39 Chapter 2
Exercise 2.7
Q6. The area of a rectangular field is 252 square meter. The length of its side is 9 meter longer than its width. Find its sides Sol: Consider dimension of rectangular field
x 2+ 3x - 108 =
0 x 2+ 12x - 9x - 108 =
0 x ( x + 12 ) - 9 ( x + 12 )
=
0 Width of the hall = x Length of the hall = x + 9
( x - 9 )( x + 12 )
=
0
Then according to the given condition Area = 252 m2
Either Or
We take only positive integer because distance between any two points should be positive Base = 9 Perpendicular = 9 + 3=12 Q9. The sides of a right triangle in cm are
Find the sides of triangle. Sol: Since in right angled triangle the longest
Either Or
side is hypotenuses, so Hypotenuses = x + 1 Base = x Perpendicular = x – 1 We take only positive integer because distance
Using Pythagoras theorem between any two points should be positive Base2 + perpendicular2= hypotenuses2 Width of the rectangular field = 12 meter Length of the rectangular field = 12 +9=21 meter Q7. One side of a rectangle is 3 centimeter less than twice the other. If the area of the rectangle is 54 square centimeters, then find the sides of rectangle. Sol: Let width of rectangle = x Than according to condition Length of the rectangle = 2 x- 3 Given that Area = 54 cm2
Either Or
When x = 0 Perpendicular = 0 – 1 = -1 Base = 0 Hypotenuses = 0 + 1 = 1 We take only positive measurement because distance between any two points should be positive, so we neglect the above dimensions When x = 4
Either Or
Perpendicular =4 – 1 = 3 Base = 4 Hypotenuses = 4 + 1 = 5 Q10. A farmer bought some goats for Rs.9000. If he had paid Rs. 100 less for each,
We take only positive integer because distance between any two points should be positive Width of the rectangular field = 6 centimeter Length of the rectangular field =
he would have got 3 goats more for the same amount of money. How many goats did he buy, when the rate in each case is uniform? Sol: since cost each goat is uniform Let the cost of each goat = x Number of goats = y Former bought some goats for Rs. 9000 i.e., xy = 9000 Q8. The length of one side of right triangle exceeds the length of other by 3 centimeters. If the hypotenuse is 15 centimeters, then find
x = 9000 y 1000 less for each, he got 3 the length of the sides of triangle Sol: since the triangle has three sides
goats ( 100 more )( for same 3 ) amount 9000 Let Base = x Perpendicular = x + 3 And hypotenuses = 15 centimeter
3 100 300 9000
Using Pythagoras theorem Base+ ( 2 + + perpendicular) =
x - 9 =
0
x + 12 =
0
x ( x + 9 )
=
252
x =
9
x = -
12
x 2+ 9x - 252 =
0 x 2+ 21x - 12x - 252 =
0 x ( x + 21 ) - 12 ( x + 21 )
=
0
( x - 1 ) ,x, ( x + 1 ) . ( x - 12 )( x + 21 )
=
0
x - 12 =
0
x + 21 =
0 x =
12
x = -
21
x 2+ ( x - 1 ) 2 = ( x +
1
) 2 x 2 + x 2 - 2.x.1 + 1 2 = x 2 + 2.x.1 +
1
2
2x 2 - 2x + 1 - x 2
- 2x - 1 =
0 2x 2 - x 2
- 2x - 2x + 1 - 1 =
0 x 2- 4x =
0 x ( x - 4 )
=0 x = 0 x - 4 =
0 x =
4
x ( 2x - 3 )
=
54
2x 2- 3x - 54 =
0
2x 2
- 12x + 9x - 54 =
0 2x ( x - 6 ) + 9 ( x - 6 )
=
0 ( 2x + 9 )( x - 6 )
=
0
2x + 9 =
0
x - 6 =
0 2x = -9x =
6
x =
- 29 = 2 ( 6 ) -
3
= 12 -
3 =
9 cm
x - y
+ = xy + x - y
- = Putting the values of x and xy 2= hypotenuses2
9000 + 3 ⎛ │ ⎝ 9000 y + + + = + + - = + - = ÷
⎞ │ ⎠ x 2 x 3 2 15
2 - 100y - 300 = 9000 x 2 x 2 2.x.3 3 2
225 2x 26x 9 225 0
2x 2
6x 216 0 by 2
27000 y
- 100y - 300 = 0 ÷ by 100 270 y - y - 3 = 0 × by "y"
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