Question

D32.5
15: A 5m long ladder is placed leaning
towards a vertical wall such that it
reaches the wall at a point 4m high.If foot
of ladder is moved 1.6m towards the
wall,then the distamce by which the top of
the ladder would slide upwards on the
wall is​

Answer 1

Answer:

distance the ladder slide upwards=7m

Step-by-step explanation:

length of ladder= 5m

height of wall till the ladder reaches= 4m

base= let it be x.

distance from wall to foot = x -1.6

distance the ladder slide upwards= let it be h .

by Pythagoras theorem

5²=x²+4²

25-16= x²

9= x²

√9= x

x= 3m.

therefore... now .

height = h+4m

base= 3-1.6=1.4m

length of ladder= 5m

by Pythagoras theorem

5²=( h+4)²+ 1.4²

25= h²+16+8h+1.96

25-16-1.96= h²+8h

7.04= h²+8h

h²+8h-7.04

h²+8h-7

h²+7h+h-7

h(h-7)+1(h-7)

h+1 and h-7

therefore h=7m

distance the ladder slide upwards=7m