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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.

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.
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

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

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

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.