Question

from a square of length 5 cm the four triangular pieces from the four corners are cut down to form a regular octagon find the area cut from all the corners in total . ​

Answer 1

$\huge{\bold{\star{\fcolorbox{black}{red}{Given...}}}}$

• Area of the square = (5 × 5) sq. units = 25 sq. units.☃️✔

$\huge{\bold{\fcolorbox{black}{lime}{To\:Find...}}}$

• Area of the left (cut) portion...?

______________________________

✔Let, x units of each side be removed from each corner to form the regular octagon.

✔So, each side of the regular octagon shall be (5 - 2x) units long.

$\tt\large{Now, \:if \:we\: consider\: any \:right\: triangle \:formed\: at\: any\: corner:}$

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$\rm{(x^2) + (x^2) = {(5 - 2x)^2}}$

$\rm{2(x^2) = 25 - 20x + 4(x^2)}$

$\rm{2(x^2) - 20x + 25 = 0}$

$\rm{x = \div{20 ± sqrt{400 - 200}}{ / 4}}$

$\rm{x = 5 ± \sqrt{12.5}}$

__________________

$\rm\large{\bold{\underline{Now,\: x \:can't \:be\: larger\: than\: 5.}}}$

✔So, $\rm\large{x = (5 - \sqrt{12.5} = (5 - 3.5355) = 1.4645.}$☃️

_

$\tt\large\red{\underline{So,\: total\: area\: removed\: from \:four \:corners}}$

$\rm\large{= 4*(1 / 2)*(x^2) = 2*(x^2) ≈ 2*{(1.4645)^2} sq\: units\: ≈\: 4.2895 \:sq \:units}$.☃️

_

$\sf\large{\underline{\pink{If\: this\: is\: rounded\: off \:to \:nearest\: integer, \:the\: total \:area\: removed \:is\: 4 \:sq \:units.}}}$☃️✔

Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

Area of the square = (5 × 5) sq. units = 25 sq. units.

$\large\underline\bold{TO\:FIND,}$

• Area of the left (cut) portion.

$\large\underline\bold{SOLUTION,}$

$Let, \:x \:units\: of\: each\: side \:be\: removed\: from \:each \:corner\: to \:form \:the \:regular\: octagon.$

$So,\: each\: side \:of\: the \:regular\: octagon \:shall \:be\: (5 - 2x)\: units \: long.$

$\blue{\text{Now, \:if \:we\: consider\: any \:right\: triangle \:formed\: at\: any\: corner}}$

$\sf{(x^2) + (x^2) = {(5 - 2x)^2}}$

$\sf{2(x^2) = 25 - 20x + 4(x^2)}$

$\rm{2(x^2) - 20x + 25 = 0}$

$\sf x=20 \±\sqrt{400}-\frac{200}{4}$

$\sf{x = 5 \± \sqrt{12.5}}$

$\sf\mathfrak\red{now,}$

$\sf\implies x\:is \:not\:greater\:than\:length\:5cm$

$\sf\mathfrak\red{so,}$

$\sf\large{x =(5 -\sqrt{12.5} )= (5 - 3.5355) = 1.4645.}$

$\large{\boxed{\bf{x =(5 -\sqrt{12.5} = (5 - 3.5355) = 1.4645 }}}$

$\purple{\text{So,\: total\: area\: removed\: from \:four \:corners}}$

$\sf\large{= 4×\frac{1}{2}×(x^2) = 2×(x^2) =2×{(1.4645)^2} sq\: units\: =\: 4.2895 \:sq \:units}$

$\sf\therefore\red{If\: this\: is\: rounded\: off \:to \:nearest\: integer, \:the\: total \:area\: removed \:is\: 4 \:sq \:units.}$

$\large{\boxed{\bf{ answer\:4sq.units}}}$

_________________

✯ADDITIONAL INFORMATION,

$\setlength{\unitlength}{1.6mm}\begin{picture}(5,6)\put(0,35){\line(1,0){35}}\put(0,0){\line(1,0){35}}\put(0,0){\line(0,1){35}}\put(35,0){\line(0,1){35}}\put(17,-2){5cm}\put(18,-2){}\put(35,15){5cm}\put(18,0){ }\end{picture}$

✯DIAGRAM FOR RIGHT ANGLED TRIANGLE,

$\setlength{\unitlength}{1.6mm}\begin{picture}(30,20)\linethickness{0.1mm}\put(0,0){\line(0,1){37.5}}\put(0,0){\line(1,0){25}}\put(25,0){\line(-2,3){25}}\end{picture}\put(-32,18){x}\put(-18,-1.5){x}\put(-18.2,20){(5-2x^2)}$