Question

Will give medal! The altitude to the hypotenuse of a right triangle divides the hypotenuse into segments of lengths 6 and 9. What is the length of the altitude?
A. 9 sqrt (2)
B. 6 sqrt (6)
C. 3 sqrt (6)
D. 6 sqrt (3)

Answer 1

I think that:

from theorem of Euclide, we have:
x^2=6*15,
y^=9*15
now we can write:
x*y=k*15=2*area of triangle,
squaring both sides of the above equality, we can write:
\[x ^{2}y ^{2}=k ^{2}*15^{2}\]
please now continue

Answer 2

Umm..I'm afraid I don't quite understand.

Answer 3

please substitute in the above equation, x^2, y^2

Answer 4

start with a diagram



you know a^2 + b^2 = 15^2

then if you look at the 2 smaller right triangles you will see

9^2 + d^2 = a^2 and 6^2 + d^2 = b^2

so subsituting you get

9^2 + d^2 + 6^2 + d ^2 = 15^2

now solve for d

Answer 5

using a theorem of Euclide we can write:
x^2=6*15,
and analogously:
y^2=9*15
do you agree?

Answer 6

or

81 + 36 + 2d^2 = 225

117 + 2d^2 = 225

solve for d

hope it helps

Answer 7

Sorry, this is so confusing! I'm really bad at this stuff...

Answer 8

ok... so do you know pythagoras' theorem...?

Answer 9

No, I don't remember learning that

Answer 10

wow... that makes it extremely hard

Answer 11

yea...sorry

Answer 12

I'll leave my method there... so perhaps you can go and research pythagoras theorem and then comeback and study the solution to see if it makes sense

Answer 13

area of our right triangle is:
A=(x*y)/2,
or
A=(hypotenuse*altitude)/2
so we can write:
\[\frac{ x*y }{ 2 }=\frac{ hypotenuse*altitude }{ 2 }\]
and then:
\[x*y=hypotenuse*altitude\]
now I called k our altitude, so substituting, we have:
\[x*y=k*hypotenuse\]
squaring both sides:
\[x ^{2}y ^{2}=k ^{2}*(hypotenuse)^{2}\]
since hypotenuse = 15:
\[x ^{2}*y ^{2}=k ^{2}*15^{2}\]
now inserting x^2 and y^2, we have:
\[(6*15)*(9*15)=k ^{2}*15^{2}\]
now, please, solve for k

Answer 14

Okay, I will try that then.

Answer 15

Pythagoras's theorem is a Year 7 or 8 topic and is fundamental well working with right triangles.