Gujarat BoardEnglish MediumSTD 8MATHSUnderstanding Quadrilaterals and Practical Geometry3 Marks
Question
$ABCDE$ is a regular pentagon. The bisector of angle A meets the side $CD$ at $M$. Find $\angle\text{AMC}.$
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Answer
Given, a pentagon $ABCDE.$ The line segment $AM$ is the bisector of the $\angle\text{A}$.
Now, since the measure of each interior angle of a regular pentagon is $108^\circ $
$\therefore\angle\text{BAM}=\frac{1}{2}\times108^\circ=54^\circ$ By the angle sum property of a quadrilateral, we have (in quadrilateral ABCM) $\angle\text{BAM}+\angle\text{ABC}+\angle\text{BCM}+\angle\text{AMC}=360^\circ$
$\Rightarrow54^\circ+108^\circ+108^\circ+\angle\text{AMC}=360^\circ$
$\Rightarrow\angle\text{AMC}=360^\circ-270^\circ$
.$\Rightarrow\angle\text{AMC}=90^\circ.$
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