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This mock test of Test: Construction Of Triangles for Class 9 helps you for every Class 9 entrance exam.
This contains 10 Multiple Choice Questions for Class 9 Test: Construction Of Triangles (mcq) to study with solutions a complete question bank.
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QUESTION: 1

In triangle ABC, altitude BE = altitude CF. Then triangle ABC is

Solution:

QUESTION: 2

To construct a ΔABC in which BC = 10 cm and ∠B= 60 degrees and AB + AC = 14 cm, then the length of BD used for construction.

Solution:

QUESTION: 3

Which of these triangles are possible to construct by knowing only its altitude?

Solution:

QUESTION: 4

The construction of a triangle ABC, given that BC = 6 cm, B = ∠45° is not possible when difference of AB and AC is equal to :

Solution:

Given, BC = 6 cm and ∠B= 45°

We know that, the construction of a triangle is not possible, if sum of two sides is less than or equal to the third side of the triangle.

i.e., AB + BC < AC

⇒ BC < AC – AB

⇒ 6 < AC-AB

So, if AC – AB= 6.9 cm, then construction of ΔABC with given conditions is not possible.

QUESTION: 5

Choose the correct statement

Solution:

QUESTION: 6

The point of concurrence of the three angle bisectors of a triangle, is called

Solution:
The point of interection of

medians - centroid

altitude - orthocentre

angle bisector - incentre

perpendicular bisectors - circumcentre

medians - centroid

altitude - orthocentre

angle bisector - incentre

perpendicular bisectors - circumcentre

QUESTION: 7

In triangle ABC, side AB is produced to D so that BD = BC. If angle B = 60° and angle A = 70°, then

Solution:

in this particular given problem angle

corresponding to the sides

ad : angle A

CDcd : angle CAB

as, angle ACD > angle CAB, So

ad>cd

QUESTION: 8

The point of intersection of the perpendicular bisectors of the sides of a triangle is called

Solution:

The perpendicular bisectors of the sides of a triangle intersect at a point called the circumcenter of the triangle, which is equidistant from the vertices of the triangle.

QUESTION: 9

To construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. We draw line segment AB of 12 cm. Draw a ray AX making 90° with AB. The next step is:

Solution:

**The below given steps will be followed to construct the required triangle.**

Step I:

Draw line segment AB of 12 cm. Draw a ray AX making 90deg with AB.

Step II:

Cut a line segment AD of 18 cm (as the sum of the other two sides is 18) from ray AX.

Step III:

Join DB and make an angle DBY equal to ADB.

Step IV:

Let BY intersect AX at C. Join AC, BC.

ΔABC is the required triangle.

QUESTION: 10

The lengths of the sides of some triangles are given, which of them is not a right angled triangle?

Solution:
If a triangle is right angle then it must satisfy Pythagoras theorem

Now let us check whether all options satisfy it or not....

a) 5^2+12^2=169

5^2+12^2=13^2

b) 5^2+8^2 is not equal to 10^2

c) 3^2+4^2=5^2

d) 7^2+24^2=25^2

It is clear that 2nd option doesn't satisfy Pythagoras theorem...soo option 2 is the answer

Now let us check whether all options satisfy it or not....

a) 5^2+12^2=169

5^2+12^2=13^2

b) 5^2+8^2 is not equal to 10^2

c) 3^2+4^2=5^2

d) 7^2+24^2=25^2

It is clear that 2nd option doesn't satisfy Pythagoras theorem...soo option 2 is the answer

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