Question

select the correct answer. a company makes the forks for forklift trucks. each fork has the dimensions shown. the top of each fork is painted. how much area is painted on the top of each fork? 90 mm 900 mm 40 mm 50 mm 40 mm 60° a. 83,425 mm² b. 2,425 mm² c. 84,118 mm² d. 81,000 mm²

Explanation:

Step1: Calculate area of rectangle

The rectangular part has length $l = 900$ mm and width $w=90$ mm. The area of a rectangle $A_{rect}=l\times w$. So $A_{rect}=900\times90 = 81000$ $mm^{2}$.

Step2: Calculate area of triangle

The triangle at the end has base $b = 50$ mm and height $h$ which can be found using trigonometry. Given the angle $\theta = 60^{\circ}$ and the non - base side of the right - triangle formed (adjacent to the $60^{\circ}$ angle) is $40$ mm. The height $h = 40\times\tan60^{\circ}=40\sqrt{3}$ mm. The area of a triangle $A_{tri}=\frac{1}{2}\times b\times h=\frac{1}{2}\times50\times40\sqrt{3}=1000\sqrt{3}\approx1732$ $mm^{2}$.

Step3: Calculate total area

The total area $A = A_{rect}+A_{tri}=81000 + 1732=82732$ $mm^{2}$. But if we consider another way of calculating the triangle area with base as the side of length 50 mm and height calculated from the right - triangle formed by the 40 - mm side and the 60 - degree angle.
We have two congruent right - triangles at the end. The combined area of the two right - triangles:
The base of the combined triangle is 50 mm and the height (from the 40 - mm side and 60 - degree angle) is $h = 40\times\tan60^{\circ}=40\sqrt{3}$ mm. Area of the combined triangle $A_{tri}= \frac{1}{2}\times50\times40\sqrt{3}\times2 = 2000\sqrt{3}\approx3464$ $mm^{2}$.
The area of the rectangle is $A_{rect}=90\times900 = 81000$ $mm^{2}$.
The total area $A=A_{rect}+A_{tri}=81000 + 3464=84464$ $mm^{2}$.
If we assume a different approach for the triangular part:
The triangular part at the end can be considered as an isosceles triangle with base 50 mm and the two equal sides of length 40 mm. Using Heron's formula, first find the semi - perimeter $s=\frac{40 + 40+50}{2}=65$ mm.
The area of the triangle $A_{tri}=\sqrt{s(s - a)(s - b)(s - c)}=\sqrt{65(65 - 40)(65 - 40)(65 - 50)}=\sqrt{65\times25\times25\times15}=\sqrt{65\times9375}=\sqrt{609375}\approx780$ $mm^{2}$.
The area of the rectangle is $A_{rect}=90\times900=81000$ $mm^{2}$.
The total area $A = A_{rect}+A_{tri}=81000+780 = 81780$ $mm^{2}$.
A more accurate way:
The rectangular part has area $A_{1}=90\times900 = 81000$ $mm^{2}$.
The triangular part: We have a triangle with base $b = 50$ mm and height $h$ calculated from the right - triangle with side 40 mm and angle $60^{\circ}$. The height $h = 40\times\sqrt{3}$ mm. The area of the triangle $A_{2}=\frac{1}{2}\times50\times40\times\sqrt{3}=1000\sqrt{3}$ mm².
The total area $A=A_{1}+A_{2}=81000 + 1000\sqrt{3}\approx81000+1732=82732$ $mm^{2}$.
If we consider the two right - triangles at the end more precisely:
The area of the rectangle $A_{r}=90\times900 = 81000$ $mm^{2}$.
The two right - triangles together: base $b = 50$ mm, height $h = 40\times\sqrt{3}$ mm. Area of the two - triangle combination $A_{t}=2\times\frac{1}{2}\times\frac{50}{2}\times40\times\sqrt{3}=2000\sqrt{3}\approx3464$ $mm^{2}$.
The total area $A = 81000+3464 = 84464$ $mm^{2}$.
The closest answer to our calculations is C. 84,118 $mm^{2}$.

Answer:

C. 84,118 $mm^{2}$