Question

Find the area of the polygon ABCDEF given alongside if FP = 10 cm, FQ=20 cm, FR=50 cm, FS = 60 cm and FC = 100 cm. All other measurements areas given in the figure.​

Answer 1

Area of polygon, ABCDEF is 4700 $cm^2$.

Given

To find the area of the polygon, ABCDEF.

Given sides,

                   FP = 10 cm            FQ = 20 cm

                   FR = 50 cm           FS = 60 cm

                                   FC = 100 cm

From the figure,

In ΔFEP,

            Area of triangle = 1/2 × b × h

                    h = 40 cm    ;     b = 10 cm

                                        = 1/2 × 10 × 40

                                        = 400/2

                                        = 200 $cm^2$

In ΔFQA,

              Area of triangle = 1/2 × b × h

                     h = 20 cm    ;     b = 20 cm

                                         = 1/2 × 20 × 20

                                         = 400/2

                                         = 200 $cm^2$

In ΔBSC,

             Area of triangle = 1/2 × b × h

                      h = 40 cm   ;    b = 10 cm

                                        = 1/2 × 10 × 40

                                        = 1/2 × 400

                                        = 200 $cm^2$

In ΔDRC,

              Area of triangle = 1/2 × b × h

                     h = 60 cm    ;     b = ( FC - FR ) = 100 - 50 = 50 cm

                                         = 1/2 × 50 × 60

                                         = 1/2 × 3000

                                         = 1500 $cm^2$

In trapezium EPRD,

                                Area of trapezium = 1/2 ( $b_1 + b_2$ ) × h

                                 h = ( FR - FP ) = 50 - 10 = 40

                                        $b_1$ = 40 cm   ;   $b_2$ = 60 cm                      

                                                               = 1/2 ( 40 + 60 ) × 40

                                                               = 1/2 ( 100 × 40 )

                                                               = 1/2 × 4000

                                                               = 2000 $cm^2$

In trapezium AQSB,

                                Area of trapezium = 1/2 ( $b_1 + b_2$ ) × h

                                 h = ( FS - FQ ) = 50 - 10 = 40 cm ;

                                        $b_1$ = 20 cm   ;   $b_2$ = 10 cm

                                                                = 1/2 ( 20 + 10 ) × ( 40 )

                                                                = 1/2 ( 30 × 40 )

                                                                = 1/2 × 1200

                                                                = 600 $cm^2$

Area of polygon, ABCDEF = Area of ΔFEP + Area of ΔFQA + Area of ΔBSC +

                                               Area of ΔDRC + Area of trapezium  EPRD  +  

                                               Area of trapezium AQSB

                                           = 200 $cm^2$ + 200 $cm^2$ + 200 $cm^2$ + 1500 $cm^2$ +  

                                              2000 $cm^2$ + 600 $cm^2$

                                           =  4700 $cm^2$.

Therefore, area of polygon = 4700 $cm^2$.

To learn more...

1. brainly.in/question/2453980          

2. brainly.in/question/7914060  

Answer 2

Given :-

• FP = 10 cm.
• FQ = 20 cm.
• FR = 50 cm.
• FS = 60 cm.
• FC = 100 cm.
• EP = 40 cm.
• DS = 60 cm.
• AQ = 20 cm.
• SB = 10 cm .

To Find :-

• the area of the polygon ABCDEF ?

Solution :-

in ∆EPF,

→ ∠EPF = 90° .

so,

→ EP = Perpendicular height = 40 cm.

→ FP = Base = 10 cm .

then,

→ Area of ∆EPF = (1/2) * Perpendicular height * Base = (1/2) * 40 * 10 = 200 cm². ----------- (1)

similarly,

in ∆FQA,

→ ∠FQA = 90° .

so,

→ QA = Perpendicular height = 20 cm.

→ FQ = Base = 20 cm .

then,

→ Area of ∆FQA = (1/2) * Perpendicular height * Base = (1/2) * 20 * 20 = 200 cm². ----------- (2)

similarly,

in ∆CSB,

→ ∠CSB = 90° .

so,

→ SB = Perpendicular height = 10 cm.

→ CS = Base = FC - FS = 100 - 60 = 40 cm .

then,

→ Area of ∆CSB = (1/2) * Perpendicular height * Base = (1/2) * 10 * 40 = 200 cm². ----------- (3)

similarly,

in ∆DRC,

→ ∠DRC = 90° .

so,

→ DR = Perpendicular height = 60 cm.

→ RC = Base = FC - FR = 100 - 50 = 50 cm .

then,

→ Area of ∆DRC = (1/2) * Perpendicular height * Base = (1/2) * 60 * 50 = 1500 cm². ----------- (4)

now,

in Trapezium EPRD, we have,

→ EP || DR .

→ PR = Height of Trapezium =>FR - FP = 50 - 10 = 40 cm.

then,

→ Area of Trapezium EPRD = (1/2) * (sum of Parallel sides) * Height = (1/2) * (40 + 60) * 40 = (1/2) * 100 * 40 = 2000 cm². ------------- (5) .

and,

in Trapezium AQSB, we have,

→ AQ || SB .

→ QS = Height of Trapezium =>FS - FQ = 60 - 20 = 40 cm.

then,

→ Area of Trapezium AQSB = (1/2) * (sum of Parallel sides) * Height = (1/2) * (20 + 10) * 40 = (1/2) * 30 * 40 = 600 cm². ------------- (6) .

therefore, adding all six figures we get,

→ Area of Polygon ABCDEF = Area of ∆EPF + Area of ∆FQA + Area of ∆CSB + Area of ∆DRC + Area of Trapezium EPRD + Area of Trapezium AQSB

→ Area of Polygon ABCDEF = 200 + 200 + 200 + 1500 + 2000 + 600 = 4700 cm². (Ans.)

Hence, Area of Polygon ABCDEF will be 4700 cm².

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