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And, shape B can be a rectangle with dimensions:

10 units × 2 units or 5 units × 4 units

III IV
Perimeter of III shape = 2(10 + 2) = 2 × 12 Perimeter of IV shape = 2(5 + 4) = 2 × 9
= 24 units = 18 units
Since, shape A has a longer perimeter than the shape B. So, we choose rectangles I and IV, satisfying
the conditions.

Example 17: Find the dimensions of a rectangle whose area is the sum of the areas of two rectangles
with measurements: 5 m × 10 m and 2 m × 7 m respectively.
Solution: Dimensions of rectangle I : 5 m × 10 m

Dimensions of rectangle II: 2 m × 7 m
Area of rectangle I = (5 × 10) sq. m = 50 sq. m
Area of rectangle II = (2 × 7) sq. m = 14 sq. m

Now, the area of the required rectangle = sum of areas of rectangles I and II

= 50 sq. m + 14 sq. m = 64 sq. m
Considering the dimensions in whole numbers, the rectangle with an area of 64 sq. m can be of
dimensions: 1 m × 64 m; 2 m × 32 m; 4 m × 16 m and 8 m × 8 m.
Example 18: Gurucharan, an architect, has made a house plan as shown below. Some of the
measurements in the plan are given and some are missing. Look at the plan and answer the
following questions:

48 ft

Toilet
Kitchen
(...... ft × (20 ft × 10 ft)
...... ft) Area = 80 sq. ft (...... ft × ...... ft) Utility
Master Bedroom Area = Area = 200 sq. ft
(14 ft × 16 ft) ......
Area = 224 sq. ft

Hall
(26 ft × ........ ft)
Small Bedroom Area = ................ Area = ......... (...... ft × ...... ft) Entrance
(14 ft × 12 ft)
Area = ...............

(a) Find the missing measurements.

(b) Find out the total area of the house.

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