Question

14. In the below figure, the line segment XY is parallel to side AC of ABC and it divides the triangle into two equal parts of equal areas. Find the ratio AX AB .

Answer 1

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HIGH RATED GABRU---HARSH ♦️♦️♦️♦️♦️♦️♦️♦️♦️♦️Step-by-step explanation:Sol: In ΔABC, XY || AC and area of ΔBXY = area of quadrilateral XYCA ⇒ ar (ΔABC) = 2.ar (ΔBXY) ----------------(1) XY || AC and BA is a transversal. ⇒ ∠BXY = ∠BAC --------------------------- (2) So, In ΔBAC and ΔBXY, ∠XBY = ∠ABC (common angle) ∠BXY = ∠BAC [from equation (2)] ⇒ ΔBAC ~ ΔBXY ⇒ ar(ΔBAC) / ar(ΔBXY) = BA2 / BX2 ⇒ BA = √2 BX ⇒ BA = √2 (BA – AX) ⇒ (√2 – 1) BA = √2 AX ⇒ AX/XB = (√2 – 1) / √2

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Answer 2

Yooooo!!!!!!

Ur answer is in the attachment above....

Hope it helps!!!

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