Question

In the given figure , ∠ = ∠ , then prove that ∠ = ∠ (3)





P

S Q R T​

Answer 1

Answer:

Sum of angles on a straight line=180 ∘

⟹∠PAE=180 ∘ −∠1

Similarly,∠PEA=180 ∘ −∠5

In△PAE

∠PAE+∠PEA+∠APE=180 ∘

⟹180−∠1+180−∠5+∠APE=180 ∘

⟹∠APE=∠1+∠5−180 ∘ =∠P

Similarly,∠BSC=∠2+∠3−180 ∘ =∠S

∠DRC=∠3+∠4−180 ∘ =∠R

∠DQE=∠4+∠∠5−180 ∘ =∠Q

∠ATS=∠1+∠2−180 ∘ =∠T

⟹∠P+∠Q+∠R+∠S+∠T

=∠1+∠2+∠3+∠4+∠2+∠3+∠4+∠5+∠1+∠5−(180×5)

=2(∠1+∠2+∠3+∠4+∠5)−900−(1)

From pentagon ABCDE,

Sum of angles of pentagon=((2×5)−4)×90 ∘

(∵ Sum of angles=(2n−4)×90 ∘ )

=6×90=540 ∘

⟹∠P+∠Q+∠R+∠S+∠T=2(540 ∘ )−900 ∘

=1080 ∘ −900 ∘

=180 ∘

=2×90

=2Rightangles

Answer 2

Answer:

Pressure exerted by water on the bottom of a deep dam(Hydrostatic pressure) is 120000 Pa.

Step-by-step Explanation :

GivEn :

Density of water, ϱ = 10³kg/m³

Height = 12 m

g = 10m/s²

To find :

Pressure exerted by water on the bottom of a deep dam(Hydrostatic pressure) =

SoluTion :

We know that,

\begin{gathered}\begin{gathered}\sf \longrightarrow \: Hydrostatic \: pressure = \rho \times g \times h \\ \\\end{gathered}\end{gathered}

⟶Hydrostaticpressure=ρ×g×h

Substituting the values,

\begin{gathered}\begin{gathered}\sf \longrightarrow \: Hydrostatic \: pressure = {10}^{3} \times 10 \times 12 \\ \\\end{gathered}\end{gathered}

⟶Hydrostaticpressure=10

3

×10×12

\begin{gathered}\begin{gathered}\sf \longrightarrow \: Hydrostatic \: pressure = {10}^{4} \times 12 \\ \\\end{gathered}\end{gathered}

⟶Hydrostaticpressure=10

4

×12

\begin{gathered}\begin{gathered}\sf \longrightarrow \: Hydrostatic \: pressure = 10000 \times 12 \\ \\\end{gathered}\end{gathered}

⟶Hydrostaticpressure=10000×12

\begin{gathered}\begin{gathered}\sf \therefore { \blue{Hydrostatic \: pressure = 120000 \: Pa}} \\ \\\end{gathered}\end{gathered}

∴Hydrostaticpressure=120000Pa

Hence, Pressure exerted by water on the bottom of a deep dam(Hydrostatic pressure) is 120000 Pa.