Problem 1 :
A girl is in the beach with her father. She spots a swimmer drowning. She shouts to her father who is 50 m due west of her. Her father is 10 m nearer to a boat than the girl. If her father uses the boat to reach the swimmer, he has to travel a distance 126 m from the boat. At the same time, the girl spots a man riding a water craft who is 98 m away from the boat. The man n the water craft is due east of the swimmer. How far must the man travel to rescue the swimmer?
Solution :
Let “A” be the place where her father is standing
Let “C” be the place where a girl is standing
Let “B” be the place where the boat is
Let D” be the place where water craft is
Let “E” be the place where the swimmer is
BC = x m
AB = (x - 10) m
By considering he triangles ∆ ABC, ∆ DBE
∠ABC = ∠DBE (vertically opposite angles)
∠BAC = ∠BDE (alternate angles)
By using AA similarity criterion ∆ ABC ~ ∆ DBE
(AB/DB) = (BC/BE) = (AC/DE)
(AB/DB) = (BC/BE)
(x – 10)/98 = x/126
126 (x – 10) = 98 x
126 x – 1260 = 98 x
126 x – 98 x = 1260
28 x = 1260
x = 1260/28
x = 45
BC = 45 m
Also, (BC/BE) = (AC/DE)
DE = (AC x BE)/BC
= (50 x 126)/45
= 6300/45
= 140 m
The man has to travel 140 m to rescue the swimmer.
Problem 2 :
P and Q are points on sides AB and AC respectively, of triangle ABC. If AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ.
Solution :
AP/PB = AQ/AC
3/9 = 5/15
1/3 = 1/3
In triangles APQ, and ABC we get
(AP/AB) = (AQ/AC)
∠A = ∠A
By using SAS criterion ∆ APQ ~ ∆ ABC
(AP/AB) = (Q/AC) = (PQ/BC)
(AP/PB) = (PQ/BC)
PQ/BC = 3/9
PQ/BC = 1/3
3PQ = BC
BC = 3PQ
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