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Long Answer Type

351.

Find the largest possible area of a right-angled triangle whose hypotenuse is 5 cm long.

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352. A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the maximum length of the hypotenuse is

open parentheses straight a to the power of 2 over 3 end exponent plus straight b to the power of 2 over 3 end exponent close parentheses to the power of 3 over 2 end exponent.

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353. Find the maximum-area of an isosceles triangle inscribed in the ellipse $straight x squared over straight a squared plus straight y squared over straight b squared space equals space 1$ with its vertex at one end of the major axis.

Let APQ be the isosceles triangle inscribed in the ellipse with centre at C. A is (a, 0).
Let P and Q be (a cos ө , b sin ө) and (a cos ө, – b sin ө) respectively.
Let ∆ be area of ∆ APQ

$therefore space space space space increment space equals space 1 half left parenthesis PQ right parenthesis space left parenthesis AM right parenthesis space equals space 1 half left parenthesis 2 straight b space sin space straight theta right parenthesis thin space left parenthesis straight a minus acosθ right parenthesis space equals space ab space sin space straight theta space left parenthesis 1 minus cos space straight theta right parenthesis space space space space space space space space space space space space space space space space space space space equals ab space left parenthesis sin space straight theta space minus space sin space straight theta space cos space straight theta right parenthesis space equals space ab open parentheses sin space straight theta space minus space 1 half sin space 2 straight theta close parentheses therefore space space space space space space fraction numerator straight d increment over denominator dθ end fraction space equals space ab space left parenthesis cos space straight theta space minus space cos space 2 straight theta right parenthesis space space space space space space space space space space space fraction numerator straight d increment over denominator dθ end fraction space equals space 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space ab space left parenthesis cos space straight theta space minus space cos space 2 straight theta right parenthesis space equals space 0 rightwards double arrow space space space space space cos space straight theta minus space cos space 2 straight theta space equals space 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space cos space 2 straight theta space equals space cos space straight theta rightwards double arrow space space space space space space space 2 straight theta space equals space 2 straight pi minus straight theta space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space 3 straight theta space equals space 2 straight pi space space space space space rightwards double arrow space space space straight theta space equals space fraction numerator 2 straight pi over denominator 3 end fraction space space space space$
$fraction numerator straight d squared increment over denominator dθ squared end fraction space equals space ab space left parenthesis negative sinθ space plus space 2 space sin 2 straight theta right parenthesis$
$When space space space straight theta equals space fraction numerator 2 straight pi over denominator 3 end fraction comma space space fraction numerator straight d squared increment over denominator dθ squared end fraction space equals space ab open parentheses negative sin space fraction numerator 2 straight pi over denominator 3 end fraction plus space 2 space sin fraction numerator 4 straight pi over denominator 3 end fraction close parentheses space equals space ab space open parentheses negative sin space straight pi over 2 space minus space 2 space sin space straight pi over 3 close parentheses space space space space space space space space space space space space space space space space space equals space ab open parentheses negative space 3 space sin space straight pi over 3 close parentheses space equals space ab space open parentheses negative 3 space cross times space 3 over 2 close parentheses space equals space minus fraction numerator 3 square root of 3 over denominator 2 end fraction ab less than 0 therefore space space space space space space space space increment space is space maximum space when space straight theta space equals space fraction numerator 2 straight pi over denominator 3 end fraction therefore space space space space space space maximum space area space equals space ab space open parentheses sin space fraction numerator 2 straight pi over denominator 3 end fraction space minus space 1 half sin space fraction numerator 4 straight pi over denominator 3 end fraction close parentheses space equals space ab space open parentheses sin space straight pi over 3 plus 1 half sin space straight pi over 3 close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals ab open parentheses 3 over 2 sin space straight pi over 3 close parentheses space equals space ab space open parentheses 3 over 2 cross times fraction numerator square root of 3 over denominator 2 end fraction close parentheses equals space fraction numerator 3 square root of 3 over denominator 4 end fraction space ab space space sq. space units.$

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Short Answer Type

354.

Let AP and BQ be two vertical poles at points A and B, respectively. If AP = 16 m. BQ = 22 m and AB = 20 m, then find the distance of a point R on AB from the point A such that RP² + RQ² is minimum.

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