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## Geometry and Mensuration: Level 3 Test 9

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*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%%
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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?

2•5 | |

3•5 | |

3•8 | |

4•0 |

Question 1 Explanation:

½ of the lawn would be 800/2 = 400 sq. m.

Area left after 1

Area left after 2

Area left after 3

There is still some area left.

Area left after 4

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

Thus fraction of area to be mowed is 0.82

Thus total number of times = 3.8 (approx)

Correct option is (c)

Area left after 1

^{st}time = 18x38Area left after 2

^{nd}time=16x36Area left after 3

^{rd}time= 14x34= 476There 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)

Question 2 |

In the figure below, AB =BC =CD= DE= EF =FG= GA. Then, ∠DAE is approximately:

15 ^{o} | |

20 ^{o} | |

30 ^{o} | |

25 ^{o} |

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.)

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

$ \displaystyle 1:\sqrt{3}$ | |

$ \displaystyle 1:\sqrt{2}$ | |

$ \displaystyle 1:2\sqrt{3}$ | |

$ \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?

7.5 | |

7 | |

7.75 | |

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 P

_{0}and AB be a diameter of C. Suppose P^{1}is the mid-point of the line segment P_{0}B, P_{2}is the mid-point of the line segment P_{1}B and so on. Let C_{1},C_{2},C_{3},… be circles with diameters P_{0}P_{1},P_{2},P_{2}P_{3}……. respectively. Suppose the circles C_{1},C_{2},C_{3 }….. respectively. Suppose the circles C_{1},C_{2},C_{3}…. Are all shaded. The ratio of the area of the unshaded portion of C to that of the original circle C is:8: 9 | |

9: 10 | |

10: 11 | |

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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