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A rectangle has height h and base b
The area of the rectangle is 54cm^2
The perimeter of the rectangle is 33cm
h=54/b
Work out the height and base of the rectangle
*
hb=54
2h+2b=33
h=54/b
therefore,
2(54/b)+b=33
(108/b) + b = 33
I’ve got a feeling I’ve gone down a blind alley here.
Any hints?
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A rectangle has height h and base b
The area of the rectangle is 54cm^2
The perimeter of the rectangle is 33cm
h=54/bWork out the height and base of the rectangle
*
hb=54
2h+2b=33
h=54/btherefore,
2(54/b)+b=33
(108/b) + b = 33I’ve got a feeling I’ve gone down a blind alley here.
Any hints?
A = 54
P = 33
h = 54/b
P = 2h + 2b
33 = 2(54/b) + 2b
2b^2 - 33b + 108 = 0
I get b = 12 and b = 4.5.
This is as far as I got.
The greatest truth about the Rapture is not its timing but it's reality.
Dr. David Jeremiah
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I get b = 12 and b = 4.5.
Well done.
if Ax^2+Bx+C=0
x=[-B±√(B^2-4AC)]/2A
Here:
2b^2 - 33b + 108 = 0
A=2
B=-33
C=108
Therefore, if b=12 cm, h=54/12=4.5 cm
And, if b=4.5 cm, h=54/4.5=12 cm
Last edited by KerimF (2024-05-05 11:31:45)
Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.
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nycguitarguy wrote:I get b = 12 and b = 4.5.
Well done.
if Ax^2+Bx+C=0
x=[-B±√(B^2-4AC)]/2AHere:
2b^2 - 33b + 108 = 0
A=2
B=-33
C=108Therefore, if b=12 cm, h=54/12=4.5 cm
And, if b=4.5 cm, h=54/4.5=12 cm
Perfect.
The greatest truth about the Rapture is not its timing but it's reality.
Dr. David Jeremiah
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Thanks, guys.
How do we know which is the height, and which is the base?
h=54/b
h=54/12=4.5
h=54/4.5=12
??
Also, although this answer comes from a quadratic equation and the quadratic formula can we get it from a simultaneous equation? I ask because someone in class said it was a sim.eq.
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Thanks, guys.
How do we know which is the height, and which is the base?
Good remark.
In general, a problem may have more than one solution (even more than two as it is the case here).
In real life, one usually chooses the solution which seems be the best one for his application.
For example, in this exercise, we use to see the base of a rectangle be longer than its height. So, our preference is likely to say that h=4.5 cm and b=12 cm. But this is valid as long no one complains (mainly a math teacher at one's school).
Also, although this answer comes from a quadratic equation and the quadratic formula can we get it from a simultaneous equation? I ask because someone in class said it was a sim.eq.
At the start we had indeed two simultaneous equations to solve. But, from them, we got a quadratic one to be solved.
I mean, let us consider the following two simultaneous equations.
3x + 2y - 7 = 0 [eq. 1]
x + 4y - 9 = 0 [eq. 2]
Multiplying [eq. 2] by 3
3x + 12y - 27 = 0 [eq. 3]
From [eq. 3] - [eq. 1], we get
0x - 10y - 20 = 0
10y - 20 = 0 [eq. 4]
Here, we derived a linear equation from the two simultaneous ones instead of a quadratic equation.
In both cases, we had two simultaneous equations to solve
Anyway, perhaps someone knows better what solving simultaneous equations does mean.
Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.
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paulb203 wrote:Thanks, guys.
How do we know which is the height, and which is the base?Good remark.
In general, a problem may have more than one solution (even more than two as it is the case here).
In real life, one usually chooses the solution which seems be the best one for his application.
For example, in this exercise, we use to see the base of a rectangle be longer than its height. So, our preference is likely to say that h=4.5 cm and b=12 cm. But this is valid as long no one complains (mainly a math teacher at one's school).paulb203 wrote:Also, although this answer comes from a quadratic equation and the quadratic formula can we get it from a simultaneous equation? I ask because someone in class said it was a sim.eq.
At the start we had indeed two simultaneous equations to solve. But, from them, we got a quadratic one to be solved.
I mean, let us consider the following two simultaneous equations.
3x + 2y - 7 = 0 [eq. 1]
x + 4y - 9 = 0 [eq. 2]Multiplying [eq. 2] by 3
3x + 12y - 27 = 0 [eq. 3]From [eq. 3] - [eq. 1], we get
0x - 10y - 20 = 0
10y - 20 = 0 [eq. 4]Here, we derived a linear equation from the two simultaneous ones instead of a quadratic equation.
In both cases, we had two simultaneous equations to solve
Anyway, perhaps someone knows better what solving simultaneous equations does mean.
A very informative reply. I will post my problems throughout the week UNLESS I land a new job soon.
The greatest truth about the Rapture is not its timing but it's reality.
Dr. David Jeremiah
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Thanks guys.
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