# Perimeter and Area of Mixed Figures

Here we will discuss about the Perimeter and area of mixed figures.

1. The length and breadth of a rectangular field is 8 cm and 6 cm respectively. On the shorter sides of the rectangular field two equilateral triangles are constructed outside. Two right-angled isosceles triangles are constructed outside the rectangular field, with the longer sides as the hypotenuses. Find the total area and perimeter of the figure.

Solution:

The figure consists of the following.

(i) The rectangular field ABCD, whose area = 8 × 6 cm$$^{2}$$ = 48 cm$$^{2}$$

(ii) Two equilateral triangles BCG and ADH. For each, area = $$\frac{√3}{4}$$ × 6$$^{2}$$ cm$$^{2}$$ = 9√3 cm$$^{2}$$

(iii) Two isosceles right-angled triangles CDE and ABF, whose areas are equal.

IF CE = ED = x then x$$^{2}$$ + x$$^{2}$$ = 8$$^{2}$$ cm$$^{2}$$ (by Pythagoras’ theorem)

or, 2x$$^{2}$$ = 64 cm$$^{2}$$

or, x$$^{2}$$ = 32 cm$$^{2}$$

Therefore, x = 4√2 cm

Therefore, area of the ∆CDE = $$\frac{1}{2}$$ CE × DE

= $$\frac{1}{2}$$ x$$^{2}$$

= $$\frac{1}{2}$$ (4√2)$$^{2}$$ cm2

= $$\frac{1}{2}$$ 32 cm$$^{2}$$

= 16 cm$$^{2}$$

Therefore, area of the figure = area of the rectangular field ABCD + 2 × area of the ∆BCG + 2 × area of the ∆CDE

= (48 + 2 × 9√3 + 2 × 16) cm$$^{2}$$

= (80 + 18√3) cm$$^{2}$$

= (80 + 18 × 1.73) cm$$^{2}$$

= (80 + 31.14) cm$$^{2}$$

= 111.14 cm$$^{2}$$

Perimeter of the figure = length of the boundary of the figure

= AF + FB + BG + GC + CE + ED + DH + HA

= 4 × CE + 4 × BG

= (4 × 4√2 + 4 × 6) cm

= 8(3 + 2√2) cm

= 8(3 + 2 × 1.41) cm

= 8 × 5.82 cm

= 46.56 cm

2. The dimension of a field are 110 m × 80 m. The field is to be converted into a garden, leaving a path 5 m broad around the garden. Find the total cost of making the garden if the cost per square metre is Rs 12.

Solution:

For the garden, length = (110 – 2 × 5) m = 100 m, and

Breadth = (80 – 2 × 5) m = 70 m

Therefore, area of the garden = 100 × 70 m$$^{2}$$ = 7000 m$$^{2}$$

Therefore, total cost of making the garden = 7000 × Rs 12 = Rs 84000

3. A square-shaped piece of paper is cut into two pieces along a line joining a corner and a point on an opposite edge. If the ratio of the areas of the two pieces be 3:1, find the ratio of the perimeters of the smaller piece and the original piece of paper.

Solution:

Let PQRS be the square-shaped piece of paper. Let its side measure a units.

It is cut along PM. Let SM = b units

Area of the ∆MSP = $$\frac{1}{2}$$ PS × SM = $$\frac{1}{2}$$ ab square units.

Area of the square PQRS = a$$^{2}$$ square units.

According to the question,

$$\frac{\textrm{area of the quadrilateral PQRM}}{\textrm{area of the ∆MSP}}$$ = $$\frac{3}{1}$$

⟹ $$\frac{\textrm{area of the quadrilateral PQRM}}{\textrm{area of the ∆MSP}}$$ + 1 = 4

⟹ $$\frac{\textrm{area of the quadrilateral PQRM + area of the ∆MSP}}{\textrm{area of the ∆MSP}}$$ = 4

⟹ $$\frac{\textrm{area of the square PQRS}}{\textrm{area of the ∆MSP}}$$ = 4

⟹ $$\frac{a^{2}}{\frac{\textrm{1}}{2} ab} = 4$$

⟹$$\frac{2a}{b}$$ = 4

⟹ a = 2b

⟹ b = $$\frac{1}{2}$$a

Now, PM2 = PS2 + SM2; (by Pythagoras’ theorem)

Therefore, PM2 = a2 + b2

= a2 + ($$\frac{1}{2}$$a )2

= a2 + $$\frac{1}{4}$$a2

= $$\frac{5}{4}$$a2.

Therefore, PM2 = $$\frac{√5}{2}$$a.

Now, $$\frac{\textrm{perimeter of the ∆MSP}}{\textrm{perimeter of the square PQRS}}$$ = $$\frac{\textrm{MS + PS + PM}}{\textrm{4a}}$$

= $$\frac{\frac{1}{2}a + a +\frac{\sqrt{5}}{2}a}{4a}$$

= $$\frac{(\frac{3 + \sqrt{5}}{2})a}{4a}$$

= $$\frac{3 + √5}{8}$$

= (3 + √5) : 8.

4. From a 20 cm × 10 cm plywood board an F-shaped block is cut out, as shown in the figure. What is the area of a face of the remaining board? Also find the length of the boundary of the block.

Solution:

Clearly, the block is a combination of three rectangular blocks, as shown in the below figure.

Therefore, area of a face of the block = 20 × 3 cm$$^{2}$$ + 3 × 2 cm$$^{2}$$ + 7 × 3 cm$$^{2}$$

= 60 cm$$^{2}$$ + 6 cm$$^{2}$$ + 21 cm$$^{2}$$

= 87 cm$$^{2}$$

Area of a face of the uncut board = 20 × 10 cm$$^{2}$$

= 200 cm$$^{2}$$

Therefore, area of a face of the remaining board = 200 cm$$^{2}$$ - 87 cm$$^{2}$$

= 113 cm$$^{2}$$

Required length of the boundary = (20 + 3 + 11 + 2 + 3 + 2 + 3 + 7 + 3 + 10) cm

= 64 cm

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