Question

A triangle and a parallelogram stand
on the same base and are equal in area. If the sides of the triangle 40cm,24cm
and 32cm and the base of the parallelogram is 40cm, find the
corresponding height of the parallelogram​

Answer 1

Answer :-

$\:\\\underbrace{\underline{\sf{Let's\;understand\;to\;the\;concept\;:-}}}$

Here the concept Equality in Areas of Triangle and Areas of Parallelogram has been used. We are given that the Parallelogram and Triangle stand on same base and have equal area. This means, that Area of Triangle will be Equal to Area of Parallelogram. We are already given the sides of Triangle, so we can find its area using Heron's Formula and then find the height of Parallelogram.

Let's do it !!

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★ Formula Used :-

$\\\;\large{\boxed{\sf{Semi\:-\:Perimeter\;of\;\Delta,\;s\;=\;\bf{\dfrac{Sum\;of\;all\;sides,\;a\:+\:b\:+\:c}{2}}}}}$

$\\\;\large{\boxed{\sf{Area\;of\;\Delta\;=\;\bf{\sqrt{s(s\:-\:a)(s\:-\:b)(s\:-\:c)}}}}}$

$\\\;\large{\boxed{\sf{Area\;of\;Parallelogram\;=\;\bf{Base\;\times\;Height}}}}$

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★ Solution :-

Given,

» First side of Triangle = a = 40 cm

» Second Side of Triangle = b = 24 cm

» Third Side of Triangle = c = 32 cm

» Base of Parallelogram = 40 cm

Since here all the sides of Triangle are of different length, then its a Scalene Triangle. We can find its area by using Heron's Formula.

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~ For the Semi - Perimeter of Triangle :-

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Semi\:-\:Perimeter\;of\;\Delta,\;s\;=\;\bf{\dfrac{Sum\;of\;all\;sides,\;a\:+\:b\:+\:c}{2}}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Semi\:-\:Perimeter\;of\;\Delta,\;s\;=\;\bf{\dfrac{40\:+\:24\:+\:32}{2}}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Semi\:-\:Perimeter\;of\;\Delta,\;s\;=\;\bf{\dfrac{96}{2}}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Semi\:-\:Perimeter\;of\;\Delta,\;s\;=\;\underline{\underline{\bf{48\;\;cm}}}}$

$\\\;\underline{\boxed{\tt{Semi\:-\:Perimeter\;of\;the\;Triangle\;=\;\bf{48\;\;cm}}}}$

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~ For the Area of Triangle :-

$\\\;\;\;\;\;\sf{:\Longrightarrow\;\;\;Area\;of\;\Delta\;=\;\bf{\sqrt{s(s\:-\:a)(s\:-\:b)(s\:-\:c)}}}$

$\\\;\;\;\;\;\sf{:\Longrightarrow\;\;\;Area\;of\;\Delta\;=\;\bf{\sqrt{(48)(48\:-\:40)(48\:-\:24)(48\:-\:32)}}}$

$\\\;\;\;\;\;\sf{:\Longrightarrow\;\;\;Area\;of\;\Delta\;=\;\bf{\sqrt{(48)(8)(24)(16)}}}$

$\\\;\;\;\;\;\sf{:\Longrightarrow\;\;\;Area\;of\;\Delta\;=\;\bf{\sqrt{147456}}}$

$\\\;\;\;\;\;\sf{:\Longrightarrow\;\;\;Area\;of\;\Delta\;=\;\underline{\underline{\bf{384\;\;cm^{2}}}}}$

$\\\;\underline{\boxed{\tt{Area\;of\;the\;Triangle\;=\;\bf{384\;\;cm^{2}}}}}$

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~ For the Height of the Triangle :-

Clearly, its given that,

Area of Triangle = Area of Parallelogram

Then,

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Area\;of\;Parallelogram\;=\;\bf{Base\;\times\;Height}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Area\;of\;Triangle\;=\;\bf{Base\;\times\;Height}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Base\;\times\;Height\;\;=\;\;\bf{384\;\;cm^{2}}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;40\;\times\;Height\;\;=\;\;\bf{384}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Height\;\;=\;\;\bf{\dfrac{384}{40}}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Height\;\;=\;\;\underline{\underline{\bf{9.6\;\;cm}}}}$

$\\\:\large{\underline{\underline{\rm{Thus,\;corresponding\;Height\;of\;Parallelogram\;is\;\;\boxed{\bf{9.6\;\;cm}}}}}}$

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★ More Formulas To know :-

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

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

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

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

$\\\:\sf{\leadsto\;\;\;Perimeter\;of\;Square\;=\;4\;\times\;(Side)}$

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

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

Answer 2

Answer:

