The diagonal of a rectangular room is exactly 17 feet and the area of the room is 120 sq. feet. what is the length of the room ?

Question

no additional details 

Answer 1

Hi Channaraju,

If you happen to be familiar with Pythagorean triples like 3,4,5 and 5,12,13 for right triangle side lengths, then you might know what Pythagorean triple has a hypotenuse of 17. If not, here is how you can work the problem.

 

Set the length of the rectangle = x making the width = 120/x.

Using those as the legs of a right triangle with the hypotenuse = 17ft, use the Pythagorean theorem and the quadratic formula to solve for x.

x² +(120/x)²  = 17² which simplifies to (14400/x²) + x² = 289

Then multiplying by x² gives 14400 + x⁴ = 289 x²

Convert this equation to a quadratic form by substituting z for x²

Then z² -289z +14400 = 0

By the quadratic formula, z = 225 or 64, but remember that z = x², so x is the square root of z making x = 15 or x = 8.

Length is usually considered longer than width, so 15 ft. is the length of the rectangle.

 

By the way, that Pythagorean triple is 8,15,17.

 

Answer 2

Hey Channaraju,

 

First, let's go and visit (special) right triangle.  Do you remember any of them?

(last numbers are hypotenuse (diagonal).)

 

3   4  5

12 7 13

15 8 17

24 7 25.

 

It will be 15 ft by 8 ft room. 

 

How? 15² + 8² = 17² --> 225 + 64 = 289    289^1/2 = 17

 

And lastly, 15 x 8 = 120  (for an area)

 

Good luck to you.

 

If you want this in old Algebraic way, here it is:

 

L x W = 120   &   L² + W² = 17² = 289

 

solve for one of the variable and substitute:

 

L = 120/W       

 

(120)² + W2 = 289

    W²          

 

Multiply by W2 both sides to cancel the denominator.

 

(120)² + W4 = 289W²      

 

14400 - 289W² + W4 = 0

 

Use quadratic formula: a = 1, b = -289, c = 14400

 

you will get 289 (+/-) 161

                          2 

 

So, you will have 450/2 & 128/2

225 & 64.  But since it is W², you must take square root of them.

 

Finally, you will get 15 & 8.

 

Good Luck to you.  I hope you understand how I did it.

 

 

 

P.S.  Just in case, here is the factor form (below):

(225 - W²) (64 - W²) = 0

 

W2 = 225  W2 = 64.  Treat one of them as length.

 

 

 

Answer 3

Call the sides of the rectangle x and y

You can then use the area formula to write xy = 120

 

Since the diagonal of the rectangle is 17, you canuse the Pythagorean theorem to write x^2 + y^2 = 17^2

 

From the first equation, y = 120/x, substituting that in eq 2 you get

 

x^2 + (120/(x^2)) = 289

x^2 + 14400/x^2 = 289    multiply both sides by x^2 to get

x^4 + 14400 = 289x^2    This is called a biquadratic and is factored like a quadratic

x^4 - 289x^2 + 14400 = 0

(x^2 - 64)(x^2 - 225) = 0

x^2 = 64 or x^2 = 225

x=8, x=-8, x=15, x=-15    The negatives cannot be lengths so x=8 or x=15

 

Substituting back in,

When x=8, y=15

When x=15, y=8

 

So either way, your dimensions are 8 and 15.

 

Hope this helps,