2. In A ABC, DE II BC and AD = 4cm, AB = 9 cm, AC = 13.5 cm, then the value of EC=
a) 6 cm b) 7.5 cm c) 9 cm d) none
Answers
Answer:
b 7.5 cm
Step-by-step explanation:
AD/AB=AE/AC
4/9=AE/13.5
AE=13.5*4/9=6cm
EC=AC-AE
=13.5-6= 7.5cm
Value of EC= 7.5 cm, if in ΔABC, DE II BC and AD = 4cm, AB = 9 cm, AC = 13.5 cm
Given : In ΔABC, DE II BC
AD = 4cm, AB = 9 cm, AC = 13.5 cm,
To Find : the value of EC
a) 6 cm b) 7.5 cm c) 9 cm d) none
Solution:
Thales Theorem / BPT ( Basic Proportionality Theorem)
"if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio"
In ΔABC, DE II BC
=> AD/ DB = AE / EC
AD = 4 cm
AB = AD + DB ( Segment Addition postulate)
=> 9 = 4 + DB
=> DB = 5 cm
4/ 5 = AE/EC
=> AE = 4EC/5
AE + EC = AC ( Segment Addition postulate)
4EC/5 + EC = 13.5
=> 9EC = 5 * 13.5
=> EC = 5 * 1.5
=> EC = 7.5 cm
value of EC= 7.5 cm
Correct option is b) 7.5 cm
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