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  • These tests focus on geometry and mensuration and are meant to indicate your preparation level for the subject.
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Geometry and Mensuration: Test 13

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Question 1
If the median drawn on the base of a triangle is half its base, the triangle will be:
A
right-angled
B
acute-angled
C
obtuse-angled
D
equilateral
Question 1 Explanation: 
geometry-and-mensuration-test-13-qestion-1-pic-1
Since the median is = ½ the base therefore, the triangle has been divided into 2
Isosceles congruent triangles. So each of the triangles is 450-450 -900.
Therefore the sum of the base angles of the original triangle is 450+ 450 = 900.
So the vertex angle must be 90 degrees.
Question 2
In a right-angled triangle ABC, ∠ABC= 90o, AB=5 cm and BC=12 cm. The radius of the circumcircle of the triangle ABC is
A
7.5 cm
B
6 cm
C
6.5 cm
D
7 cm
Question 2 Explanation: 
7
We know that the radius of the circumcircle = ½ the length of the hypotenuse. The hypotenuse = 13 cm ( by Pythagoras theorem) Thus the required length of the radius = ½ X 13 = 6.5 cm.
Question 3
The exterior angles obtained on producing the base BC of a triangle ABC in both ways are 120o and 105o, then the vertical ∠A of the triangle is of measure
A
36o
B
40o
C
45o
D
55o
Question 3 Explanation: 
8
The base angles are = (180o-120o) = 60o
and 180-105 = 75o .
Thus the vertical angle = 180o- 60o-75o = 45o.
Question 4
If AD, BE and DF are medians of ΔABC, then which one of the following statements is correct?
A
$ \displaystyle \left( AD+BE+CF \right)
B
$ \displaystyle AD+BE+CF>AB+BC+CA$
C
$ \displaystyle AD+BE+CF=AB+BC+CA$
D
$ \displaystyle AD+BE+CF=\sqrt{2}\,\left( AB+BC+CA \right)$
Question 4 Explanation: 
9
In an Equilateral triangle ,
The Height = side x (√3/2) = 0.85 of the side
Thus we can conclude that the length of the median < length of the side.
Sum of the length of the medians < sum of length of the sides.
Question 5
In a regular polygon, the exterior and interior angles are in the ratio 1: 4. The number of sides of the polygon is
A
10
B
12
C
15
D
16
Question 5 Explanation: 
The number of sides of the polygon is
$ \begin{array}{l}\frac{\frac{360}{n}}{\left( n-2 \right)\times \frac{180}{n}}=\frac{1}{4}\\n=10\end{array}$
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