Tuesday, October 20, 2020

ABCD is a cyclic quadrilateral Find the angles of the cyclic quadrilateral.

10th ncert cbse mathematics  chapter 3 miscellaneous / optional exercise

pair of linear equations in two variables

 

 ABCD is a cyclic quadrilateral  Find the angles of the cyclic quadrilateral,

if angles are A =(4y+20) , B =(3y-5) , C=(-4x), D=(-7x+5)

 

We know that the opposite angles of a cyclic quadrilateral are supplementary, 

which means they add up to 180 degrees.

 

A+C=180 degrees

and 

B+D = 180 degrees

 

A+C=180 

gives 

(4y+20) +(-4x)=180

or

 -4x +4y =180-20

or

-4x +4y =160 

dividing by 4 makes it

-x+y=40--------------(1)


 

 B+D = 180

gives

 (3y-5)+(-7x+5) =180

or

-7x+3y=180-------------(2)


eliminating y

-x+y=40--------------(1)  *3

-7x+3y=180-------------(2)

 

-3x+3y =120

-7x+3y =180 

--------------------------subtracting

4x     =(-60)

x = (-60)/4

 

x=(-15)

 

substitute in

-x+y=40

-(-15) +y =40

15+y-40

y=40-15

 

y=25

 

use  x=(-15) and y=25

A =(4y+20) =4*25+20=120 degrees

 B =(3y-5)=3*25-5=75-5=70 degrees

 C=(-4x) =(-4)*(-15) = 60 degrees

 D=(-7x+5) =(-7)*(-15) +5 =105+5=110 degrees

 

exercise 3.6

2

(iii) 

Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

 

let x = speed of train in km/hr

and y = speed of bus in km/hr

 

plan 1

60 km by train and remaining 300-60 = 240 km by bus

time = distance / speed

t1=(60/x) in the train 

and 

t2=(240/y) hours in the bus


total of 4 hours means

t1 +t2 =4 hours

(60/x) +(240/y) =4--------------------(1)

 

plan 2

100 km by train and remaining 300-100 = 200 km by bus

time = distance / speed

t3=(100/x) in the train 

and 

t4=(200/y) hours in the bus

10 minutes longer means 4hours 10minutes

This has to be changed in term of hours 

4hours 10minutes =4+(10/60)=4+(1/6) = (25/6)hours

so 

t3 +t4 = (25/6)hours

(100/x)+(200/y) =(25/6)----------------(2)

 

equations are

 

(60/x) +(240/y) =4--------------------(1)

(100/x)+(200/y) =(25/6)----------------(2)

 

we have to use substitution (1/x)=u , (1/y)=v

 

equations change to

 

60u   + 240v =4   ----------------------(3)  *5

100u + 200v =(25/6)-------------------(4)  *3 

--------------------------------------------------------------------eliminate u


300u + 1200v =20

300u +600v =(25/2)

------------------------------------------subtracting

600v =  20 -(25/2)

600v = (15/2)

v = [ 15 / (2*600) ]

v = [1/80]

resubstitute v = [1/80] in 60u   + 240v =4  

60u+{240/80} =4

60u+3=4

60u=4-3

60u=1

u=(1/60)

 

u=(1/60)

v = [1/80]

 

take reciprocal

x=60 km/hr 

y=80 km/hr 


speed of train =60km/hr

speed of bus =80km/hr


=================================================

ncert cbse 10th mathematics chapter 3 optional exercise 3.7 

 The ages of two friends Ani and Biju differ by 3 years. Ani’s father  is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Ani’s father differ by 30 years. Find the ages of Ani and Biju

solution

 

2. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? 

solution

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. 

solution  

 

4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

solution  


5. In a ∆ ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

solution

 

Solve the following pair of linear equations:

 px + qy = p – q 

 qx – py = p + q

solution

 

(ii) ax + by = c
     bx + ay = 1 + c

solution  

 

(iii) 

(x/a) -(y/b) = 0

ax +by = (a^2)  + (b^2)

solution  

 

(iv)

(a – b)x + (a + b) y = (a^2) – 2ab – (b^2)


(a + b)(x + y) = (a^2) + (b^2 )

solution

 

(v)

152x – 378y = – 74

–378x + 152y = – 604

solution   


 ABCD is a cyclic quadrilateral  Find the angles of the cyclic quadrilateral,

if angles are A =(4y+20) , B =(3y-5) , C=(-4x), D=(-7x+5)

solution

 

exercise 3.6

2

(iii) 

Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

 

solution

 

 

  ncert cbse 10th  mathematics chapter 2 optional exercise
 If the zeroes of the polynomial (x^3) – 3(x^2) + x + 1 are a – b, a, a + b, find a and b.
solution
 
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively
solution



4. If two zeroes of the polynomial (x^4) – 6(x^3) – 26(x^2) + 138x – 35 are 
 [2 ±sqrt(3) ] , find other zeroes

 solution
 
5. If the polynomial (x^4) – 6(x^3) + 16(x^2) – 25x + 10 is divided by another polynomial (x^2) – 2x + k, the remainder comes out to be x + a, find k and a
solution  
 
exercise 2.3
 

3. obtain all other zeroes of 3(x^4)+6(x^3)-2(x^2)-10x-5 if two of its zeroes are sqrt(5/3) and [sqrt(5/3)]
solution
 
4. On dividing (x^3) – 3(x^2) + x + 2 by a polynomial g(x), the quotient and remainder were x – 2
and –2x + 4, respectively. Find g(x). 
  

solution
 
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