X cm and CF = y CMAS we KNOW,BE = BD [tangents]DC = CF [tangents]AE = AF [tangents]AB + BE = AC + CF⇒ 10 + x = 8.6 + y⇒ x – y = 8.6 – 10⇒ x – y = - 1.4 ----(1)BD + DC = 6.4⇒ x + y = 6.4 ----(2)From equation (1) and equation (2)⇒ x = 2.5⇒ BE = 2.5 cm

"> X cm and CF = y CMAS we KNOW,BE = BD [tangents]DC = CF [tangents]AE = AF [tangents]AB + BE = AC + CF⇒ 10 + x = 8.6 + y⇒ x – y = 8.6 – 10⇒ x – y = - 1.4 ----(1)BD + DC = 6.4⇒ x + y = 6.4 ----(2)From equation (1) and equation (2)⇒ x = 2.5⇒ BE = 2.5 cm

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In a ΔABC. D is point on BC. AB is produce to E and AC is produce to F. A circle is passes through DEF, where AB = 10 cm, AC = 8.6 cm and BC = 6.4, then BE = ?

SRMJEEE Analytical Geometry in SRMJEEE 9 months ago

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Given, AB = 10 cm, AC = 8.6 cm and BC = 6.4 cmLet BE = X cm and CF = y CMAS we KNOW,BE = BD [tangents]DC = CF [tangents]AE = AF [tangents]AB + BE = AC + CF⇒ 10 + x = 8.6 + y⇒ x – y = 8.6 – 10⇒ x – y = - 1.4 ----(1)BD + DC = 6.4⇒ x + y = 6.4 ----(2)From equation (1) and equation (2)⇒ x = 2.5⇒ BE = 2.5 cm

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Posted on 26 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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