D is a point on BC such that ∠ABD = ∠CAD. If AB = 5 cm, AD = 4 cm and AC = 3 cm. Find (i) BC (ii) DC (iii) A(∆ ACD) : A( ∆ BCA).
Answers
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.
Solution :-
In ∆ACD and ∆BCA we have,
→ ∠DCA = ∠ACB { common. }
→ ∠DAC = ∠ABC { given. }
so,
→ ∆ACD ~ ∆BCA { By AA similarity. }
then,
→ AC/BC = CD/CA = AD/BA { when two ∆'s are similar corresponding sides are in same proportion. }
putting given values now, we get,
→ AC/BC = AD/BA
→ 3/BC = 4/5
→ 4•BC = 3 * 5
→ BC = (15/4)
→ BC = 3.75 cm (i) (Ans.)
again,
→ CD/CA = AD/BA
→ CD/3 = 4/5
→ 5•CD = 3 * 4
→ CD = (12/5)
→ CD = DC = 2.4 cm (ii) (Ans.)
also, when two ∆'s are similar, the ratio of the areas is equal to the square of the ratio of their corresponding sides.
therefore,
→ A(∆ACD) : A(∆BCA) = AD² : BA² = 4² : 5² = 16 : 25 (iii) (Ans.)
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