Question 17

Let us find the missing measurements and areas of Sharan's house.

Step 1 — Find the total height of the house

Look at the left side of the house. The Master Bedroom is 15 ft high. The Small Bedroom is 10 ft high. The total height of the house is the sum of these heights.

$= 15 \text{ ft} + 10 \text{ ft}$

$= 25 \text{ ft}$

$\boxed{\text{Total height} = 25 \text{ ft}}$

Step 2 — Find the total area of the house

The total width of the house is given as 42 ft. The total height of the house is 25 ft (from Step 1). The area of the house is width multiplied by height.

$= 42 \text{ ft} \times 25 \text{ ft}$

$= 1050 \text{ sq ft}$

$\boxed{\text{Total area of house} = 1050 \text{ sq ft}}$

Step 3 — Calculate Small Bedroom area

The dimensions of the Small Bedroom are 12 ft x 10 ft. The area is length multiplied by width.

$= 12 \text{ ft} \times 10 \text{ ft}$

$= 120 \text{ sq ft}$

$\boxed{\text{Small Bedroom Area} = 120 \text{ sq ft}}$

Step 4 — Calculate Utility room dimensions and area

Look at the top right corner. The Kitchen is 10 ft high. The Utility room is next to it. So, the Utility room is also 10 ft high. The area of the Utility room is given as 70 sq ft. To find the width, we divide the area by the height.

$\text{Width} = \frac{\text{Area}}{\text{Height}}$

$= \frac{70 \text{ sq ft}}{10 \text{ ft}}$

$= 7 \text{ ft}$

So, the Utility room dimensions are 7 ft x 10 ft. The area is 70 sq ft.

$\boxed{\text{Utility: 7 ft} \times 10 \text{ ft, Area} = 70 \text{ sq ft}}$

Step 5 — Calculate Toilet dimensions and area

Look at the top section of the house. The total width is 42 ft. The Master Bedroom width is 12 ft. The Kitchen width is 18 ft. The Utility width is 7 ft (from Step 4). The Toilet is between the Master Bedroom and the Kitchen. Let us find the Toilet's width.

$\text{Toilet Width} = \text{Total Width} - (\text{Master Bedroom Width} + \text{Kitchen Width} + \text{Utility Width})$

$= 42 \text{ ft} - (12 \text{ ft} + 18 \text{ ft} + 7 \text{ ft})$

$= 42 \text{ ft} - 37 \text{ ft}$

$= 5 \text{ ft}$

Now, let us find the Toilet's height. The Kitchen and Utility rooms are 10 ft high. The Toilet is next to the Kitchen and has the same top and bottom lines as the Kitchen. So, the Toilet's height is also 10 ft. The area of the Toilet is width multiplied by height.

$= 5 \text{ ft} \times 10 \text{ ft}$

$= 50 \text{ sq ft}$

$\boxed{\text{Toilet: 5 ft} \times 10 \text{ ft, Area} = 50 \text{ sq ft}}$

Step 6 — Calculate Hall dimensions and area

Look at the bottom section of the house. The Hall is to the right of the Small Bedroom. The Small Bedroom is 10 ft high. It is below the Master Bedroom, which is 15 ft high. So, the Small Bedroom's top is 15 ft from the top of the house. Its bottom is 25 ft from the top. The Toilet, Kitchen, and Utility rooms are 10 ft high. Their bottom edge is 10 ft from the top of the house. The Hall is below the Toilet and Kitchen. Its top edge is 10 ft from the top of the house. Its bottom edge is at the bottom of the house, which is 25 ft from the top. So, the Hall's height is the difference between these distances.

$\text{Hall Height} = 25 \text{ ft} - 10 \text{ ft}$

$= 15 \text{ ft}$

The Hall's width is given as 23 ft. The area of the Hall is width multiplied by height.

$= 23 \text{ ft} \times 15 \text{ ft}$

$= 345 \text{ sq ft}$

$\boxed{\text{Hall: 23 ft} \times 15 \text{ ft, Area} = 345 \text{ sq ft}}$

Step 7 — Calculate Entrance dimensions and area

Look at the bottom right corner. The Entrance is to the right of the Hall. The Hall's height is 15 ft (from Step 6). The Entrance is next to the Hall and has the same top and bottom lines. So, the Entrance's height is also 15 ft. Now, let us find the Entrance's width. The total width of the house is 42 ft. The Small Bedroom width is 12 ft. The Hall width is 23 ft (given). The Entrance is to the right of the Hall.

$\text{Entrance Width} = \text{Total Width} - (\text{Small Bedroom Width} + \text{Hall Width})$

$= 42 \text{ ft} - (12 \text{ ft} + 23 \text{ ft})$

$= 42 \text{ ft} - 35 \text{ ft}$

$= 7 \text{ ft}$

The area of the Entrance is width multiplied by height.

$= 7 \text{ ft} \times 15 \text{ ft}$

$= 105 \text{ sq ft}$

$\boxed{\text{Entrance: 7 ft} \times 15 \text{ ft, Area} = 105 \text{ sq ft}}$

Step 8 — List all room dimensions and areas

We have found all the missing measurements. Master Bedroom: 12 ft x 15 ft, Area = 180 sq ft. Toilet: 5 ft x 10 ft, Area = 50 sq ft. Kitchen: 18 ft x 10 ft, Area = 180 sq ft. Utility: 7 ft x 10 ft, Area = 70 sq ft. Small Bedroom: 12 ft x 10 ft, Area = 120 sq ft. Hall: 23 ft x 15 ft, Area = 345 sq ft. Entrance: 7 ft x 15 ft, Area = 105 sq ft.

Let us check the total area by adding all room areas.

$180 + 50 + 180 + 70 + 120 + 345 + 105$

$= 1050 \text{ sq ft}$

This matches the total area of the house we found in Step 2.

Step 9 — Find the perimeter of Sharan's house

The house is a large rectangle. Its total width is 42 ft. Its total height is 25 ft. The perimeter is $2 \times (\text{width} + \text{height})$.

$= 2 \times (42 \text{ ft} + 25 \text{ ft})$

$= 2 \times 67 \text{ ft}$

$= 134 \text{ ft}$

$\boxed{\text{Perimeter of Sharan's house} = 134 \text{ ft}}$

Step 10 — Compare areas and perimeters

The area of Sharan's house is 1050 sq ft. The problem states that Charan's house has the same area, 1050 sq ft. The perimeter of Sharan's house is 134 ft. The problem states that the perimeter of Charan's house is 130 ft. So, Sharan's house has a greater perimeter than Charan's house.

Answer

a. The missing measurements are as follows:

• Master Bedroom: 12 ft × 15 ft, Area = 180 sq ft
• Toilet: 5 ft × 10 ft, Area = 50 sq ft
• Kitchen: 18 ft × 10 ft, Area = 180 sq ft
• Utility: 7 ft × 10 ft, Area = 70 sq ft
• Small Bedroom: 12 ft × 10 ft, Area = 120 sq ft
• Hall: 23 ft × 15 ft, Area = 345 sq ft
• Entrance: 7 ft × 15 ft, Area = 105 sq ft b. The area of his house = 25 ft × 42 ft = 1050 sq ft The area of Sharan's house and Charan's house are same and equal to 1050 sq ft. The perimeter of Sharan's house (134 ft) is greater than Charan's house (130 ft).