Perimeter and Area | IT

Question 17

Now, find out the missing dimensions and area of Sharan's home. Below is the plan:

Some of the measurements are given.

a. Find the missing measurements.

b. Find out the area of his house.

What are the dimensions of all the different rooms in Sharan's house? Compare the areas and perimeters of Sharan's house and Charan's house.

Question diagram 1
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Solution
Understand the Question

The floor plan of Sharan's house is a large rectangle with a total width of 42 ft42\text{ ft}. By observing how different rooms line up horizontally and vertically:

  • The total height is the sum of the heights along the left side (Master Bedroom + Small Bedroom).
  • Missing room widths are found by subtracting known room widths along a horizontal row from the total width (42 ft42\text{ ft}).
  • Missing room heights are found using the heights of adjacent aligned rooms.
  • Room Area =Length×Width= \text{Length} \times \text{Width}, and House Perimeter =2×(Total Width+Total Height)= 2 \times (\text{Total Width} + \text{Total Height}).

a Find the missing measurements and dimensions of all the different rooms in Sharan's house.

Step 1 · Find Total Height of the House

From the left edge of the plan, the Master Bedroom has height 15 ft15\text{ ft} and the Small Bedroom has height 10 ft10\text{ ft}.Diagram 1

Total height=15 ft+10 ft=25 ft\begin{aligned} \text{Total height} &= 15\text{ ft} + 10\text{ ft} \\[0.6em] &= 25\text{ ft} \end{aligned}

Step 2 · Calculate Utility Room Dimensions

The Kitchen has height 10 ft10\text{ ft}, so the adjacent Utility room also has height 10 ft10\text{ ft}. Given Area=70 sq ft\text{Area} = 70\text{ sq ft}:Diagram 2

Width=AreaHeight=70 sq ft10 ft=7 ft\begin{aligned} \text{Width} &= \dfrac{\text{Area}}{\text{Height}} \\[0.8em] &= \dfrac{70\text{ sq ft}}{10\text{ ft}} \\[0.8em] &= 7\text{ ft} \end{aligned}

Dimensions of Utility room =7 ft×10 ft= 7\text{ ft} \times 10\text{ ft}, Area =70 sq ft= 70\text{ sq ft}.

Step 3 · Calculate Toilet Dimensions and Area

Along the top row, the total width is 42 ft42\text{ ft}. The Toilet lies between the Master Bedroom (12 ft12\text{ ft}) and Kitchen (18 ft18\text{ ft}), with Utility width =7 ft= 7\text{ ft}.Diagram 3

Toilet Width=Total Width−(Master Bedroom Width+Kitchen Width+Utility Width)=42 ft−(12 ft+18 ft+7 ft)=42 ft−37 ft=5 ft\begin{aligned} \text{Toilet Width} &= \text{Total Width} - (\text{Master Bedroom Width} + \text{Kitchen Width} + \text{Utility Width}) \\[0.6em] &= 42\text{ ft} - (12\text{ ft} + 18\text{ ft} + 7\text{ ft}) \\[0.6em] &= 42\text{ ft} - 37\text{ ft} \\[0.6em] &= 5\text{ ft} \end{aligned}

Height of Toilet =10 ft= 10\text{ ft} (aligned with Kitchen).

Area=5 ft×10 ft=50 sq ft\begin{aligned} \text{Area} &= 5\text{ ft} \times 10\text{ ft} \\[0.6em] &= 50\text{ sq ft} \end{aligned}

Dimensions of Toilet =5 ft×10 ft= 5\text{ ft} \times 10\text{ ft}, Area =50 sq ft= 50\text{ sq ft}.

Step 4 · Calculate Hall Dimensions and Area

The rooms above the Hall (Toilet and Kitchen) have height 10 ft10\text{ ft}. Since the total house height is 25 ft25\text{ ft}:Diagram 4

Hall Height=25 ft−10 ft=15 ft\begin{aligned} \text{Hall Height} &= 25\text{ ft} - 10\text{ ft} \\[0.6em] &= 15\text{ ft} \end{aligned}

Given Hall width =23 ft= 23\text{ ft}:

Area=23 ft×15 ft=345 sq ft\begin{aligned} \text{Area} &= 23\text{ ft} \times 15\text{ ft} \\[0.6em] &= 345\text{ sq ft} \end{aligned}

Dimensions of Hall =23 ft×15 ft= 23\text{ ft} \times 15\text{ ft}, Area =345 sq ft= 345\text{ sq ft}.

Step 5 · Calculate Entrance Dimensions and Area

The Entrance is aligned with the Hall, so its height is 15 ft15\text{ ft}. Along the bottom row:Diagram 5

Entrance Width=Total Width−(Small Bedroom Width+Hall Width)=42 ft−(12 ft+23 ft)=42 ft−35 ft=7 ft\begin{aligned} \text{Entrance Width} &= \text{Total Width} - (\text{Small Bedroom Width} + \text{Hall Width}) \\[0.6em] &= 42\text{ ft} - (12\text{ ft} + 23\text{ ft}) \\[0.6em] &= 42\text{ ft} - 35\text{ ft} \\[0.6em] &= 7\text{ ft} \end{aligned} Area=7 ft×15 ft=105 sq ft\begin{aligned} \text{Area} &= 7\text{ ft} \times 15\text{ ft} \\[0.6em] &= 105\text{ sq ft} \end{aligned}

Dimensions of Entrance =7 ft×15 ft= 7\text{ ft} \times 15\text{ ft}, Area =105 sq ft= 105\text{ sq ft}.

