Let PQ, a side of equilateral triangle be b
By symmetry QT=ST=z (say)
A ∆ PQT = x PQ x QT
= x a x z
A ∆ TRU = x RT x OR
= x (a-z) x (a-z)
∴ =
= --------------(p)
Since PQT and PTO are right angled triangles
PQ^{2} + QT^{2} = PT^{2}
RT^{2} +RU^{2} = UT^{2}
a^{2} + z^{2} = b^{2} --------- (1)
And, (a-z)^{2} + (a-z)^{2} = b^{2}--------- (2)
=) a^{2} + z^{2} = 2(a-z)^{2}
=) a^{2} + z^{2} = 2a^{2} + 2z^{2} – 4az
=) a^{2} + z^{2}- 4az = 0
=) a^{2} + z^{2} – 2az = 2az (Please note how the solution is being managed here. You must always be aware of what you are looking for. Here, as equation -℗ we are looking for (a-z)2 in terms of az)
=) (a-z)^{2} =2az
Putting in equation (p) =
Choice (D) is therefore, the correct answer.
Correct Answer: 1 : 2