Question

please answer this question please its urgent no fake answer who will answer the correct I will mark as brainliest answer

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Answer 1

Answer:

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Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question}}}$

Here the Concept of Areas of Trapezium has been used. We see that the given figure has been divided into two Trapeizums. We already see that we are given the dimensions of both the Trapeziums. So firstly we can find the areas of them separately and then add them both to find the answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{Area\;of\;Trapezium\;=\;\bf{\dfrac{1}{2}\:\times\:(Sum\:of\:Parallel\:sides)\:\times\:Height}}}$

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★ Solution :-

Given,

» For Trapezium ABFC ::

• Parallel sides = AB and FC = 10 cm and 16 cm

• Height = 3 cm

» For Trapezium EDFC ::

• Parallel Sides = ED and FC = 12 cm and 16 cm

• Height = 5 cm

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~ For the Area of Trapezium ABFC ::

$\\\;\;\sf{:\rightarrow\;\;Area\;of\;Trapezium\;=\;\bf{\dfrac{1}{2}\:\times\:(Sum\:of\:Parallel\:sides)\:\times\:Height}}$

By applying values, we get,

$\\\;\;\sf{:\Longrightarrow\;\;Area\;of\;Trapezium\:ABFC\;=\;\bf{\dfrac{1}{2}\:\times\:(AB\;+\;FC)\:\times\;3}}$

$\\\;\;\sf{:\Longrightarrow\;\;Area\;of\;Trapezium\:ABFC\;=\;\bf{\dfrac{1}{2}\:\times\:(10\;+\;16)\:\times\;3}}$

$\\\;\;\sf{:\Longrightarrow\;\;Area\;of\;Trapezium\:ABFC\;=\;\bf{\dfrac{1}{2}\:\times\:(26)\:\times\;3}}$

$\\\;\;\sf{:\Longrightarrow\;\;Area\;of\;Trapezium\:ABFC\;=\;\bf{13\;\times\;3}}$

$\\\;\;\bf{:\Longrightarrow\;\;Area\;of\;Trapezium\:ABFC\;=\;\bf{\green{39\;\:cm^{2}}}}$

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~ For the Area of Trapezium EDFC ::

$\\\;\;\sf{:\rightarrow\;\;Area\;of\;Trapezium\;=\;\bf{\dfrac{1}{2}\:\times\:(Sum\:of\:Parallel\:sides)\:\times\:Height}}$

By applying values, we get,

$\\\;\;\sf{:\Longrightarrow\;\;Area\;of\;Trapezium\:EDFC\;=\;\bf{\dfrac{1}{2}\:\times\:(ED\;+\;FC)\:\times\;5}}$

$\\\;\;\sf{:\Longrightarrow\;\;Area\;of\;Trapezium\:EDFC\;=\;\bf{\dfrac{1}{2}\:\times\:(12\;+\;16)\:\times\;5}}$

$\\\;\;\sf{:\Longrightarrow\;\;Area\;of\;Trapezium\:EDFC\;=\;\bf{\dfrac{1}{2}\:\times\:(28)\:\times\;5}}$

$\\\;\;\sf{:\Longrightarrow\;\;Area\;of\;Trapezium\:EDFC\;=\;\bf{14\;\times\;5}}$

$\\\;\;\bf{:\Longrightarrow\;\;Area\;of\;Trapezium\:EDFC\;=\;\bf{\green{70\;\:cm^{2}}}}$

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~ For the Area of the Polygon ::

$\\\;\sf{:\mapsto\;\;Area\;of\;Polygon\;=\;\bf{Sum\;of\;areas\;both\;Trapeziums}}$

$\\\;\sf{:\mapsto\;\;Area\;of\;Polygon\;=\;\bf{Area\;of\;Trapezium\;ABFC\;+\;Area\;of\;Trapezium\;EDFC}}$

$\\\;\sf{:\mapsto\;\;Area\;of\;Polygon\;=\;\bf{39\;+\;70}}$

$\\\;\sf{:\mapsto\;\;Area\;of\;Polygon\;=\;\bf{\orange{109\;\:cm^{2}}}}$

$\\\;\underline{\boxed{\tt{Area\;\:of\;\:Polygon\;=\;\bf{\purple{109\;\;cm^{2}}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;Area\;of\;Square\;=\;(Side)^{2}}$

$\\\;\sf{\leadsto\;\;Area\;of\;Rectangle\;=\;Length\;\times\;Breadth}$

$\\\;\sf{\leadsto\;\;Area\;of\;Circle\;=\;\pi r^{2}}$

$\\\;\sf{\leadsto\;\;Area\;of\;Triangle\;=\;\dfrac{1}{2}\;\times\;Base\;\times\;Height}$

$\\\;\sf{\leadsto\;\;Area\;of\;Parallelogram\;=\;Base\;\times\;Height}$