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Answers
a+b+c=0
b+c=-a
a+b=-c
c+a=-b
Putting these in the question
a²/bc+b²/ac+c²/ab
a³+b³+c³/abc
It is an identity that if a+b+c=0 then a³+b³+c³=3abc
3abc/abc = 3
(c) Parallel sides are 25 cm and 46 cm
Let x be the distance between the height and the apex of the trapezium
The other side the distance will be 46 - ( 25 + x )
= 21 - x
Let height be h
By Pythagoras theorem :
h² = 20² - ( 21 - x )²
==> h² = 400 - ( 441 + x² - 42 x )
==> h² = 400 - 441 - x² + 42 x
==> h² = - 41 - x² + 42 x ......................(1)
Also :
h² = 13² - x²
==> h² = 169 - x²........................(2)
From (1) and (2) :
==> - 41 - x² + 42 x = 169 - x²
==> 42 x = 169 + 41
==> 42 x = 210
==> x = 210/42
==> x = 5
By Pythagoras theorem :
h² = 13² - 5²
==> h² = 169 - 125
==> h² = 144
==> h = 12
Area = 1/2 × (25 + 46 ) × 6
==> 3 × 71
==> 213
The height is 12 cm and area is 213 cm²
( d ) 4/3 cot²30° + 3 sin²60° - 2 cosec²60° - 3/4 tan²30° + sin 15°/cos 75°
==> 4/3 × (√3)² + 3 × (√3)/2² - 2 × 2²/(√3)² - 3/4 × 1/(√3)² + sin 15°/sin 15°
==> 4/3 × 3 + 3 × 3/4 - 2 × 4/3 - 3/4 × 1/3 + 1
==> 4 + 9/4 - 8/3 - 1/4 +1
==> 4 + 8/4 - 8/3 + 1
==> 4 + 2 - 8/3 + 1
==> 5 - 8/3
==> ( 15 - 8 )/3
==> 7/3
The value is 7/3
( c ) Let the value that B gets be x
A gets 33+1/3 % more
= 100/3 % more
A gets x + 100/3% × x
= x + 100 x / ( 3 × 100 )
= x + x/3
= 4 x/3
C gets 60% more than A and B
= ( 4 x/3 + x )+ 60/100 × ( 4 x/3 + x )
==> 7 x/3 + 3/5 × 7 x/3
==> 7 x/3 + 7 x/5
==> ( 35 x + 21 x ) / 15
==> 56 x / 15
A + B + C = Rs 45,500
==> x + 4 x/3 + 56 x/15 = Rs 45,500
==> ( 15 x + 20 x + 56 x ) / 15 = Rs 45,500
==> 91 x = Rs 45,500 × 15
==> 91 x = Rs 682500
==> x = Rs 682500/91
==> x = Rs 7500
B gets Rs 7500
A gets 4/3 × Rs 7500
==> Rs 10,000
C gets 56 x/15
==> Rs 56 × 7500/15
==> Rs 28,000
A gets Rs 10,000 , B gets Rs 7,500 and C gets Rs 28,000
Question - 7
(a)
( b ) Let the first boy be A
A got 30% and failed for 12 marks .
Let the total marks be x
Let the pass marks be y
30 x/100 = y - 12
==> 30 x = 100 y - 1200
==> 3 x = 10 y - 120
==> 3 x - 10 y + 120 = 0 .......................(1)
For B :
B got 40 % marks
B got 28 more than the pass marks
40 x/100 = y + 28
==> 40 x = 100 ( y + 28 )
==> 4 x = 10 ( y + 28)
==> 4 x = 10 y + 280
==> 2 x = 5 y + 140
==> 2 x - 5 y - 140 = 0 ..............................(2)
Multiplying (2) by 2 we get :
==> 4 x - 10 y - 280 = 0 ........................(3)
Subtracting (3) and (2) we get :
==> x = 280 + 120
==> x = 400
Put this value in 2 we get :
==> 2×400 - 5 y - 140 = 0
==> 800-140 = 5 y
==> 5 y = 660
==> y = 660/5
==> y = 132
The pass mark is 132
The full marks is 400
( c ) a + b + c = 0
==> a + b = - c ....................(1)
==> a + c = - b ...................(2)
==> b + c = - a....................(3)
Now :
( a + b )² / ab + ( b + c )²/bc + ( a + c )² / ac
From (1) , (2) and (3) we get :
( - c )² / ab + ( - a )² /bc + ( - b )² / ac
==> c² / ab + a²/ bc + b² / ac
==> ( c³ + a³ + b³ ) / abc
==> ( a³ + b³ + c³ ) / abc..........................(4)
Now :
a + b + c= 0
==> ( a + b ) = - c .................(5)
Cubing both sides :
==> ( a + b )³ = -c³
==> a³ + b³ + 3 ab ( a + b ) = -c³
==> a³ + b³ - 3 abc = -c³ [ From ( 5 ) ]
==> a³ + b³ + c³ = 3 abc ...............................(6)
Putting this value of (6) in (4) we get :
==> ( a³ + b³ + c³ ) /abc
==> 3 abc / abc
==> 3
Answer : 3