8
c
I AABC-ADEF such that DE-5cm,EF2cm, DF = 2.5cm, BC = 4cm, then
perimeter of the AABC is
(a) 18cm
(b) 20cm (c) 19cm
(d)15cm
Answers
Answered by
0
Answer:
b
Explanation:
Answered by
0
Answer:
please mark me as brainliest
Explanation:
Given : ΔABC ~ ΔDEF, DE = 3 cm, BC = 4 cm , DF = 2.5 cm , EF = 2 cm
Solution :
AB/DE = BC/EF = CA/FD
[corresponding sides of two similar triangles are in proportional]
AB/DE = BC/EF
AB/3 = 4/2
AB/3 = 2/1
AB = 6 cm …………..(1)
Similarly ,
BC/EF = CA/FD
4/2 = CA/2.5
2 AC = 4 × 2.5
2 AC = 10
AC = 10/2
AC = 5 cm .. …………(2)
Perimeter of ∆ ABC = AB + BC + AC
= 6 + 4 + 5
[From equation 1 and 2]
Perimeter of ∆ ABC = 15 cm
Hence, the Perimeter of ∆ABC is 15 cm.
Among the given options Option (d) 15 cm is the correct answer.
HOPE THIS ANSWER WILL HELP YOU……
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