# perimeter and areas urgent check please

• Jul 22nd 2007, 12:22 PM
sanee66
perimeter and areas urgent check please
I think I have these right finally could someone please check and let me know? Will be adding more in a bit. Thanks in advance!!
• Jul 22nd 2007, 12:25 PM
Jhevon
Quote:

Originally Posted by sanee66
I think I have these right finally could someone please check and let me know? Will be adding more in a bit. Thanks in advance!!

A is incorrect. how did you get 4?

assume you are correct and each of the sides are 4, what would the diagonal be? (use Pythagoras' theorem). It won't be 8, I can tell you that

Quote:

Originally Posted by sanee66
I think I have these right finally could someone please check and let me know? Will be adding more in a bit. Thanks in advance!!

B, C and D is ok

E is incorrect. you have to find the other side, it is not given. use Pythagoras' theorem
• Jul 22nd 2007, 01:21 PM
sanee66
So E is
Asquared+Bsquared=Csquared
asquared+36squared=39squared
asquared+1296=1521
asquared=225
a=15
If this is right then what, I thought it would be 15+15+36+36 but apparently not. I don't understand A yet either. The answers For E I have to choose from are 520
540
450
440

And A answers are 8,16,24,and 32
• Jul 22nd 2007, 01:27 PM
Jhevon
Quote:

Originally Posted by sanee66
So E is
Asquared+Bsquared=Csquared
asquared+36squared=39squared
asquared+1296=1521
asquared=225
a=15
If this is right then what, I thought it would be 15+15+36+36 but apparently not.

you were asked for the area, why are you adding? it is length times width

the answer is 540, you got it, but i don't get what you did, it seems to me you just tried to put the numbers together to get one of the answers. the area you have for the triangle is incorrect, but you just added 36 twice to correct it
• Jul 22nd 2007, 01:34 PM
sanee66
[FONT='Times New Roman','serif']Here is how I did it and I had 540 the firwst time I did it but you said it was wrong so I thought I did something [/FONT]
[FONT='Times New Roman','serif']Triangle ABC area is .5*36*39=702/3 sides =234 as the length of AB and DC Area of ABCD=234+234+36+36=540[/FONT]
[FONT='Times New Roman','serif']For A I took 8square root 2 and added 4 as the lenght +4 sides to get 16[/FONT]
• Jul 22nd 2007, 01:34 PM
Jhevon
Quote:

Originally Posted by sanee66
I don't understand A yet either.

i see what the problem is, you have the question incorrect somehow. they would not put the diagonal in squared units, that is the unit for area. revise your post
• Jul 22nd 2007, 01:39 PM
Jhevon
Quote:

Originally Posted by sanee66
[FONT='Times New Roman','serif']Here is how I did it and I had 540 the firwst time I did it but you said it was wrong so I thought I did something [/font]
[FONT='Times New Roman','serif']Triangle ABC area is .5*36*39=702/3 sides =234 as the length of AB and DC Area of ABCD=234+234+36+36=540[/font]
[FONT='Times New Roman','serif']For A I took 8square root 2 and added 4 as the lenght +4 sides to get 16[/font]

you typed the diagonal is "8 squared 2 units" are you telling me you meant "8* squareroot(2) units"??

the area of a triangle is half base times height. by Pythagoras, the height of either triangle is 15.

so the area of one triangle is 0.5*15*36 = 270

but there are two such triangles in the rectangle, so the area of the rectangle is twice that. so the area of the rectangle is 2*270 = 540

you got the right answer, but your method is wrong. i believe were it not for the answer choices guiding you, your answer would have been 468 (which is twice the area of the triangle you found) which would be wrong
• Jul 22nd 2007, 01:46 PM
sanee66
Question A is 8*square root symbol 2
I just am not getting this stuff today
And having a real hard time with the second posting lol.
• Jul 22nd 2007, 02:12 PM
Jhevon
Quote:

Originally Posted by sanee66
Question A is 8*square root symbol 2
I just am not getting this stuff today
And having a real hard time with the second posting lol.

that's fine, we all have those days

Let the side length of the square be $\displaystyle x$

Note that the diagonal will form a right-triangle with two of the sides

Thus, by Pythagoras, $\displaystyle (8 \sqrt {2})^2 = x^2 + x^2$

$\displaystyle \Rightarrow 128 = 2 x^2$

$\displaystyle \Rightarrow 64 = x^2$

$\displaystyle \Rightarrow x = 8$

Thus the perimeter of the square is: $\displaystyle 4x = 32$