SOLUTION: Help I am in college and studying for a test. I don't understand how to do this question??????????/ you have two different triangles. the hypotenuse of the first is 6 inches

Algebra ->  Triangles -> SOLUTION: Help I am in college and studying for a test. I don't understand how to do this question??????????/ you have two different triangles. the hypotenuse of the first is 6 inches      Log On


   



Question 53907: Help I am in college and studying for a test. I don't understand how to do this question??????????/

you have two different triangles.
the hypotenuse of the first is 6 inches and the second if 14 inches.
which would have the greater area and why?

Answer by venugopalramana(3286) About Me  (Show Source):
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Help I am in college and studying for a test. I don't understand how to do this question??????????/
you have two different triangles.
the hypotenuse of the first is 6 inches and the second if 14 inches.
which would have the greater area and why?
the 2 triangles given are right angled as it mentions the word hypotenuse,which means the side opposite 90 degree angle.hence we have one side = hypotenuse =6" or 14" and an angle=90 deg.but we need 3 dimensions to fix a triangle.hence the given data does not firm up a triangle.
area of right triangle = 0.5*a*b....where a and b are 2 legs or sides which are perpendicular to each other.
from pythogarus theorem,we have
a^2+b^2=6^2=36...in one triangle and
p^2+q^2=14^2=196..in the second triangle..
but this in no way fixes a&b for first triangle or p&q for the second triangle.
CASE 1
for example in triangle 1....
a=b could be sqrt(18) each giving an area of 0.5*sqrt(18)*sqrt(18)=9 sq.inches
on the other hand in the second triangle...
p could be 0.1...q = sqrt(195.99)...giving an area of 0.5*0.1*sqrt(195.99)=0.7 about.....which is much less than the area of first triangle...
CASE 2
on the other hand it could be otherway round ....
a=0.1....b=sqrt(35.99)......area=0.5*0.1*sqrt(35.99)=0.3 about
where as
p=sqrt(98)=q.....with an area = 0.5*sqrt(98)sqrt(98)=49 sq.inches
which gives higher area for second triangle..
SO THE ANSWER IS ... EITHER TRIANGLE COULD BE BIGGER in AREA DEPENDING ON THE
3 RD.DIMENSION WHICH IS NOT GIVEN.THE GIVEN DATA IS INSUFFICIENT TO CONCLUDE THAT ONE TRIANGLE IS BIGGER IN AREA THAN THE OTHER.