In the given figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ∼ ∆ADE and hence find the length of AE and DE.
Answers
SOLUTION :
Given : ΔACB is right angled triangle at C = 90°.
From the figure : BC = 12 cm , AD=3 cm, DC = 2 cm.
AC = AD + DC = 3 +2= 5 cm
In ∆ACB,
AB² = AC² + BC² (by pythagoras theorem)
AB² = 5² + 12²
AB² = 25 + 144 = 169
AB= √169 = 13
AB = 13 cm
In ΔABC & ΔADE
∠BAC = ∠DAE (common)
∠ACB = ∠AED (each 90°)
ΔABC∼ΔADE (by A-A similarity criterion),
AB/AD = BC/DE = AC/AE
[Since corresponding sides of two similar triangles are proportional]
13/3 = 12/ DE = 5/AE
13/3 = 12/DE
13 DE = 12×3
DE = 36/13
13/3 = 5/AE
13 AE = 5×3
AE = 15/13
Hence, the length of DE= 36/13 & AE = 15/13
HOPE THIS ANSWER WILL HELP YOU...
Answer:
Step-by-step explanation:
In triangle ΔADE and ΔABC
∠BAC = ∠EAD.............(common angle)
∠AED = ∠ACB.................(each 90 degree)
hence, ΔADE is similar to ΔABC
In ΔABC , ∠C=90°
Therefore, AB²=AC²+BC²...............pythagoras theorem
AB=√5²+12²
=√25+144
=√169
=13 cm
Therefore AB/AD=BC/DE=AC/AE..................ΔABC is similar to ΔADE
⇒13/3=12/DE=5/AE
⇒DE=12×3/13
=36/13
and AE=5×3/13
=15/13