In ΔADC, E and B are the points on the sides AD and AC respectively such that ∠ABE = ∠ADC. If AE = 6 cm, BC = 2 cm, BE = 3 cm and CD = 5 cm, then (AB + DE) is equal to:
Show answer & explanation
Correct answer: A
Explanation
Correct Answer is option 1 i.e. 46/3 cm
If all three angles of 2 triangles are equal, then these two triangles are similar triangles.
Given ∠ABE = ∠ADC
AE = 6, BC = 2, EB = 3 and DC = 5 cm
[AB + DE] = ?
In triangles ADC and ABE:
Since ∠ABE = ∠ADC
∠A is common in both
3rd angle will also be equal
Hence, triangles ADC and ABE are similar triangles.
Hence,
AE/AC = BE/DC
⇒ 6/AC = 3/5
⇒ AC = 10
Now,
AB = AC – BC = 10 – 2 = 8 cm
Similarly,
AB/AD = = BE/DC
⇒ 8/AD = 3/5
⇒ AD = 40/3
Now,
DE = AD – DE = 40/3 – 6 = 22/3 cm
Hence,
[AB + DE] = 8 + 22/3 = 46/3 cm