Pythagor triangle exercice pls

Exercise 1: AB = √39

Exercise 2: Perimeter = 68

Step-by-step explanation:

Let's solve for CD using the Pythagorean Theorem

Let CD = b

a² + b² = c²

5² + b² = 9

25 + b² = 81

b² = 81 - 25

b = √56

We can simplify the radical: √56

                                           = √8 * √7

                                           = 2√2 * √7

                                           = 2√14

CB = 1/2 CD

CB =

CB =

Use the pythagorean theorem, let c = AB

a² + b² = c²

5² + (√14)² = c²

25 + 14 = c²

c =

AB =

Let's image a rhombus so that the diagonals are vertical and horizontal. You can see that the point of intersection of the two diagonals from 4-90° angles, and the diagonals form 4 congruent right triangles.

We can find the hypotenuse of one triangle, and multiply by 4 to find the perimeter.

a² + b² = c²

a = 1/2(30) = 15

b = 1/2(16) = 8

15² + 8² = c²

225 + 64 = c²

289 = c²

c = √289

c = 17

Now we must multiply by 4 to find the perimeter

          17 x 4 = 68

Perimeter = 68

-Chetan K

its kind of blurry. can you put a clearer pic?

step-by-step explanation: