Finding Dimensions and Maximum Area for Geometric Shapes

[jess1515]

Hey I have two questions that I do not know how to answer...help? Please try to answer at a grade 10 level! And, this isn't a homework question.

1. A yard is to be enclosed by 40 meters of fencing. If all of the fencing is used, what dimensions will result in a yard with an area of 75m^2?

2. What is the maximum area of a right angle triangle whose hypotenuse is 10 cm and perimeter is 26 cm?

 

[symbolipoint]

#1:
You are saying that the perimeter is 40 meters. You should be able to develop this information:
$$\[<br /> \begin{array}{l}<br /> p = perimeter \\ <br /> 75 = theArea \\ <br /> p = 2x + 2y \\ <br /> 75 = xy \\ <br /> \end{array}<br /> \]$$

 

[symbolipoint]

Please note that in the above, I assumed that the yard is rectangular shaped.

 

[Math Jeans]

Exactly, and if you use that method for #1, you end up with two equations and two variables which is easily solved.

For #2:
So you know that c^2=a^2+b^2
You also know that p=a+b+c
c and p are known variables, and b and a are your unknowns. Therefore, you once again have two equations and two unknowns: solvable.