Constructions Test

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Constructions Test - 10

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Question 1
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You are asked to "construct" an angle whose measure is $$30^\circ$$. Which of the following methods would be considered as an acceptable construction?

A
Using a compass and straightedge, construct two parallel lines and label one of the angles $$30^\circ$$.

B
Using a compass and straightedge, construct an equilateral triangle and then bisect one of its angles.

C
Using a compass and straightedge, copy an angle that appears to be close to $$30^\circ$$ from a diagram in your textbook.

D
None of these

Solution

We only need to construct an angle of $$30^{o}$$.

We have not been provided with any side or point.

So, its best to construct an equilateral triangle by drawing $$60^{\circ}$$ angles and then extending and joining them with the straighedges

Since every angle is of measure $$60^{o}$$, so bisecting one of the angle will form an angle of measure $$30^{o}$$.

Hence, the acceptable construction is given by option B.
Question 2
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Which diagram below shows a correct mathematical construction using only a compass and a straightedge to bisect an angle?

A

B

C

D

Solution

To bisect an angle, we need to draw an arc from vertex which intersects all the two sides.
Then draw an arc from the point where the first arc(from vertex) intersects the lines.
Draw a line by joining vertex of an angle and the point of intersection of two arcs. The new line formed will bisect the angle.
This construction is shown by option D.
Hence, option D is correct.
Question 3
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Which of the following angles is possible to construct using a compass?

A
$$60^\circ$$

B
$$32^\circ$$

C
$$51.25^\circ$$

D
$$49^\circ$$

Solution

An angle of $$60^{\circ}$$ can be constructed using a compass.
Step 1 : Make a compelete arc on a straight line.
Step 2 : Make an arc on the previous arc with the same opening of compass from the start point of previous arc.
Step 3: Draw a line through the centre of arc and point of intersection.
Question 4
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With the help of a ruler and a compass it is not possible to construct an angle of.

A
$$37.5^{\circ}$$

B
$$40^{\circ}$$

C
$$22.5^{\circ}$$

D
$$67.5^{\circ}$$

Solution

$$ {\textbf{Step -1: Find angle by ruler and compass}} $$

$$ \angle {37.5^ \circ }{\text{can be constructed by bisecting }} {150^ \circ } {\text{twice which can be done by compass}}. $$

$$ \angle {22.5^ \circ }{\text{is the bisector of }}{90^ \circ }{\text{ which can be also constructed using compass}}{\text{.}} $$

$$ \angle {67.5^ \circ }{\text{is the bisector of }} {135^ \circ } {\text{which can also be drawn using compass}}{\text{.}} $$
$${\textbf{Step -2: Find 40}^{\circ} {\textbf{why not done by ruler}}}$$

$$ {\text{But }}40^ \circ {\text{can not be drawn using ruler and compass}}{\text{.}} $$

$$ {\textbf{Hence, option B is correct}}{\text{.}} $$
Question 5
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For constructing a triangle whose perimeter and both base angles are given, the base length is equal to:

A
the length of the perimeter

B
the length of the largest side

C
the difference between the largest and the shortest side

D
None of these

Solution

To construct a triangle when the perimeter and both base angles are given, we first draw the base with length equal to the perimeter of the triangle. After that we draw the complete triangle with proper method.
Question 6
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The steps of construction of an $$\angle AOB=45^{o}$$ is given in jumbled order below:
1. Place compass on intersection point.
2. Place ruler on start point and where arc intersects perpendicular line.
3. Adjust compass width to reach start point.
4. Construct a perpendicular line.
5. Draw $$45$$ degree line.
6. Draw an arc that intersects perpendicular line.
The third step in process is:

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$
Question 7
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Directions For Questions

The steps for construction for an $$\angle PQR$$ of measure $$90^\circ$$ are given in jumbled order below:
1. Mark a point $$B$$ on the same arc with the same radius from point $$A$$. Similarly, mark a point $$C$$ from $$B$$.
2. Join $$Q-D$$ and extend it to obtain ray $$QP$$.
3. Draw ray $$QR$$.
4. Place the pointed end of the compass on $$Q$$ and draw a semi-circular arc with an arbitrary radius.
5. Draw two intersecting arcs from $$B$$ and $$C$$ and mark the intersection point as $$D$$.

...view full instructions

The last step in the process is:

A
$$1$$

B
$$2$$

C
$$4$$

D
$$5$$

Solution

Correct sequence is
Step 1. Draw a ray $$QR$$
Step 2. Place the pointed end of the compass on $$Q$$ and draw a semi circular arc with arbitrary radius.
Step 3.Mark a point $$B$$ on the same arc with the same radius from point $$A$$. Similarly , mark a point $$C$$ from $$B.$$
Step 4. Draw two intersecting arcs from $$B$$ and $$C$$ and mark the intersection as point $$D$$.
Step 5. Join $$Q-D$$ and extend it to obtain $$QP.$$
So the last step is $$2$$
Option $$B$$ is correct.
Question 8
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Each angle of equilateral triangle is $$ 60^\circ$$. The angles are bisected then each angle will be of:

A
$$60^o$$

B
$$30^o$$

C
$$90^o$$

D
$$120^o$$

Solution

Angle bisector divide the angle in two equal parts.
$$\therefore $$ bisected angle $$=\dfrac{60^{\circ}}{2}=30^{\circ}$$
So option $$B$$ is correct.
Question 9
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An architect needs a staircase attached to a wall.The angle between stair and ground needs to be 30.
His plan will look like:

A

B

C

D

Solution

Only option $$(B)$$ shape have an angle equivalent to $$30^o$$.
So architect plan will look like shape $$(B)$$.
Question 10
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The steps for constructing an $$\angle ABC$$ of measure $$120^\circ$$ are given below in jumbled order:
1. From the point $$R$$, mark a point $$P$$ on the same arc with the same radius.
2. Place the pointed end of the compass on $$B$$ and draw a semi-circular arc with arbitrary radius and name its intersection with ray $$BC$$ as $$Q$$.
3. Draw a ray $$BC$$.
4. From point $$Q$$, mark a point $$R$$ on the arc with the same radius.
5. Join $$B-P$$ and extend it to obtain ray $$BA$$

$$The \ fifth \ step \ in \ the \ process \ is:$$

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

Correct sequence is :
Step 1. Draw a ray $$BC$$.
Step 2.Place the pointed end of the compass on $$B$$ and draw a semi-circular arc with arbitrary radius and name its intersection with the ray $$BC$$ as $$Q$$.
Step 3. From $$Q$$ mark a point $$R$$ on the arc with the same radius.
Step 4. From point $$R$$, mark a point $$P$$ on the same arc with same radius.
Step 5. Join $$B-P$$ and extend it to obtain ray $$BA$$

So the fifth step is $$5$$
Option $$D$$ is correct.