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We know that the sum of interior angles of any polygon (convex or concave) having n sides is $(n-2)\times {180}^{\circ }$.
$\therefore$ The sum of angles of a concave quadrilateral is $(4-2)\times {180}^{\circ }={360}^{\circ }$.

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Sticks can be treated as the diagonals of a quadrilateral.
Now, since the diagonals (sticks) are bisecting each other each other at right angles, therefore the shape formed by joining their end points can be a rhombus or a square.
Both the sticks are of the same length. This means that the diagonals of the quadrilateral formed are equal and also bisect each other perpendicularly.
$\therefore$ The shape formed is that of a square.

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We know that, a kite has two pairs of congruent adjacent sides and we can observe that figure R resembles a kite.

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True
A rhombus can be constructed uniquely, if both diagonals are given.

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Line segments.
A closed curve made up of only line segments is called a polygon.

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Given: BEST is a rhombus, so,
$TS\parallel BE$
and BS is a transversal.
$\therefore \angle SBE=\angle TSB={40}^{\circ }$ [alternate interior anges]
$Also,\angle y={90}^{\circ }$ [diagonals of a rhombus bisect at ${90}^{\circ }$]
In $\Delta TSO$,
$\angle STO+\angle TSO+\angle SOT={180}^{\circ }$ [Angle sum property of a triangle]
$x+{40}^{\circ }+{90}^{\circ }={180}^{\circ }\Rightarrow x={180}^{\circ }-{90}^{\circ }-{40}^{\circ }={50}^{\circ }$
$\therefore y-x={90}^{\circ }-{50}^{\circ }={40}^{\circ }$

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True.
A square a quadrilateral with allfour angles right angles. Therefore, we can say that all squares are rectangles but vice-versa is not true.

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Given, $\angle FIS={60}^{\circ }$
Now, $\angle FTS=\angle FIS={60}^{\circ }$ [$\because$ opposite angles of a parallelogram are equal]
Now, $FT\parallel IS$ and TI is a transversal, therefore $\angle FIO=\angle STO={35}^{\circ }$ [alternate angles]
Also, $\angle FOT+\angle SOT={180}^{\circ }$ [linear pair]
$\Rightarrow {110}^{\circ }+\angle SOT={180}^{\circ }$
$\Rightarrow \angle SOT={70}^{\circ }$
In $\Delta TOS,\angle TSO+\angle OTS+\angle TOS={180}^{\circ }$ [angle sum property of triangle]
$\therefore \angle OST={180}^{\circ }-({70}^{\circ }+{35}^{\circ })={75}^{\circ }$
In $\Delta FST,\angle SFT+\angle FTS+\angle TSF={180}^{\circ }$ [angle sum property of triangle]
$\Rightarrow \angle SFT={180}^{\circ }-({60}^{\circ }+{75}^{\circ })\therefore \angle SFT={45}^{\circ }$

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The number of non-overlapping triangles in an n-gon = n - 2,
i.e. 2 less than the number of sides.

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Let ABCD be a parallelogram,
where $\angle C={45}^{\circ }$

Since, ABCD is a parallelogram,
$\angle A=\angle C$ [opposite angles of parallelogram are equal]
Again, since PRQA is a parallelogram,
$\angle A=\angle R$ [opposite angles of parallelogram are equal]
$\Rightarrow \angle R={45}^{\circ }$ [$\therefore \angle A=\angle C={45}^{\circ }$]