Math, asked by tanikutu6344, 4 months ago

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.​

Answers

Answered by supriyagoswami242001
0

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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