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प्रश्न
In the given figure, ABC is a right angled triangle with ∠BAC = 90°.
- Prove that : ΔADB ∼ ΔCDA.
- If BD = 18 cm and CD = 8 cm, find AD.
- Find the ratio of the area of ΔADB is to area of ΔCDA.
उत्तर
i. Let ∠CAD = x
`=>` m ∠DAB = 90° – x
`=>` m ∠DBA = 180° – (90° + 90° – x) = x
`=>` ∠CDA = ∠DBA ...(1)
In ΔADB and ΔCDA,
∠ADB = ∠CDA ...[Each 90°]
∠ABD = ∠CAD ...[From (1)]
∴ ΔADB ∼ ΔCDA ...[By A.A]
ii. Since the corresponding sides of similar triangles are proportional, we have.
`(BD)/(AD) = (AD)/(CD)`
`=> (18)/(AD) = (AD)/(8)`
`=>` AD2 = 18 × 8 = 144
`=>` AD = 12 cm
iii. The ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.
`=> (Ar(ΔADB))/(Ar(ΔCDA)) = (AD^2)/(CD^2)`
= `12^2/8^2`
= `144/64`
= `9/4`
= 9 : 4
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