Question

Q.1. The difference between the length and breadth of a rectangle is
33 m. If its perimeter is 134 m, then its area is:​

Answer 1

Given,

Perimeter of the rectangle = 2[l + b] = 134m

==> 2[l + b] = 134

==> l + b = 134/2 = 67 -------(1)

The difference between the length and breadth of a rectangle = 33m

==> l - b = 33 ------------(2)

Solve eq - (1) & (2)

==> 2L = 100

==> l = 100/2

==> l = 50

Length = 50m

Substitute l = 50 in eq - (1)

==> (50) + b = 67

==> b = 67 - 50

==> b = 17

Breadth = 17m

We know that,

Area of Rectangle = length × breadth

==> = 50 × 17

==> = 850m²

Therefore, the area of Rectangle is 850cm².

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline{ \mathfrak{ \: \: Given : \: \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \text{Difference between the length and breadth } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text{of the rectangle = 33 m} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \text{Perimeter of the rectangle = 134 m} \\ \\ \underline{ \mathfrak{ \: \: To \: \: find : \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \text{ Area of the rectangle = ? }$

$\underline{ \mathfrak{ \: \: Suppose, \: \: }} \\ \\ \: \: \: \: \: \: \: \: \: \bold{Length \: \: of \: \: the \: \: rectangle = l } \\ \\ \: \: \: \: \: \: \: \: \: \bold{Breadth \: \: of \: \: the \: \: rectangle = b}$

$\underline{ \mathfrak{ \: \: We \: \: know \: \: that, \: \: }} \\ \\ \boxed{ \bold{ \pink{ \: Perimeter \: \: of \: \: a \: \: rectangle = 2 (Length + Breadth) \: }}}$

$\underline{ \mathfrak{ \: According \: \: to \: \: the \: \: first \: \: condition, \: }} \\ \\ \: \: \: \: \: \: \: \: \: \sf{l - b = 33 \: m \: \: - - - - - - - - > (1) } \\ \\ \underline{ \mathfrak{ \: According \: \: to \: \: the \: \: second \: \: condition, \: }} \\ \\ \: \: \: \: \: \: \: \: \: \sf{2 ( l + b) = 134 \: m} \\ \\ \sf{ \implies l + b =( \frac{134}{2} )\: m} \\ \\ \: \: \: \: \: \: \sf{ \therefore \: l + b = 67 \: m\: \: - - - - - - - - > (2)}$

$\tt{(1) + (2) \implies 2 \: l = 100} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{ \implies l =\frac{100}{2}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{ \therefore \: \: l=50 } \\ \\ \tt{(2)-(1)\implies 2 \: b = 34 } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{\implies b =\frac{34}{2} }\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{\therefore \: \: b =17} \\ \\ \\ \: \: \: \: \: \: \: \: \tt{ \therefore \: Length \: \: of \: \: the \: \: rectangle,l = 50 \: m} \\ \bold{And, } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{Breadth \: \: of \: \: the \: \: rectangle,b = 17 \: m }$

$\underline{ \mathfrak{ \: \: We \: \: know \: \: that, \: \: }} \\ \\ \boxed{ \bold{ \pink{ \: \: Area \: \: of \: \: a \: \: rectangle = Length \times Breadth \: \: }}}$

$\rm{\therefore \: Area \: \: of \: \: the \: \: rectangle = l \times b } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{= (50 \times 17) \: \: m{}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{= 850 \: \: m {}^{2} }$

$\bold{Hence, } \\ \: \: \: \: \: \: \: \: \bold{Area \: \: of \: \: the \: \: rectangle = \red{850 \: m {}^{2} }}$