Question

There is a rectangle of length 22 cm. Find its area if its perimeter is 75 cm.​

Answer 1

Answer:

• given length is 22 now perimter = 2(l+b) =75
• it gives breadth as 31/2
• now area=length×breadth=341

Step-by-step explanation:

please mark me as brainlist

Answer 2

Given:

• There is a rectangle of length 22 cm
• its perimeter is 75 cm.

To Find:

• it's area

Solution:

◐  Here ,we have got the perimeter and the length of the rectangle so, with the help of the given clues let's find the breadth and then the area.

${ \purple{ \underline{ \mathfrak{Given \: clues \dag}}}}$

• Perimeter: 75cm
• length: 22cm

${\purple{ \underline{ \mathfrak{As \: we \: know \: that \dag}}}} \:$

${\longrightarrow}\blue{ \underline{ \boxed{ \pink{ \mathfrak{ \: perimeter = 2(length + breadth)}}} \bigstar}}$

❍ Now, let's substitute the values and find out the breadth

${ : \implies} \sf \: perimeter = 2(l + b) \\ \\ \\ { : \implies} \sf75 = 2(22 + b) \\ \\ \\ { : \implies} \sf75 = 44 + 2b \\ \\ \\ { : \implies} \sf2b = 75 - 44 \\ \\ \\ { : \implies} \sf \: 2b = 31 \\ \\ \\ { : \implies} \sf \: b = \frac{31}{2} \\ \\ \\ { : \implies} \sf \: \ b = 15.5 \bigstar$

${\therefore} \: \: { \rm {\underline{the \: breadth \: of \: the \: rectangle \: is \: 15.5cm}}}$

$\rule{300}{2}$

✦Now, let's find the area of the rectangle:

${\purple{ \underline{ \mathfrak{As \: we \: know \: that \dag}}}}$

${ \longrightarrow }\blue{ \underline{ \boxed{ \pink{ \mathfrak{ area = lenght \times breadth}}} \bigstar }}$

❍ let's substitute the values again now:

${ : \implies} \sf \: area \: = l \times b \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf \: area = 22 \times 15.5 \\ \\ \\ { : \implies} \sf \: area = 341 {cm}^{2} \bigstar$

❶hence , the area of the given rectangle is 341cm²

___________________________________________

More to know:

$\leadsto \sf \: area \: of \: a \: triangle = \frac{1}{2} \times bh \\ \\ \leadsto \sf \: area \: of \: a \: rhombus = \frac{ d_{2} \times d_{1} }{2 \: } \\ \\\leadsto \sf \: area \: of \: a \: square = {side}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \leadsto \sf \: area \: of \: a \: paralellogram = b \times h$