This is Not a Square

Geometry Level 3

A quadrilateral having sides of length 3, 4, 5 and 6 units is inscribed in a circle.

The area of this quadrilateral can be expressed as 2 A 2 B 2A \sqrt{2B} , where A A and B B are coprime positive integers. Find A + B A+B .


The answer is 8.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Relevant wiki: Brahmagupta's Formula

Area of a cyclic quadrilateral with lengths of four sides given is

A r e a = ( s a ) ( s b ) ( s c ) ( s d ) Area=\sqrt{(s-a)(s-b)(s-c)(s-d)}

s = a + b + c + d 2 = 3 + 4 + 5 + 6 2 = 9 s=\frac{a+b+c+d}{2}=\frac{3+4+5+6}{2}=9

s a = 9 3 = 6 s-a=9-3=6

s b = 9 4 = 5 s-b=9-4=5

s c = 9 5 = 4 s-c=9-5=4

s d = 9 6 = 3 s-d=9-6=3

A r e a = ( 6 ) ( 5 ) ( 4 ) ( 3 ) = 2 ( 3 ) 2 ( 5 ) Area=\sqrt{(6)(5)(4)(3)}=2(3)\sqrt{2(5)}

A + B = 3 + 5 = 8 A+B=3+5=8

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...