BC is tangent to circle A at B and to circle D at C. AB= 10 BC= 21 DC= 8 find AD to the nearest tenth
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Answer:
The length of river frontage for each lot are 96.55 ft. 98.85 ft, 101.15 ft and 103.45 ft.
Step-by-step explanation:
See the attached diagram.
The river frontage of 400 ft will be divided into 84 : 86 : 88 : 90 for each lot as AP, BQ, CR, DS and ET all are parallel.
Therefore, PQ : QR : RS : ST = 84 : 86 : 88 : 90 = 42 : 43 : 44 : 45
Let, PQ = 42x, QR = 43x, RS = 44x and ST = 45x
So, (42x + 43x + 44x + 45x) = 400
⇒ 175x = 400
⇒ x = 2.2988.
So, PQ = 42x = 96.55 ft.
QR = 43x = 98.85 ft.
RS = 44x = 101.15 ft and
ST = 45x = 103.45 ft
(Answer)
