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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:

Perimeter and Area of Mixed Figures

The figure consists of the following.

(i) The rectangular field ABCD, whose area = 8 × 6 cm2 = 48 cm2

(ii) Two equilateral triangles BCG and ADH. For each, area = 34 × 62 cm2 = 9√3 cm2

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

IF CE = ED = x then x2 + x2 = 82 cm2 (by Pythagoras’ theorem)

or, 2x2 = 64 cm2

or, x2 = 32 cm2

Therefore, x = 4√2 cm

Therefore, area of the ∆CDE = 12 CE × DE

                                         = 12 x2

                                         = 12 (4√2)2 cm2

                                         = 12 32 cm2

                                         = 16 cm2

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) cm2

                                          = (80 + 18√3) cm2

                                          = (80 + 18 × 1.73) cm2

                                          = (80 + 31.14) cm2

                                          = 111.14 cm2

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:

Rectangular Field Problem

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 m2 = 7000 m2

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.

Area of Square-shaped Piece of Paper

It is cut along PM. Let SM = b units

Area of the ∆MSP = 12 PS × SM = 12 ab square units.

Area of the square PQRS = a2 square units.

According to the question,

area of the quadrilateral PQRMarea of the ∆MSP = 31

area of the quadrilateral PQRMarea of the ∆MSP + 1 = 4

area of the quadrilateral PQRM + area of the ∆MSParea of the ∆MSP = 4

area of the square PQRSarea of the ∆MSP = 4

a212ab=4

2ab = 4

⟹ a = 2b

⟹ b = 12a

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

Therefore, PM2 = a2 + b2

                       = a2 + (12a )2

                       = a2 + 14a2

                       = 54a2.

Therefore, PM2 = 52a.

Now, perimeter of the ∆MSPperimeter of the square PQRS = MS + PS + PM4a

                                         = 12a+a+52a4a

                                        = (3+52)a4a

                                        = 3+58

                                        = (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.

Application Problem on Plane Figure

Solution:

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

Plane Figure Problem

Therefore, area of a face of the block = 20 × 3 cm2 + 3 × 2 cm2 + 7 × 3 cm2

                                                      = 60 cm2 + 6 cm2 + 21 cm2

                                                      = 87 cm2

Area of a face of the uncut board = 20 × 10 cm2

                                                = 200 cm2

Therefore, area of a face of the remaining board = 200 cm2 - 87 cm2

                                                                      = 113 cm2

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

                                               = 64 cm







9th Grade Math

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