arhumsid wrote:
What is the perimeter of ∆PQS ?
(1) x = 45
(2) w = 15
Because angle w is outside of ∆PQS, it is tempting to think that angle w is not required to construct ∆PQS,
but it is. The second paragraph below shows that w can determine x. Hence, if Statement (1) alone can
answer the question, then Statement (2) can as well.
The minimum details needed to construct a triangle are one side and two angles, or two sides and one
angle, or all the three sides. In ∆PQS, we have ∠PSQ = 30° (from the figure) and the side PS = 1. So, we
are short of knowing one angle or one side to construct ∆PQS. Statement (1) would help with an angle in
the triangle. So, Statement (1) is sufficient.
Construct line PS = 1. Draw an angle making 30 degrees with PS at S and name the line l. Extend the line
PS 2 units further to the right of the point S to locate the point R. Now, draw another line measuring w (=
15) degrees with PR from the point R, and name the line m. So, the point of intersection of l and m is the
point Q. Now, measure ∠PQS to find x.
The answer is (D).
I will try,
The Question is whether we can draw a fixed Triangle ∆PQS or not.
As we know to get a fixed Triangle, we need atleast two sides and one included angle or two angles and one included side.Statement 1): We are given ∠x=45 then ∠p=180-45-30=105.
Now we have Two angles and one included side so we have fixed triangle.
we will have fixed perimeter for ∆PQS.
Hence , Sufficient.
Statement 2): We are given ∠w=15 and also we know ∠y=150, so ∠(q-x)=15=∠w.
now ∆SQR is Isosceles trainlge, if side opposite to ∠(q-x) ,SR, is 2 mm then side opposite to ∠w, QS, is also 2 mm.
Now in ∆PQS, we have two sides and one included angle , so we have one fixed trainlge ∆PQS.
we will have fixed perimeter for ∆PQS.
Hence, Suffiecient.
And) D
please correct me if i am wrong.