Question

In the figure,D is a point on BC such that angle ABD=angle CAD. If AB=5 CM,AD=4cm and AC=3cm find BC and A(triangle ACD) :A(triangleBCA) Note:it is not given that angle ACD is right angle ​

Answer 1

Answer:
figure,D is a point on BC such that angle ABD=angle CAD. If AB=5 CM,AD=4cm and AC=3cm find BC
and A(triangle ACD) :A(triangleBCA) Note:it is not given that angle ACD is right angle ​ngle ACD is right angle ​

Step-by-step explanation:

It is given that D is a point on BC such that ∠ABD=∠CAD.

Now, In the figure,D is a point on BC such that angle ABD=angle CAD. If AB=5 CM,AD=4cm and AC=3cm find BC and A(triangle ACD) :A(triangleBCA) Note:it is not given that angle ACD is right angle ​From ΔACD and ΔBCA, we have

∠ACD=BCA (Common angle)

∠ABD=∠CAD (Given)

By AA similarity, ΔACD is similar to ΔBCA.

Thus, \frac{AC}{BC}=\frac{AD}{BA}BCAC​=BAAD​

\frac{3}{BC}=\frac{4}{5}BC3​=54​

15=4BC15=4BC

BC=3.75BC=3.75

Now, \frac{a{\triangle}ACD}{a{\triangle}BCA}=\frac{AD^{2}}{BA^{2}}=\frac{4^{2}}{5^{2}}=\frac{16}{25}a△BCAa△ACD​=BA2AD2​=5242​=2516​

(Because of theorem of area of similar triangles)

Therefore, areaΔACD=16cm^{2}[/tex] and areaΔBCA=25cm^{2}[/tex]
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Answer 2

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