solving systems by substitution?

Question

the length of the diagonal of a rectangle is 13 cm and its area is 10cm^2 determine the length and width of the rectangle.

Answer 1

a^2+b^2=169
ab=10.

This is what you know.

b=10/a
a^2+100/(a^2)=169.
a^4-169a^2+100=0.

a^2=(169+/-sqrt(169^2-400))/2

a^2=(169+/-sqrt(28161))/2

Calculator says: a=12.977 or a=.771.

b is the other one.

Hope that helps--it was a neat problem.

Answer 2

X * Y = 10 cm*2
X*2 + Y*2 = (13cm)*2 > Pythagorean
X*2 + Y*2 = 169
In your question the rectangle has much less area than the hypothenue thus the area suggested is wrong!
Theoretically based on the length of the hypothenue (13cm), the values of the width and the lengths could have been 12cm and 5 cm [12*2 + 5*2 = 144 + 25 = 169]
Verify your question!

Answer 3

The length is 12 and the width is 5.
It's a 5,12,13 triangle, that's a Pythagorean Triple.

Just write that it's a Pythagoran Triple and prove it by doing the Pythagorean Theroem.
a²+b²=c²
5²+12²=13²
25+144=169
169=169

Answer 4

Area= 10cm^2
Diagonal = 13 cm

Diagonal is the longest side ==> d^2=l^2 + b^2
L * B = 10
B = 10/L

13^2 = l^2 + (10/L)^2

Answer 5

we know L*w = 10
and also L^2 + w^2 = 13^2 by pythagorean theorem
so let w = 10/L and substitute it into the 2nd equation
Hey China the area is supposed to be 10

Answer 6

a.b = 10
a^2 + b^2 = 13x13 = 169

substituting one variable into another equation qe have:

Or a=12.977 and b=.771.
or b = 12.977 and a=.771.