# is this triginometry or algebra?

• August 25th 2009, 06:17 AM
cliste09
is this triginometry or algebra?
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
• August 25th 2009, 06:33 AM
gtbiyb
Hello,
You basically have got a simultaneous equation here. Im struggling to see what you have done wrong but i think its in your pythagorous, probaly in the use of brackets. See if this helps-

Firstly the area of the rectangle is length x bredth so

area = (2√x)(√x) = 2x

Then using pythagorous like you said to form a second equation thus:

(√45)^2 = (√x)^2 + (2√x)^2

45 = x + 4x

45 = 5x

x = 9

The area then equals 2 x 9 = 18cm2

Hope that helps! Not sure about part i as there is no mention in the question of a square?

• August 25th 2009, 06:43 AM
cliste09
ohhh now i see ...:D but where did u get the 2 when u multiplied it by 9?????
• August 25th 2009, 07:23 AM
$area = (2\sqrt{x})(\sqrt{x}) = 2x$
$area = (2\sqrt{x})(\sqrt{x}) = 2x$