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)\(^{2}\) cm^{2}

= \(\frac{1}{2}\) 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, PM^{2} = PS^{2} + SM^{2}; (by Pythagoras’ theorem)

Therefore, PM^{2} = a^{2} + b^{2}

= a^{2} + (\(\frac{1}{2}\)a )^{2}

= a^{2} + \(\frac{1}{4}\)a^{2}

= \(\frac{5}{4}\)a^{2}.

Therefore, PM^{2} = \(\frac{√5}{2}\)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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