and please give the numbers of questions with them so I know, I appreciate you if u help!!

fatimahalnabulsi55 Apr 28, 2020

#1**0 **

Hello! I tried to explain a little so that you can understand. If you have questions, let me know! I hope this helps!

1)

a)The legs are* g, p,* and *j*. The hypotenuse is the leg opposite of the right angle, so the hypentenuse in this problem is *g*.

b) The legs are *n, d*, and *r*. The hypotenuse is *d*.

auxiarc Apr 28, 2020

#3**0 **

1. a) & b) remember that the hypotenuse is the longest side of a right triangle or the side that isn't next to the right angle, and the rest are legs.

2. a), b), and c) I'm assuming that you know the pythagorean theorem. remember that it is \(a^2+b^2=c^2\), when c is the hypotenuse. Now you can solve them :)

Hints: a) \(?\) isn't next to the right angle, so it's the hypotenuse. Therefore, \(c^2= 8^2+5^2\), and \(c= \sqrt{8^2+5^2}\).

b) and c) both of the \(?\) are next to the right angle, so it's a leg. Thus, \(b= \sqrt{c^2-a^2}\) for both. (it doesn't really matter whether the \(?\) is a or b)

3. a) \(h^2=7^2+3^2\), so \(h= \sqrt{7^2+3^2}\).

b) \(b=\sqrt{8^2-6^2}\)

c)\(h=\sqrt{12^2+10^2}\)

d) it's actually a triangle you should memorize, which is \(5-12-13\), so you know what \(l\) is.

4. a), b), and c) again, it's almost the same as above.

a) \(?=\sqrt{9^2-6^2}\)

b) \(?=\sqrt{10^2+12^2}\)

c) \(?=\sqrt{25^2-22^2}\)

Hope this helps!

Guest Apr 28, 2020

#4**0 **

2)

a) To solve this problem, you'd use a^{2}+b^{2}=c^{2}, where c is the hypotenuse.

8^{2}+5^{2}=c^{2 } solve

64+25=c^{2 }add

c^{2}=89 square root each side

c=√89

b) Use a^{2}+b^{2}=c^{2} again.

a^{2}+3^{2}=4^{2} solve

a^{2}+9=16 subtract

a^{2}=7 square root each side

a=√7

c) Use a^{2}+b^{2}=c^{2} again.

a^{2}+8^{2}=15^{2} solve

a^{2}+64=225 subtract

a^{2}=161 square root each side

a=√161

auxiarc Apr 28, 2020