Question

$\to very\:important$-what is a square.theorms for squares.draw a square of perimeter 36 and find its area

Answer 1

$\mathfrak{dear\:user}$

$\textsf{question-what \:is \:a\: square\:and write\:about\:it}$

$\mathfrak{here\:is\:the\:solution}$

$\mathbb{ANSWER}$

$SQUARE$

$\textsc{a \:square\:is\:a\:quadrilateral\:with\:4\:sides}$

$\textsc{it is a rectangle in which two adjacent sides have equal length}$

$\textsc{Number of vertices: 4}$

$\textsc{Number of edges}-4$

$\textsc{internal angle- 90}$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D}\put(-0.5,4.2){\bf A}\put(4.2,-0.5){\bf C}\put(4.2,4.2){\bf B}\put(1.5,-0.6){\bf\large }\put(4.4,2){\bf\large \ }\end{picture}$

$\textsf{prove that diagonals of square are equal}$

Given : $\textsf{ square ABCD}$

$\textsf{diagonals AC and BD}$

To Prove :$AC=BD$

Proof:$\textsf{In triangles BAD and ABC}$

$\textsf{AB = AB (COMMON)}$

$\textsf{ AD = BC (ALL SIDES IN A SQUARE ARE EQUAL)}$

$\angle BAD=\angle ABC \textsf{(VERTEX ANGLES, SO EQUAL)}$

$\textsf{BY SAS, }$$\triangle BAD\cong\triangle ABC$

$\textsf{ BY CPCT, AC = BD}$

$\textsf{ HENCE PROVED}$

$\mathbb{TO\:FIND\:AREA}$

${Area=side\times side}$

$\text{A-area,S-side}$

$A=S\times S$

$A=S^2$

$\boxed{A=S^2}\to1$

$\mathbb{TO\:FIND\:PERIMETER}$

$\textit{Perimeter=side+side+side+side}$

$\text{P-perimeter,S-sides}$

$\text{P=side+side+side+side}$

$\text{P=4S}$

$\boxed{\text{P=4S}}\to2$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D}\put(-0.5,4.2){\bf A}\put(4.2,-0.5){\bf C}\put(4.2,4.2){\bf B}\put(1.5,-0.6){\bf\large side}\put(4.4,2){\bf\large side}\end{picture}$

$\textsf{given perimeter=36}$

$\textsf{from \boxed{\to2} perimeter =4s}$

$36=4S\\4S=36\\S=36\div4\\S=9$

$\boxed{S=9}\to3$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D}\put(-0.5,4.2){\bf A}\put(4.2,-0.5){\bf C}\put(4.2,4.2){\bf B}\put(1.5,-0.6){\bf\large 9\ cm}\put(4.4,2){\bf\large 9\ cm}\end{picture}$

$\textsf{from\boxed{\to\:1} area=s x s}$

$\textsf{from\boxed{\to 3}substitute the value of s=9 }$

$A=S^2$

$A=9^2$

$A=9\times9$

$A=81$

$\boxed{A=81}$

$\boxed{\therefore\textsf{area}=81 cm^{2}}$

$\mathbf{EXTRA \:INFORMATION}$

$\textsf{1) proporties of squares}$

$\to\textsc{The diagonals of a square bisect each other and meet at 90 }$

$\to\textsc{The diagonals of a square bisect its angles.}$

$\to\textsc{Opposite sides of a square are both parallel and equal in length.}$

$\to\textsc{All four sides of a square are equal.}$

$\to\textsc{The diagonals of a square are equal.}$

$\textsf{2) angles of a square add up to?}$

Proof: $\textsf{In the square ABCD,}$

$\angle ABC, \angle BCD, \angle CDA, and \angle DAB\: \textsf{are the internal angles.}$

$\textsf{AC is a diagonal}$

$\textsf{AC divides the quadrilateral into two triangles, }\triangle ABC\: and\: \triangle ADC$

