Can sum1 pleaz help me with this geometry extra credit problem?

Question

In triangle PRQ, angle PRQ is a right angle; segment PR is 3 units and RQ segment is 5 units. What is the perimeter of the triangle?

Answer 1

Too bad the legs aren't length 3 and 4 so the hypotenuse is 5.
That would be an easy perimeter of 12.

In this case the legs are 3 and 5
3^2 + 5^2 = H^2
H^2 = 34
so Hypotenuse = "square root of 34"

So Perimeter = 3 + 5 + "square root of 34"
= 8 + "square root of 34" units

Answer 2

3+5+6=14

Answer 3

Wait did you write the question in correctly?
You need to use the pythagorean theorem.
so its (3x3)+(5x5)=cXc
9+25=34
34=cXc
so you have to find the square root.
there is no square root for 34 so just put the square root of 34
(in numbers of course, it doesnt have it on this keyboard)

But if you did type it in wrong, it's:
Its a 3, 4, 5 triangle so the perimeter would be 3+4+5=12
=)

Answer 4

if's a special triangle with sides 3, 4, 5
you can get the missing side using the pythagorean ther.
a^2 +b^2 =c^2
so 3^2 +b^2 =5^2
so 9 + b^2 = 25
so b^2 = 16
so b = 4
add the sides to find perimeter 3+4+5=12

Answer 5

P
I\
I \
5 I \ root34
I \
I_ _ _\
R 3 Q


PQ= root(5X5+3X3) = root34
Therefore the perimeter is 5+3+root34 = 8+root34

Answer 6

all you need to do is Pythagoras, so 3^2 +5^2 =34
the square root of 34 is 5.83 so add 3 5 and 5.83 to get 13.83 as a perimeter.

Answer 7

a^2+b^2=c^2 [or (aXa)+(bXb)=(cXc) ]

PR=a=3
RQ=b=5
PQ=c=x

3^2+5^2=x^2
9+25=x^2
34=x^2

This next part depends on your teacher...some teachers want the actual number, and others don't.

So, the answer is either
P=a+b+c
P=3+5+the square root of 34
P=8+the square root of 34
or
P=3+5+the square root of 34 (5.83)
the square root of 34=5.83...so
x 5.83
3+5+5.83=13.83

Answer 8

3 in....but you cheated