AC x AD + AC x DC.........[Adding equations (i) and (ii)]

              = AC [AD + DC]

               = AC x AC

               =AC2

∴ AC= AB+ BC2

 

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