# sorry if this isnt directly algebra,

• Aug 25th 2009, 05:54 AM
cliste09
sorry if this isnt directly algebra,
im sorry if this isnt directly algebra but in my exam papers its in the algebra Q

a rectangle has a length 2√x cm and width √x cm
the length of the diagonal of the triangle is √45 cm

i) find the area of the rectangle
ii) the area of the square is twice the area of the rectangle
find the length of a side of the square.

for part i) i did the a^2 + 0^2 = h^2

then in the calc. i did 2√x got 2√5 then sq it and got 10 ,...then i
did √x got √5 then sq it and got 5 , added them and got 15 so then i did the hypotenuse thing and √45 and got 20, yeah i know you are prob wondering how i got that and to be honest so am i lol because i used a new calc. for this and when i repeated it on an older calc. of mine i got a different answer so the CORRECT answer was :

2x=18cm^2

part 2 was= 6cm, but i got the answer wrong for i) so i was confused
• Aug 25th 2009, 07:38 AM
HallsofIvy
Quote:

Originally Posted by cliste09
im sorry if this isnt directly algebra but in my exam papers its in the algebra Q

a rectangle has a length 2√x cm and width √x cm
the length of the diagonal of the triangle is √45 cm

You mean "diagonal of the rectangle".

Quote:

i) find the area of the rectangle
ii) the area of the square is twice the area of the rectangle
find the length of a side of the square.

for part i) i did the a^2 + 0^2 = h^2
?? Ah! a= "adjacent side" and "0= Opposite side". A very good reason NOT to use "O" as a variable is that it is easy to mistake it for zero: "0". (And, in fact, you used "0" rather than "O" or "o"!). Now, what do you mean by "i did" that? It would be far better to say that you recognized that the diagonal of the rectangle forms the hypotenuse of a right triangle with the two sides of the rectangle as legs.

Quote:

then in the calc. i did 2√x got 2√5 then sq it and got 10 ,...then i did √x got √5
I assume you mean that you took one leg to be $2\sqrt{x}$ but how do you know that x= 5?

[/quote] then sq it and got 5 , added them and got 15 so then i did the hypotenuse thing and √45 and got 20, yeah i know you are prob wondering how i got that and to be honest so am i lol because i used a new calc. for this and when i repeated it on an older calc. of mine i got a different answer[/quote]
45= 9(5). [tex]]\sqrt{45}= 3\sqrt{5}= 6.708...[tex]. Even without a calculator you should have been able to see that $20^2$= (20)(20)= (2)(2)(10)(10)= 400, nowhere near 45!
Quote:

so the CORRECT answer was :
2x=18cm^2
The correct answer to what problem? You appear to be saying that x= 9 cm^2 which is impossible because "x" is a length, not an area. The first question asked for the area of the rectangle not "2x" nor "x".

part 2 was= 6cm, but i got the answer wrong for i) so i was confused[/QUOTE]
It is close to impossible to figure out what you are saying here but I presume that you used the Pythagorean theorem to say that $(2\sqrt{x})^2+ (\sqrt{x})^2= (\sqrt{45})^2$ or [tex] 4x+ x= 5x= 45[tex] and so x= 45/5= 9, NOT 5. Then $\sqrt{x}= \sqrt{9}= 3$ so the two sides have length $2\sqrt{x}= 2(3)= 6$ and [tex]sqrt{x}= 3[tex]. Now, what is the area of a rectangle with side lengths 6 cm and 3 cm?