Question

IN PQR , (,) S is midpiint of side qr . then pq = 11 ,pr= 17 , ps=13 ,,since find the length of qr. ​

Answer 1

SOLUTION :-

=> in ΔPQR , seg PS is median.

=> .°. Applying Appollonius's theorem.

=> PQ²+PR²= 2PS²+2QS²

=> .°. 11²+17² = 2(13)² + 2(QS)²

=> .°. 121 + 289 = 2(169) + 2QS²

=> .°. 410 = 338 + 2QS²

=> .°. 2QS² = 410-338

=> .°. 2QS² = 72

=> .°. QS² = 72/2

=> .°. QS² = 36

=> .°. √(QS)² = √36 ----(take square root on both side)

=> .°. QS = 6.

_________________

=> QS= ½×QR -------[(point S, midpoint of side QR)].

=> 6 = ½ × QR

=> 6×2 = QR

=> 12 = QR

=> .°. QR = 12.

Hence, the length of side QR is 12.

Answer 2

Given :

• In Δ PQR, S is midpoint of side QR.
• Side PQ = 11 units.
• Side PS = 13 units.

To Find :

• Length of QR.

Solution :

In Δ PQR,

QS = ½QR (1)

$\sf{\red{\Big[\because\:S\:is\:the\:midpoint\:of\:seg\:QR\:\Big]}}$

In Δ PQR, PS is the median.

$\sf{\purple{\big[\because\:PS\:is\:the\:line\:segment\:joining\:vertex\:of\:opposite\:side\:to\:the\:midpoint}\big]}$

We know that the relation between the sides and median of a triangle is given by the Appollonius theorem.

Applying the Appollonius thm;

$\longrightarrow$ $\sf{PQ^2+PR^2=2PS^2+2QS^2}$

$\longrightarrow$ $\sf{11^2+17^2=2(13)^2+2QS^2}$

$\longrightarrow$ $\sf{121+289=2\:\times\:169+2QS^2}$

$\longrightarrow$ $\sf{410=338+2QS^2}$

$\longrightarrow$ $\sf{410-338=2QS^2}$

$\longrightarrow$ $\sf{72=2QS^2}$

$\longrightarrow$ $\sf{\dfrac{72}{2}=QS^2}$

$\longrightarrow$ $\sf{36=QS^2}$

$\longrightarrow$ $\sf{\sqrt{36}=QS}$

$\longrightarrow$ $\sf{QS=6}$

Now, we have to calculate QR and as per the figure, Q-S-R, so QR is double (twice) of QS since S is the midpoint.

Substitute, QS = 6 in equation (i),

$\longrightarrow$ $\sf{QS=\dfrac{1}{2}QR}$

$\longrightarrow$ $\sf{6=\dfrac{1}{2}\:\times\:QR}$

$\longrightarrow$ $\sf{6\:\times\:2=QR}$

$\longrightarrow$ $\sf{12=QR}$

$\large{\boxed{\sf{\purple{Length\:of\:QR\:is\:12\:units}}}}$