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Home » In the figure (2) given below, CA || BD, the lines AB and CD meet at G. (i) Prove that ∆ACO ~ ∆BDO. (ii) If BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm and AC = 3.6 cm, calculate OA and OC.
In the figure (2) given below, CA || BD, the lines AB and CD meet at G.
(i) Prove that ∆ACO ~ ∆BDO.
(ii) If BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm and AC = 3.6 cm, calculate OA and OC.
CBSE | Class 10 | maths | Maths | ML Aggarwal | Similarity
In the figure (2) given below, CA || BD, the lines AB and CD meet at G.
(i) Prove that ∆ACO ~ ∆BDO.
(ii) If BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm and AC = 3.6 cm, calculate OA and OC.

Solution:-

(i) We have to prove that, ∆ACO ~ ∆BDO.

So, from the figure

Consider ∆ACO and ∆BDO

Then,

∠AOC = ∠BOD [from vertically opposite angles]

∠A = ∠B

Therefore, ∆ACO = ∆BDO

Given, BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm, AC = 3.6 cm,

∆ACO ~ ∆BOD

So, AO/OB = CO/OD = AC/BD

Consider AC/BD = AO/OB

3.6/2.4 = AO/3.2

AO = (3.6 × 3.2)/2.4

AO = 4.8 cm

Now, consider AC/BD = CO/OD

3.6/2.4 = CO/4

CO = (3.6 × 4)/2.4

CO = 6 cm

i

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