In a triangle Pqr right angled at q, Xand y are two points on pq and qr such that qy:Yr =1:2 and px :xq =1:2. Prove that 9( Py2 + Xr2) = 13 pr2
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Given that X divides PQ in 2:1
= QX = 2/3*PQ. ....(1)
Similarly, Y divides QR in 2:1
= RY = 2/3*QR ......(2)
In ∆PYQ,
PY^2 = PQ^2 + QY^22
PY^2 = PQ^2 + (2/3*QR)^2
9PY^2 = 9PQ^2 + 4QR^2 .......(3)
In ∆XQR,
XR^2 = XQ^2 + QR^2
XR^2 = (2/3*PQ)^2 + QR^2
XR^2 = 4/9*PQ^2 + QR^2
9XR^2 = 4PQ^2 + 9QR^2. ........(4)
On adding (3)+(4), we get
9(PY^2 + XR^2) = 13(PQ^2 + QR^2)
9(PY^2 + XR^2) = 13PR^2
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= QX = 2/3*PQ. ....(1)
Similarly, Y divides QR in 2:1
= RY = 2/3*QR ......(2)
In ∆PYQ,
PY^2 = PQ^2 + QY^22
PY^2 = PQ^2 + (2/3*QR)^2
9PY^2 = 9PQ^2 + 4QR^2 .......(3)
In ∆XQR,
XR^2 = XQ^2 + QR^2
XR^2 = (2/3*PQ)^2 + QR^2
XR^2 = 4/9*PQ^2 + QR^2
9XR^2 = 4PQ^2 + 9QR^2. ........(4)
On adding (3)+(4), we get
9(PY^2 + XR^2) = 13(PQ^2 + QR^2)
9(PY^2 + XR^2) = 13PR^2
Please mark as brainliest if it helps you...
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