Question

Saiya medicating card of length 10 cm and with in 8 cm find the perimeter of the greeting card
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Answer 1

Answer:

The perimeter of the greeting card is 22 cm

Step-by-step explanation:

Length(l) = 7 cm.

Breadth(b) = 4 cm.

∴ Perimeter = 2(l+b).

= 2 (7+4).

= 2 × 11.

= 22 cm.

Answer 2

Question:-

• Saiya medicating card of length 10 cm and breadth 8 cm . find the perimeter of the greeting card .

Given:-

• From the Question we can see that it is a rectangle as length and breadth is given
• Length = 10cm
• Breadth = 8cm

To Find:-

• Perimeter of rectangle greeting card

Solution:-

$\begin{gathered}\begin{gathered}: \implies \underline{ \boxed{\displaystyle \sf \bold{\: Perimeter\: of \: Rectangle \: = \: 2 \: ( \: L \: + \: B\: ) }} }\\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}: \implies \displaystyle \sf \: 2 \: ( \: 10cm \: + \: 8cm\: ) \\ \\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}: \implies \displaystyle \sf \: 2 \: × 18cm \\ \\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}: \implies \underline{ \boxed{\displaystyle \sf \bold{\: 36cm }} }\\ \\\end{gathered}\end{gathered}$

So, the perimeter of greeting card = 36cm

Note:-

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Diagram:-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 10 cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 8 cm}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$