Question

Please find the solution ​

Answer 1

Answer:

the solution of this question is 257

Answer 2

$\sf\huge\bold{Answer:}$

Given :

A right angled triangle ∆ ABF

where

A=Standby of Batsman

B=point where ball is hit.

F=fielder's standby

AF=50m

AB=40m

To find :

Distance the fielder has to cover to pick the ball from point B i.e BF

Solution :

As we know, ∆AFB is right angled at B

$\fbox\purple{Using Pythagoras Property }$

$\sf \: :⟶ H²=B²+P²$

where, H = 50m

P = 40m

B =?

$\sf:⟶ (50m)²=B²+(40m)²$

$\sf:⟶ 2500m {}^{2} =B²+1600m²$

$\sf:⟶ B² = 2500m {}^{2} - 1600 {m}^{2}$

$\sf:⟶ B {}^{2} = 900m {}^{2}$

$\sf:⟶ B = \sqrt{900 {m}^{2} }$

$\sf:⟶ B = 30m$

Hence, BF = 30m

Distance the fielder has to cover to pick the ball from point B is 30 m

$\sf\huge\purple{30m}$