Step 6 · Summary of All Room Dimensions and Areas

Calculating area of Small Bedroom:

Small Bedroom Area=12 ft×10 ft=120 sq ft\begin{aligned} \text{Small Bedroom Area} &= 12\text{ ft} \times 10\text{ ft} \\[0.6em] &= 120\text{ sq ft} \end{aligned}

Summary of all room dimensions and areas:

  • Master Bedroom: 12 ft×15 ft12\text{ ft} \times 15\text{ ft}, Area=180 sq ft\text{Area} = 180\text{ sq ft}
  • Toilet: 5 ft×10 ft5\text{ ft} \times 10\text{ ft}, Area=50 sq ft\text{Area} = 50\text{ sq ft}
  • Kitchen: 18 ft×10 ft18\text{ ft} \times 10\text{ ft}, Area=180 sq ft\text{Area} = 180\text{ sq ft}
  • Utility: 7 ft×10 ft7\text{ ft} \times 10\text{ ft}, Area=70 sq ft\text{Area} = 70\text{ sq ft}
  • Small Bedroom: 12 ft×10 ft12\text{ ft} \times 10\text{ ft}, Area=120 sq ft\text{Area} = 120\text{ sq ft}
  • Hall: 23 ft×15 ft23\text{ ft} \times 15\text{ ft}, Area=345 sq ft\text{Area} = 345\text{ sq ft}
  • Entrance: 7 ft×15 ft7\text{ ft} \times 15\text{ ft}, Area=105 sq ft\text{Area} = 105\text{ sq ft}

Verifying total sum of room areas:

Total Area=180+50+180+70+120+345+105=1050 sq ft\begin{aligned} \text{Total Area} &= 180 + 50 + 180 + 70 + 120 + 345 + 105 \\[0.6em] &= 1050\text{ sq ft} \end{aligned}
Answer

a The room dimensions and areas are:

  • Master Bedroom: 12 ft×15 ft12\text{ ft} \times 15\text{ ft}, Area =180 sq ft= 180\text{ sq ft}
  • Toilet: 5 ft×10 ft5\text{ ft} \times 10\text{ ft}, Area =50 sq ft= 50\text{ sq ft}
  • Kitchen: 18 ft×10 ft18\text{ ft} \times 10\text{ ft}, Area =180 sq ft= 180\text{ sq ft}
  • Utility: 7 ft×10 ft7\text{ ft} \times 10\text{ ft}, Area =70 sq ft= 70\text{ sq ft}
  • Small Bedroom: 12 ft×10 ft12\text{ ft} \times 10\text{ ft}, Area =120 sq ft= 120\text{ sq ft}
  • Hall: 23 ft×15 ft23\text{ ft} \times 15\text{ ft}, Area =345 sq ft= 345\text{ sq ft}
  • Entrance: 7 ft×15 ft7\text{ ft} \times 15\text{ ft}, Area =105 sq ft= 105\text{ sq ft}

b Find out the area of his house and compare the areas and perimeters of Sharan's house and Charan's house.

Step 1 · Calculate Total Area of Sharan's House

Using total width =42 ft= 42\text{ ft} and total height =25 ft= 25\text{ ft}:

Total area=42 ft×25 ft=1050 sq ft\begin{aligned} \text{Total area} &= 42\text{ ft} \times 25\text{ ft} \\[0.6em] &= 1050\text{ sq ft} \end{aligned}

Step 2 · Calculate Perimeter of Sharan's House

Using the perimeter formula for a rectangle:

Perimeter=2×(width+height)=2×(42 ft+25 ft)=2×67 ft=134 ft\begin{aligned} \text{Perimeter} &= 2 \times (\text{width} + \text{height}) \\[0.6em] &= 2 \times (42\text{ ft} + 25\text{ ft}) \\[0.6em] &= 2 \times 67\text{ ft} \\[0.6em] &= 134\text{ ft} \end{aligned}

Step 3 · Compare Area and Perimeter with Charan's House

Comparing values:

  • Area: Sharan's house area =1050 sq ft= 1050\text{ sq ft}, Charan's house area =1050 sq ft= 1050\text{ sq ft} (both are equal).
  • Perimeter: Sharan's house perimeter =134 ft= 134\text{ ft}, Charan's house perimeter =130 ft= 130\text{ ft}.
Answer

b The area of Sharan's house is 1050 sq ft1050\text{ sq ft}. Both houses have equal area (1050 sq ft1050\text{ sq ft}), but Sharan's house has a larger perimeter (134 ft>130 ft134\text{ ft} > 130\text{ ft}).

Common Mistakes
  • Row vs Total Width: Forgetting that widths across any complete horizontal strip must add up to the total width of 42 ft42\text{ ft}.
  • Determining Room Height: Assuming all lower rooms have height 10 ft10\text{ ft}; the Hall and Entrance have a height of 25 ft−10 ft=15 ft25\text{ ft} - 10\text{ ft} = 15\text{ ft}.
  • Area vs Perimeter: Confusing perimeter with area when comparing the two houses—shapes with equal area can have different perimeters.

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Q17

Now, find out the missing dimensions and area of Sharan's home. Below is the plan:

Some of the measurements are given.

a. Find the missing measurements.

b. Find out the area of his house.

What are the dimensions of all the different rooms in Sharan's house? Compare the areas and perimeters of Sharan's house and Charan's house.

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