PQR and SQR are two triangles on same base QR. if PS intersect atO then show that area triangle pQR/area of triangle SQR =PO/SO
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see diagram..
PA and SB are perpendiculars from P and S respectively onto QR.
ΔPQR = 1/2 * QR * PA
ΔSQR = 1/2 * QR * SB
So we have ΔPQR / ΔSQR = PA/SB --- (1)
ΔPAO and ΔSBO are similar as ∠PAO = ∠SBO = 90°
PA || SB and ∠POA = ∠SOB (vertical angles).
So, PA/ SB = PO / SO ---(2)
compare (1) and (2) , we have the answer.
PA and SB are perpendiculars from P and S respectively onto QR.
ΔPQR = 1/2 * QR * PA
ΔSQR = 1/2 * QR * SB
So we have ΔPQR / ΔSQR = PA/SB --- (1)
ΔPAO and ΔSBO are similar as ∠PAO = ∠SBO = 90°
PA || SB and ∠POA = ∠SOB (vertical angles).
So, PA/ SB = PO / SO ---(2)
compare (1) and (2) , we have the answer.
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