+561 \(a^2=b^2+c^2\)
\(a^2=17^2+8.5^2\)
You'll have to use whatever method you prefer for multiplication without calculator. I like to use this method.
\(a^2=289+72.25\)
\(a^2=361.25\)
\(a=\sqrt{361.25}\)
This is an answer for root form. To convert it to decimal, we can do some trial and error.
\(20^2=400\)
\(19^2=361\)
That's probably the closest we're going to get, so we give the decimal answer as:
\(a\approx19\)
+561 \(a^2=b^2+c^2\)
\(a^2=17^2+8.5^2\)
You'll have to use whatever method you prefer for multiplication without calculator. I like to use this method.
\(a^2=289+72.25\)
\(a^2=361.25\)
\(a=\sqrt{361.25}\)
This is an answer for root form. To convert it to decimal, we can do some trial and error.
\(20^2=400\)
\(19^2=361\)
That's probably the closest we're going to get, so we give the decimal answer as:
\(a\approx19\)
+118608 I am assuming that A,c and E are collinear.
Use pythagorean theorum :)
\(AE^2=8.5^2+17^2\\\)
AE^2=8.5^2+17^2\\
Here is a nefty trick for finding the square of numbers ending in 5
85^2
= 8^2+8 with 25 tacked on the end.
= 64+8 with 25 tacked on the end.
= 72 with 25 tacked on the end.
=7225
so 8.5^2=72.25
17^2=289 (I just know that)
But I could do it as (10+7)^2 = 100+140+49=289
so
\(AE^2=8.5^2+17^2\\ AE^2=72.25+289\\ AE^2=361.25\\ AE^2=\frac{36125}{100}\\ AE^2=\frac{25*1445}{100}\\ AE^2=\frac{25*5*289}{100}\\ AE=\frac{5*\sqrt5*17}{10}\\ AE=\frac{85\sqrt5}{10}\\ AE=8.5\sqrt5\\\)
That is in 8.5sqrt5 cm of course.
Will has given you a good decimal approximation. Thanks Will
