Question

In the line D is a point on BC such that, <ABD =<CAD. If AB= 5cm , AD=4cm and AC=3cm find: (i) BC, (ii)DC (iii)A(∆ACD): A(∆BCA) ​

Answer 1

Answer:

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Step-by-step explanation:

From the question it is given that,

∠ABD=∠CAD

AB=5cm,AC=3cm and AD=4cm

Now, consider the △ABC and △ACD

∠C=∠C … [common angle for both triangles]

∠ABC=∠CAD … [from the question]

So, △ABC∼△ACD

Then, AB/AD=BC/AC=AC/DC

(i) Consider AB/AD=BC/AC

5/4=BC/3

BC=(5×3)/4

BC=15/4

BC=3.75cm

(ii) Consider AB/AD=AC/DC

5/4=3/DC

DC=(3×4)/5

DC=12/5

Dc=2.4cm

(iii) Consider the △ABC and △ACD

∠CAD=∠ABC … [from the question]

∠ACD=∠ACB … [common angle for both triangle]

Therefore, △ACD∼△ABC

Then, area of △ACD/area of △ABC=AD

2

/AB

2

=4

2

/5

2

=16/25

Therefore, area of △ACD : area of △BCA is 16:25.