Question

In a right-angled triangle, AB = 6cm, BC = 8cm, what is the length of AC ?


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Answer 1

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

Here the Concept of Pythagoras Theorem has been used . We see that in a Right Triangle, according to Pythagoras Theorem, the square of the length of Hypotenuse that is the longest side is equal to the sum of squares of other two sides . Using this, let's do the question .

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}}}$

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★ Solution :-

Given,

» AB = Height = 6 cm

» BC = Base = 8 cm

» AC = Hypotenuse

We see here we have to find the length of AC . AC is the Hypotenuse because it is the opposite side to the right angle that is ∠ABC = 90° .

Using the Pythagoras Theorem here, we get,

$\\\;\sf{:\rightarrow\;\;(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}$

By applying values here, we get

$\\\;\sf{:\Longrightarrow\;\;(AC)^{2}\;=\;\bf{(BC)^{2}\;+\;(AB)^{2}}}$

$\\\;\sf{:\Longrightarrow\;\;(AC)^{2}\;=\;\bf{(8)^{2}\;+\;(6)^{2}}}$

$\\\;\sf{:\Longrightarrow\;\;(AC)^{2}\;=\;\bf{64\;+\;36}}$

$\\\;\sf{:\Longrightarrow\;\;(AC)^{2}\;=\;\bf{100}}$

$\\\;\sf{:\Longrightarrow\;\;(AC)\;=\;\bf{\sqrt{100}}}$

$\\\;\bf{:\Longrightarrow\;\;AC\;=\;\bf{\green{10\:\;cm}}}$

✒ AC = 10 cm

$\\\;\underline{\boxed{\tt{Length\;\;of\;\;AC\;=\;\bf{\purple{10\;\;cm}}}}}$

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★ More to know :-

• Why is this answer correct ?

We know that a triangle is formed only when the sum of two sides is more than the third side . Here (8 + 6) > 10 . This is means this combination is correct. Even if AC is Hypotenuse its the longest side which is proved as AC > 6,8

So our answer is correct .

• Other Formulas Related to Triangle ::

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

$\\\:\sf{\leadsto\;\;Area\;of\;Triangle\;=\;\sqrt{s(s\;-\;a)(s\:-\:b)(s\:-\:c)}}$

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

$\\\;\sf{\leadsto\;\;Perimeter\;of\;Triangle\;=\;Sum\;of\;all\;sides}$

• Formulas related to other plane (2 D) figures ::

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

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

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

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

Answer 2

Step-by-step explanation:

$\text{\Large\underline{\red{Given:-}}}$

$\sf Given = \begin{cases} \sf{●\ Length\ of\ AB\ =\ 6cm.} \\ \\ \sf{●\ Length\ of\ BC\ =\ 8cm.} \\ \\ \sf{●\ And\ it\ is\ Right\ angled\ at\ B.} \end{cases}$

$\text{\Large\underline{\orange{To\ find:-}}}$

$\sf To\ find = \begin{cases} \sf{●\ In\ this\ question\ we\ have\ to\ find\ the\ length\ of\ AC.} \end{cases}$

$\text{\Large\underline{\purple{Solution:-}}}$

According to Pythagoras Theorem:-

● Addition of square of side AB and BC will equal to square of third side AC that is:-

$\sf{(AB)²\ +\ (BC)²\ =\ (AC)²}$

$\text{\large\underline{\blue{By\ putting\ the\ values:-}}}$

$\Longrightarrow$ $\sf{(8)²\ +\ (6)²\ =\ (AC)²}$

$\Longrightarrow$ $\sf{64\ +\ 36\ =\ (AC)²}$

$\Longrightarrow$ $\sf{100\ =\ (AC)².}$

Now here:-

$\Longrightarrow$ $\sf{(AC)²\ =\ 100}$

$\Longrightarrow$ $\sf{AC\ =\ \sqrt{100}}$

$\Longrightarrow$ $\sf{AC\ =\ 10cm.}$

Hence:-

● Length of the side AC is 10cm.

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