presc04
Hi,
Which side are you taking about when you say "If side opposite to 30º is x" ? ... Little lost here.. where is x ?
thks !!
Hi
presc04,
Please look at the attached fig. below:
As we know that angles of a \(\Delta\) add upto \(180.\)
So in case we have a \(\Delta\) in which the angles are \(30\) \(60 \) \(90\) then the sides obey a rule:
Based on this rule if we know the value one side of this \(30\) \(60 \) \(90 \) triangle we can find the value of the other two sides.
For example: suppose if side opposite to \(30\) angle is \(x\) then opposite \(60\) will be \(x\sqrt3\) and opposite \(90\) we will be \(2x \)
Attachment:
GMAT t 1.png [ 10.02 KiB | Viewed 1305 times ]
By the same rule if the side opposite \(60\) is \(h \) then side opposite \(30\) will be \(\frac{h}{\sqrt3}\) and side opposite \(90\) will be \(2h\)
Coming back to the original question:
Attachment:
GMAT t 2.jpg [ 8.08 KiB | Viewed 1304 times ]
Here also we have two \(\Delta \)'s in which the angles are \(30\) \(60\) and \(90\)
We can see how the side opposite to \(30\) is \(\frac{h}{\sqrt3} \) if the side opposite to \(60 \) is \(h\) as explained above.
Also the two \(\Delta\)'s in green are congruent so their corresponding sides are same.
so we can write s = \(\frac{h}{\sqrt3}+r +\frac{h}{\sqrt3}\)
\(\frac{2h}{\sqrt 3} = s-r\)
\(h= {\frac{(s-r)* \sqrt{3}}{2}\)
Let me know which part is still unclear.
( Also note : Answer choices A and C look same to me , probably a typo )