Two different ways to calculate triangle's area

[yyasin1061]

Triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR (hypotenuse) and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS?


Question 1: would it be correct to solve this problem by coming up with the sides 15-20-25(15+20+25= 60) of a right triangle?

Question 2: an area of a right triangle =1/2 (leg1)*(leg2), is there a way to calculate an area of a right triangle if only hypotenuse is given to us?

[Guest]

Is it 4:3?

[tali62]

my 2 cents:

Question 1: would it be correct to solve this problem by coming up with the sides 15-20-25(15+20+25= 60) of a right triangle?

I think it's very quick way of solving any right triangle. All you need is to find the right pythagorean triplet. I've used this trick in some questions and the trick has never failed me but remember to back substitute your answer in the question and verify when you use this trick.


Question 2: an area of a right triangle =1/2 (leg1)*(leg2), is there a way to calculate an area of a right triangle if only hypotenuse is given to us?

I think it's not possible. You need either an angle or another side along with hypotnuse. Consider this : you have a semicircle with diameter as hypotnuse. Now, you can draw several right triangles by choosing different points on the circumference of the circle and they'll different areas.

Thanks,
Tauqueer.

[yyasin1061]

actually the correct answer is 16/9

in reference to Question2, i don't understand the following:

(1) a^2 + b^2 = c^2 Pythagorean Theorem for right triangle PQR
(2) ab/2 = 12c/2 Two different ways to calculate triangle PQR's area????

what is 12c/2?

[Guest]

I disagree. Given the rt angled triangle 20-15-25, the areas of the triangles bisected by a perpendicular to the hypotenuse can assume a 16:9 split of the hypotenuse and therefore the areas of the subtriangles.

In this case, PQ=20, QR=15. The question stem says a perpendicular to hyp is dropped with length 12. As the area of a triangle is prop to the lengths of the legs, knowing that the hypotenuse is divided by 16:9 ratio, will divide the areas of the triangles in the same ratio.

In this example, the area of the full triangle is 1/2*15*20=150. The area of PQS = 1/2*12*16 (triangle with larger side PQ)=96 and the area of QRS = 1/2*12*9=54. After reducing the areas you will find that they are divided in 16:9 ratio.

Please let me know this doesnt make sense.

[StaceyKoprince]

Almost perfect. Dropping a perpendicular to the hypotenuse does not necessarily bisect the triangle, though - bisect means to cut in half, so that will only happen if you have a 45-45-90 triangle. Otherwise, you will split the triangle into two different pieces.

One useful rule to know: if you have a right triangle and drop a perpendicular from the right triangle to the hypotenuse, you will create 3 similar triangles - the big one and the two interior ones. So, as our last guest said, I can figure out the ratio of one triangle's area to another's area by applying either the ratios of the sides or the actual sides, if I know them.

[yyasin1061]

Stacey,

it is still not clear if there are two ways to find an area of a right triangle?
"ab/2 = 12c/2 Two different ways to calculate triangle PQR's area"?????

can you please clarify?

thanks

p.s. this question is taken from Computer Adaptive Exam

[StaceyKoprince]

Oh, sorry - yeah, that could've been explained better in the explanation. It doesn't actually use a "different" way to calculate the area - it's still using the standard formula (1/2*b*h) - but using different bases and heights.

So, I can use base = leg a and height = leg b to get Area of PQR = ab/2
I can also use base = hypotenuse c and height = 12 (given) to get Area of PQR = 12c/2

The area of PQR is the same in both cases, so I can set the two equal to each other: ab/2 = 12c/2.

And you cannot find the area of a right triangle given ONLY the hypotenuse and nothing else. In this case, though, we were given the hypotenuse and the height drawn to that hypotenuse.

[shaan17]

Hi Stacey,

Can I please have the complete answer to this question, I am unable to understand the approach to this problem. Do we first find the sides of the triangle PQR ? How do we get the sides as 20-15-25 ?

Thanks in advance
Shaan

[StaceyKoprince]

This question is from our CAT exams and includes a diagram, which I can't post here without a lot of work (and I have too many questions to answer to do that!). I'll give you the text but you may have to rely on the kindness of another poster to give you the diagram.
* * *
In the diagram to the right, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS?

A) 3/2
B) 7/4
C) 15/8
D) 16/9
E) 2
* * *
We are given a right triangle PQR with perimeter 60 and a height to the hypotenuse QS of length 12. We're asked to find the ratio of the area of the larger internal triangle PQS to the area of the smaller internal triangle RQS.

First let's find the side lengths of the original triangle. Let c equal the length of the hypotenuse PR, and let a and b equal the lengths of the sides PQ and QR respectively. First of all we know that:

(1) a2 + b2 = c2.....Pythagorean Theorem for right triangle PQR
(2) ab/2 = 12c/2.....Triangle PQR's area computed using the standard formula (1/2*b*h) but using a different base-height combination:
- We can use base = leg a and height = leg b to get Area of PQR = ab/2
- We can also use base = hypotenuse c and height = 12 (given) to get Area of PQR = 12c/2
- The area of PQR is the same in both cases, so I can set the two equal to each other: ab/2 = 12c/2.

(3) a + b + c = 60....The problem states that triangle PQR's perimeter is 60
(4) a > b..................PQ > QR is given

(5) (a + b)2 = (a2 + b2) + 2ab.....Expansion of (a + b)2
(6) (a + b)2 = c2 + 24c................Substitute (1) and (2) into right side of (5)
(7) (60 - c)2 = c2 + 24c...............Substitute (a + b) = 60 - c from (3)
(8) 3600 - 120c + c2 = c2 + 24c
(9) 3600 = 144c
(10) 25 = c

Substituting c = 25 into equations (2) and (3) gives us:

(11) ab = 300
(12) a + b = 35

which can be combined into a quadratic equation and solved to yield a = 20 and b = 15. The other possible solution of the quadratic is a = 15 and b = 20, which does not fit the requirement that a > b.

Remembering that a height to the hypotenuse always divides a right triangle into two smaller triangles that are similar to the original one (since they all have a right angle and they share another of the included angles), therefore all three triangles are similar to each other. Therefore their areas will be in the ratio of the square of their respective side lengths. The larger internal triangle has a hypotenuse of 20 (= a) and the smaller has a hypotenuse of 15 (= b), so the side lengths are in the ratio of 20/15 = 4/3. You must square this to get the ratio of their areas, which is (4/3)2 = 16/9.

The correct answer is D.