How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Understand the problem

Show that among all quadrilaterals of a given perimeter the square has the largest area.

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.23.3" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]Start with a quadrilateral with sides $a,b,c$ and $d$. Divide it into two triangles and write its area as the sum of the areas of the triangles.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]Show that the area $A$ satisfies $A\le\frac{ab+cd}{2}$ and $A\le\frac{ad+bc}{2}$.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]Using hint 2, derive that $A\le \frac{(a+c)(b+d)}{2}$.[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]From the AM-GM inequality, we can write that $(a+c)(b+d)\le\frac{(a+b+c+d)^2}{4}$. Hence $\text{Area}\le\frac{(\text{Perimetre)^2}{8}$. Equality is achieved iff all the angles are right angles (this follows from hint 1) and $a+c=b+d$. If all the angles are right angles then the quadrilateral is a rectangle and hence $a=c$ and $b=d$. Finally, $a=b=c=d$. Thus the area is maximised for a square. [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

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