Question

In figure, S and T trisect the side QR of a right triangle PQR, prove that $8P{T^2} = 3P{R^2} + 5P{S^2}$

Answer 1

Hint: Follow the step by step approach. First find the relation of individual hypotenuse so formed three right angled triangles. Also, suppose the variable “x” for the base is formed by the trisection. Trisection means three equal sections or portions.

Complete step-by-step answer:
Given that S and T trisects the side QR.
Therefore the three line-segments so formed are equal in length.
$ \Rightarrow QS = TS = RT$
Let us suppose that all the equal lengths are of measure “x”
$\therefore QS = TS = RT = x$
The above supposition implies that –
$QR = 3x$
And $QT = 2x$
Now, using Pythagora's theorem which states that in any right angled triangle the square of the hypotenuse is the sum of the square of the adjacent side and the square of the opposite side.
Therefore in the given right angled triangle PQR,
$P{R^2} = P{Q^2} + Q{R^2}$
Place the values we know in terms of “x”
$ \Rightarrow P{R^2} = P{Q^2} + {(3x)^2}$
Simplify the above equation –
$ \Rightarrow P{R^2} = P{Q^2} + 9{x^2}\;{\text{ }}.....{\text{(i)}}$
Similarly, in the given right angled triangle PQS,
$P{S^2} = P{Q^2} + Q{S^2}$
Place the values we know in terms of “x”
$ \Rightarrow P{S^2} = P{Q^2} + {(x)^2}$
Simplify the above equation –
$ \Rightarrow P{S^2} = P{Q^2} + {x^2}\;{\text{ }}.....{\text{(ii)}}$
Similarly in the given right angled triangle PQT,
$P{T^2} = P{Q^2} + Q{T^2}$
Place the values we know in terms of “x”
$ \Rightarrow P{T^2} = P{Q^2} + {(2x)^2}$
Simplify the above equation –
$ \Rightarrow P{T^2} = P{Q^2} + 4{x^2}\;{\text{ }}.....{\text{(iii)}}$
Now, take Left hand side of the equation –
LHS $ = P{T^2}$
Place values by using equation (iii)
LHS $ = 8(P{Q^2} + 4{x^2})$
Simplify the above equation –
LHS $ = 8P{Q^2} + 32{x^2}$
In order to get the required RHS in terms of and , split the terms of “x” and
We can write $32 = 27 + 5$ and $8 = 3 + 5$ in the above terms-
LHS $ = \underline {3P{Q^2} + 5P{Q^2}} + \underline {27{x^2} + 5{x^2}} $
We can rearrange the above terms –
LHS $ = \underline {3P{Q^2} + 27{x^2}} + \underline {5P{Q^2} + 5{x^2}} $
 Take common terms from the above pair of terms –
LHS $ = 3(P{Q^2} + 9{x^2}) + 5(P{Q^2} + {x^2})$
Simplify by using the equations (i) and (ii) –
LHS $ = 3P{R^2} + 5P{S^2}$
LHS$ = $ RHS
$ \Rightarrow 8P{T^2} = 3P{R^2} + 5P{S^2}$
Hence, the required equation is proved.

Note: Remember the concepts of Pythagoras theorem and apply accordingly identifying the right angled triangles in the given figure. Simplify the equation so formed as per the required answer.