The rectangular front windshield of the bus has the length of 1.5√3 m and the wi...

amitnrw

Given : The rectangular front windshield of the bus has the length of 1.5√3 m and the width of 1.5 m

The wipers are attached to the bottom corners.

length of each wiper is equal to the width of the windshield

To Find : the percentage of the windshield that wipers can reach

(Note : Solve this task without using Trigonometry)

Solution:

Method 1 : Using Heron formula :

$\text{Area of Rectangle} = \text{Length} \times \text{width}$

Substitute length= $1.5\sqrt{3}$ and width=1.5 to find area of windshield.

$\text{Area of windshield} = 1.5\sqrt{3} \times 1.5=2.25\sqrt{3}=3.897$

Area covered by wipers is area covered by sector AOD, Sector BOC and Triangle AOB.

Area of Triangle AOD using heron formula.

Heron Formula is: $\sqrt{s(s-a)(s-b)(s-c)}$ where s is semi perimeter and a, b, c are sides of triangle.

AO=1.5, BO=1.5, AB=1.5√3

Substituting values in Formula and simplify to find Area of triangle AOB = $\dfrac{(1.5)(1.5)\sqrt{3}}{4}.$

Area of triangle AOB using formula $\dfrac{base \times height}{2}.$

Equate both area to find OM = $\dfrac{1.5}{2}.$

Hence O is the center of Rectangle.

Hence DO=AO= BO=CO (As Diagonals of rectangle are equal and bisect each other )

Triangle AOD and Triangle BOC are Equilateral triangle (As all sides are Equal)

Hence angle DAO is 60° and angle CBO is 60°

Area of Circular sector is: $\dfrac{\text{sector angle}}{360} \pi (radius)^2.$

Area of Circular sector AOD = BOC = $\dfrac{60}{360} \pi (1.5)^2.$

$\dfrac{60}{360} \pi (1.5)^2=\dfrac{1}{6} \pi (2.25)$

Total area covered by wipers is : Area of Circular sector AOD + Area of Circular sector BOC + Area of Triangle AOB

$\dfrac{1}{6} \pi (2.25)+\dfrac{1}{6} \pi (2.25)+ \dfrac{(1.5)(1.5)\sqrt{3}}{4}=3.33$

the percentage of the windshield that wipers can reach is :

$\dfrac{3.33}{3.897} \times100 = 85.46 \%$

Another Method Using Pythagoras theorem :

AC² = AB² + BC²

=> AC² = (1.5√3)² + 1.5²

=> AC² = 1.5²(3 + 1)

=> AC = 1.5(2)

=> AC = 3

AO = 1.5 = AC/2

Hence O is the center of rectangle

=> AO = BO = CO = DO = 1.5

Rest solution is same.

using trigonometry :

Cosine rule in ΔAOB

can find angle ∠OAB = 30° or ∠OBA = 30°

=> angle DAO is 60° and angle CBO is 60°

area of triangle can be found using sine rule.

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TheEmeraldBoyy

${\huge{\boxed{\sf{\green{❥✰Question✰}}}}}$

The rectangular front windshield of the bus has the length of 1.5√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)

${\huge{\underline{\bf{\pink{❥✰➵Given:-✰}}}}}$

Length of Rectangular front of Windshield = $\sqrt1.5[]{3}$ m
Breadth of Rectangular front of Windshield = 1.5 m
The wipers are attached to the bottom corners

${\huge{\underline{\bf{\pink{❥✰➵ANSWER :-✰}}}}}$

Percentage of windshield that the wipers can reach is 85.5 %

${\huge{\underline{\bf{\pink{❥✰➵Calculation :-✰}}}}}$

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 $\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}(x−h) 2 +(y−k) 2 =r 2$

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,

$\sf{(x-0)^{2}+(y-0)^{2}=(1.5)^{2}}(x−0) 2 +(y−0) 2 =(1.5) 2 \to\sf{x^2+y^2=2.25\:\:\:\:\bf{..1}}→x 2 +y 2 =2.25...1$

