Question

2. In the given figure ABC is a right angled triangle with ZBAC = 90' and
AD BC. .
[
a Prove AADB - ACDA
b.ID=18, CD= 8cm, find AD.
Find the ratio of the area of AADB to area of ACDA
​

Answer 1

Answer:

(i) let ∠CAD = x

⇒ m ∠dab = 90° - x

⇒ m ∠DBA = 180° - (90° + 90° - x) = x

⇒ ∠CDA = ∠DBA ……… (1)

In ΔADB and ΔCDA,

∠ADB = ∠CDA …… [each 90°]

∠ABD = ∠CAD …… [From (1)]

∴ΔADB ~ ΔCDA …….. [By A.A]

(ii) Since the corresponding sides of similar triangles are proportional, we have.

(iii) The ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

Step-by-step explanation:

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