Question

Find the length of side a in the following right-
angled triangles.
​

Answer 1

☯ We need to find side a in both right - angled triangles A and B.

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$\underline{\bigstar\:\boldsymbol{By\:Using\:Pythagoras\:Theorem\::}}$

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

• In First triangle A)

⠀⠀⠀

$\frak{Here} \begin{cases} & \sf{Base = \frak{6\:cm}} \\ & \sf{Perpendicular = \frak{8\:cm}} \\ & \sf{Hypotenuse = \frak{a}} \end{cases}$

$\because$ We know that, The the side opposite the right angle is Hypotenuse.

⠀⠀⠀

$\bf{\dag}\;{\underline{\frak{Putting\:given\:\:values\:,}}}$

$:\implies\sf a^2 = 6^2 + 8^2\\ \\ \\ :\implies\sf a^2 = 36 + 64\\ \\ \\ :\implies\sf a^2 = 100$

$:\implies\sf \sqrt{a^2} = \sqrt{100}\qquad\qquad\bigg[\:Taking\:Sqrt.\:Both\;sides\:\bigg]$

$:\implies{\underline{\boxed{\frak{\purple{a = 10}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Hence,\:the\:length\:of\;side\:a\:{\textsf{\textbf{10\:cm}}}.}}}$

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• In First triangle B)

⠀⠀⠀

$\frak{Here} \begin{cases} & \sf{Base = \frak{5\:cm}} \\ & \sf{Perpendicular = \frak{12\:cm}} \\ & \sf{Hypotenuse = \frak{a}} \end{cases}$

$\bf{\dag}\;{\underline{\frak{Putting\:given\:values\:,}}}$

$:\implies\sf a^2 = 5^2 + 12^2\\ \\ \\ :\implies\sf a^2 = 25 + 144\\ \\ \\ :\implies\sf a^2 = 169$

$:\implies\sf \sqrt{a^2} = \sqrt{169}\qquad\qquad\bigg[\:Taking\:Sqrt.\:Both\;sides\:\bigg]$

$:\implies{\underline{\boxed{\frak{\purple{a = 13}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Hence,\:the\:length\:of\;side\:a\: {\textsf{\textbf{13\:cm}}}.}}}$

Answer 2

$\huge\colorbox{Orange}{Answer}$

Part(A)-

$\huge\large\mathcal\red{We\:have\:to\:find\:the\:length\:of\:(a)}$

Given-

$\huge\large\color{purple}{Base=6cm/:Perpendicular=8cm}$

Solution)-

$:\:(H)²=(P)²+(B)²$

$:\:(H)²=(8)²+(6)²$

$:\:(H)²=64+36$

$:\:(H)²=100$

$:\:(H)=√100$

$:\:(A)=10$

$\huge\large\mathcal\pink{(a)=10\:cm}$

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Part(B)-

$\huge\large\color{purple}\boxed{We\:have\:to\:again\:find\:the\:value\:of\:(a)}$

Given:

$\huge\large\mathcal\pink{Base=5cm\:Perpendicular=12cm}$

Solution:

$:\:(H)²=(P)²+(B)²$

$:\:(H)²=(12)²+(5)²$

$:\:(H)²=(144)+(25)$

$:\:(H)²=169$

$:\:(H)=√169$

$:\:(H)=13$

$\huge\large\mathcal\pink{(a)=13cm}$

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Hope it helps:)