Question

In the adjoining figure, AB || DC, AO = 10 cm, OC = 5 cm,
AB = 6.5 cm and OD = 2.8 cm.
(i) Prove that AOAB ~ AOCD.
(ii) Find CD and OB.
(iii) Find the ratio of areas of triangle OAB and triangle OCD.​

Answer 1

Answer:

Step-by-step explanation:

(i) ∠AOB ≅ ∠COD (Vertical angles)

∠ODC ≅ ∠OBA and ∠OCD ≅ ∠OAB (Alternate interior angles)

Δ AOB ~ Δ COD (AAA similarity theorem)

(ii) $\frac{OC}{OA}$ = $\frac{DC}{BA}$ ==> $\frac{5}{10}$ = $\frac{DC}{6.5}$ ==> DC = (5 × 6.5) ÷ 10 ==> DC = 3.25 cm

   $\frac{5}{10}$ = $\frac{2.8}{OB}$ ==> OB = (2.8 × 10) / 5 ==> OB = 5.6 cm

(iii) $P_{AOB}$ = 10 + 5.6 + 6.5 = 22.1 cm; $P_{COD}$ = 2.8 + 3.25 + 5 = 11.05 cm;

Semiperimeters: 11.05 cm and 5.525 cm

$A_{AOB}$ = $\sqrt{11.05(11.05-5.6)(11.05-6.5)(11.05-10)}$ ≈ 16.96 cm²

$A_{COD}$ = $\sqrt{5.525(5.525-3.25)(5.525-5)(5.525-2.8)}$ ≈ 4.241 cm²

Ratio:

$A_{AOB}$ : $A_{COD}$ = $\frac{16.96221}{4.241}$ ≈ 4