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

$\sf{(x-1.5 \sqrt{3})^{2}+(y-0)^{2}=1.5^{2}}(x−1.5 3 ) 2 +(y−0) 2 =1.5 2 \to\sf{(x-1.5 \sqrt{3})^{2}+y^{2}=2.25\:\:\:\:\bf{..2}}→(x−1.5 3 ) 2 +y 2 =2.25..2$

Subtract Equation 2 from Equation 1

$\sf{(x-1.5 \sqrt{3})^{2}-x^{2}=0}(x−1.5 3 ) 2 −x 2 =0\to\sf{(x-1.5 \sqrt{3}+x)(x-1.5 \sqrt{3}-x)=0}→(x−1.5 3 +x)(x−1.5 3 −x)=0\to\sf{(2 x-1.5 \sqrt{3})(-1.5 \sqrt{3})=0}→(2x−1.5 3 )(−1.5 3 )=0\to\sf{2 x-1.5 \sqrt{3}=0}→2x−1.5 3 =0\to\sf{2 x=1.5 \sqrt{3}}→2x=1.5 3 \to\sf{x=1.299}→x=1.299\to\sf{x\approx 1.3}→x≈1.3$

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

$\to\sf{(1.3)^{2}+y^{2}=2.25}→(1.3) 2 +y 2 =2.25\to\sf{y^{2}=2.25-1.69}→y 2 =2.25−1.69\to\sf{y^{2}=0.56}→y 2 =0.56\to\sf{y=\sqrt{0.56}}→y= 0.56 \to\sf{y\approx 0.75}→y≈0.75$

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 $\bf{dA=dydx}dA=dydx ,$

(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}AreaunderPR=∬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}→AreaunderPR=∫ 1.31.5 ∫ 02.25−x 2 dy dx\to\sf{\displaystyle Area\: under\:PR=\int_{1.3}^{1.5}[y]_{0}^{\sqrt{1.5^{2}-x^{2}}} \mathrm{~d} x}→AreaunderPR=∫ 1.31.5 [y] 01.5 2 −x 2 $

$\to\sf{\displaystyle Area\: under\:PR=\int_{1.3}^{1.5} \sqrt{1.5^{2}-x^{2}} \mathrm{~d} x}→AreaunderPR=∫ 1.31.5$

Then the Area under the curve PQR is ,

$\sf{Area\:under\:PQR=2 \times Area \:under\: PR}AreaunderPQR=2×AreaunderPR\to\sf{Area\:under\:PQR=2 \times 0.1011}→AreaunderPQR=2×0.1011\to\sf{Area\:under\:PQR=0.2022}→AreaunderPQR=0.2022The area swept by the two wipers is calculated as,\sf{A_{w}=\textsf{Area of left quadrant}+\textsf{Area of right quadrant }-\textsf{Common Area}}A w$

The area swept by the two wipers is calculated as,

$\sf{A_{w}=\textsf{Area of left quadrant}+\textsf{Area of right quadrant }-\textsf{Common Area}}A w\\\to\sf{\displaystyle A_{w}=\frac{1}{4}\left\{\pi(1.5)^{2}\right\}+\frac{1}{4}\left\{\pi(1.5)^{2}\right\}-0.2022}→A w = 41 {π(1.5) 2 }+ 41 {π(1.5) 2 }−0.2022\to\sf{\displaystyle A_{w}=3.332}→A w =3.332$

Total Area of the rectangular windshield is given by,

(Refer to Attachment - 3)

The percentage area covered by the wipers is calculated as,

$\sf{\displaystyle\left(A_{w}\right) \%=\frac{A_{w}}{A} \times 100}(A w )%= AA w ×100\to\sf{\displaystyle\left(A_{w}\right) \%=\frac{3.332}{3.897} \times 100}→(A w )%= 3.8973.332 ×100\to\sf{\displaystyle\left(A_{w}\right) \%=\frac{3.332\times 100}{3.897}}→(A w )%= 3.8973.332×100 \to\sf{\displaystyle\left(A_{w}\right) \%=\frac{333.2}{3.897}}→(A w )%= 3.897333.2 \large\boxed{\boxed{\to\sf{\displaystyle\left(A_{w}\right) \%=85.5}}} →(A w )%=85.5$

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

TheEmeraldBoyy

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