Answer:
- AB = √39 ≈ 6.245
- P = 68
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.