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Ex.3.1 Q3 Understanding Quadrilaterals Solution - NCERT Maths Class 8

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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!)

What is known?

Quadrilateral \(ABCD\)

What is unknown?

Sum of the measures of the angles of a convex quadrilateral.


Let \(ABCD\) be a convex quadrilateral. Then, we draw a diagonal \(AC\) which divides the Quadrilateral into two triangles. We know that the sum of the angles of a triangle is \(180\) degree, so by calculating the sum of the angles of a \(∆ABC \) and , we can measure the sum of angles of convex quadrilateral.

 Video Solution
Understanding Quadrilaterals
Ex 3.1 | Question 3

Text Solution

\(ABCD\) is a convex quadrilateral made of two triangles \(∆ABC\) and \(∆ADC.\) We know that the sum of the angles of a triangle is \(180\) degree. So:

\[\begin{align}\!\angle 6\! +\! \angle 5 \!+ \!\angle 4\! =\! {180^ \circ }\left[ \begin{array}{l}{\rm{Sum}}\,\,{\rm{of}}\,{\rm{the}}\,{\rm{angles}}\,\\{\rm{of}}\,{\rm{\Delta ABC}}\,{\rm{ = 18}}{{\rm{0}}^{\rm{o}}}\end{array} \right]\\
\angle 1\!+ \!\angle2\!+\!\angle 3\!=\!{180^ \circ }\left[ \begin{array}{l}{\rm{Sum}}\,\,{\rm{of}}\,{\rm{the}}\,{\rm{angles}}\,\\{\rm{of}}\,{\rm{\Delta ADC}}\,{\rm{ = 18}}{{\rm{0}}^{\rm{o}}}\end{array} \right]

Adding we get

\[\begin{align}\angle 6 &+ \angle 5 + \angle 4 + \angle 1 + \angle 2 + \angle 3\\ &= {{180}^{\rm{o}}} + {{180}^{\rm{\circ}}}\\ & = {{360}^{\rm{\circ}}}\end{align}\]

Hence, the sum of measures of the triangles of a convex quadrilateral is \(360^\circ\). Yes, even if quadrilateral is not convex then, this property applies. Let \(ABCD\) be a non-convex quadrilateral; join \(BD\), which also divides the quadrilateral in two triangles

Using the angle sum property of triangle, again\(ABCD\) is a concave quadrilateral, made of two triangles \(\Delta \text{ABD}\) and \(\Delta \text{BCD}\) .Therefore, the sum of all the interior angles of this quadrilateral will also be,

\[180^\circ + 180^\circ=360^\circ\]

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