Geometry and Mensuration: Level 3 Test 9

½ of the lawn would be 800/2 = 400 sq. m.
Area left after 1^st time = 18x38
Area left after 2^nd time=16x36
Area left after 3^rd time= 14x34= 476
There is still some area left.
Area left after 4^th time=12x32=384.
Already half of area is crossed.
So the correct answer should be less than 4.
Area to be scrapped out for leaving ½ park is = 476-400=76
Area to be scraped out between 3^rd and 4^th = 476-384=92.
Thus fraction of area to be mowed is 0.82
Thus total number of times = 3.8 (approx)
Correct option is (c)

∠DAE=x (say).
Since AB = BC, ∠BCA = ∠CAB = x
Hence, ∠CBD = ∠CAB + ∠BCA = x + x = 2x
BC = CD => ∠CBD = ∠CDB = 2x
Hence, ∠DCE = ∠DAE + ∠CDA = x + 2x = 3x
Since, CD = DE => ∠DCE = ∠DEC = ∠AED = 3x
Similarly, ∠ADE = ∠EFD = ∠AEF +∠ DAE
= ∠EGF + ∠DAE =(∠DAE + ∠GFA) + ∠DAE = ∠DAE + ∠DAE + ∠DAE = 3x
Hence, in triangle ∠ADE, ∠ADE +∠ DAE + ∠AED = 3x +3x+x =7x
Thus ∠DAE= 180/7 = 25(approx.)

$ \begin{array}{l}Let\,\,the\,\,radius\,\,of\,\,the\,\,circle=R\\We\,\,have\,\,\frac{Area\,\,of\,\,\bigcirc }{Area\,\,of\,\,\square }=\frac{\pi {{R}^{2}}}{lb}=\frac{\pi }{\sqrt{3}}\\=>\sqrt{3}\,{{R}^{2}}=lb................................a\\l=\frac{\sqrt{3}\,{{R}^{2}}}{b}\,..........................................1\\Now\,\,we\,\,are\,\,given\,\,by\,\angle ODC\text{ }=\angle ADE\\Therefore\,\,we\,\,can\,\,say\,\,that\,\,\vartriangle AED\sim \vartriangle CBD\\\therefore \frac{\text{AE}}{\text{CB}}\text{= }\frac{\text{AD}}{\text{DC}}\\\text{ }\\\therefore \frac{\text{AE }}{\text{AD}}\text{=}\frac{\text{BC}}{\text{ DC}}\\\therefore \,\frac{\text{AE }}{\text{AD}}=\frac{b}{l}.....................................2\\Now\,\,From\,\,triangle\,\,\,DBC,\,\,\,\,\,\,\,\,\,\,\,\\4{{R}^{2}}={{l}^{2}}+{{b}^{2}}\\4{{R}^{2}}=\frac{\sqrt{3}\,{{R}^{4}}}{{{b}^{2}}}+{{b}^{2}}\\\Rightarrow {{b}^{4}}-4{{R}^{2}}{{b}^{2}}+3{{R}^{4}}=0\\\Rightarrow {{b}^{4}}-3{{R}^{2}}{{b}^{2}}-{{R}^{2}}{{b}^{2}}+3{{R}^{4}}=0\\\Rightarrow {{b}^{2}}\left( {{b}^{2}}-3{{R}^{2}} \right)-{{R}^{2}}\,\left( {{b}^{2}}-3{{R}^{2}} \right)=0\\\Rightarrow {{b}^{2}}={{R}^{2}}\,\,\,\,and\,\,\,{{b}^{2}}=3{{R}^{2}}\\=>b=R\\and\,\,b=\sqrt{3}R\\Now\,we\,\,have\,\,two\,\,values\,\,of\,\,b\\whenb=\sqrt{3}R\\therfore\,\,required\,\,ratio=\sqrt{3}:1\\and\,\,when\,\,b=r\\therfore\,\,\,\,required\,\,ratio=1:\sqrt{3}:\\but\,\,form\,\,the\,\,option\,\,a\,\,is\,\,the\,\,right\,\,option\end{array}$

In any circle, an inscribed angle that intercepts the diameter is a RIGHT ANGLE.
Thus, ∠ACD is a right angle, and ∆ACD is a right triangle.
In any right triangle, a height drawn through the right angle forms SIMILAR triangles.
Thus, in ∆ACD, height CE forms the following similar triangles:
∆ACE, ∆CED, and ∆ACD.
As shown in the figure above, each of these triangles has the same combination of angles: x-y-90.
Corresponding sides of similar triangles are always in the SAME RATIO.
Thus, in ∆ACD and ∆CED:
(leg opposite angle x)/hypotenuse = (leg opposite angle x)/hypotenuse
2/8 = DE/2
DE = 1/2.
The same reasoning can be used to determine that AF = 1/2.
Thus, FE = AD - AF - DE = 8 - 1/2 - 1/2 = 7.
Since BC || AD, quadrilateral BCEF is a rectangle, implying that BC=FE.
Thus, BC=FE=7.

$ \displaystyle \begin{array}{l}Let\text{ }the\text{ }total\text{ }area\text{ }of\text{ }circle\text{ }be\text{ }A.\\Thus\text{ }{{C}_{1}}=~\frac{A}{16}\\{{C}_{2}}=\frac{A}{64}\\~{{C}_{3}}=~Total\text{ }area\text{ }of\text{ }shaded\text{ }circles\text{ }\\=~\frac{A}{16}+\frac{A}{64}+\frac{A}{256}...\\=\frac{\frac{A}{16}}{1-\frac{1}{4}}=\frac{A}{12}\\\text{Thus the required ratio = }\frac{A-\frac{A}{12}}{A}=\frac{11}{12}\end{array}$