Question

What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try?)

Answer 1

Hint: We use the concept of convex and non-convex quadrilateral and break each quadrilateral into two halves. Use the property of the sum of interior angles of a triangle to find the sum of interior angles in each triangle. Add the angles and write the sum of angles of the quadrilateral.

Complete step-by-step answer:
Let us draw a convex quadrilateral ABCD having a diagonal AC.

In quadrilateral we have four angles $$\mathrm{\angle }A,\mathrm{\angle }B,\mathrm{\angle }C,\mathrm{\angle }D$$
So, the sum of interior angles of quadrilateral ABCD is given as $$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C+\mathrm{\angle }D$$.................… (1)
We have diagonal AC which divides quadrilateral ABCD into two triangles, $$△ABC$$ and $$△ADC$$
We apply the property of the sum of interior angles in each triangle.
In$$△ABC$$,
$$\mathrm{\angle }BAC+\mathrm{\angle }B+\mathrm{\angle }BCA={180}^{\circ }$$..............… (2)
In$$△ADC$$,
$$\mathrm{\angle }CAD+\mathrm{\angle }D+\mathrm{\angle }ACD={180}^{\circ }$$................… (3)
Add equations (3) and (4)
$$\Rightarrow \mathrm{\angle }BAC+\mathrm{\angle }B+\mathrm{\angle }BCA+\mathrm{\angle }CAD+\mathrm{\angle }D+\mathrm{\angle }ACD={180}^{\circ }+{180}^{\circ }$$
$$\Rightarrow \left(\mathrm{\angle }BAC+\mathrm{\angle }CAD\right)+\mathrm{\angle }B+\left(\mathrm{\angle }BCA+\mathrm{\angle }ACD\right)+\mathrm{\angle }D={360}^{\circ }$$
Since, we know $$\left(\mathrm{\angle }BAC+\mathrm{\angle }CAD\right)=\mathrm{\angle }A;\left(\mathrm{\angle }BCA+\mathrm{\angle }ACD\right)=\mathrm{\angle }B$$
$$\Rightarrow \mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C+\mathrm{\angle }D={360}^{\circ }$$
From equation (1), LHS is the sum of all interior angles of quadrilateral ABCD
$$\therefore$$Sum of interior angles of a convex quadrilateral is $${360}^{\circ }$$.
Now let us assume PQRS as a non-convex quadrilateral, having diagonal PR

In quadrilateral we have four angles $$\mathrm{\angle }P,\mathrm{\angle }Q,\mathrm{\angle }R,\mathrm{\angle }S$$
So, the sum of interior angles of quadrilateral ABCD is given as $$\mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R+\mathrm{\angle }S$$............… (4)
Again we divide the quadrilateral PQRS into two triangles, $$△PQR$$ and $$△PSR$$
We apply the property of the sum of interior angles in each triangle.
In $$△PQR$$,
$$\mathrm{\angle }PRQ+\mathrm{\angle }Q+\mathrm{\angle }RPQ={180}^{\circ }$$............… (5)
In$$△PSR$$,
$$\mathrm{\angle }PRS+\mathrm{\angle }S+\mathrm{\angle }RPS={180}^{\circ }$$...........… (6)
Add equations (5) and (6)
$$\Rightarrow \mathrm{\angle }PRQ+\mathrm{\angle }Q+\mathrm{\angle }RPQ+\mathrm{\angle }PRS+\mathrm{\angle }S+\mathrm{\angle }RPS={180}^{\circ }+{180}^{\circ }$$
$$\Rightarrow \left(\mathrm{\angle }PRQ+\mathrm{\angle }PRS\right)+\mathrm{\angle }Q+\left(\mathrm{\angle }RPQ+\mathrm{\angle }RPS\right)+\mathrm{\angle }S={360}^{\circ }$$
Since, we know $$\left(\mathrm{\angle }PRQ+\mathrm{\angle }PRS\right)=\mathrm{\angle }R;\left(\mathrm{\angle }RPQ+\mathrm{\angle }RPS\right)=\mathrm{\angle }P$$
$$\Rightarrow \mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R+\mathrm{\angle }S={360}^{\circ }$$
From equation (2), LHS is the sum of all interior angles of quadrilateral PQRS
$$\therefore$$Sum of interior angles of a non-convex quadrilateral is also $${360}^{\circ }$$.
So, the property holds true even if the quadrilateral is non-convex.

Note: Convex Quadrilateral: A quadrilateral which has each interior angle less than$${180}^{\circ }$$and both the diagonals lie inside the quadrilateral is called a convex quadrilateral.
Non-convex quadrilateral: A quadrilateral having one interior angle greater than $${180}^{\circ }$$ and diagonal lies outside the closed shape of the quadrilateral is called a non-convex of concave quadrilateral.
* In any triangle ABC, with angles A, B and C the sum of interior angles adds up to $${180}^{\circ }$$
$$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$$