A straight line drawn parallel to the side BC of AABC meet AB and AC
at points P and Q respectively. If AQ = 3AP, find PQ : BC.
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Here , AB = 12 units and AQ = 2 units
∠P = ∠B
∠Q = ∠C
∴ By AA — similarity,
∆ APQ ~ ∆ ABC
∴
AP
=
AQ
AB AC
⇒
AB
=
AC
AP AQ
⇒
AB
- 1 =
AC
- 1 =
AC - AQ
AP AQ AQ
⇒
AB
- 1 =
QC
AP AQ
⇒
12
- 1 =
QC
QC 2
⇒
12
- 1 =
y
y 2
⇒
12 - y
=
y
y 2
⇒ y² + 2y – 24 = 0
⇒ y² + 6y – 4y – 24 = 0
⇒ y (y + 6) – 4(y + 6) = 0
⇒ (y – 4) (y + 6) = 0
⇒ y = 4 because y ≠ –6
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