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sergey [27]
2 years ago
11

Question 10 The hypotenuse of a right triangle is I m longer than the longer leg. The other leg is 7 m shorter than the longer l

eg. Determine the lengths of the three sides of the triangle. (3 marks)​
Mathematics
1 answer:
bonufazy [111]2 years ago
5 0

Answer:

5, 12, 13

Step-by-step explanation:

let x be the longer leg then x + 1 is the hypotenuse and x - 7 the shorter leg

Using Pythagoras' identity in the right triangle

x² + (x - 7)² = (x + 1)² ← expand using FOIL

x² + x² - 14x + 49 = x² + 2x + 1

2x² - 14x + 49 = x² + 2x + 1 ( subtract x² + 2x + 1 from both sides )

x² - 16x + 48 = 0 ← in standard form

(x - 4)(x - 12) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 4 = 0 ⇒ x = 4

x - 12 = 0 ⇒ x = 12

x = 4 , then x - 7 = 4 - 7 = - 3 ← not possible

x = 12, then x - 7 = 12 - 7 = 5 and x + 1 = 12 + 1 = 13

The lengths of the 3 sides are

longer leg = 12 m , shorter leg = 5 m and hypotenuse = 13 m

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\large\boxed{-9y^5}

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What two numbers multiply to 54 and add up to 7
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Xy = 54
x + y = 7

xy = 54
\frac{xy}{x} = \frac{54}{x}
y = \frac{54}{x}

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\frac{x^{2}}{x} + \frac{54}{x} = 7
\frac{x^{2} + 54}{x} = 7
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x^{2} - 7x + 54 = 0
x = \frac{-(-7) \± \sqrt{(-7)^{2} - 4(1)(54)}}{2(1)}
x = \frac{7 \± \sqrt{49 - 216}}{2}
x = \frac{7 \± \sqrt{-167}}{2}
x = \frac{7 \± i\sqrt{167}}{2}
x = \frac{7 \± 12.9i}{2}
x = 3.5 \± 6.45i
x = 3.5 + 6.45i    or    3.5 - 6.45i

                  x + y = 7
   3.5 + 6.45i + y = 7
- (3.5 + 6.45i)       - (3.5 + 6.45i)
                        y = 3.5 - 6.45i
                  (x, y) = (3.5 + 6.45i, (3.5 - 6.45i)

                          or

                  x + y = 7
   3.5 - 6.45i + y = 7
- (3.5 - 6.45i)     - (3.5 - 6.45i)
                       y = 3.5 + 6.45i
                 (x, y) = (3.5 - 6.45i, 3.5 + 6.45i)

The two numbers that add up to 7 and can multiply to 54 is 3.5 ± 6.45i.
5 0
3 years ago
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