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Let ABC be a triangle; a parallelogram ADEF is completed where D,E,F are on sides AB,BC,CA respectively. Prove that maximum area of parallelogram ADEF will be attained if D,E,F be the mid points of respective sides.(use calculus)

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4 months ago

```							Parallelogram :A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram.  In a parallelogram diagonal divides it into two triangles of equal areas.  Mid point theorem:The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.  SOLUTION : Given:ABC is a Triangle in which the midpoints of sides BC ,CA and AB are  D, E and F. To show:(i) BDEF is a parallelogram. (ii) ar (DEF) = 1/4ar (ABC) (iii) ar (BDEF) =1/2 ar (ABC) Proof: i)Since E and F are the midpoints of AC and AB. BC||FE & FE= ½ BC= BD (By mid point theorem) BD || FE & BD= FE  Similarly, BF||DE & BF= DE  Hence, BDEF is a parallelogram .[A pair of opposite sides are equal and parallel]  (ii) Similarly, we can prove that FDCE & AFDE are also parallelograms. Now, BDEF is a parallelogram so its diagonal FD divides its into two Triangles of equal areas. ∴ ar(ΔBDF) = ar(ΔDEF) — (i)  In Parallelogram AFDE ar(ΔAFE) = ar(ΔDEF) (EF is a diagonal) — (ii)  In Parallelogram FDCE ar(ΔCDE) = ar(ΔDEF) (DE is a diagonal) — (iii)  From (i), (ii) and (iii) ar(ΔBDF) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF).....(iv) ar(ΔBDF) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = ar(ΔABC) 4 ar(ΔDEF) = ar(ΔABC)(From eq iv) ar(∆DEF) = 1/4 ar(∆ABC)........(v)  (iii) Area (parallelogram BDEF) = ar(ΔDEF) + ar(ΔBDF)ar(parallelogram BDEF) = ar(ΔDEF) + ar(ΔDEF) ar(parallelogram BDEF) = 2× ar(ΔDEF)(From eq iv)  ar(parallelogram BDEF) = 2× 1/4  ar(ΔABC)(From eq v) ar(parallelogram BDEF) = 1/2 ar(ΔABC)
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4 months ago
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