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Geometry and Mensuration: Test 10
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Question 1 
The length of a rectangle is three fifth of the side of a square. The radius of a circle is equal to side of the square. The circumference of the circle is 132 cm. What is the area of the rectangle if the breadth of the rectangle is 8 cm?
112.4 sq. cm.  
104.2 sq. cm.  
100.8 sq. cm.  
Cannot be determined

Question 1 Explanation:
$ \begin{array}{l}2\pi r=132\\\Rightarrow r=21cm\\Side\,\,of\,\,the\,\,square=21cm\\Length\,\,of\,\,the\,\,rec\tan gle=\frac{3}{5}\times 21=12.6cm\\Breadth=8cm\\Area=12.6\times 8=100.8c{{m}^{2}}\end{array}$
Question 2 
Smallest side of a right angled triangle is 8 cm less than the side of a square of perimeter 56 cm. Second largest side of the right angled triangle is 4 cm less than the length of rectangle of area 96 sq. cm and breadth 8 cm. What is the largest side of the right angled triangle?
20 cm  
12 cm  
10 cm  
15 cm 
Question 2 Explanation:
$ \displaystyle \begin{array}{l}Side\text{ }of\text{ }{{1}^{st}}square\text{ }=\text{ }56/4\text{ }=\text{ }14\text{ }cm.\\Smallest\text{ }side\text{ }of\text{ }right\text{ }angled\text{ }triangle=\text{ }14\text{ }8~=\text{ }6\text{ }cm.\\length\text{ }of\text{ }{{2}^{nd}}rectangle\text{ }=\text{ }96/8\text{ }=\text{ }12\text{ }cm.\\Second\text{ }largest\text{ }side\text{ }of\text{ }the\text{ }{{1}^{st}}rectangle\text{ }=~124\text{ }=\text{ }8\text{ }cm.\\largest\text{ }side\text{ }=\text{ }hypotenuse=\sqrt{{{8}^{2}}+{{6}^{2}}}=10cm~\end{array}$
Question 3 
In a swimming pool measuring 90 m by 40 m. 150 men take a dip. If the average displacement of water by a man is 8 cu m, what will be the rise in water level?
33.33 cm  
30 cm
 
20 cm  
25 cm 
Question 3 Explanation:
Total volume displaced by men = 150X 8 = 1200 cu. m.
Area of the surface of swimming pool = 90m X 40m = 3600 sq.m.
Rise in water level = 1200/3600 m = 1/3 th of a metre = 33.33 cm
Area of the surface of swimming pool = 90m X 40m = 3600 sq.m.
Rise in water level = 1200/3600 m = 1/3 th of a metre = 33.33 cm
Question 4 
A 4 cm cube is cut into 1 cm cubes. Find the percentage increase in surface area.
200%  
100%  
400%  
300% 
Question 4 Explanation:
Total surface area of 4 cm cube= 6a^{2}
=6 x 4 x 4 = 96cm^{2}
Number of new cubes = 4^{3}/1^{3 }= 64
Total surface area of new cube = 64x6x1^{2} = 384cm^{2}
Increase = 38496 = 288
Increase Percentage = 288/96 x 100 = 300%
=6 x 4 x 4 = 96cm^{2}
Number of new cubes = 4^{3}/1^{3 }= 64
Total surface area of new cube = 64x6x1^{2} = 384cm^{2}
Increase = 38496 = 288
Increase Percentage = 288/96 x 100 = 300%
Question 5 
The radius of the base of a conical tent is 5 m. If the tent is 12 m high, then the area of the canvas required in making the tent is
300Â Î ^{2 }  
60Â Î ^{2}  
90Â Î ^{2 }  
None of these 
Question 5 Explanation:
$ \displaystyle \begin{array}{l}The\text{ }lateral\text{ }length=\sqrt{{{12}^{2}}+{{5}^{2}}}=13m\\Lateral\text{ }surface\text{ }area=\pi \times r\times l\\=\pi \times 5\times 13=65\pi {{m}^{2}}\end{array}$
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