- This is an assessment test.
- 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:
right-angled | |
acute-angled | |
obtuse-angled | |
equilateral |
Question 1 Explanation:
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
7.5 cm | |
6 cm | |
6.5 cm | |
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 120o and 105o, then the vertical ∠A of the triangle is of measure
36o | |
40o | |
45o | |
55o |
Question 3 Explanation:
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?
$ \displaystyle \left( AD+BE+CF \right) | |
$ \displaystyle AD+BE+CF>AB+BC+CA$ | |
$ \displaystyle AD+BE+CF=AB+BC+CA$ | |
$ \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
10 | |
12 | |
15 | |
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}$
$ \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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