Question

If S is any point in the interior of \[\Delta PQR\]then, prove that \[SQ+SR

Answer 1

Hint: First of all, extend the point S such that the line QS intersects the side PR at point T. Now, join the points Q and S, and points R and S. We know the property that in any triangle the sum of two sides is always greater than the third side. Now, use this property for \[\Delta PQT\] and \[\Delta TSR\] , and get the equations \[PQ+PT > QT\] and \[ST+TR > SR\] . The side QT is the summation of the line segment QS and ST. Now, add both equations. The side PR is the summation of the line segment PT and TR. Now, solve it further and get the required result.

Complete step-by-step answer:
First of all, let us take a point S in the interior of the triangle \[\Delta PQR\] .

Now, extending the point S such that the line QS intersects the side PR at point T.
Similarly, join the points Q and S, and points R and S.
We know the property that in any triangle the sum of two sides is always greater than the third side …………………………………………..(1)
Now, in the \[\Delta PQT\] , we have
\[PQ+PT>QT\] (using the property shown in equation (1))
Now, from the figure of \[\Delta PQR\] , we can see that
\[QT=QS+ST\] ……………………………………(2)
Now, from equation (1) and equation (2), we get
\[PQ+PT>QS+ST\] ……………………………………..(3)
Now, in \[\Delta TSR\] , we have
\[ST+TR>SR\] (using the property shown in equation (1)) …………………………………….(4)
Now, on adding equation (3) and equation (4), we get
\[PQ+PT+ST+TR>SR+QS+ST\]
\[PQ+PT+TR>QS+SR\] ………………………………………………(5)
Now, from the figure of \[\Delta PQR\] , we can see that the side PR is the summation of the line segment PT and TR. So, we can say that
\[PR=PT+TR\] ………………………………….(6)
Now, replacing \[PT+TR\] by \[PR\] in equation (5), we get
\[PQ+PR>QS+SR\] .
Therefore, \[PQ+PR>QS+SR\] .
Hence, proved.

So, the correct answer is “Option A”.

Note: Here, one might think to solve this question without extending the point S to intersect the side PR at point T. If we don’t extend the point our figure will look like

We know the property that in any triangle the sum of two sides is always greater than the third side.
Now, using this property in \[\Delta PQR\] and \[\Delta SQR\] , w have
\[PQ+PR>QR\] ……………………………….(1)
\[QS+SR>QR\] …………………………………(2)
But, the equation (1) and equation (2) is not sufficient to prove \[PQ+PR>QS+SR\] .
So, to get sufficient equations to prove \[PQ+PR>QS+SR\] we need to extend the point S such that the line QS intersects the line PR at T.