Question

A thin converging lens forms a (i) real magnified image (ii) virtual magnified image:- (a) write the position of object in each case. (b) draw labelled diagram for each case.
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Answer 1

Answer:

$\huge{\underline{\underline{Given→}}}$

$\sf\green{Length\:of\:rectangel=x+2\:units}$

$\sf\green{Breadth\:of\:rectangel=2x\:units}$

$\huge{\underline{\underline{To\:Find→}}}$

$\sf\purple{Perimeter\:of\:the\:rectangle}$

$\sf\color{purple}{Area\:of\:the\:rectangle}$

$\huge{\underline{\underline{Answer→}}}$

Perimeter:-

As perimeter =2(Lenght+Breadth)

$\sf\blue{→P=2(l+b)}$

$\sf\blue{→P=2(2x+x+2)}$

$\sf\blue{→P=4x+2x+4}$

$\sf\blue{→P=6x+4}$

$\mathrm{\boxed{\boxed{\pink{→P=6x+4✔}}}}$

Area:-

As area =Length x Breadth

$\sf\color{blue}{→Ar=l\times b}$

$\sf\color{blue}{→Ar=2x\times (x+2)}$

$\sf\color{blue}{→Ar=2x(x+2)}$

$\sf\color{blue}{→Ar=2x^2+4x}$

${\boxed{\boxed{\pink{→Ar=2x^2+4x✔}}}}$

☞ Hence the Perimeter of the rectangle is 6x+4 units and it's area is $\mathrm{2x^2+4x}$ units which is the required answer.

HOPE IT HELPS.