Question

Sanya, Aarushi, Manav and Kabir made greeting cards. Complete the table for their cards:

Whose card	Length	Width	Perimeter	Area
Sanya	10 cm	8cm
Manav	11 cm		44 cm
Aarushi		8 cm		80 square cm
Kabir			40 cm	100 square cm

Answer 1

Hint: In this question, we are given cards of four children and we have to complete its missing measurements. For Sanya, we are given the length and breadth of the card and we have to find the perimeter and area. So, we will use formula as $\text{Perimeter}=2\left( l+b \right)$ and $\text{Area}=l\times b$ where l is length and b is breadth. For Manav, we are given length and perimeter. Using $\text{Perimeter}=2\left( l+b \right)$ we will find the breadth of the card and then use l and b to find the area. For Aarushi, we are given breadth and area. Using $\text{Area}=l\times b$ we will find the length and then find the perimeter. For Kabir, we are given perimeter and area only. So, we will form two equations for perimeter and area and then find length and breadth. After finding all values we will fill the card.

Complete step-by-step solution:
Here, four children have made cards and we have to complete their measurements. Since cards are rectangular in shape, we will use the formula of rectangle only.
Perimeter of the rectangle is given by $\text{P}=2\left( l+b \right)$ where, P is perimeter, l is length and b is breadth/width. Area of the rectangle is given by $\text{A}=l\times b$ where, A is area, l is length and b is breadth.
Sanya's card.
Length of the card is 10cm. Therefore, l = 10cm
The Width of the card is 8cm. Therefore, b = 8cm.
Perimeter of the card will be given by,
\[\begin{align}
  & \text{P}=2\left( l+b \right) \\
 & \Rightarrow \text{P}=2\left( 10+8 \right) \\
 & \Rightarrow P=2\times 18 \\
 & \Rightarrow P=36cm \\
\end{align}\]
Hence, the perimeter of Sanya's card is 36cm.
Area of the card will be given by,
\[\begin{align}
  & \text{A}=l\times b \\
 & \Rightarrow A=10\times 8 \\
 & \Rightarrow A=80c{{m}^{2}} \\
\end{align}\]
Hence, the area of Sanya's card is $80c{{m}^{2}}\Rightarrow 80$ square cm.
Manav's card.
Length of the card is 11cm. Therefore, l = 11cm.
Perimeter of the card is 44cm. Therefore, P = 44cm.
Since, perimeter is given by $\text{P}=2\left( l+b \right)$.
Therefore, $44=2\left( 11+b \right)$.
Dividing by 2 both sides, we get:
\[\begin{align}
  & 11+b=22 \\
 & \Rightarrow b=11cm \\
\end{align}\]
Hence, the breadth of Manav's card is 11cm.
Area of card is given by,
\[\begin{align}
  & \text{A}=l\times b \\
 & \Rightarrow A=11\times 11 \\
 & \Rightarrow A=121c{{m}^{2}} \\
\end{align}\]
Hence, the area of Manav's card is $121c{{m}^{2}}\Rightarrow 121$ square cm.
Aarushi's card.
Breadth of card is given as 8cm. Therefore, b = 8cm.
Area of the card is given as 80 square cm. Therefore, A = $80c{{m}^{2}}$.
Since, area is given by $\text{A}=l\times b$.
Therefore, $\text{80}=l\times 8$.
Dividing both sides by 8, we get:
\[\begin{align}
  & l=\dfrac{80}{8} \\
 & \Rightarrow l=10cm \\
\end{align}\]
Hence, the length of Aarushi's card is 10cm.
Perimeter of the card is given by $\text{P}=2\left( l+b \right)$.
Therefore,
\[\begin{align}
  & \text{P}=2\left( 10+8 \right) \\
 & \Rightarrow P=2\times 18 \\
 & \Rightarrow P=36cm \\
\end{align}\]
Hence, the perimeter of Aarushi's card is 36cm.
Kabir's card.
Perimeter of the card is 40cm. Therefore, P = 40cm.
Area of the card is given as 100 square cm. Therefore, A = $100c{{m}^{2}}$.
Since, $\text{P}=2\left( l+b \right)$ therefore
\[\begin{align}
  & 2\left( l+b \right)=40 \\
 & \Rightarrow l+b=20cm\cdots \cdots \cdots \cdots \cdots \left( 1 \right) \\
\end{align}\]
Since, $\text{A}=l\times b$ therefore,
\[l\times b=100c{{m}^{2}}\cdots \cdots \cdots \cdots \cdots \left( 2 \right)\]
From (1) l = 20-b, Putting this in (2) we get:
\[\begin{align}
  & \left( 20-b \right)\times b=100 \\
 & \Rightarrow 20b-{{b}^{2}}=100 \\
 & \Rightarrow {{b}^{2}}-20b+100=0 \\
\end{align}\]
Which is a quadratic equation in b. By splitting middle term, we get:
\[\begin{align}
  & {{b}^{2}}-20b+100=0 \\
 & \Rightarrow b\left( b-10 \right)-10\left( b-10 \right)=0 \\
 & \Rightarrow \left( b-10 \right)\left( b-10 \right)=0 \\
 & \Rightarrow b=10,10 \\
\end{align}\]
Hence, the breadth of Kabir's card is 10cm.
Putting value of b in (1) we get:
\[\begin{align}
  & l+10=20cm \\
 & \Rightarrow l=10cm \\
\end{align}\]
Hence, the length of Kabir's card is 10cm.
Putting all measurements found in the table, we get:

Whose card	Length	Width	Perimeter	Area
Sanya	10 cm	8cm	36 cm	80 square cm
Manav	11 cm	11 cm	44 cm	121 square cm
Aarushi	10 cm	8 cm	36 cm	80 square cm
Kabir	10 cm	10 cm	40 cm	100 square cm

Note: Students should always write units of measurement for length, breadth, perimeter, use cm and for area use $c{{m}^{2}}$. Students should be careful while calculating length and breadth for Kabir's card as a quadratic equation is involved. They can directly use the hit and trial method to find two numbers having a sum of 20 and a product as 100.