PKN wrote:

The length of sides of triangle PQR are p,q and r units. The length of sides of triangle STU are s,t and u units. The following relations hold true:

\(p(q+r-p)=s^2\)

\(q(p+r-q)=t^2\)

\(r(p+q-r)=u^2\)

The type of triangle STU is

(A) Acute angled

(B) Right angled

(C) Obtuse angled

(D) Either A or B

(E) Either A or C

Difficult to understand ..

Then \(s^2, t^2 ,u^2\) should tell you something

Now the observation when you square the three sides..1) RIGHT angle- If it is right angled triangle, the sum of square of two sides will be equal to square of third(hypotenuse)

\((side_1)^2+(side_2)^2=(side_3)^2\)

2) ACUTE angle-If you keep the two sides same and REDUCE the hypotenuse, what will happen??

Sum of square of the two sides will be more than the square of third side.

\((side_1)^2+(side_2)^2>(side_3)^2\)

3) OBTUSE angle-If you keep the two sides same and INCREASE the hypotenuse, what will happen??

Sum of square of the two sides will be less than the square of third side.

\((side_1)^2+(side_2)^2<(side_3)^2\)

Now back to the question\(p(q+r-p)=s^2\)..(i)

\(q(p+r-q)=t^2\)...(ii)

\(r(p+q-r)=u^2\)...(iii)

Add any two

\(q(p+r-q)+p(q+r-p)=s^2+t^2.............qr+rp-(p^2+q^2-2pq)=s^2+t^2...........rp+rq-(p-q)^2=s^2+t^2\)...

from (iii)

\(r(p+q-r)=u^2......rp+rq-r^2=u^2.......rp+rq=u^2+r^2\)..

substitute this in addition above

\(u^2+r^2-(p-q)^2=s^2+t^2\)

since p,q and r are three sides of a triangle and as per triangle INEQUALITY rule, any side of the triangle is more than the difference of other two sides. Hence r>|p-q| ..

therefore r^2>(p-q)^2.....r^2-(p-q)^2>0 and let that be equal to Athis further tells us that \(u^2+r^2-(p-q)^2=s^2+t^2........u^2+A=s^2+t^2\)...

thus \(s^2+t^2>r^2\)

also all equations are similar so we will get similar results for other sides THAT is SUM of square of two sides is greater than the third side..

hence it is ACUTE angled triangle

A

Should we solve such a question in GMAT or take a guess to save time? Because this will definitely eat up a lot of time.