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Let the point of intersection of CP and AQ be R. If you draw a diagram you would see that,
Consider APR and CQR traingles
`Ahat(P)R = Chat(Q)R` (Given)
`Chat(R)Q = Ahat(R)P`
`Phat(A)R = Chat(R)Q`
Also you can further deduce considering BCP and ABQ triangles,
`Ahat(Q)B = Chat(P)B`
`Phat(B)C = Chat(B)P` (Common angle)
This makes the two triangles, ABQ and BCP are similar in shape.
Therefore their respective sides has the same ratios.
`(AB) / (BC) = (AQ)/(CP) = (BQ)/(BP)`
AB = 100 and AQ = 120 and CP = 150.
`100/(BC) = (120)/(150)`
BC = 125
`120/150 = (BQ)/30`
`BQ = 24`
`BC = BQ + QC `
`125 = 24 + QC`
`QC = 125 -24 = 101`
Therefore the length of CQ is 101 cm
Given : AP=70, BP=30, AQ= 120, PC=150 angle APC= angle AQC, Let the interecting points of the two lines AQ and CP be R. We take the two triangles BPC and ABQ. The triangle BPC similar to triangle ABQ by test A.A.A [ angle BPC(suplimentary to APC)= angle AQB(suplimentary to AQC), angle PBC=angle ABQ (common angles),Since two angles of one triangle = corresponding two angles of the another triangle therefore the remaining angle BCP=remaining angle BAQ a Using properties (side proportionality)of the similar triangle we get, AB/BC = AQ/PC = BQ/BP
100/BC = 120/150 [ taking AB/BC = AQ/PC and AB=AP+PB=70+30 ]
BC= (100*150)/120 = 125
Next, taking the ratio AB/BC = BQ/BP
100/125 = BQ/30
BQ=(100*30)/125 = 24
QC= BC - BQ = 125 - 24 = 101
Yeah that's correct, nothing to worry about.
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