Question

in triangle ABC, if angle ADE=angle B, then prove that triangle ADE triangle ABC Also, if AD =7.6 cm, AE =7.2 cm, BE=4.2 cm and BC=8.4 cm, then find DE

Answer 1

Given : AD= 7.6 cm, AE= 7.2 cm, BE= 4.2 cm, BC= 8.4 cm.
In ∆ADE and ∆ABC
∠A = ∠A (COMMON)
∠ADE = ∠ABC (GIVEN)
∆ADE~∆ABC (AA Similarity)
AD/AB = DE /BC
[Ratios of the corresponding sides of the similar triangles are equal]
AD/(AE+BE) = DE/BC
7.6 / (7.2+4.2)= DE/8.4
7.6 /11.4 = DE/8.4
DE= (7.6 × 8.4) /11.4
DE= (7.6×8.4)/11.4
DE = 63.84/11.4= 5.6 cm
Hence, DE= 5.6 cm

HOPE THIS WILL HELP YOU...

Answer 2

In ΔADE and ΔABC,  ∠ADE = ∠ABC    (given)

 ∠A = ∠A      (common)

ΔADE ~ ΔABC   (AA similarity)

⇒$\frac{AD}{AB}$  = $\frac{DE}{BC}$  = $\frac{AE}{AC}$   [Corresponding sides of similar △'s are proportional]

⇒ $\frac{AD}{AB}$ = $\frac{DE}{BC}$

⇒ 3.83.6 + 2.1 = $\frac{DE}{4.2}$

⇒ 3.85.7 = $\frac{DE}{4.2}$

⇒ 23 = $\frac{DE}{4.2}$

⇒ DE = 2.8 cm