For traffic control, a CCTV camera is fixed on a straight and vertical pole. The camera can see 113m distance straight from the top. If the area visible by the camera around the pole is 39424${m^2}$, then find the height of the pole.
Answer
335.1k+ views
Hint: In this particular question use the concept that the area visible by the camera around the pole is always in circular shape, so equate the given area to the area of the circle and calculate the radius of the circle, later on in the solution use the concept of Pythagoras theorem so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Let OA be the vertical pole as shown in the above figure, at point A, CCTV camera is fixed and the reach of the CCTV camera is up to 113m as shown in the above figure.
Let the reach is up to point B, so AB = 113m (see figure).
Let the height of the vertical pole be h meters.
Therefore, OA = h meters.
Now it is given that the area visible by the camera around the pole is 39424${m^2}$.
As we all know that the visible area is always around a circle as shown in the figure, and the area of the circle is given as $\pi {r^2}$ square units, where r is the radius of the circle.
Therefore, $\pi {r^2} = 39424$
$ \Rightarrow {r^2} = \dfrac{{39424}}{\pi } = \dfrac{{39424}}{{\dfrac{{22}}{7}}} = 12544 = {\left( {112} \right)^2}$, $\left[ {\because \pi = \dfrac{{22}}{7}} \right]$
$ \Rightarrow r = 112$Meters.
As the pole is vertical so it makes a 90 degree with the radius of the circle.
So in triangle OAB apply Pythagoras theorem we have,
Now as we know that according to the Pythagoras theorem that in a right triangle the sum of the square of the base and perpendicular is equal to the square of the hypotenuse.
$ \Rightarrow {\left( {{\text{base}}} \right)^2} + {\left( {{\text{perpendicular}}} \right)^2} = {\left( {{\text{hypotenuse}}} \right)^{^2}}$................. (1)
Now in triangle OAB, base = OB, perpendicular = OA and hypotenuse = AB
Now from equation (1) we have,
$ \Rightarrow {\left( {{\text{OB}}} \right)^2} + {\left( {{\text{OA}}} \right)^2} = {\left( {{\text{AB}}} \right)^{^2}}$
Now substitute the values we have,
$ \Rightarrow {\left( {{\text{112}}} \right)^2} + {\left( {\text{h}} \right)^2} = {\left( {{\text{113}}} \right)^{^2}}$
$ \Rightarrow {\left( {\text{h}} \right)^2} = {\left( {{\text{113}}} \right)^{^2}} - {\left( {{\text{112}}} \right)^2} = 225 = {\left( {15} \right)^2}$
$ \Rightarrow h = 15$ Meters.
So the height of the vertical pole is 15 meters.
Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of the area of the circle which is given as $\pi {r^2}$ square units, and always recall the definition of the Pythagoras theorem which is stated above then simplify substitute the values as above and simplify we will get the required height of the pole.
Complete step-by-step answer:

Let OA be the vertical pole as shown in the above figure, at point A, CCTV camera is fixed and the reach of the CCTV camera is up to 113m as shown in the above figure.
Let the reach is up to point B, so AB = 113m (see figure).
Let the height of the vertical pole be h meters.
Therefore, OA = h meters.
Now it is given that the area visible by the camera around the pole is 39424${m^2}$.
As we all know that the visible area is always around a circle as shown in the figure, and the area of the circle is given as $\pi {r^2}$ square units, where r is the radius of the circle.
Therefore, $\pi {r^2} = 39424$
$ \Rightarrow {r^2} = \dfrac{{39424}}{\pi } = \dfrac{{39424}}{{\dfrac{{22}}{7}}} = 12544 = {\left( {112} \right)^2}$, $\left[ {\because \pi = \dfrac{{22}}{7}} \right]$
$ \Rightarrow r = 112$Meters.
As the pole is vertical so it makes a 90 degree with the radius of the circle.
So in triangle OAB apply Pythagoras theorem we have,
Now as we know that according to the Pythagoras theorem that in a right triangle the sum of the square of the base and perpendicular is equal to the square of the hypotenuse.
$ \Rightarrow {\left( {{\text{base}}} \right)^2} + {\left( {{\text{perpendicular}}} \right)^2} = {\left( {{\text{hypotenuse}}} \right)^{^2}}$................. (1)
Now in triangle OAB, base = OB, perpendicular = OA and hypotenuse = AB
Now from equation (1) we have,
$ \Rightarrow {\left( {{\text{OB}}} \right)^2} + {\left( {{\text{OA}}} \right)^2} = {\left( {{\text{AB}}} \right)^{^2}}$
Now substitute the values we have,
$ \Rightarrow {\left( {{\text{112}}} \right)^2} + {\left( {\text{h}} \right)^2} = {\left( {{\text{113}}} \right)^{^2}}$
$ \Rightarrow {\left( {\text{h}} \right)^2} = {\left( {{\text{113}}} \right)^{^2}} - {\left( {{\text{112}}} \right)^2} = 225 = {\left( {15} \right)^2}$
$ \Rightarrow h = 15$ Meters.
So the height of the vertical pole is 15 meters.
Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of the area of the circle which is given as $\pi {r^2}$ square units, and always recall the definition of the Pythagoras theorem which is stated above then simplify substitute the values as above and simplify we will get the required height of the pole.
Last updated date: 26th Sep 2023
•
Total views: 335.1k
•
Views today: 4.35k
Recently Updated Pages
What do you mean by public facilities

Slogan on Noise Pollution

Paragraph on Friendship

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

What is the Full Form of ILO, UNICEF and UNESCO

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

What is the IUPAC name of CH3CH CH COOH A 2Butenoic class 11 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

The dimensions of potential gradient are A MLT 3A 1 class 11 physics CBSE

Define electric potential and write down its dimen class 9 physics CBSE

Why is the electric field perpendicular to the equipotential class 12 physics CBSE
