Respuesta :

chetankachana

Answer:

Exercise 1: AB = √39

Exercise 2: Perimeter = 68

Step-by-step explanation:

Exercise 1

Let's solve for CD using the Pythagorean Theorem

Solve for CD:

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

Solve for AB

CB = 1/2 CD

CB = [tex]\frac{2\sqrt{14}}{2}[/tex]

CB = [tex]\sqrt{14}[/tex]

Use the pythagorean theorem, let c = AB

a² + b² = c²

5² + (√14)² = c²

25 + 14 = c²

c = [tex]\sqrt{39}[/tex]

AB = [tex]\sqrt {39}[/tex]

Exercise 2

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.

Find the hypotenuse

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

Find the Perimeter

Now we must multiply by 4 to find the perimeter

          17 x 4 = 68

Perimeter = 68

-Chetan K

sqdancefan

Answer:

  1. AB = √39 ≈ 6.245
  2. P = 68

Step-by-step explanation:

Exercise 1

Let's try it this way:

The Pythagorean theorem tells you the square of the hypotenuse is the sum of the squares of the sides. Then we have ...

  AB² = AC² +BC²

  BC² = AB² -AC² . . . . . for the smaller right triangle

  AD² = AC² +BD² . . . . . for the larger triangle

  AD² = AC² +(2BC)² = AC² +4BC² . . . use BD=2BC

  AD² = AC² +4(AB² -AC²) = 4AB² -3AC² . . . substitute for BC²

  4AB² = AD² +3AC² = 9² +3(5²) = 81 +75 = 156

  AB² = 156/4 = 39

  AB = √39 ≈ 6.245

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Exercise 2

The sides of a rhombus are each the hypotenuse of a right triangle that has legs equal to half the diagonals. The perimeter is 4 times the side length, so we have

  one side = √((16/2)² +(30/2)²) = √(64 +224) = 17

  Perimeter = 4×one side = 4×17

  Perimeter = 68

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Additional comment

The more usual approach to Exercise 1 might be to find the length of CD as √(81-25) = √56 = 2√14, then use that to find AB as √(14 +25) = √39. That is, the playing would be with numbers, rather than symbolic side lengths.

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