The perimeter of a rectangle is 22, and its diagonal is sqrt{65}. Find its dimensions and area.

Question

longer side:

shorter side:

area:

Answer 1

perimeter = 2a+2b= 22, in other words 2(a+b)=22 or a+b= 11, then the diagonal is sqrt of 65 then you know that a^2+b^2=sqrt 65^2, or a^2+b^2= 65. I think you can solve it from there, just plug the first equation in the second one.

Answer 2

Let l be the length of the rectangle, and let w be the width. From definition of the perimeter of a rectangle, 2*l + 2*w = 22 Rearranging to solve for l in terms of w, l = 11 - w From Pythagorean Theorem, applied to the diagonal: 65 = l^2 + w^2 Using our definition of l in term of w: 65 = (11 - w)^2 + w^2 Expanding and collecting terms: 65 = 121 - 22w + w^2 + w^2 = 121 - 22w + 2w^2 Subtracting 65 from both sides: 2w^2 - 22w + 56 = 0 Dividing through by 2: w^2 - 11w + 28 = 0 Factoring to solve for w: (w - 7)(w - 4) = 0 So, in one solution: Width w = 7 Length l = 11 - 7 = 4 Area = wl = 7*4 = 28 square units If we use the second zero at w = 4: Width w = 4, Length l = 7, Area = 28 square units. Whether the length or width is 7 or 4, the longer side is 7 and the shorter side is 4. The area is the same in both solutions.

Answer 3

The perimeter of a rectangle is the sum of twice the length and twice the width.  When a diagonal is drawn across the rectangle, two congruent triangles are formed.  The legs of the triangle are the sides of the rectangle.  Right triangles indicates that we can use the Pythagorean theorem.

 

Let length = x

Let width = y

 

 

Set the equation based on what we know.

 

2x + 2y = 22

 

x + y = 11                   eq1

 

 

x² + y² = [√65]²               eq2

 

 

 

We have two equations to work with.

 

x + y = 11                eq1

 

x² + y² = 65               eq2

 

 

Substitute eq1 into eq2.  From eq1,

 

x = 11 - y

 

x² = (11 - y)²

 

x² = (11 - y)(11 - y)

 

x² = 121 - 22y + y²

 

 

(121 - 22y + y²) + y² = 65

 

121 - 22y + 2y² = 65

 

2y² - 22y + 56 = 0

 

2(y² - 11y + 28) = 0

 

2(y - 7)(y - 4) = 0

 

y = 7       and        y = 4

 

 

Substitute these y values into eq1 to solve for x.

 

x = 11 - 7                and              x = 11 - 4

 

x = 4                       and              x = 7

 

 

Each pair of solutions has the same dimensions. 

 

Shorter side = 4

Longer side = 7