Here we will solve different types of examples on establishing riders based on Pythagoras’ theorem.
1. In the quadrilateral PQRS the diagonals PR and QS intersects at a right angle. Prove that PQ^{2}+ RS^{2} = PS^{2} + QR^{2}.
Solution:
Let the diagonals intersect at O, the angle of intersection being a right angle.
In the right-angle ∆POQ, PQ^{2} = OP^{2} + OQ^{2}.
In the right-angle ∆ROS, RS^{2} = OR^{2} + OS^{2}.
Therefore, PQ^{2} + RS^{2} = OP^{2} + OQ^{2} + OR^{2} + OS^{2} ................. (i)
In the right-angle ∆POS, PS^{2} = OP^{2} + OS^{2}.
In the right-angle ∆QOR, QR^{2} = OQ^{2} + OR^{2}.
Therefore, PS^{2} + QR^{2} = OP^{2} + OS^{2} + OQ^{2} + OR^{2} ................. (ii)
From (i) and (ii), PQ^{2}+ RS^{2} = PS^{2} + QR^{2}. (Proved).
2. In ∆XYZ, ∠Z = 90° and ZM ⊥ XY, where M is the foot of the perpendicular. Prove that \(\frac{1}{ZM^{2}}\) = \(\frac{1}{YZ^{2}}\) + \(\frac{1}{XZ^{2}}\).
Solution:
In ∆XYZ and ∆ZYM,
∠XZY = ∠ZMY = 90°,
∠XYZ = ∠ZYM (Common Angle)
Therefore, by AA criterion of similarity, ∆XYZ ∼ ∆ZYM.
\(\frac{XY}{YZ}\) = \(\frac{XZ}{ZM}\)
⟹ YZ ∙ XZ = XY ∙ ZM
Therefore, ZM = \(\frac{YZ ∙ XZ}{XY}\)
Therefore, \(\frac{1}{ZM^{2}}\) = \(\frac{XY^{2}}{YZ^{2} ∙ XZ^{2}}\) = \(\frac{XZ^{2} + YZ^{2}}{YZ^{2} ∙ XZ^{2}}\); [By Pythagoras’ theorem)
Therefore, \(\frac{1}{ZM^{2}}\) = \(\frac{1}{YZ^{2}}\) + \(\frac{1}{XZ^{2}}\). (Proved)
3. In ∆XYZ, ∠Z is acute and XM ⊥ YZ, M being the foot of the perpendicular. Prove that 2YZ ∙ ZM = YZ^{2} + ZX^{2} - XY^{2}.
Solution:
From the right-angled ∆XMY,
XY^{2} = XM^{2} + YM^{2}
= XM^{2} + (YZ - ZM)^{2}
= XM^{2} + YZ^{2} + ZM^{2} - 2YZ ∙ ZM (from algebra)
= YZ^{2} - 2YZ ∙ ZM + (XM^{2} + ZM^{2})
= YZ^{2} - 2YZ ∙ ZM + XZ^{2} (from right-angled ∆XMZ)
Therefore, 2YZ ∙ ZM = YZ^{2} + ZX^{2} – XY^{2}. (Proved)
4. Let PQRS be a rectangle. O is a point inside the rectangle. Prove that OP^{2} + OR^{2} = OQ^{2} + OS^{2}.
Solution:
PQRS is a rectangle for which PQ = SR = length and QR = PS = breadth.
Join OP, OQ, OR and OS.
Draw XY through O, parallel to PQ.
As ∠QPS and ∠RSP are right angles, ∆PXO, ∆SXO, ∆RYO and ∆QYO are right-angled triangles.
Therefore, by Pythagoras’ theorem,
OP^{2} = PX^{2} + OX^{2},
OR^{2} = RY^{2} + OY^{2},
OQ^{2} = QY^{2} + OY^{2} and
OS^{2} = SX^{2} + OX^{2}
Therefore, OP^{2} + OR^{2} = PX^{2} + OX^{2} + RY^{2} + OY^{2} ......... (i)
OQ^{2} + OS^{2} = QY^{2} + OY^{2} + SX^{2} + OX^{2} ......... (ii)
But in the rectangle XSRY, SX = RY = breadth
and in the rectangle PXYQ, PX = QY = breadth.
Therefore, from (i) and (ii), OP^{2} + OR^{2} = OQ^{2} + OS^{2}.
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