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Nostrana [21]
3 years ago
14

Given a right triangle with legs a, b and hypotenuse c, solve for b if a= 72 and c= 75

Mathematics
1 answer:
kotykmax [81]3 years ago
7 0

Answer:

The answer is a. 21.

Step-by-step explanation:

We solve this question by using Pythagoras theorem to relate the sides of a right angled triangle.

Here,

Hypotenuse of the given triangle(c)=75

Two sides of right angled triangle are:

a=72

b=?

Then,

for a given right angled triangle abc,

Using Pythagoras theorem,

c=\sqrt{a^{2} +b^{2} }

Squaring on both sides,

c^{2} =a^{2}+b^{2}

or,75^{2}=72^{2}+b^{2}

or, b^{2}=75^{2}-72^{2}

or,b^{2}=5625- 5184

or,b^{2} =441

∴b=21

So, the value of b is obtained to 21 by the use of Pythagoras theorem.

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How to get answer of 5/7 of 22??
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Presume we have this sack of marbles. There are 4 green marbles, 3 red marbles, 2 blue marbles, and 1 yellow marble.
aleksley [76]

Answer:

There is a 33.3% chance you will pull a red marble after pulling a blue one.

Step-by-step explanation:

There are a total of 10 marbles in the sack. A blue marble is removed from the bag without being replaced, leaving 9 in the sack. 3 of the marbles left are red. To find the probability of pulling a red marble, we divide the number of red marbles by the number of total marbles.

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Prove that the sum of the squares of the lengths of the medians of a triangle is three fourths the sum of the squares of the len
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Answer:

3[|AB|² + |BC|² + |AC|²] = 4[|AD|² + |BE|² + |CF|²]

Step-by-step explanation:

I've attached an image showing a triangle divided into it's medians.

Now, from the attached image, we see that AB, BC & CD are the lengths of sides of triangle while AD, BE & CF are lengths of the 3 medians of the triangle.

Now, to prove our question, we will use appolonius theorem which states that "the sum of the squares of any of the two sides of a triangle is equal to twice the square on half the third side including twice the square on the median that bisects the third side.

Applying this theorem to the image attached, we have the following;

|AB|² + |AC|² = 2[|AD|² + |BC/2|²]

|AB|² + |BC|² = 2[|BE|² + |AC/2|²]

|AC|² + |BC|² = 2[|CF|² + |AB/2|²]

Adding the 3 equations above gives us;

2|AB|² + 2|BC|² + 2|AC|² = 2|AD|² + |BC|²/2 + 2|BE|² + |AC|²/2 + 2|CF|² + |AB|²/2

Collecting like terms;

(2|AB|² - |AB|²/2) + (2|BC|² - |BC|²/2) + (2|AC|² - |AC|²/2) = 2|AD|² + 2|BE|² + 2|CF|²

Thus gives;

(3/2)[|AB|² + |BC|² + |AC|²] = 2[|AD|² + |BE|² + |CF|²]

Multiply both sides by 2 to give;

3[|AB|² + |BC|² + |AC|²] = 4[|AD|² + |BE|² + |CF|²]

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3 years ago
Someone please please help me??
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Use photomath it works better with functions rather than brainly
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