The vertices P, Q, R, and S of a parallelogram are at (3,-5), (-5,-4), (7,10) and (15,9) respectively The length of the diagonal PR is

\sqrt{241}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\sqrt{124}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{200}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{180}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\sqrt{241}$
PR = $\sqrt{{(3 - 7)}^{2} + {(- 5 - 10)}^{2}} = \sqrt{16 + 225} = \sqrt{241}$

Suggest Corrections

Similar questions

In parallelogram PQRS, P(-3,2), Q(2,7) and S(1,9) are three vertices. Then the length of diagonal PR is

Q. If

P (1, 2), Q (a, b), R (5, 7)

and

S (2, 3)

are the vertices of a parallelogram, then sum of the squares of the lengths of its diagonals is

Draw a rectangle PQRS in which vertices P, Q, R and S are (1, 4), (–5, 4), (–5, –3) and (1, – 3) respectively.

P (–10, 2) and Q (–5, –1) are the vertices of a parallelogram PQRS. The coordinates of the point where the diagonals intersect are (–3, 2). Find the coordinates of R and S.

Join BYJU'S Learning Program

The vertices P, Q, R, and S of a parallelogram are at (3,-5), (-5,-4), (7,10) and (15,9) respectively The length of the diagonal PR is

The correct option is A √241PR = √(3−7)2+(−5−10)2=√16+225=√241

The correct option is A $\sqrt{241}$
PR = $\sqrt{{(3 - 7)}^{2} + {(- 5 - 10)}^{2}} = \sqrt{16 + 225} = \sqrt{241}$