Question

Solve this question using Phythagoras Theorem​

Answer 1

$\huge\tt\underline{Answer!!!} \\ \\ \mathtt{Hypotenuse \: =25 } \: \: \: \: \\ \mathtt{perpendicuar = 24} \\ \mathtt{base = \: \: ? } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ { \underline{ \boxed{ \mathtt{Pythagoras \: theoram \colon {h}^{2} = {p}^{2} + {b}^{2} }}}} \\ \\ \rightarrow \sf \: (25) ^{2} = (24) ^{2} + {b}^{2} \\ \\ \rightarrow \sf \: 625 = 576 + {b}^{2} \: \: \: \: \: \\ \\ \rightarrow \sf \: 625 - 576 = {b}^{2} \: \: \: \\ \\ \rightarrow \sf \: 49 = {b}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rightarrow \: \sf b = 7 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{Base = 7 \: cm} \: \: \: \: \: \: \: \: \:$

$\mathbb{♡♡ \: SABRINA \: ♡♡}$

Answer 2

Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.1,2){\sf{\large{24}}}\put(9,0.7){\sf{\large{?}}}\put(9.4,1.9){\sf{\large{25 }}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

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❍ Let's consider measure of Base , Perpendicular and Hypotenuse be in units .

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$\frak{Given\:} \begin{cases} \sf{Base = ?}\\\sf{Perpendicular \:= 24 \:unit}\\\sf{Hypotenuse \:=\:25\:unit }\end{cases}$

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Need To Find : Base of Triangle.

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$\bigstar \underline {\bf{ By\:Pythagoras \:Theorem \::}}$

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$\qquad\qquad:\implies {\sf{ (Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2} }}$

Or ,

$\qquad\qquad:\implies {\sf{ (AC)^{2} = (CB)^{2} + (AB)^{2} }}$

$\qquad\qquad:\implies {\sf{ (AC)^{2} - (AB)^{2} = (CB)^{2} }}$

$\qquad\qquad:\implies {\sf{ (25)^{2} - (24)^{2} = (CB)^{2} }}$

$\qquad\qquad:\implies {\sf{ 625 - 576 = (CB)^{2} }}$

$\qquad\qquad:\implies {\sf{ 49 = (CB)^{2} }}$

$\qquad\qquad:\implies {\sf{ \sqrt {49} = CB }}$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  CB \:or \:Base \: = 7\: unit}}}}\:\bf{\bigstar}$

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Therefore,

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⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Hence ,\:CB\:or\:Base \:of\:Right \:Angled \:Triangle \:is\:\bf{7\:}}}}$

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