Question

prove that in the following figure ∆AOB≈∆COD​

Answer 1

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ \implies \triangle \text{AOB and} \: \triangle \text{COD \: are \: right \: angled \: triangle} \\ \\ \implies \text{AO = CO = 4 \: cm} \\ \\ \implies \text{BO = DO = 3 \: cm} \\ \\ \implies \angle ABO = \angle BDO = 90 \degree \\ \\ \red{ \underline \bold{To \: Prove : }} \\ \implies \triangle AOB \cong \triangle COD$

• According to given question :

$\text{Proof : } \\ \bold{In \triangle AOB \: and \: \triangle COD} \\ \implies \text{AO = CO = 4 \: cm \: \: \: (Given )} \\ \\ \implies \text{BO = DO = 3 \: cm \: \: \:(Given )} \\ \\ \implies \angle B = \angle D = 90 \degree \: \: \: \: \: (Right \: angle \: triangle) \\ \\ \therefore \triangle AOB \: \cong \triangle COD\: \: \: \: (By \: R.H.S \: property) \\ \: \: \: \: \: \: \: \: \: \: \huge\green{Proved }$