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ritula
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mrsmarthi
There is one formula that is worth remembering.

In a right triangle, when a perpendicular(altitude) is drawn from the vertex opposite to hypotenuse on the hypotenuse, then

(length of the altitude)^2 = product of the segments that it divides on the hypotenuse.

Based on the given figure, applying the above formula,
\((QS)^2 = PS * SR\) ==> \(QS = sqrt(16 * 9) = 12.\)

Now height is known, base is known. Applying the formula we get area of the trainagle to be 150.

great!!!
How did you derive that?

+1 for you.
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excellent formula.Thank you very much
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x2suresh
mrsmarthi
There is one formula that is worth remembering.

In a right triangle, when a perpendicular(altitude) is drawn from the vertex opposite to hypotenuse on the hypotenuse, then

(length of the altitude)^2 = product of the segments that it divides on the hypotenuse.

Based on the given figure, applying the above formula,
\((QS)^2 = PS * SR\) ==> \(QS = sqrt(16 * 9) = 12.\)

Now height is known, base is known. Applying the formula we get area of the trainagle to be 150.

great!!!
How did you derive that?

+1 for you.

I got this formula from one of the forums / net and made it's entry into my flashcard list. :)

But if you are interested in the derivation, here is the link

https://www.cliffsnotes.com/WileyCDA/Cli ... 18818.html

Enjoy.........
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mrsmarthi
x2suresh
mrsmarthi
There is one formula that is worth remembering.

In a right triangle, when a perpendicular(altitude) is drawn from the vertex opposite to hypotenuse on the hypotenuse, then

(length of the altitude)^2 = product of the segments that it divides on the hypotenuse.

Based on the given figure, applying the above formula,
\((QS)^2 = PS * SR\) ==> \(QS = sqrt(16 * 9) = 12.\)

Now height is known, base is known. Applying the formula we get area of the trainagle to be 150.

great!!!
How did you derive that?

+1 for you.

I got this formula from one of the forums / net and made it's entry into my flashcard list. :)

But if you are interested in the derivation, here is the link

https://www.cliffsnotes.com/WileyCDA/Cli ... 18818.html

Enjoy.........


Thats real cool. 8-)

+1 for the formula.
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FN
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wow...really neat +1
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The other easy and quick way to do this problem is to recognize that the problem states the large triangle is a right triangle and it gives you the length of its hypotenuse which is 25 (the addition of the two segments....16+9). One of the rules of a right triangle is that if the hypotenuse is 5, then the other two sides have lengths of 3 and 4. (Perhaps you have heard of a 3:4:5 triangle?). Since 25 is a multiple of 5, you can recognize that this is a multiple of a 3:4:5 triangle. So the other two sides (the base and height) are 15 and 20.

A = (1/2)b*h
A = (1/2)*20*15
A = 10*15
A = 150



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