AD||BC, AB = 5 cm, AD = 8 cm, BC = 14 സെന്റീമീറ്റർ എന്നിങ്ങനെയുള്ള ഒരു സമപാർശ്വ ലംബകമാണ് ABCD. ലംബകത്തിന്റെ വിസ്തീർണ്ണം (cm2 ൽ) എത്രയാണ്?

This question was previously asked in
SSC CGL Previous Paper 13 (Held On: 11 August 2017 Shift 1)
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  1. 36
  2. 44
  3. 88
  4. 144

Answer (Detailed Solution Below)

Option 2 : 44
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മുകളിലുള്ള ചിത്രത്തിൽ നിന്ന്, AOB ത്രികോണത്തിൽ പൈഥഗോറസ് സിദ്ധാന്തം ഉപയോഗിക്കുമ്പോൾ 

⇒ AB2 = BO2 + OA2

⇒ 52 = 32 + OA2

⇒ OA = 4cm ഇത് ലംബകത്തിന്റെ ഉയരത്തിന് തുല്യമാണ്

⇒ വിസ്തീർണ്ണം = 1/2 × (ഉയരം) × (സമാന്തര വശങ്ങളുടെ ആകെത്തുക)

⇒ വിസ്തീർണ്ണം = 1/2 × 4 × (8 + 14) = 44

∴ വിസ്തീർണ്ണം 44cm2 ആണ്.

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