9th Grade > Mathematics
CONSTRUCTIONS MCQs
:
A
The correct procedure to construct the given triangle is:
1. Draw the base BC of Δ ABC as given and construct ∠XBC of the required measure at B as shown.
2. Keeping the compass at point B cut an arc from the ray BX such that its length equals to AB + AC at point P and join it to C as shown.
3. Now measure ∠BPC and from C draw an angle equal to ∠BPC as shown.
Thus, option A is correct.
:
A, B, and C
For constructing an angle of 60∘, we draw an arc of arbitrary radius from A cutting the base of the angle at C. From C we draw another arc of the same radius cutting the previous arc at B. Since, B and C are on arcs of same radii centred at A, AB = AC.
Also, B is on arc centred at C having the same radius, so, BC = AC. So, AB = BC = AC.
The steps to draw a triangle with the base BC, a base angle ∠B and the difference of other two sides is given below.
1. Draw the base BC and at point B make an angle say XBC equal to the given angle.
2. Cut the line segment BD equal to AB - AC on the reflection of ray BX (i.e. BX').
3. Join DC and draw the perpendicular bisector, say PQ of DC.​​
The next step will be:
:
B
The perpendicular bisector is a line that divides a line segment into two equal parts and also makes a right angle with the line segment. Here CD divides AB into two equal parts but is not perpendicular to it. Hence it is not a perpendicular bisector.Â
:
A
With a ruler and compass we can construct 15∘, 30∘, 45∘, 60∘, 90∘, 105∘ angles. We cannot construct 80∘.
:
A
A 105∘ angle can be constructed by drawing a bisector between 120∘ angle arm and 90∘ arm as shown in the figure below. But it can be seen clearly that for making 120∘ and 90 ∘ angles, one needs to make a 60∘ angle first. In the image given, ∠DAB will be 60∘ .
:
A
An angle bisector divides a given angle into two equal halves. So bisecting a given 80∘ angle will give two 40∘ angles. Hence, the given statement is true.
:
B
We know that sum of the length of any two sides of a triangle must be greater than the length of the third side, but the condition in this question defies this law. So the given triangle can not be formed.
The steps to construct a triangle with given base angles ∠ B and ∠ C and BC + CA + AB, are:.
A. Draw a line segment, say XY equal to BC + CA + AB.
B. Make ∠ LXY equal to ∠  B and MYX equal to ∠ C.
C. Bisect  ∠ LXY and  ∠ MYX. Let these bisectors intersect at a point A.
D. Draw perpendicular bisectors PQ of AX and RS of AY.
E. Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC.
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