Question

The rectangular front windshield of the bus has the length of $1.5\sqrt{3}$ m and the width of 1.5 m . The wipers are attached to the bottom corners. If the length of each wiper is equal to the width of the windshield, find the percentage of the windshield that wipers can reach.
(Solve this task without using Trigonometry)

Answer 1

Area covered by wiper=area of arc(A0D)+area of traingleAOB+area of arc(BOC)

We know AO and OB are wipers,whose lengths are 1.5

In ∆ABC

AB^2+BC^2=AC^2

(1.5√3)^2+(1.5)^2=(AC)^2

2.25×3+2.25=(AC)^2

2.25×4=AC^2

AC=3

Also AO=1.5

similarly using Pythagoras theorem in ∆OBC

we get

OC=1.5

this shows that point O is the bisector of diagonal AC

so,AO=OC

also from property of rectangle we know that, diagonal bisects each other equally

so,O will also be the bisector of diagonal BD

hence OB=OD=1.5

AC=BD (property of triangle)

AC/2=BD/2

OB=OC,hence OB=OC=OD=OA=1.5=AD=BC

Hence OAD will be an equilateral triangle

so

angle OAD=60

similarly angleOBC=60

area of sector=angle of that sector ×area of circle/360

area of arc(AOD)+arc(BOC)=60/360 ×3.14 ×1.5×1.5 +60/360 ×3.14×1.5×1.5

=2×(60/360)×1.5×1.5

area of triangleAOB=area of rectangle/4 (diagonal divides the rectangle in 4 equal parts)

area of triangle=1.5×1.5√3/4

=>1.5×2.6/4=3.90/4

=0.975 or 0.98(approx)

Area covered by wiper=.98+2×60/360 ×3.14×1.5×1.5

=.98+0.75×3.14=2.35+0.97

=2.35+.98

Percentage of area of windshield covered by wiper=3.3×100/3.9=84.61

hence percentage of area of windshield covered by wiper is 85percent(approx)

Answer 2

$\underline{\underline{\sf{\maltese\:\:Question}}}$

• The rectangular front windshield of the bus has the length of $\sf{1.5\sqrt{3}}$ m and the width of 1.5 m. The wipers are attached to the bottom corners. If the length of each wiper is equal to the width of the windshield, Find the Percentage of the windshield that wipers can reach.

$\underline{\underline{\sf{\maltese\:\:Given}}}$

• Length of Rectangular front of Windshield = $\sf{1.5\sqrt{3}}$ m

• Breadth of Rectangular front of Windshield = 1.5 m

• The wipers are attached to the bottom corners

$\underline{\underline{\sf{\maltese\:\:Answer}}}$

• Percentage of windshield that the wipers can reach is 85.5 %

$\underline{\underline{\sf{\maltese\:\:Calculations}}}$

Consider the windshield diagram as given below plotted on the 1st quadrant of xy-plane, (Refer To Attachment - 1)

Also consider that the equation of a circular curve with center (h, k) and radius 'r' is given by $\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}$

The area formed by each wipers is a sector with radius equal to length of wiper, Therefore r = 1.5 m

The left circular quadrant has the center (0,0) and radius 1.5 m, so its equation is given by,

$\sf{(x-0)^{2}+(y-0)^{2}=(1.5)^{2}}$

$\to\sf{x^2+y^2=2.25\:\:\:\:\bf{........1}}$

Similarly the right circular quadrant has the center $\sf{(1.5 \sqrt{3}, 0)}$ and radius 1.5 m, so its equation is given by,

$\sf{(x-1.5 \sqrt{3})^{2}+(y-0)^{2}=1.5^{2}}$

$\to\sf{(x-1.5 \sqrt{3})^{2}+y^{2}=2.25\:\:\:\:\bf{........2}}$

Subtract Equation 2 from Equation 1

$\sf{(x-1.5 \sqrt{3})^{2}-x^{2}=0}$

$\to\sf{(x-1.5 \sqrt{3}+x)(x-1.5 \sqrt{3}-x)=0}$

$\to\sf{(2 x-1.5 \sqrt{3})(-1.5 \sqrt{3})=0}$

$\to\sf{2 x-1.5 \sqrt{3}=0}$

$\to\sf{2 x=1.5 \sqrt{3}}$

$\to\sf{x=1.299}$

$\to\sf{x\approx 1.3}$

Put this value in Equation 1, and solve for 'y'

