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Citrus2011 [14]
2 years ago
13

ABC is a triangle right angled at A and D is a point on BC such that AD Ʇ BC. Show that AD2 = BD x DC. please answer fast URGENT

Mathematics
1 answer:
postnew [5]2 years ago
6 0

Answer:

The answer is below

Step-by-step explanation:

Pythagoras theorem states that for a right angled triangle, the square of the hypotenuse side is equal to the sum of the square of the remaining sides. The hypotenuse is the longest side (that is side opposite to the 90° angle).

In right angle triangle ABD:

AB² = AD² + BD²     (1)

In right angle triangle ACD:

AC² = AD² + CD²      (2)

Also:

AC² + AB² = BC²       (3)

But BC = BD + CD

AC² + AB² = (BD + CD)²     (4)

Adding equation 1 and 2 gives:

AB² + AC² = (AD² + BD²) + (AD² + CD²)  

AB² + AC² = 2AD² + BD² + CD²

substituting AC² + AB² = (BD + CD)²:

(BD + CD)² = 2AD² + BD² + CD²

BD² + 2(BD)(CD)+ CD² = 2AD² + BD² + CD²

2AD² = 2(BD)(CD)

AD² = BD * CD

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7 0
3 years ago
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Select the rational expression that is equivalent to the given expression below. 4/x-3
Aliun [14]
To find the correct option we need to calculate each of the given options.
Option (1):
\frac{x+2}{x-3} /  \frac{4}{x+2} =  \frac{x+2}{x-3} *  \frac{x+2}{4} =  \frac{(x+2)^2}{4(x-3)}

Option (2):
\frac{x+2}{x-3} *  \frac{4}{x+2} =  \frac{4}{x-3}

Option (3):
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So, the correct answer is option (2)
7 0
3 years ago
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Is 3(x + 1)2 = (3x + 3)2 an identity? Explain and show your reasoning.
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Answer:

YES

Step-by-step explanation:

The equation, 3(x + 1)2 = (3x + 3)2, would be an identity if the equation remains true regardless of the value of x we choose to plug in into the equation.

Let's find out if we would always get a true statement using different value of x.

✍️Substituting x = 1 into the equation:

3(x + 1)2 = (3x + 3)2

3(1 + 1)2 = (3(1) + 3)2

3(2)2 = (3 + 3)2

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✍️Substituting x = 2 into the equation:

3(x + 1)2 = (3x + 3)2

3(2 + 1)2 = (3(2) + 3)2

3(3)2 = (6 + 3)2

18 = 18 (TRUE)

✍️Substituting x = 3 into the equation:

3(x + 1)2 = (3x + 3)2

3(3 + 1)2 = (3(3) + 3)2

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Answer:

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