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Re: Which is greater b^2+2ad or (a+d)^2 - c^2
[#permalink]
10 Jan 2022, 21:30
10 Jan 2022, 21:30
1
Because a and d form 88degrees (which is less than 90degrees), and b and c form an angle of 92degrees (greater than 90degrees), (a^2 + d^2) < (b^2 + c^2) < e^2 (1)
B= (a + d)^2 - c^2 = (a^2 + d^2) + 2ad - c^2 < (b^2 + c^2) + 2ad - c^2 = b^2 + 2ad = A
Thus A is greater
B= (a + d)^2 - c^2 = (a^2 + d^2) + 2ad - c^2 < (b^2 + c^2) + 2ad - c^2 = b^2 + 2ad = A
Thus A is greater
Which is greater b^2+2ad or (a+d)^2 - c^2
[#permalink]
10 Aug 2022, 17:00
10 Aug 2022, 17:00
1
Step 1: Let's simplify what we're comparing here.
A: b^2 + 2ad and B: expands out to a^2 + 2ad + d^2 - c^2
Since both contain the term 2ad, we can subtract it from both sides.
Now we're comparing A: b^2 and B: a^2 + d^2 - c^2
Step 2: Let's look at the figure. These are almost right triangles, where one of them is slightly less than 90 degrees and one is slightly more.
So let's imagine these were exactly right triangles. Then we would know b^2 + c^2 = e^2 and a^2 + d^2 = e^2.
However, since they're not exactly 90 degrees, let's think about how this affects the above equations.
Since 88 degrees is slightly less than 90, it would slightly shorten the length of the "hypotenuse" e relative to the sides a and d.
Thus a^2 + d^2 > e^2
Also, since 92 degrees is slightly more than 90, it would slightly lengthen the length of the "hypotenuse" e relative to the sides b and c.
Thus b^2 + c^2 < e^2
From here, we can see that since one side of the one equation is greater than e^2 and one side of the second equation is less than e^2, we get the following relationship:
a^2 + d^2 > b^2 + c^2
If we subtract c^2 from both sides, we have
a^2 + d^2 - c^2 > b^2
Let's look back at Step 1. This is exactly the two quantities we are trying to compare. Thus, quantity B is greater.
A: b^2 + 2ad and B: expands out to a^2 + 2ad + d^2 - c^2
Since both contain the term 2ad, we can subtract it from both sides.
Now we're comparing A: b^2 and B: a^2 + d^2 - c^2
Step 2: Let's look at the figure. These are almost right triangles, where one of them is slightly less than 90 degrees and one is slightly more.
So let's imagine these were exactly right triangles. Then we would know b^2 + c^2 = e^2 and a^2 + d^2 = e^2.
However, since they're not exactly 90 degrees, let's think about how this affects the above equations.
Since 88 degrees is slightly less than 90, it would slightly shorten the length of the "hypotenuse" e relative to the sides a and d.
Thus a^2 + d^2 > e^2
Also, since 92 degrees is slightly more than 90, it would slightly lengthen the length of the "hypotenuse" e relative to the sides b and c.
Thus b^2 + c^2 < e^2
From here, we can see that since one side of the one equation is greater than e^2 and one side of the second equation is less than e^2, we get the following relationship:
a^2 + d^2 > b^2 + c^2
If we subtract c^2 from both sides, we have
a^2 + d^2 - c^2 > b^2
Let's look back at Step 1. This is exactly the two quantities we are trying to compare. Thus, quantity B is greater.
Re: Which is greater b^2+2ad or (a+d)^2 - c^2
[#permalink]
15 Aug 2023, 05:11
15 Aug 2023, 05:11
As SherpaPrep mentioned, this can be simplified to comparing b^2 +c^2 and a^2 + d^2. If you don't believe their graph, you can use the cos theorem (Pythagorean for all triangles). By this theorem: e^2 = a^2 + d^2 - 2ad * cos(88), or e^2 = b^2 + c^2 - 2bc* cos(92). Substituting for the two quantities, we are left to compare: 2bc*cos(92) and 2ad*cos(88). You can just google the values, but it's clear that cos(92)<0, cos(88)>0. So the comparison of A and B is equivalent to the comparison of a negative and a positive number since b,c,d,a are all positive. So, B>A as positive is larger than negative. Hope that helps somebody.
