Stuck on a geometry question.
Post Fiend
Joined: Apr 2004
Posts: 6,473
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From: Lafayette, IN
Wow. Did ANY of you pass geometry? All of you whipped out the calculator way too soon.
Traderjoe came pretty close by identifying the right isosceles triangle. That was step one, but then he went to the calculator.
Since the hypotenuse (usually labelled 'c') is 27 units, each of the legs would be 27/sqrt(2). The area of a triangle is 0.5xbasexheight. In this case base and height are the same and are the two legs. So, if you multiply base times height, you get 27/sqrt(2) squared, which is 27 squared over 2. Multiply that by 1/2 to get the answer, so it's 27 squared over 4, which works out to exactly 729/4, or 182.25 units.
Your answer is EXACTLY 182.25.
Jason (who didn't even touch a calculator for this)
On edit: my apologies to buckarcher, who also had the correct answer, but didn't show how he got it (and also appeared to have whipped out the calculator pretty quick
)
Traderjoe came pretty close by identifying the right isosceles triangle. That was step one, but then he went to the calculator.
Since the hypotenuse (usually labelled 'c') is 27 units, each of the legs would be 27/sqrt(2). The area of a triangle is 0.5xbasexheight. In this case base and height are the same and are the two legs. So, if you multiply base times height, you get 27/sqrt(2) squared, which is 27 squared over 2. Multiply that by 1/2 to get the answer, so it's 27 squared over 4, which works out to exactly 729/4, or 182.25 units.
Your answer is EXACTLY 182.25.
Jason (who didn't even touch a calculator for this)
On edit: my apologies to buckarcher, who also had the correct answer, but didn't show how he got it (and also appeared to have whipped out the calculator pretty quick
)
Last edited by jroehl; Oct 16, 2007 at 08:47 PM.
Senior User
Joined: Oct 2007
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My answer
I like jroehl's answer best. It uses only given information and geometric knowledge.
My answer:
Per the pythagorean theorem:
for a right triangle the hypotenuse is equal to the square root of the sum of the squares of the two opposite sides.
given: that sides b and c are equal
27 squared is 729
729/2 is 364.5
Square root of 364.5 is approximately 19.1 (19.1 squared is 364.8+)
The area of a triangle equals half the area of the area of the square described by its two sides, therefore:
19.1 x 19.1 / 2 is approx. is 364.8/2 equals 182.4 square units
C'mon, this was not "to figure something for his truck." It was for his kid's geometry homework question that he didn't know the answer to. Fess up.
My answer:
Per the pythagorean theorem:
for a right triangle the hypotenuse is equal to the square root of the sum of the squares of the two opposite sides.
given: that sides b and c are equal
27 squared is 729
729/2 is 364.5
Square root of 364.5 is approximately 19.1 (19.1 squared is 364.8+)
The area of a triangle equals half the area of the area of the square described by its two sides, therefore:
19.1 x 19.1 / 2 is approx. is 364.8/2 equals 182.4 square units
C'mon, this was not "to figure something for his truck." It was for his kid's geometry homework question that he didn't know the answer to. Fess up.
Posting Guru
Joined: Dec 2002
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From: Princeton, MN
Originally Posted by jroehl
On edit: my apologies to buckarcher, who also had the correct answer, but didn't show how he got it (and also appeared to have whipped out the calculator pretty quick )
)
Post Fiend
Joined: Apr 2004
Posts: 6,473
Likes: 4
From: Lafayette, IN
My point was that in geometry (and other mathematical disciplines in general) it is easier, quicker and more accurate to work through the problem with variables and radicals (like the square root of two) than to plug in numbers right away. Plugging in numbers makes you lose track of what you are doing--you lose reference to the problem you are trying to solve.
Ronwill's answer is a great example. He takes the square root of 364.5, coming up with 19.1, then turns right around and and squares 19.1 to come up with 364.8 (approximately), when 364.5 is exactly correct.
It's not rocket science, folks.
Jason (800 SAT math, junior year)
Ronwill's answer is a great example. He takes the square root of 364.5, coming up with 19.1, then turns right around and and squares 19.1 to come up with 364.8 (approximately), when 364.5 is exactly correct.
It's not rocket science, folks.
Jason (800 SAT math, junior year)
Post Fiend
Joined: Jun 2003
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How's this for a different angle: any 'right' tirangle's area is half the area of a rectangle with the same two sides. I like Jason's answer best though, as the lengths of the legs are not provided. I just thought i would offer different perspective.
Senior User
Joined: Mar 2003
Posts: 288
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From: Northeast Wisconsin
Originally Posted by jroehl
Wow. Did ANY of you pass geometry? All of you whipped out the calculator way too soon.
Traderjoe came pretty close by identifying the right isosceles triangle. That was step one, but then he went to the calculator. )
Traderjoe came pretty close by identifying the right isosceles triangle. That was step one, but then he went to the calculator.
)
