For each triangle find the missing length. Round your answer to the nearest tenth. Then find the area and the perimeter.
Problem 1 :
Solution :
Let the missing side be x.
hypotenuse = 19
x^{2} + 17^{2} = 19^{2}
x^{2} + 289 = 361
x^{2} = 361 - 289
x^{2} = 72
x = √72
x = 8.48
Problem 2 :
Solution :
Let the missing side be x.
hypotenuse = x
5^{2} + 13^{2} = x^{2}
25 + 169 = x^{2}
x^{2} = 194
x = √194
x = 13.92
Problem 3 :
Find a third number so that the three numbers form a right triangle:
i) 9 , 41
Solution :
Let x be the unknown.
In the given measure 41 is the greatest.
41^{2} = 9^{2} + x^{2}
1681 = 81 + x^{2}
Subtracting 81 on both sides.
1681 - 81 = x^{2}
x^{2} = 1600
x = √1600
x = 40
Problem 4 :
Ms. Green tells you that a right triangle has a hypotenuse of 13 and a leg of 5. She asks you to find the other leg of the triangle. What is your answer?
Solution :
Hypotenuse = 13, one leg = 5, other leg = x ?
x^{2} + 5^{2} = 13^{2}
x^{2} + 25 = 169
x^{2} = 169 - 25
x^{2} = 144
x = √144
x = 12
Problem 5 :
The sides of a triangle have lengths x, x + 5, and 25.
If the length of the longest side is 25, what value of x makes the triangle a right triangle?
Solution :
Longest side = 25
25^{2} = x^{2} + (x + 5)^{2}
625 = x^{2} + x^{2} + 2x(5) + 5^{2}
625 = 2x^{2} + 10x + 25
2x^{2} + 10x + 25 - 625 = 0
2x^{2} + 10x - 600 = 0
Dividing by 2, we get
x^{2} + 5x - 300 = 0
(x - 15)(x + 20) = 0
x = 15 and x = -20
Since x is one of the side of the triangle, we ignore -20.
So, the value of x is 15.
Problem 6 :
A 22 foot ladder lean against a shed reaching a height of x feet. The base of the ladder is 10 feet from the shed.
Solution :
22^{2} = x^{2} + 10^{2}
484 = x^{2} + 100
x^{2} = 484 - 100
x^{2} = 384
x = √384
x = 19.59
Approximately height of the wall is 19.6 feet.
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