# In rectangle ABCD, the diagonal BD is 13 and the perimeter is 34. Find the area of this rectangle.

Let AB = CD = a, BC = AD = b.

1. Triangle ABD – rectangular, hence: BD ^ 2 = AB ^ 2 + AD ^ 2

BD ^ 2 = AB ^ 2 + AD ^ 2, by condition BD = 13, hence: a ^ 2 + b ^ 2 = 169.

2. The perimeter of the rectangle is found by the formula P = (a + b) * 2

By the condition P = 34, hence (a + b) * 2 = 34, a + b = 17.

3. The result is a system of equations:

a ^ 2 + b ^ 2 = 169

a + b = 17

4. Let us express the variable a from the second equation and substitute it into the first equation.

a = 17 – b

(17 – b) ^ 2 + b ^ 2 = 169

5. Let’s open the brackets according to the formula of the square of the difference.

289 – 34b + b ^ 2 + b ^ 2 – 169 = 0

2b ^ 2 – 34b + 120 = 0

6. Divide the equation by 2.

b ^ 2 – 17b + 60 = 0

7. Solve the quadratic equation through the discriminant.

D = 289 – 240 = 49

b1 = (17 + 7) / 2 = 12

b2 = (17 – 7) / 2 = 5

Since a = 17 – b, then a1 = 17 – 12 = 5, a2 = 17 – 5 = 12.

That is, the sides of the rectangle are 12 and 5, which means the area is S = a * b = 5 * 12 = 60.