Answer:
It is proved that 2PQ + 2SR + 2QR + 2 PS = 4(2(OP)+2(OQ))
Step-by-step explanation:
In a rhombus, all sides are equal.
Thus, in this question;
PQ = SR = QR = PS
By inspection, QS is the same dimension as the four sides. So, QS = PQ
Thus, OQ = PQ/2
For OP, we can find it using Pythagoreas theorem since the angle that divides the diagonals is 90°
Thus;
|OP|² + |OQ|² = |PQ|²
Earlier, we saw that;OQ = PQ/2
Thus;
|OP|² + |PQ/2|² = |PQ|²
|OP|² = |PQ|² - |PQ/2|²
|OP|² = |PQ/2|²
Taking square root of both sides, we have;
OP = PQ/2
So,going back to the question, on the right hand side, we have;
4(2(OP) + 2(OQ))
Let's put,
PQ/2 for OQ and OP as gotten earlier
So,
4(2(PQ/2) + 2(PQ/2)) = 4(PQ + PQ)
Since PQ = SR = QR = PS, we can rewrite as;
4PQ + 4PQ = (PQ + SR + QR + PS) + (PQ + SR + QR + PS) = 2PQ + 2SR + 2QR + 2 PS
This is equal to the left hand side, so the equation is proved correct.
