Question

Area of trapezium of height 4 cm is 32 cm^2 One of the parallel sides of the trapezium is cm. Find the length of the other parallel side​

Answer 1

Answer:

Answer :

h=2A

a+b.... ............

Answer 2

Answer :-

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

Here the concept of Areas of Trapezium has been used. We are given the dimensions of the Trapezium. Its area is half the product of sum of parallel sides of Trapezium multiplied by perpendicular distance between them. We are given the height, area and one parallel side. We can apply the value and find the answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{Area\;of\;Trapezium\;=\;\bf{\dfrac{1}{2}\;\times\;(Sum\;of\;Parallel\;Side)\;\times\;Distance\;Between\;them}}}$

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★ Correct Question :-

Area of trapezium of height 4 cm is 32 cm².One of the parallel sides of the trapezium is 6 cm. Find the length of the other parallel side.

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★ Solution :-

Given,

» Perpendicular distance between the parallel sides of the Trapezium = height = 4 cm

» Area of the Trapezium = 32 cm²

» One of the Parallel side = 6 cm

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~ For the length of other Parallel sides of Parallelogram ::

• Let the other parallel side of the Trapezium be x .

Then,

$\\\;\;\;\sf{:\mapsto\;\;Area\;of\;Trapezium\;=\;\bf{\dfrac{1}{2}\;\times\;(Sum\;of\;Parallel\;Side)\;\times\;Distance\;Between\;them}}$

$\\\;\;\;\sf{:\mapsto\;\;32\;=\;\bf{\dfrac{1}{2}\;\times\;(6\;+\;x)\;\times\;4}}$

By transposing this to other side, we get,

$\\\;\;\;\sf{:\mapsto\;\;(6\;+\;x)\;=\;\bf{\dfrac{32\;\times\;2}{4}}}$

$\\\;\;\;\sf{:\mapsto\;\;(6\;+\;x)\;=\;\bf{8\;\times\;2}}$

$\\\;\;\;\sf{:\mapsto\;\;(6\;+\;x)\;=\;\bf{16}}$

$\\\;\;\;\sf{:\mapsto\;\;x\;=\;\bf{16\;-\;6}}$

$\\\;\;\;\bf{:\mapsto\;\;x\;=\;\bf{10\;\;cm}}$

$\\\;\large{\underline{\underline{\rm{Hence,\;length\;of\;other\;side\;is\;\;\boxed{\bf{10\;\;cm}}}}}}$

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★ More to know :-

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

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

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

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

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

$\\\;\sf{\leadsto\;\;Perimeter\;of\;Rectangle\;=\;2\;\times\;(Length\;+\;Breadth)}$

$\\\;\sf{\leadsto\;\;Perimeter\;of\;Cube\;=\;2\;\times\;(Side)}$

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