I have a rectangle the L 10 feet and W 5 feet . I want increase the L and W by the same amount to area of 75?

Answer 1

You have to increase the L and W by the 1.515 ft to get the area of 75 ft^2.

1. Area of a rectangle
= L*W
= ( L * W )
= (10 *5 )
= 50)
= 50 FT^2.
2. Now, you want to increase the L and w by same amount means L by x ft. and W by x feet.
Area of a rectangle
75= L* W
75= L *W
75= {(10+x)*(5+x)}
75=50+5x+10x+x^2
75-50=5x+10x+x^2
25=x^2+15x
x^2+15x-25=0
Let us compare this with ax^2+bx+c=0
a=1,b=15 and c=-25.

x= -b+sqrt(b^2-4.a.c)divided by 2a
x= -15+sqrt{(-15)^2 - 4*1*(-25)} divided by 2*1
x= {-15+sqrt(225+100)}/2
x= {-15+sqrt(325)}/2
x= (-15+18.03)/2
x= 3.03/2
x= 1.515 ft. is the answer.
or
x= -b - sqrt(b^2-4.a.c)divided by 2a
x= {-15 - sqrt(325)}/2.1
x=(-15-18.03)/2
x=-16.515
(now, x^2+15x-25=0
2.275+22.725-25=0)

Answer 2

The original area of the rectangle with L 10 feet and W 5 feet would be 5 x 10 = 50 ft.

Now, utilizing the same formula for area( L X W) we add the variable "the same amount" to the formula giving us the equation:
(5+x)(10+x) = 75(Since this is the desired area.)
50 + 15x + x^2 = 75

Now, Subtract 75 from both sides.
-25 + 15x + x^2 = 0

Use of the qradratic equation gives the result:
x = [-15 +- 5(13)^(1/2)] / 2

Answer 3

Yes, use algebra:
A = L*W
Let x = how much you want to increase both sides by

A = L*W
Original rectangle: 50 = 10 * 5
Enlarged rectangle: 75 = (10+x)*(5+x)

(10+x)(5+x) = 75
x^2+15x+50 = 75
x^2+15x-25 = 0

Now you can solve for x using the quadratic formula.
Have fun with that.

Answer 4

Area = LW

Increasing the length to 15 feet

A = 15 x 5 =

A = 75 feet

- - - - - - -

Incresing th width to 7.5 feet

a = 10 x 7.5

a = 75 feet

- - - - - -

Increasing the length to 25 feet and decreasing the width to 3 feet

a = 25 x 3 =

a = 75

- - - - - - s-

Answer 5

enable the size of the rectangle be x and y, so the size of fencing required is given by potential of: L = 3x + 2y And the section, A is given by potential of A = xy = 1500000 --> y = 1500000/x so L = 3x + 3000000/x dL/dx = 3 - 3000000/x^2 = 0 (at maximus) x^2 = a million x = 1000 --> y = 1500 So your answer is 1500 feet, 1000 feet

Answer 6

so the length l will be 10+x and W = 5+x

the surface = (10+x)* (5+x) = 75

50 + x^2 + 15 x = 75

x^2 + 15 x -25 =0 you find x = 1.5

Answer 7

Increase length and breadth by x
New length is 10+x and new breadth is 5+x
Thus (10+x)(5+x)=75
50 + 10x+5x+x²=75
x² + 15x -25 = 0
Solve using quadratic formula
x=-15+-(15² + 4x25)^1/2)/2
x=(-15+-(325^1/2))/2
x=(-15+-18.03)/2
Consider the positive value of 18.03
Thus x = (-15+18.03)/2
x=3.03/2
x=1.515
So new dimensions are10+1.515 and 5+1.515
giving length 11.515 and breadth 6.515
Check 11.515 x 6.515 =75.0 as required

Answer 8

(x+a)*(y+a)=A
A=75
x=10
y=5
(10+a)*(5+a)=75
50+15a+a²=75
pq-formula:
a1/2= -15/2 +/- the root of ((15/2)²+25)
is round about -71/2 +/-9,015
a1= round about 1,5095

Answer 9

According to statement of the problem
L=10 + a
W=5 + a
and we know that
LW=75
then
(10 + a)(5 + a)=75

a^2+15a + 50 = 75
a^2+15a - 25=0
Solve for a

a1,2=(-15(+/-)(sqrt(225+100))/2
a1=(-15+18.03)/2=1.515
a2= Who cares? - it is negative.

Answer 10

Use algebra.

(10 + X) * (5 + X) = 75

X^2 + 15X + 50 - 75 = 0

X^2 + 15X - 25 = 0

Solve for X.