Question

the foot of a ladder is 6m away from its wall and its top reaches a window 8m above the ground find the lenght of the ladder.

Answer 1

Given : The foot of a ladder is 6m away from its wall and it's top reaches a window 8m above the ground .

Exigency To Find : The lenght of the ladder.

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❍ Let's Consider AB as the Distance from the wall to foot of ladder & BC as it's top that reaches a window above the ground .

⠀⠀⠀⠀⠀ SO ,

⠀AC will be the length of ladder & consider it as x m .

[ Note : Kindly refer the given attachment ( image ) ]

Diagram :

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\:\large \bf 6m\:\: \:}\put(2.8,.3){\large\bf 8\:m}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C} \end{picture}\\\\ \qquad \qquad \maltese \:\bf Question \: Diagram\:$

Now ,

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\bigstar\:\bf By \:Pythagoras\:Theorem\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ (Hypotenuse)^2 = (\:Side\:_1\:)\:^2 + (\:Side\:_2 \:)\:^2 }\bigg\rgroup$

Now ,

⠀⠀⠀⠀⠀In $\triangle$ ABC :

$\qquad:\implies \sf (AC)^2 = (AB)^2 + (BC)^2$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad:\implies \sf (AC)^2 = (AB)^2 + (BC)^2$

$\qquad:\implies \sf (x)^2 = (6)^2 + (8)^2$

$\qquad:\implies \sf (x)^2 = 36 + 64$

$\qquad:\implies \sf (x)^2 = 100$

$\qquad:\implies \sf x = \sqrt {100}$

$\qquad:\implies \bf x = 10$

$\qquad :\implies \frak{\underline{\purple{\:x = 10 \:m }} }\:\:\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The\:Length \:of\:Ladder \:is\:\bf{10\:m}}}}$

[ Note : For Diagram see answer from web ( brainly.in ) ]

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Answer 2

${\large{\pmb{\sf{\underline{RequirEd \: Solution}}}}}$

${\bigstar \:{\pmb{\sf{\underline{Understanding \: the \: question}}}}}$

⋆ This question says that we have to find out the length of the ladder if the foot of a ladder is 6 metres away from its wall and its top reaches a window 8 metres above the ground.

${\bigstar \:{\pmb{\sf{\underline{Given \: that}}}}}$

⋆ The foot of a ladder is 6 metres away from its wall.

⋆ The top of ladder reaches a window 8 metres above the ground.

${\bigstar \:{\pmb{\sf{\underline{To \: find}}}}}$

⋆ The length of the ladder

${\bigstar \:{\pmb{\sf{\underline{Solution}}}}}$

⋆ The length of the ladder = 10 metres

${\bigstar \:{\pmb{\sf{\underline{Using \: concept}}}}}$

⋆ Phythagoras theorm

${\bigstar \:{\pmb{\sf{\underline{Using \: formula}}}}}$

${\small{\underline{\boxed{\sf{\star \: (Hypotenuse)^{2} \: = (Base)^{2} + (Perpendicular)^{2}}}}}}$

${\bigstar \:{\pmb{\sf{\underline{Knowledge \: requied}}}}}$

⋆ Phythagoras Theorm: Phythagoras Theorm is applied only in right angled triangles. Phythagoras Theorm states that in a right angled triangle, the square of the hypotenuse is always equal to the sum of the square of base and the sum of the square of perpendicular of the right angled triangle. Perpendicular is also termed as height

⋆ A triangle is right angled triangle if the square of the hypotenuse is equal to the sum of the square of base and the sum of the square of perpendicular.

• Kindly see the attachment 1st for understanding about phythagoras theorm.

⋆ Right angled triangle is always equal to 90° and the other two triangle's are acute.

⋆ The side that is opposite to the right angle is known as hypotenuse.

⋆ The other two sides are known as base or perpendicular/height or legs of the right angled triangle.

⋆ Hypotenuse is the longest side of a right angled triangle.

${\bigstar \:{\pmb{\sf{\underline{Full \: Solution}}}}}$

~ To solve this question we have to use phythagoras theorm, just have to put the values according to the formula.

$:\implies \sf (Hypotenuse)^{2} \: = (Base)^{2} + (Perpendicular)^{2} \\ \\ :\implies \sf (Hypotenuse)^{2} \: = (6)^{2} + (8)^{2} \\ \\ :\implies \sf (Hypotenuse)^{2} \: = 36 + 64 \\ \\ :\implies \sf (Hypotenuse)^{2} \: = 100 \\ \\ :\implies \sf Hypotenuse \: = \sqrt{100} \\ \\ :\implies \sf Hypotenuse \: = 10 \: metres$

• You can see the diagram regards this question from attachment 2nd too.

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf 8 \: m}\put(2.8,.3){\large\bf 6 \: m}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf Z}\put(.8,.3){\large\bf X}\put(5.8,.3){\large\bf Y}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf \Theta }\end{picture}$