For this case we have the following equation:
w = || F || • || PQ || costheta
Where,
|| F ||: force vector module
|| PQ ||: distance module
costheta: cosine of the angle between the force vector and the distance vector.
Substituting values:
w = (60) * (100) * (cos (45))
w = 4242.640687 lb.ft
Answer:
The work done pushing the lawn mower is:
w = 4242.640687 lb.ft
Step-by-step explanation:
it filled up half the circle (up to the center point) - if we had a full circle. but a little bit is cut off (below AB).
what we see is that the shaded area is the sum of the area of the triangle AOB and 2 equally sized circle segment areas left and right of AOB.
since we are dealing with a half-circle, we have 180° in total. 120° are taken by AOB, so, that leaves us with 180-120 = 60° for both circle segments (so, one has an angle of 30°).
and 2×30° = 1×60°, so we can calculate the area of one 60° segment instead of two 30° segments.
AOB is an isoceles triangle (the legs are equally long, and therefore also the 2 side angles are equal).
the area of this triangle AOB is
1/2 × a × b × sin(C) = 1/2 × 3 × 3 × sin(120) =
= 3.897114317... m²
a circle segment area of 60° is 60/360 = 1/6 of the full circle area (as a full circle = 360°).
so, it's area is
pi×r² × 1/6 = pi×3²/6 = pi×3/2 = 4.71238898... m²
so, the total area of the shaded area is
3.897114317... m² + 4.71238898... m² =
= 8.609503297... m²
£1400
A reduction of 40% means that £840 is 60% of the original price
Divide £840 by 60 to find 1% then multiply by 100 to find original price
original price = £840 × 
= £1400