$\large ❀ \bf\red {\underline{Question}}❀$

A ᴛʀɪᴀɴɢʟᴇ ᴀɴᴅ ᴀ ᴘᴀʀᴀʟʟᴇʟᴏɢʀᴀᴍ ꜱᴛᴀɴᴅ ᴏɴ ᴛʜᴇ ꜱᴀᴍᴇ ʙᴀꜱᴇ ᴀɴᴅ ᴀʀᴇ ᴇϙᴜᴀʟ ɪɴ ᴀʀᴇᴀ. Iғ ᴛʜᴇ ꜱɪᴅᴇꜱ ᴏғ ᴛʜᴇ ᴛʀɪᴀɴɢʟᴇ 40ᴄᴍ,24ᴄᴍ ᴀɴᴅ 32ᴄᴍ ᴀɴᴅ ᴛʜᴇ ʙᴀꜱᴇ ᴏғ ᴛʜᴇ ᴘᴀʀᴀʟʟᴇʟᴏɢʀᴀᴍ ɪꜱ 40ᴄᴍ, ғɪɴᴅ ᴛʜᴇ ᴄᴏʀʀᴇꜱᴘᴏɴᴅɪɴɢ ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ᴘᴀʀᴀʟʟᴇʟᴏɢʀᴀᴍ

$\large ❀ \bf\red {\underline{Given}}❀$

Tʀɪᴀɴɢʟᴇ'ꜱ ꜱɪᴅᴇꜱ -

• ◾40 ᴄᴍ
• ◾24 ᴄᴍ
• ◾32 ᴄᴍ

Pᴀʀᴀʟʟᴇʟᴏɢʀᴀᴍ'ꜱ ʙᴀꜱᴇ :-

• ◼️ 40 ᴄᴍ

$\large ❀ \bf\red {\underline{To \: Find}}❀$

• ♦️ Hᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ᴘᴀʀᴀʟʟᴇʟᴏɢʀᴀᴍ = ?

$\large ❀ \bf\red {\underline{Formula}}❀$

$\\ \large \bigstar\bf {\green{s = } \frac{a + b + c}{2} } \red \bigstar \\ \\ \large \bigstar \bf {\color{orange}{ \sqrt{s(s - a)(s - b)(s - c) \: \: } }} \bigstar \\ \\ \\ \large \bigstar \bf {\red{area \: of \: the \: parallelogram} = base \: \times \: height} \blue \bigstar$

$\large ❀ \bf\red {\underline{Solution}}❀$

Lᴇᴛ

• ♡ ᴀ = 40 cm ♡
• ♡ ʙ = 24 cm ♡
• ♡ ᴄ = 32 cm ♡

$\\ \therefore \: \: \large \bf \bold \green{s = \frac{a +b +c }{2} } \\ \\ \\ \\ \: \: \: \: \: \: \: \: \large \bf \bold {➲ \: s = \frac{40 +24 +32 }{2} } \\ \\ \\ \large \bf \bold {➲ \: s = \cancel\frac{96 }{2} } \\ \\ \\ \large \bf \bold {➲ \: s = 48 \: cm {}^{2} }$

$Nᴏᴡ , \\ \\ \red\bigstar\bf s - ᴀ = 48 - 40 \\ = 8 \\ \\ \red \bigstar\bf \: s- ʙ = 48 - 24 \\ = 24 \\ \\ \red \bigstar \bf \: s \: - ᴄ = 48 - 32 \\ = 16$

$\\ \blue\therefore \bf \bold{area \: of \: the \: triangle} \\ \\ \bf \green{➲ \: \sqrt{s(s - a)(s - b)(s - c)} } \\ \\ \bf \green{➲ \sqrt{48(8)(24)(16)} } \\ \\ \bf \green{➲ \sqrt{147456} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \green{➲ \: 384 \: cm {}^{2} }$

Aʀᴇᴀ ᴏғ ᴛʜᴇ ᴘᴀʀᴀʟʟᴇʟᴏɢʀᴀᴍ = Aʀᴇᴀ ᴏғ ᴛʜᴇ ᴛʀɪᴀɴɢʟᴇ

Aʀᴇᴀ ᴏғ ᴛʜᴇ ᴘᴀʀᴀʟʟᴇʟᴏɢʀᴀᴍ = ʙᴀꜱᴇ × ʜᴇɪɢʜᴛ

Aʀᴇᴀ ᴏғ ᴛʜᴇ ᴛʀɪᴀɴɢʟᴇ = ʙᴀꜱᴇ × ʜᴇɪɢʜᴛ

➔ 384 = 40 × ʜᴇɪɢʜᴛ

➔ 40 × ʜᴇɪɢʜᴛ = 384

➔ ʜᴇɪɢʜᴛ = $\large\cancel\frac{384}{40}$

➔ ʜᴇɪɢʜᴛ = 9.6 ᴄᴍ