Geometry and Mensuration: Test 12

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Geometry and Mensuration: Test 12

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Question 1

A man standing in one corner of a square football field observes that the angle subtended by a pole in the corner just diagonally opposite to this corner is 60^o. When he retires 80 m from the corner, along the same straight line, he finds the angle to be 30^o. The length of the field, in m, is:

A	40
B	20√2
C	20
D	40√2

Question 1 Explanation:

$ \displaystyle \begin{array}{l}From\text{ }the\text{ }given\text{ }description\text{ }if\text{ }the\text{ }diagonal\text{ }is\text{ }x\text{ }m\text{ },\\we\text{ }can\text{ }derive,\\\sqrt{3}x=\frac{x+80}{\sqrt{3}}\\\Rightarrow 3x=x+80,\\\Rightarrow 2x=80\\\Rightarrow x=40\\Thus\text{ }the\text{ }length\text{ }of\text{ }the\text{ }football\text{ }field\\=\frac{40}{\sqrt{2}}=20\sqrt{2}m\end{array}$

Question 2

Each interior angle of a regular polygon is three times its exterior angle, then the number of sides of the regular polygon is:

A	9
B	8
C	10
D	7

Question 2 Explanation:

geometry-and-mensuration-test-12-qestion2-pic-1

Question 3

$ \displaystyle ~O\text{ }is\text{ }the\text{ }in\text{ }center\text{ }of\vartriangle ABC\text{ }and\angle A=\text{ }{{30}^{o}},\text{ }then\angle BOC\text{ }is$

A	100^o
B	105^o
C	110^o
D	90^o

Question 3 Explanation:

geometry-and-mensuration-test-12-qestion4-pic-1

Question 4

$ \displaystyle \begin{array}{l}Let\text{ O }the\text{ }in-centre\text{ }of\text{ }a\text{ }triangle\text{ }ABC\text{ }and\text{ }\\D\text{ }be\text{ }a\text{ }point\text{ }on\text{ }the\text{ }side\text{ }BC\text{ }of\vartriangle ABC,\text{ }\\such\text{ }that\text{ }OD\bot BC.\text{ }If\angle BOD=\text{ }{{15}^{o}},\text{ }\\then\angle ABC=.\end{array}$

A	75^o
B	45^o
C	150^o
D	90^o

Question 4 Explanation:

$ \displaystyle \begin{array}{l}\angle OBD\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }of\angle ABC.\\Thus\text{ }the\text{ }required\text{ }angle\text{ }=\text{ }2\times \left( 180-90-15 \right)\text{ }=\text{ }\\{{150}^{0}}.Correct\text{ }option\text{ }is\text{ }\left( c \right)\end{array}$

Question 5

If the circum radius of an equilateral triangle be 10 cm, then the measure of its in-radius is

A	5 cm
B	10 cm
C	20 cm
D	15 cm

Question 5 Explanation:

The circumradius: inradius of the equilateral triangle is 2:1.
Thus the inradius = 10/2 = 5 cm.

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