Question

draw any 4 figures for eg square,rectangle,triangle,etc...

Answer 1

$\mathfrak{dear\;user }$

$\mathfrak{question-}$$\textsf{draw any 4 figures for eg square,rectangle,triangle,etc...}$

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

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

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\put(0,0){\line(1,0){2.3}}\put(0.5,0.3){\bf\large x\ cm}\end{picture}$

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\qbezier(1.6,4)(1.6,4)(6,1)\put(0.7,0.5){\sf A}\put(6,0.5){\sf B}\put(1.4,4.3){\sf D}\put(6.6,4.3){\sf C}\end{picture}$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large x cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large y cm}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

$\mathcal{MY\:EXPECTATION\: FOR \:THIS \: ANSWER \:IS }10\:thanks\:and \:brainlist$

$\mathcal{BY \:ROSHAN\: A \:USER \: OF \: BRAINLY}$

                                                                                    ʙʀᴀɪɴʟʏ  ʀᴏsʜᴀɴ

Answer 2

S E N D

+M O R E

M O N E Y

★SOLUTION

• from hint one above we see that M=1;

replace M By 1

S+M sold be at least 10 so as is my mother 9 if there is no carry over from the previous column aur 8 if there is a carry over.

S+Mm can only be 10 or 11 that means O can only be 0 or 1 but O cannot be 1 as M . Hence S=9.

In the hundredth kolam E+O=N,and O=0; but E cannot be the same as N; therefore there must be carried over from the previous column and N must be 1 greater than E:E+1=N.

from the N+R Kollam we can derive the equation: N+R+(+1) =E+10. we have to interest the expression (+1) Vikas we did not know yet another one is carried from over column D+E+10 p is inserted because we do know that one has to be carried over from the column N+R TO E+O.

subtract the first equation from the second: R+(+1)=9.

R cannot be equal to 9 since S=9, therefore R=8; p hence we know that one has to be carried over kolam D+E.

Column D+E mast total at least 12 since Y I cannot be 1 or 0 what values can we give D and E to reach his total? we have already used 9.8 else where the only digits left that are high enoug are 7,6and 7,5. but one of this has to be En,and N is 1 greater than E. hence E=5 and N=6 while D=7.

Then Y=2and the puzzle is solved The solution is:9767+1085=10652.