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## Geometry and Mensuration: Level 3 Test 9

Congratulations - you have completed Geometry and Mensuration: Level 3 Test 9. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%%
 Question 1
Neeraj has agreed to mow the farm lawn, which is a 20 m by 40 m rectangle. The mower mows a 1 m wide strip. If Neeraj starts at one corner and mows around the lawn towards the centre, about how many times would he go round before he has mowed half the lawn?
 A 2•5 B 3•5 C 3•8 D 4•0
Question 1 Explanation:
½ of the lawn would be 800/2 = 400 sq. m.
Area left after 1st time = 18x38
Area left after 2nd time=16x36
Area left after 3rd time= 14x34= 476
There is still some area left.
Area left after 4th 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 3rd and 4th = 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)
 Question 2
In the figure below, AB =BC =CD= DE= EF =FG= GA. Then, ∠DAE is approximately:
 A 15o B 20o C 30o D 25o
Question 2 Explanation:
∠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.)
 Question 3
In the figure below (not drawn to scale), rectangle ABCD is inscribed in the circle with centre at O, The length of side AB is greater than that of side BC. The ratio of the area of the dircle to the area of the rectangle ABCD if Π:√3 . The line segment DE intersects AB at E such that ODC = ADE. What is the ratio AD: DC A $\displaystyle 1:\sqrt{3}$ B $\displaystyle 1:\sqrt{2}$ C $\displaystyle 1:2\sqrt{3}$ D $\displaystyle 1:2$
Question 3 Explanation:
$\begin{array}{l}Let\,\,the\,\,radius\,\,of\,\,the\,\,circle=R\\We\,\,have\,\,\frac{Area\,\,of\,\,\bigcirc }{Area\,\,of\,\,\square }=\frac{\pi {{R}^{2}}}{lb}=\frac{\pi }{\sqrt{3}}\\=>\sqrt{3}\,{{R}^{2}}=lb................................a\\l=\frac{\sqrt{3}\,{{R}^{2}}}{b}\,..........................................1\\Now\,\,we\,\,are\,\,given\,\,by\,\angle ODC\text{ }=\angle ADE\\Therefore\,\,we\,\,can\,\,say\,\,that\,\,\vartriangle AED\sim \vartriangle CBD\\\therefore \frac{\text{AE}}{\text{CB}}\text{= }\frac{\text{AD}}{\text{DC}}\\\text{ }\\\therefore \frac{\text{AE }}{\text{AD}}\text{=}\frac{\text{BC}}{\text{ DC}}\\\therefore \,\frac{\text{AE }}{\text{AD}}=\frac{b}{l}.....................................2\\Now\,\,From\,\,triangle\,\,\,DBC,\,\,\,\,\,\,\,\,\,\,\,\\4{{R}^{2}}={{l}^{2}}+{{b}^{2}}\\4{{R}^{2}}=\frac{\sqrt{3}\,{{R}^{4}}}{{{b}^{2}}}+{{b}^{2}}\\\Rightarrow {{b}^{4}}-4{{R}^{2}}{{b}^{2}}+3{{R}^{4}}=0\\\Rightarrow {{b}^{4}}-3{{R}^{2}}{{b}^{2}}-{{R}^{2}}{{b}^{2}}+3{{R}^{4}}=0\\\Rightarrow {{b}^{2}}\left( {{b}^{2}}-3{{R}^{2}} \right)-{{R}^{2}}\,\left( {{b}^{2}}-3{{R}^{2}} \right)=0\\\Rightarrow {{b}^{2}}={{R}^{2}}\,\,\,\,and\,\,\,{{b}^{2}}=3{{R}^{2}}\\=>b=R\\and\,\,b=\sqrt{3}R\\Now\,we\,\,have\,\,two\,\,values\,\,of\,\,b\\whenb=\sqrt{3}R\\therfore\,\,required\,\,ratio=\sqrt{3}:1\\and\,\,when\,\,b=r\\therfore\,\,\,\,required\,\,ratio=1:\sqrt{3}:\\but\,\,form\,\,the\,\,option\,\,a\,\,is\,\,the\,\,right\,\,option\end{array}$
 Question 4
On a semicircle with diameter AD, chord BC is parallel to the diameter. Further, each of the chord AB and CD has length 2, while AD has length 8. What is the length of BC? A 7.5 B 7 C 7.75 D None of these
Question 4 Explanation: 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.
 Question 5
Let C be a circle with center P0 and AB be a diameter of C. Suppose P1 is the mid-point of the line segment P0B, P2 is the mid-point of the line segment P1B and so on. Let C1,C2,C3 ,… be circles with diameters P0P1,P2,P2P3……. respectively. Suppose the circles C1,C2,C3 ….. respectively. Suppose the circles C1,C2,C3 …. Are all shaded. The ratio of the area of the unshaded portion of C to that of the original circle C is:
 A 8: 9 B 9: 10 C 10: 11 D 11: 12
Question 5 Explanation: $\displaystyle \begin{array}{l}Let\text{ }the\text{ }total\text{ }area\text{ }of\text{ }circle\text{ }be\text{ }A.\\Thus\text{ }{{C}_{1}}=~\frac{A}{16}\\{{C}_{2}}=\frac{A}{64}\\~{{C}_{3}}=~Total\text{ }area\text{ }of\text{ }shaded\text{ }circles\text{ }\\=~\frac{A}{16}+\frac{A}{64}+\frac{A}{256}...\\=\frac{\frac{A}{16}}{1-\frac{1}{4}}=\frac{A}{12}\\\text{Thus the required ratio = }\frac{A-\frac{A}{12}}{A}=\frac{11}{12}\end{array}$
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