Geometry and Mensuration: Test 13

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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= 90^o, 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:

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 120^o and 105^o, then the vertical ∠A of the triangle is of measure

A	36^o
B	40^o
C	45^o
D	55^o

Question 3 Explanation:

The base angles are = (180^o-120^o) = 60^o
and 180-105 = 75^o .
Thus the vertical angle = 180^o- 60^o-75^o = 45^o.

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:

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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