Question
Download Solution PDFAD||BC, AB = 5 cm, AD = 8 cm, BC = 14 സെന്റീമീറ്റർ എന്നിങ്ങനെയുള്ള ഒരു സമപാർശ്വ ലംബകമാണ് ABCD. ലംബകത്തിന്റെ വിസ്തീർണ്ണം (cm2 ൽ) എത്രയാണ്?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFമുകളിലുള്ള 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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