Question

Show that the sum of all sides of a quadrilateral is greater than the sum of the diagonals.

Answer 1

Hint:Here, we have to show that the sum of all sides of a quadrilateral is greater than the sum of the diagonals. We will consider a quadrilateral \[ABCD\] and we will use the property that the sum of two sides of triangle is greater than third side in \[\vartriangle ADB\], \[\vartriangle BCD\], \[\vartriangle ACB\] and \[\vartriangle ACD\].Then, we will add all the equations and we will simplify it.

Complete step by step answer:
Consider a quadrilateral \[ABCD\].

\[AB\], \[BC\], \[CD\] and \[DA\] are the sides of a quadrilateral and \[AC\] and \[BD\] are the diagonals of the quadrilateral.As we know, the sum of two sides of a triangle is greater than the third side.In \[\vartriangle ADB\], by triangle inequality, we can write
\[ \Rightarrow AB + DA > BD - - - (1)\]
In \[\vartriangle BCD\], by triangle inequality, we can write
\[ \Rightarrow BC + CD > BD - - - (2)\]

In \[\vartriangle ACB\], by triangle inequality, we can write
\[ \Rightarrow AB + CB > AC - - - (3)\]
In \[\vartriangle ACD\], by triangle inequality, we can write
\[ \Rightarrow DA + CD > AC - - - (4)\]
Adding equation \[(1)\], \[(2)\], \[(3)\] and \[(4)\], we get
\[ \Rightarrow AB + DA + BC + CD + AB + CB + DA + CD > BD + BD + AC + AC\]
\[ \Rightarrow 2\left( {AB + BC + CD + DA} \right) > 2\left( {BD + AC} \right)\]
As we know that inequality remains the same on dividing both the sides by a positive constant. Dividing both the sides by \[2\], we get
\[ \therefore AB + BC + CD + DA > BD + AC\]

Therefore, the sum of all sides of a quadrilateral is greater than the sum of the diagonals.

Note:Here, the question is of a quadrilateral. The sum of the measure of the angles of a quadrilateral is always \[{360^ \circ }\]. Similarly, the sum of the measure of the angles of a triangle having three sides is always \[{180^ \circ }\], the sum of the measure of the angles of a pentagon having five sides is always \[{540^ \circ }\] and the sum of the measure of the angles of a hexagon having six sides is always \[{720^ \circ }\].