use the property of similar triangles - their corresponding sides are in the same ratio
the large triangle and the smaller one containing the sides a and b are similar
So a would be 18?
how did you get that?
The triangles are similar by the AA Postulate. It appears to me that we are missing a third numerical measure that is needed to solve the problem. @Mikayla_Rose Is the outer, large triangle given to be a right triangle? |dw:1444254138067:dw|
you find the corresponding sides by looking at the equal angles - then use the sides opposite these angles.
nevermind i think its 8 because that is simplifed form of 32
I'd like to see the equation(s) you two are using. I am not seeing it. Yes, if that angle I mentioned is a right angle, the problem can be solved. But, to assume that it is a right angle is incorrect. Of course, the diagrams in these posted problems are frequently incomplete.
I agree we should not really assume its a right angle but I cant see how we can solve this unless we make that assumption
I dont know how to get the answer, that is why i posted this question to get instructions.
can you see why (assuming the big triangle is right angled whay the 2 triangles are similar - similar triangles have all 3 angles equal
We can't give instructions when the diagram is incomplete.
ok.... How do i complete the diagram?
it will be complete if the big triangle has been marked as right-angled.
- as in directrix's drawing.
do you see what i mean?
okay what is the next step?
triangles ADC and BDC are similar ( by AA postulate) therefore we can write 2 / b = b / (32+2) 2 / b = b / 34 solve for b
can you find b?
im working on it right now
b= 2 sqr root 17, -2 sqr root 17
well it has to be the positive value , of course now you can find the value of a by applying the pythagoras theorem to the small triangle.
a^2 = (2 sqrt17)^2 - 2^2
i got a= sqr root 30?
Oh - i just realised - angle ADC must be a right-angle because the 2 angles marked are equal and angle C is common.
No a^2 = 68 - 4 = 64 a = 8
oh okay i got it thanks a whole lot for helping me!