It's Geometry! 2

Calculus Level 4

If the lengths of three sides of a trapezium other than base are equal to 10 cm,then find the area of trapezium when it is maximum .

Details :-Let the area be represented as a 3 b \frac{a\sqrt{3}}{b} where a a and b b are coprime positive integers. Find a + b . a+b.


The answer is 76.

Mohit Gupta by Mohit Gupta , 20, India

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1 solution

Rishabh Jain
Feb 5, 2016

AE=FB= 100 − h 2 \sqrt{100-h^2} and EF=CD=10 Hence Area of trapezium ABCD = A ( h ) = 1 2 × ( h ) ( 10 + 10 + 2 100 − h 2 A(h)=\frac{1}{2} \times (h)(10+10+2\sqrt{100-h^2} = 10 h + h 100 − h 2 =10h+h \sqrt{100-h^2} A'(h=0) ⇒ 5 100 − h 2 = h 2 − 50 \Rightarrow 5\sqrt{100-h^2}=h^2-50 .
Squaring and solving we get h= 0 , 25 3 0,25\sqrt{3} We note since A'' ( 25 3 ) (25\sqrt3) <0 it is the point of required maximum.
Hence maximum area= 10 ( 5 ( 3 ) + 5 3 ( 100 − 75 ) = 75 3 10(5(\sqrt3) + 5\sqrt3(\sqrt{100-75})=75\sqrt3 Hence, a + b = 75 + 1 = 76 \Large a+b=75+1=\boxed{76}

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Thanks edited it...

Mohit Gupta - 5 years, 4 months ago

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Yup.... Nice problem..

Rishabh Jain - 5 years, 4 months ago

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