Question
Question: In the figure point \(T\) is the interior of the rectangle \(PQRS\). Prove that \(T{S^2} + T{Q^2} ...
In the figure point T is the interior of the rectangle PQRS.
Prove that TS2+TQ2=TP2+TR2
As shown in the figure, draw segment AB∥PS and ATB.
Solution
First of all this is a very simple problem, and basic mathematics are needed here to solve this. In order to understand how to approach this problem, we have to know about the basic properties of a rectangle, and the properties of a right angled triangle. We should have an idea about the Pythagoras theorem, where if a right-angled triangle named ABC with hypotenuse AC and the other two sides AB and BC, then the Pythagoras theorem is given by:
⇒AB2+BC2=AC2
Complete step by step answer:
Given that there is a rectangle PQRS
Now T is any interior point in the rectangle PQRS.
Now the AB line segment is parallel to PS and RQ.
The line segment AB is perpendicular to RS and PQ, as given below:
⇒AB⊥RS and AB⊥PQ
Now consider the triangle PTA, this is a right angled triangle. Hence it obeys the Pythagoras theorem, as given below:
⇒PT2=PA2+AT2
⇒PT2−PA2=AT2
Now consider the triangle ATQ, this is a right angled triangle. Hence it obeys the Pythagoras theorem, as given below:
⇒QT2=QA2+AT2
⇒QT2−QA2=AT2
From the above two expressions equating the AT2 expressions, as given below:
⇒PT2−PA2=AT2
⇒QT2−QA2=AT2
∴PT2−PA2=QT2−QA2
Now consider the triangle SBT, this is a right angled triangle. Hence it obeys the Pythagoras theorem, as given below:
⇒ST2=SB2+BT2
⇒ST2−SB2=BT2
Now consider the triangle BRT, this is a right angled triangle. Hence it obeys the Pythagoras theorem, as given below:
⇒RT2=RB2+BT2
⇒RT2−RB2=BT2
From the above two expressions equating the BT2 expressions, as given below:
⇒ST2−SB2=BT2
⇒RT2−RB2=BT2
∴ST2−SB2=RT2−RB2
From the rectangle it is clear that as AB is perpendicular to RS and PQ, which is given by:
⇒AB⊥RS and AB⊥PQ
∴SB=PA and RB=QA
Substituting these expressions in the equation ST2−SB2=RT2−RB2, as given below:
⇒ST2−PA2=RT2−QA2
Now subtracting both the equations obtained, as given below:
⇒PT2−PA2=QT2−QA2
⇒ST2−PA2=RT2−QA2
⇒PT2−ST2=QT2−RT2
Rearranging the terms gives:
⇒PT2+RT2=QT2+ST2
∴TS2+TQ2=TP2+TR2
Hence proved.
Note: While solving such kind of problems always Pythagoras theorem helps, which is important to keep that in mind. This problem can be done in the same way but slight changes, here instead of substituting the values of SB=PA and RB=QA in the second equation, and subtracting the two equations, we can substitute the values of PA=SB and QA=RB in the first equation, and subtract the resulting two equations to get the same result.