$\textsf{We have learned that the sum of internal angles of a quadrilateral is }$$\textsf{360, that is,} ABC + \angle BCD + \angle CDA +\angle DAB = 360.$

$\textsf{let's prove that the sum of all the four angles of a quadrilateral is 360 degrees.}$

$\textsf{We know that the sum of angles in a triangle is 180.}$

$\textsf{Now consider} \:\triangle ADC,$

$\angle D + \angle DAC + \angle DCA = 180$                     $\text{ (Sum of angles in a triangle)}$

$\textsf{Now consider triangle ABC,}$

$\angle B + \angle BAC + \angle BCA = 180$

$\textsf{On adding both the equations obtained above we have,}$

(∠D + ∠DAC + ∠DCA) + (∠B + ∠BAC + ∠BCA) = 180° + 180°

∠D + (∠DAC + ∠BAC) + (∠BCA + ∠DCA) + ∠B = 360°

$\textsf{We see that} (\angle DAC +\angle BAC) = \angle DAB\: and\: (\angle BCA + \angle DCA) = \angle BCD.$

$\textsf{Replacing them we have,}$

∠D + ∠DAB + ∠BCD + ∠B = 360°

$\text{That is,}$

$\angle D + \angle A + \angle C +\angle B=360$

$\textsf{Or, the sum of angles of a quadrilateral is 360. This is the angle sum property of quadrilaterals.}$

$\to PLEASE\:DONT\:COPY$

$\mathcal{HOPE\:IT\:HELPS}$

$\mathcal{BY\:BRAINLY \:ROSHAN}$

Answer 2

EXTRAINFORMATION

\textbf{How to draw a rectangle?}How to draw a rectangle?

\textsf{For example take the values of B=9cm , L=10cm}For example take the values of B=9cm , L=10cm

\bullet \textsf{draw a horizonal line AB of 10cm(length)}∙draw a horizonal line AB of 10cm(length)

\bullet\textsf{draw a right angle from the point A and draw a line AD of 9cm (breadth)}∙draw a right angle from the point A and draw a line AD of 9cm (breadth)

\bullet\textsf{From D draw a right angle and draw a line DC of 10cm(length)}∙From D draw a right angle and draw a line DC of 10cm(length)

\bullet\textsf{From C draw a right angle and draw a line CB of 9cm(breadth)}∙From C draw a right angle and draw a line CB of 9cm(breadth)

\bullet \textsf{refer to images for understanding more}∙refer to images for understanding more

\textbf{Properties of rectangle}Properties of rectangle

\textsf{The opposite sides are parallel and equal to each other}The opposite sides are parallel and equal to each other

\textsf{Each interior angle is equal to 90 degrees}Each interior angle is equal to 90 degrees

\textsf{The sum of all the interior angles is equal to 360 degrees}The sum of all the interior angles is equal to 360 degrees

\textsf{The diagonals bisect each other}The diagonals bisect each other

\textsf{Both the diagonals have the same length}Both the diagonals have the same length

\textsf{A diagonal of a rectangle is a diameter of its circumcircle}A diagonal of a rectangle is a diameter of itsEXTRAINFORMATION

\textbf{How to draw a rectangle?}How to draw a rectangle?

\textsf{For example take the values of B=9cm , L=10cm}For example take the values of B=9cm , L=10cm

\bullet \textsf{draw a horizonal line AB of 10cm(length)}∙draw a horizonal line AB of 10cm(length)

\bullet\textsf{draw a right angle from the point A and draw a line AD of 9cm (breadth)}∙draw a right angle from the point A and draw a line AD of 9cm (breadth)

\bullet\textsf{From D draw a right angle and draw a line DC of 10cm(length)}∙From D draw a right angle and draw a line DC of 10cm(length)

\bullet\textsf{From C draw a right angle and draw a line CB of 9cm(breadth)}∙From C draw a right angle and draw a line CB of 9cm(breadth)

\bullet \textsf{refer to images for understanding more}∙refer to images for understanding more