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CBSE Class 10 Answered

ymxn is a quadrilateral where p,q,r,s is the mid point of ym,mx,nx,ny respectively such that pqrs is a rhombus,show that 3pq2=sn2+nr2+qx2+xr2+py2+ys2.
Asked by rushabhjain.avv | 24 Jan, 2019, 06:49: PM
answered-by-expert Expert Answer
Let us consider a quadrilateral YMXN. P, Q, R and S are mid points of respective sides YM, MX. XN and NY.
 
Let us join PQ, QR, RS and SP. It is given that PQRS is Rhombus.
 
Since S and R are mid points of side YN and XN in ΔYXN, SR is parallel to XY and  XY = 2×SR  .....................(1)
 
Similarly we can prove, PS is parallel to MN and MN = 2×SP ..........................(2)
 
In Rhombus PQRS, we have SR = SP. Hence from (1) and (2), we can prove XY = MN.
 
If diagonals of a quadrilateral are equal, then the quardilateral is rectangle. Hence YMXN is rectangle.
 
Hence ΔNSR, ΔXRQ and ΔYPS are right triangles.
 
NS2 + NR2 = SR2 = PQ2   ..................(3)
XR2 + XQ2 = RQ2 = PQ2 .....................(4)
YS2 + YP2  = PS2 = PQ.....................(5)
 
In the above eqns.(3), (4) and (5), RHS is equated to PQ2 , because sides SR, RQ, PS and PQ are equal and they are sides of rhombus
 
By adding eqns. (3), (4) and (5), NS2 + NR2 + XR2 + XQ2 + YS2 + YP2  = 3 PQ2
 
Answered by Thiyagarajan K | 27 Jan, 2019, 12:48: PM
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