Question

In figure, 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). / आकृतीत, बाजू BC वर D हा बिंदू असा आहे की, ∠ABD = ∠CAD. जर AB = 5 सेमी, AD = 4 सेमी आणि AC = 3 सेमी, तर (i) BC (ii) DC (iii) A(∆ ACD) : A( ∆ BCA) काढा. *  2.75 cm / 2.75 सेमी 3.4 cm / 3.4 सेमी 2.4 cm / 2.4 सेमी 3.75 cm / 3.75 सेमी 16 : 25 25 : 9 BC = DC = A(∆ ACD) : A( ∆ BCA) = BC = DC = A(∆ ACD) : A( ∆ BCA) =​

Answer 1

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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