Chapter 14 Practical Geometry

Given:

Radius of the circle = 3.2 cm.

To Draw:

A circle with the given radius.

Construction:

The steps of construction are as follows:

1. Open the compass for the required radius of 3.2 cm by placing its pointed leg on the zero mark of a ruler and the pencil leg on the 3.2 cm mark.

2. Mark a point 'O' on a piece of paper. This point will be the center of the circle.

3. Place the pointed leg of the compass on the point O.

4. Turn the compass slowly to draw the circle. