Not sure how to do this

Question

In a certain rectangle, a diagonal has a length of 41. The perimeter of the rectangle is 98. Find the area of the rectangle.

I am not sure how to do this, if someone could give a solution, that would be appreciated.

Answer 1

Hi Isaac X.,

You do this problem the same as you do any word problem: you transform each bit of the problem into an equation, then put the equations together for a solution.

Let's call one side L, the other side W. Then the statement of perimeter is 2L + 2W = 98, right?

And how do you calculate a diagonal? It's a right triangle, right? And since Pythagoras, it's been known that (for the present triangle) L^2 + W^2 = 41^2 , right?

So, can you solve the first equation, for W, as an expression involving L?

Then, apply that expression into the second equation.

You'll come up with an equation involving L^2, L, and a constant, which you must solve using your quadratic roots tool: For equation aX^2 +bX + c = 0, roots (solutions) are at (-b +/- (b^2 - 4ac)^0.5)/2a , right?

That will give you two solutions for L; you only need to calculate one since the other is the value for W (the equations don't distinguish which is which, so they give you both. You need to apply your common sense to whatever a quadratic equation solves to, anyway, and check it to make sure that the solution doesn't violate any of the conditions of the original problem, such as giving you a triangle with a side of negative length, etc.)

Lastly, multiply L times W, to get your answer.

-- Cheers, -- Mr. d.

Answer 2

If you let x and y represent the length and width of the rectangle you can write the following two equations.

x² + y² = 41² (this is the Pythagorean Theorem based on the given diagonal.

2x + 2y = 98 (perimeter of the rectangle).

Now solve the 2nd equation for x or y and substitute into the other equation. Something like this:

x² + (49 - x)² = 41²

Solve the above quadratic equation by squaring the binomial, combining similar terms, set equal to zero and solve by factoring.