Question

plz do fast.........................​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{Given}}}}$

• Diameter of circular park = 17 m
• Distance difference between pole and gate A , B = 7 m

$\Large{\underline{\mathfrak{\bf{Find}}}}$

• Distance from pole to gate

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

Let, P be the position of the pole and A and B be the opposite fixed gate .

In right ∆ APB ,

• PA = a
• PB = b

So, A/c to question,

• PA - PB = 7

Or,

• a - b = 7 .........(1)

In right ∆ APB ,

$\large{\underline{\mathfrak{\bf{By\:Pythagoras\:Theorem}}}}$

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

But, here

• Hypotenuse (AB) = 17
• Perpendicular (PA) = a
• Base (PB) = b

So,

$:\implies\sf{\:(17)^2\:=\:a^2+b^2} \\ \\ :\implies\sf{\:a^2+b^2\:=\:289....(2)}$

keep value of a by equ(1),

• a = b + 7

$:\implies\sf{\:(b+7)^2+b^2\:=\:289} \\ \\ :\implies\sf{\:b^2+49+14b+b^2\:=\:289} \\ \\ :\implies\sf{\:2b^2+14b-240\:=\:0} \\ \\ :\implies\sf{\:b^2+7b-120\:=\:0} \\ \\ :\implies\sf{\:b^2+15b-8b-120\:=\:0} \\ \\ :\implies\sf{\:b(b+15)-8(b+15)\:=\:0} \\ \\ :\implies\sf{\:(b-8)(b+15)\:=\:0} \\ \\ :\implies\sf{\:(b-8)\:=\:0\:\:or\:\:(b+15)\:=\:0} \\ \\ :\implies\sf{\:b\:=\:8\:\:\:Or\:\:\:b\:=\:-15}$

But, length is always be positive

so, b = -15 negligible .

Hence, b = 8 is right value .

Keep value of b in equ(1),

$:\implies\sf{\:a-8\:=\:7} \\ \\ :\implies\sf{\:a\:=\:7+8} \\ \\ :\mapsto\sf{\:a\:=\:15}$

$\Large{\underline{\mathfrak{\bf{Hence}}}}$

• Distance from the gate A to pole P (PA) = 15 m
• Distance from the gate B to pole P (PB) = 8 m

______________________

Answer 2

Answer:

SOLUTION :-

.

Let P be the required location on the boundary of a circular park such that its distance from gate B is x metre that is BP x metres.

Then, AP = x + 7

In the right triangle ABP we have by using Pythagoras theorem

\

$AP^2 + BP^2 = AB^2 \\ (x + 7)^2 + x^2 = (13)^2 \\ x^2 + 14x + 49 + x^2 = 169 \\ 2x^2 + 14x + 49 - 169 = 0 \\ 2x^2 + 14x - 120 = 0 \\ 2(x^2 + 7x - 60) = 0 \\ x^2 + 7x - 60 = 0 \\ x^2 + 12x - 5x - 60 = 0 \\ x(x + 12) - 5(x - 12) = 0 \\ (x + 12)(x - 5) = 0 \\ x + 12 = 0 \\ x = -12 \\ or \\ x - 5 = 0 \\ x = 5$

But the side of right triangle can never be negativeTherefore, x = 5

Hence, 

P is at a distance of 5 metres from the gate B.

⇒ BP = 5m

Now,

AP = (BP + 7)m = (5 + 7)m = 12 m

∴ The pole has to be erected at a distance 5 mtrs from the gate B and 12 m from the gate A.