Question

10. A rectangular piece of land is to be fenced for an exhibition. The length of the exhibition land
is 250 m and its breadth is 200 m. We need to keep an entrance of 10 m and an emergency
exit of 5 m. If the cost of this special fencing is 175 per metre, what will be the total cost
of fencing?​

Answer 1

Answer:

✯ Given :-

Two poles of equal heights are standing opposite each other on either sides of the road, which is 80 m wide.

From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively.

✯ To Find :-

What is the height of the poles and the distance of the point from the poles.

✯ Solution :-

» Let, AD and BC be two poles of equal heights h m.

» And, P be a point on the road such that AP = x m, BP = (80 - x) m.

» Given that, ∠APD = 60°, ∠BPC = 30°

➙ In the ∆APD, we have,

⇒ tan60° = \dfrac{AD}{AP}

AP

AD

⇒ √3 = \dfrac{h}{x}

x

h

⇒ x = \dfrac{h}{√3}

√3

h

..... Equation no (1)

➙ In the ∆BPC we have,

⇒ tan30° = \dfrac{BC}{BP}

BP

BC

⇒ \dfrac{1}{√3}

√3

1

= \dfrac{h}{80 - x}

80−x

h

⇒ 80 - x = √3h

⇒ x = 80 - √3h .... Equation no (2)

➣ By solving the equation no (1) and (2) we get,

⇒ \dfrac{h}{√3}

√3

h

= 80 - √3 h

⇒ h = √3 (80 - √3h)

⇒ h = 80√3 - 3h

⇒ 4h = 80√3

⇒ h = \sf\dfrac{\cancel{80√3}}{\cancel{4}}

4

80√3

\dashrightarrow⇢ h = 20√3

➣ Putting h = 20√3 in the equation no (1) we get,

⇒ x = \dfrac{h}{√3}

√3

h

⇒ x = \dfrac{20√3}{√3}

√3

20√3

➠ x = 20 m

⋆ And,

AP = x = 20 m

BP = 80 - x = 80 - 20 = 60 m

\therefore∴ The height of the poles is 20√3 m .

\therefore∴ The distance of the point from the poles is 20 m and 60 m .

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