Help with a proof

[Red0ne]

Hello all,

I'm using this pilot preparation program which has a series of math word problems, most of it is quite easy and straight-forward (higher GCSE/A'S Level).

The reason I'm on here is that I've run into a problem and I'm sure the solution the programmers have given is wrong, would anyone be able to take a look at the problem and be able to verify their solution? Incase they are right, please can you explain why they've used their method of multiplying out the brackets rather than conventionally using the FOIL acronym.

Thanks in advance!

 

[Deleted member 4993]

Hello all,

I'm using this pilot preparation program which has a series of math word problems, most of it is quite easy and straight-forward (higher GCSE/A'S Level).

The reason I'm on here is that I've run into a problem and I'm sure the solution the programmers have given is wrong, would anyone be able to take a look at the problem and be able to verify their solution? Incase they are right, please can you explain why they've used their method of multiplying out the brackets rather than conventionally using the FOIL acronym.

Thanks in advance!
View attachment 13130

Solution given by the programmers is CORRECT.

Please share your work so that we can catch your mistake.

 

[Red0ne]

Thanks for your input Subhotosh, I tried the problem again later and got the correct result. I couldn't follow the solution that was given in the example as it seemed too complicated

0.5 (a + 2) (h + 4) - 46 = 0.5 (a - 2) (h + 8) - 40 [multiply brackets]
0.5 (ah + 4a + 2h + 8) - 46 = 0.5 (ah + 8a - 2h - 16) - 40 [remove brackets]
0.5ah + 2a + h + 4 - 46 = 0.5ah + 4a - h - 8 - 40 [basic operations]
0.5ah + 2a + h - 42 = 0.5ah + 4a - h - 48 [divide by 0.5ah]
2a + h - 42 = 4a - h - 48 [move terms]
2a - 4a = -h - 48 - h + 42 [equalize terms]
-2a = -2h - 6 [divide by -2]
a = h + 3

 

[Dr.Peterson]

What you did here is hardly different from what they did. But their version is complicated by their choice to leave units in place throughout their work. It's good to show units initially in order to verify an equation, but I would drop them before solving.

You mentioned foil; what they did is identical to foil. You don't need to use the acronym when you do the work; in fact, it gets in the way when you have more terms. But that's not an issue here.

 

[JeffM]

In support of Dr. P, using dimensional analysis to make sure your initial equations make sense dimensionally is very good practice, but I never use units in my mathematical working. Rather I use units in my definitions. Moreover, the use of deltas and subscripts in this problem is unnecessary and therefore confusing. And using a and A as variable names with no mnemonic purpose is an annoyance.

If you set things up with an eye toward simplicity, the problem is very easy.

[MATH]a = \text {area of triangle in cm}^2.[/MATH]
[MATH]b = \text {length of the base of the triangle in cm.}[/MATH]
[MATH]h = \text {height of the triangle in cm.}[/MATH]
Three unknowns require three independent equations to solve. The area of a triangle equals the product of the lengths of height and base. Yes, the units in the definitions make sense.

[MATH]a = \dfrac{bh}{2} \implies 2a = bh.[/MATH]
[MATH]a + 46 = \dfrac{(b + 2)(h + 4)}{2} \implies 2a + 92 = bh + 4b + 2h + 8 \implies 2a + 84 = bh + 4b + 2h.[/MATH]
[MATH]a + 40 = \dfrac{(b - 2)(h + 8)}{2} \implies 2a + 80 = bh + 8b - 2h - 16 \implies 2a + 96 = bh + 8b - 2h.[/MATH]
Substitute from the first equation.

[MATH]\therefore bh + 84 = bh + 4b + 2h \implies 84 = 4b + 2h.[/MATH]
[MATH]\therefore bh + 96 = bh + 8b - 2h \implies 96 = 8b - 2h.[/MATH]
Eliminate h by addition.

[MATH]84 + 96 = (4b + 2h) + (8b - 2h) \implies 180 = 12b \implies b = 15 \implies[/MATH]
[MATH]2h = 84 - 4(15) = 84 - 60 = 24 \implies h = 12 \implies a = \dfrac{15(12)}{2} = 15(6) = 90.[/MATH]
Now if you are being formal, you should indicate units again.

[MATH]\text {base} = 15 \text {cm}.[/MATH]
[MATH]\text {height} = 12 \text {cm}.[/MATH]
[MATH]\text {area} = 90 \text {cm}^2.[/MATH]
I cannot believe how they made such a simple problem obscure. It takes far more talent than I have.

 

[Deleted member 4993]

In support of Dr. P, using dimensional analysis to make sure your initial equations make sense dimensionally is very good practice, but I never use units in my mathematical working. Rather I use units in my definitions. Moreover, the use of deltas and subscripts in this problem is unnecessary and therefore confusing. And using a and A as variable names with no mnemonic purpose is an annoyance.

If you set things up with an eye toward simplicity, the problem is very easy.

[MATH]a = \text {area of triangle in cm}^2.[/MATH]
[MATH]b = \text {length of the base of the triangle in cm.}[/MATH]
[MATH]h = \text {height of the triangle in cm.}[/MATH]
Three unknowns require three independent equations to solve. The area of a triangle equals the product of the lengths of height and base. Yes, the units in the definitions make sense.

[MATH]a = \dfrac{bh}{2} \implies 2a = bh.[/MATH]
[MATH]a + 46 = \dfrac{(b + 2)(h + 4)}{2} \implies 2a + 92 = bh + 4b + 2h + 8 \implies 2a + 84 = bh + 4b + 2h.[/MATH]
[MATH]a + 40 = \dfrac{(b - 2)(h + 8)}{2} \implies 2a + 80 = bh + 8b - 2h - 16 \implies 2a + 96 = bh + 8b - 2h.[/MATH]
Substitute from the first equation.

[MATH]\therefore bh + 84 = bh + 4b + 2h \implies 84 = 4b + 2h.[/MATH]
[MATH]\therefore bh + 96 = bh + 8b - 2h \implies 96 = 8b - 2h.[/MATH]
Eliminate h by addition.

[MATH]84 + 96 = (4b + 2h) + (8b - 2h) \implies 180 = 12b \implies b = 15 \implies[/MATH]
[MATH]2h = 84 - 4(15) = 84 - 60 = 24 \implies h = 12 \implies a = \dfrac{15(12)}{2} = 15(6) = 90.[/MATH]
Now if you are being formal, you should indicate units again.

[MATH]\text {base} = 15 \text {cm}.[/MATH]
[MATH]\text {height} = 12 \text {cm}.[/MATH]
[MATH]\text {area} = 90 \text {cm}^2.[/MATH]
I cannot believe how they made such a simple problem obscure. It takes far more talent than I have.

Only thing I'll add that:

go back and confirm that the given conditions are met.

Original area = 90

1) Add 2 to base and 4 to height the new area (N) becomes = 90 + 46 = 136

N = 1/2 * (15+2) * (12+4) = 17 * 8 = 136 ....................... Checks

2) Add 8 to height and subtract 2 from base - the new area (N) becomes = 90 + 40 = 130

N = 1/2 * (15-2) * (12+8) = 10 * 13 = 130 ....................... Checks

Thus the solution is most probably correct.