i need some help checking my answers
@YoursTruly
so this triangle BAD is a right triangle ?
yes the right angle is angle A
@Vocaloid welcome back
Uh My instinct is no since there's nothing about the length of GH
* actually in the context of the problem that's yes
Let me check w/ someone on discord
@jhonyy9 sorry to bother but do you have any input on this problem?
oh. np. so my first idea was using the proportionality of corresponding sides - but i think dont work @Vocaloid
@Vocaloid i haw a new idea than we write the altitude theorem FH^2 = FE*FFG so this help without calcule the FH - yes ?
sorry FE*FG = FH^2
i already got past this problem, but you two can help me with the others i need checked
sure I found a similar solution to jhonny's|dw:1529183314718:dw|
us the altitude theorem for QS
do you know it ?
QS^2 = RS*ST from this calcule ST
and after this you need assum ST with RS and will get the right answer
i did that, and i got C
check it please i got other answer right
ST=?
"QS^2 = RS*ST" can you try plugging in the values of QS and RS into this equation? then solve for ST
looking at the diagram QS = 18 and RS = 16, right? can you try plugging these in and solving for ST?
16 + 18 = 34,
hm? the equation does not tell you to add QS and RS
QS is squared
QS^2 = RS*ST
if QS = 18 and RS = 16 we just plug them in to get 18^2 = 16 * ST can you try solving for ST now?
23?
what is 18^2 divided by 16? ^ means an exponent
20.25
good now that's the length of ST but they want the length of RT so RT = ST + RS = ?
36.25
good so 36.25 = your sol'n
@Vocaloid
hm, not quite. all of the choices except one has a proportion where short leg/hypotenuse = short leg/hypotenuse or long leg/hypotenuse = long leg/hypotenuse except one
oh there's actually one where long leg/short leg = long leg/short leg
so D?
hm, not quite forgot to mention you can flip the denominator and numerator of both sides of a proportion and the proportion will still hold so hypotenuse/long leg = hypotenuse/long leg which is the situation for D
|dw:1529185686804:dw|
c/b is the (hypotenuse of the big triangle/long leg of the big triangle) b/y is the (hypotenuse of the right triangle)/(long leg of the big triangle)
|dw:1529185769105:dw| so D is a valid proportion and thus is not the solution
have you tried A yet? are both sides of the proportion comparing corresponding parts of the triangles?
|dw:1529186019101:dw| notice how y is smaller than c right? we would expect y/c to be less than 1 notice how c is larger than x, right? we would expect c/x to be greater than 1, then. so y/c and c/x cannot be equal, thus making A the better sol'n
all of the other sol'ns are valid proportions comparing corresponding parts of the triangles
good
hm, not quite this is kind of similar to the other one with a bit of a twist if you look at the XW, XY, and the altitude WY, they make a right triangle where WY^2 + XY^2 = XW^2 can you try solving for WY?
about like, 22?
hm not quite 24^2 = 16^2 + WY^2 WY = ?
320?
almost since WY^2 = 320 then WY = sqrt(320) now using the rule from before, altitude^2 = the product of the two segments on the hypotenuse so WY^2 = YZ * XY solve for YZ then after that you will add it back to XY to get your sol'n
uh....
WY^2 = YZ * XY you have the values of WY and XY. solve for YZ.
C?
good gtg, idk if @jhonyy9 is still there but I hope they can help you w/ the rest
@princeevee you know that WY^2 = 320 yes ?
and the theorem of altitude say wy^2 = xy*yz where xy = 16 so subtituting what you know you can writing 320=16*yz from this yz = ?
divide both sides by 16 and will get 320/16 = yz
ok. ?
the right answer you get assuming yz and xy right ?
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