Question on pg 261 #11 d.
How do you arrive at the answer 2*sqrt(29)?
Pythagorean theory doesn't seem to work for me (e.g. 12^2 + (2*sqrt(5))^2 = c^2 OR (2*sqrt(5))^2 +b^2 = 12^2).
Thanks,
John
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- johnhurford
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- tommywallach
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Re: OG pg 261 #11 d
Hey John,
The diagonal is the hypotenuse of a triangle which has a height of 4 (see the dotted line all the way to the right), and a base of 10. Looks like you're saying the base is 12, but we have to subtract 2 (it's symmetrical, so we can use the 2 on the right to find the portion on the left).
10^2 + 4^2 = 116
rt. 116 = rt. 4 * rt. 29 = 2 * rt. 29
-t
The diagonal is the hypotenuse of a triangle which has a height of 4 (see the dotted line all the way to the right), and a base of 10. Looks like you're saying the base is 12, but we have to subtract 2 (it's symmetrical, so we can use the 2 on the right to find the portion on the left).
10^2 + 4^2 = 116
rt. 116 = rt. 4 * rt. 29 = 2 * rt. 29
-t
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