Question

(b) In the figure (it) given below, LABC = ZDAC and AB =8 cm, AC = AD=Som
(6) Prove that AACD is similar to ABCA () Find BC and CD
(ii) Find area of AACD: area of AABC.
(2914​

Answer 1

Answer:

∠ABC=∠DAC

AB=8cm,AC=4cm,AD=5cm

(i) Now, consider △ACD and △BCA

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

∠ABC=∠CAD … [from the question]

So, △ACD∼△BCA … [by AA axiom]

(ii) AC/BC=CD/CA=AD/AB

Consider AC/BC=AD/AB

4/BC=5/8

BC=(4×8)/5

BC=32/5

BC=6.4cm

Then, consider CD/CA=AD/AB

CD/4=5/8

CD=(4×5)/8

CD=20/8

CD=2.5cm

(iii) from (i) we proved that, △ACD∼△BCA

area of △ACB/area of △BCA=AC

2

/AB

2

=4

2

/8

2

=16/64

By dividing both numerator and denominator by 16, we get,

=

4

1

Therefore, the area of △ACD : area of △ABC is 1:4.