$\to\sf{(1.3)^{2}+y^{2}=2.25}$

$\to\sf{y^{2}=2.25-1.69}$

$\to\sf{y^{2}=0.56}$

$\to\sf{y=\sqrt{0.56}}$

$\to\sf{y\approx 0.75}$

Thus, the Point of intersection of the curves is P(1.3 , 0.75)

Now, in order to determine the area PQR (which is the common area swept by both the wipers). A better way to calculate this area is to find the area under the curve PR and double it to obtain the total area under the curve PQR. Consider a vertical elemental strip in the region PR, having width dx and height dy. Then the area of the elemental strip will be $\bf{dA=dydx}$ ,

(Refer To Attachment - 2)

Now integrate the elemental area between the limits as described in the diagram, to obtain the total area as,

$\sf{\displaystyle Area\: under\:PR=\iint dA}$

$\to\sf{\displaystyle Area\: under\:PR=\int_{1.3}^{1.5} \int_{0}^{\sqrt{2.25-x^{2}}} \mathrm{~d} y \mathrm{~d} x}$

$\to\sf{\displaystyle Area\: under\:PR=\int_{1.3}^{1.5}[y]_{0}^{\sqrt{1.5^{2}-x^{2}}} \mathrm{~d} x}$

$\to\sf{\displaystyle Area\: under\:PR=\int_{1.3}^{1.5} \sqrt{1.5^{2}-x^{2}} \mathrm{~d} x}$

$\to\sf{\displaystyle Area\: under\:PR=\left[\frac{x}{2} \sqrt{1.5^{2}-x^{2}}+\frac{1.5^{2}}{2} \sin ^{-1}\left(\frac{x}{1.5}\right)+C\right]_{1.3}^{1.5}}$

$\boxed{\bf{\displaystyle\because \int \sqrt{a^{2}-x^{2}} \mathrm{~d} x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+C}}$

$\to\sf{Area\: under\:PR=\left(\frac{1.5}{2} \sqrt{1.5^{2}-1.5^{2}}+\frac{1.5^{2}}{2} \sin ^{-1}\left(\frac{1.5}{1.5}\right)\right)-\left(\frac{1.3}{2} \sqrt{1.5^{2}-1.3^{2}}+\frac{1.5^{2}}{2} \sin ^{-1}\left(\frac{1.3}{1.5}\right)\right)}$$\to\sf{\displaystyle Area\: under\:PR=\left(0+1.125\left(\frac{\pi}{2}\right)\right)-(0.486+1.18)}$

$\to\sf{\displaystyle Area\: under\:PR=0.1011}$

Then the Area under the curve PQR is ,

$\sf{Area\:under\:PQR=2 \times Area \:under\: PR}$

$\to\sf{Area\:under\:PQR=2 \times 0.1011}$

$\to\sf{Area\:under\:PQR=0.2022}$

The area swept by the two wipers is calculated as,

$\sf{A_{w}=\textsf{Area of left quadrant}+\textsf{Area of right quadrant }-\textsf{Common Area}}$

$\to\sf{\displaystyle A_{w}=\frac{1}{4}\left\{\pi(1.5)^{2}\right\}+\frac{1}{4}\left\{\pi(1.5)^{2}\right\}-0.2022}$

$\to\sf{\displaystyle A_{w}=3.332}$

Total Area of the rectangular windshield is given by,

(Refer to Attachment - 3)

The percentage area covered by the wipers is calculated as,

$\sf{\displaystyle\left(A_{w}\right) \%=\frac{A_{w}}{A} \times 100}$

$\to\sf{\displaystyle\left(A_{w}\right) \%=\frac{3.332}{3.897} \times 100}$

$\to\sf{\displaystyle\left(A_{w}\right) \%=\frac{3.332\times 100}{3.897}}$

$\to\sf{\displaystyle\left(A_{w}\right) \%=\frac{333.2}{3.897}}$

$\large\boxed{\boxed{\to\sf{\displaystyle\left(A_{w}\right) \%=85.5}}}$

Thus The Percentage of windshield that the wipers can reach is 85.5 %