Question

The length of a rectangle is 59 inches greater than twice the width. If the diagonal is 2 inches more than the​ length, find the dimensions of the rectangle

Answer 1

Answer:

Dimensions of rectangle are 20 inches and 99 inches.

Step-by-step explanation:

Let the width be w,

then as length, which is 59 inches greater than twice the width, is 2w+59 inches

and square of diagonal is w2+(2w+59)2=w2+4w2+236w+3481

= 5w2+236w+3481

But as diagonal is 2 more than length, it is 2w+61 and hence

5w2+236w+3481=(2w+61)2=4w2+244w+3721

Hence w2−8w−240−0 or (w−20)(w+12)=0

Hence, width is 20 or −12, but as it cannot be negative, it is 20 and length is 20×2+59=99.

Hence dimensions of rectangle are 20 inches and 99 inches.

Check 202+992=400+9801=10201=1012 and diagonal is 101 and 2 more than length.

I hope this help

Answer 2

To start we need to assign variables for what were trying to find. Lets say the rectangle has width w. Now we know that the rectangle has a length 20 more than twice its width. So in algebra this would be l = 20+2w. Now we know that the diagonal is 2 more than the length. In algebra this would be d = 2+l. Substituting our length equation into the diagonal expression we get d = 2 + 20 + 2w = 22+2w.

So our equations are l = 20+2w and d = 22+2w. We also know that the angles of rectangle are right angles. So the length, width, and diagonal form a right triangle and using the Pythagorean theorem we find that:

(20+2w)^2 + w^2 = (22+2w)^2.
Now we can solve for the width.

(20+2w)^2 + w^2 = (22+2w)^2.
400 + 80w + 4w^2 + w^2 = 484 + 44w + w^2

I will leave the rest to you but just a hint, you will have to use the quadratic formula. You will eventually get it down to which if you plug into your calculator should come out to 1.9226 or -10.9226.

Since we know that a measure of a side can't be negative we know that the width has to be 1.9226. So now we go back and find the length and the diagonal. I will leave this for you to do. You should get L = 23.8452 and D = 25.8452.