A ladder is placed against a wall such that its foot is is at a distan

1.	A ladder is placed against a wall such that its foot is is at a distance of 8 m from the wall and its top reaches a window 6 metre above the ground find the length of the ladder
Answer» Step-by-step explanation: Step-by-step explanation: Answer: PR is the ladder. QR is the DISTANCE between BASE and ladder. PQ is height of the window. $\begin{gathered}\bigstar \: \underline{\sf Now,by \: using \: Phythagoras \: theorem \: we \: get : } \\\end{gathered}$ ★ Now,by using Phythagoras theorem we get: $\begin{gathered}:\implies \sf (Hypotenuse)^{2} = (Perpendicular)^{2} + (Base)^{2} \\ \\\end{gathered}$ :⟹(Hypotenuse) 2 =(Perpendicular) 2 +(Base) 2 $\begin{gathered}:\implies \sf (PR)^2 = (PQ)^2 + (QR)^2 \\ \\\end{gathered}$ :⟹(PR) 2=(PQ) 2 +(QR) 2 $\begin{gathered}:\implies \sf PR^2 = (6 \: m)^{2} + (8 \: m)^{2} \\ \\\end{gathered}$ :⟹PR 2 =(6m) 2 +(8m) 2 $\begin{gathered}:\implies \sf PR^2 = 36 + 64 \\ \\\end{gathered}$ :⟹PR 2 =36+64 $\begin{gathered}:\implies \sf PR= \sqrt{36 + 64} \\ \\\end{gathered}$ :⟹PR= 36+64 $\begin{gathered}:\implies \sf PR= \sqrt{100 \: m} \\ \\\end{gathered}$ :⟹PR= 100M $\begin{gathered}:\implies \underline{ \boxed{ \sf PR= 10 \: meter}} \\\end{gathered}$ :⟹ PR=10meter $\begin{gathered}\therefore\:\underline{\textsf{Length of the ladder is \textbf{10 meter}}}. \\\end{gathered}$ ∴ Length of the ladder is 10 meter .

A ladder is placed against a wall such that its foot is is at a distance of 8 m from the wall and its top reaches a window 6 metre above the ground find the length of the ladder

Answer»

Step-by-step explanation:

Step-by-step explanation:

Answer:

PR is the ladder.

QR is the DISTANCE between BASE and ladder.

PQ is height of the window.

$\begin{gathered}\bigstar \: \underline{\sf Now,by \: using \: Phythagoras \: theorem \: we \: get : } \\\end{gathered}$

★

Now,by using Phythagoras theorem we get:

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

:⟹(Hypotenuse)

=(Perpendicular)

+(Base)

$\begin{gathered}:\implies \sf (PR)^2 = (PQ)^2 + (QR)^2 \\ \\\end{gathered}$

:⟹(PR) 2=(PQ) 2 +(QR) 2

$\begin{gathered}:\implies \sf PR^2 = (6 \: m)^{2} + (8 \: m)^{2} \\ \\\end{gathered}$

:⟹PR 2 =(6m) 2 +(8m) 2

$\begin{gathered}:\implies \sf PR^2 = 36 + 64 \\ \\\end{gathered}$

:⟹PR

=36+64

$\begin{gathered}:\implies \sf PR= \sqrt{36 + 64} \\ \\\end{gathered}$

:⟹PR=

36+64

$\begin{gathered}:\implies \sf PR= \sqrt{100 \: m} \\ \\\end{gathered}$

:⟹PR=

$\begin{gathered}:\implies \underline{ \boxed{ \sf PR= 10 \: meter}} \\\end{gathered}$

:⟹

PR=10meter

$\begin{gathered}\therefore\:\underline{\textsf{Length of the ladder is \textbf{10 meter}}}. \\\end{gathered}$