Math, asked by ItsmysteryboyZ, 2 months ago

Answer it :- the question in figure​

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Answered by AestheticDude
93

Answer :-

  • \sf31 \dfrac{3}{4}\; cm^{2}\; \bigstar

Step-By-Step-Explanation :-

Solution :-

Area of polygon ABCDEF

= Area of Δ AFG + Area of trapezium FGIE + Area of Δ EID + Area of Δ ABH + Area of trapezium HBCJ + Area of Δ CDJ .

\sf = \dfrac12AG \times FG\;+\dfrac12(FG+EI)\times GI\;+\dfrac12ID\times EI + \;\dfrac12AH\times BH \;+\dfrac12(BH+CJ)\times HJ+\;\dfrac12JD\times CJ

\sf = \dfrac12AG\times FG+\;\dfrac12(FG+EI)\times(AI-AG)+\;\dfrac12(AD-AI)\times EI +\;\dfrac12AH\times BH+\;\dfrac12(BH+CJ)\times(AJ-AH) +\;\dfrac12(AD-AJ)\times CJ

=\bigg[ \dfrac{1}{2} \times 2.5 + \dfrac{1}{2}(2.5 + 3) (6 - 2) + \dfrac{1}{2}(9 - 6) \times 3 + \dfrac{1}{2} \times 4 \times 2.25 \sf + \dfrac{1}{2}(2.25 + 2)(7 - 4) + \dfrac{1}{2} \sf (9 - 7) \times 2\bigg] cm^2

\sf =\bigg[2.5+\dfrac12\times5.5\times4+\dfrac12\times3\times3+5+\dfrac12\times4.5\times3+\dfrac12\times2\times 2\bigg]cm^2

\sf =\bigg[\dfrac52+\dfrac{11}{2}\times2+\dfrac92+5+\dfrac12\times\dfrac92\times3+2\bigg] cm^2

\sf =\bigg[\dfrac52+11+\dfrac92+5+\dfrac{27}{4}+2\bigg]cm^2

\sf =\bf31\dfrac34 cm^2

\rm Therefore,\;31\dfrac34\; cm^2 \;is\;the\;Answer

Additional Information :-

\sf Area\;of\;Rhombus=\dfrac12\times product\; of\; its\; diagonals

\sf Area\;of\;Trapezium=\dfrac12\;sum\;of\;parallel\;sides\;\times distance\;between\;the\;parallel\;sides

\sf Area\;of\;Triangle=\dfrac12\;Base\times Height

\sf Area\;of\;Rectangle=Length\times Breadth

\sf Area\;of\;Square=a^2

a here refers to the side of the square

\sf Area\;of\;Parallelogram =Base\times Height

\sf Area\;of\;Circle=\pi r^2

\sf Perimeter\;of\;Triangle=a+b+c

A , B and C refers to three sides of triangle

\sf Perimeter\;of\;Rectangle=2(Length+ Breadth)

\sf Perimeter\;of\;Square=4a

\sf Perimeter\;of\;Parallelogram=2(Length + Breadth)

\sf Perimeter\;of\;Circle=2\pi r

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