Question

in the given figure, ABC is a right angled triangle with BAC = 90°
(i) prove that: ∆ADB ~ ∆CDA
(ii) If BD= 18 cm and CD=8 cm, find AD
(iii) Find the ratio of the area of ∆ADB is to area of ∆CDA​

Answer 1

Step-by-step explanation:

Given In the given figure, ABC is a right angled triangle with BAC = 90°(i) prove that: ∆ADB ~ ∆CDA

(ii) If BD= 18 cm and CD=8 cm, find AD

(iii) Find the ratio of the area of ∆ADB is to area of ∆CDA

• Let angle DAC = p
• So angle DAB = 90 – p
• Now angle DAB = 180 – (90 + 90 – p)
•                          = p
• So angle DAC = - angle DAB-----------------------1
• So in triangle ADB and triangle CDA we have
• angle ADB = angle ADC (each is 90 degree)
• From 1 we have,
• Angle ABD = angle CAD  
• Therefore triangle ADB is congruent to triangle CDA (from AA postulate)
• Since the corresponding sides of same triangles are proportional,
• We have CD / AD = AD / DB’
• So CD / AD = AD / DB
• So AD^2 = DB x CD
• So AD^2 = 18 x 8 = 144
•  Or AD = 12 cm  
• The ratio of two areas of same triangles is equal to the ratio of the squares of their corresponding sides.
• So Ar (triangle ADB) / Ar (triangle CDA) = AD^^2 / CD^2
•                                                                         = 144 / 64
•                                                                          =   9 /4
• So the ratio will be 9:4

Reference link will be

https://brainly.in/question/15191149

Answer 2

Answer:

9:4will be the correct answer