AD = 18 and BE = 15MD = 18 × 1/3 = 6BM = 15 × 2/3 = 10In right angled triangle BDM,USING PYTHAGORAS theorem,⇒ BM2 = BD2 + MD2⇒ BD2 = 100 – 36⇒ BD = 8⇒ BC = 2BD = 2 × 8 = 16Area of triangle ABC = 1/2 × height × base= 1/2 × AD × BC = 1/2 × 18 × 16 = 144 sq.units

"> AD = 18 and BE = 15MD = 18 × 1/3 = 6BM = 15 × 2/3 = 10In right angled triangle BDM,USING PYTHAGORAS theorem,⇒ BM2 = BD2 + MD2⇒ BD2 = 100 – 36⇒ BD = 8⇒ BC = 2BD = 2 × 8 = 16Area of triangle ABC = 1/2 × height × base= 1/2 × AD × BC = 1/2 × 18 × 16 = 144 sq.units

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AD and BE are medians and M is the centroid of ΔABC. AD = 18 and BE = 15 and AD is perpendicular bisector of side BC. Find the area of ΔABC.

SRMJEEE Analytical Geometry in SRMJEEE 11 months ago

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Given,AD = 18 and BE = 15MD = 18 × 1/3 = 6BM = 15 × 2/3 = 10In right angled triangle BDM,USING PYTHAGORAS theorem,⇒ BM2 = BD2 + MD2⇒ BD2 = 100 – 36⇒ BD = 8⇒ BC = 2BD = 2 × 8 = 16Area of triangle ABC = 1/2 × height × base= 1/2 × AD × BC = 1/2 × 18 × 16 = 144 sq.units

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Posted on 27 Oct 2024, this text provides information on SRMJEEE related to Analytical Geometry in SRMJEEE. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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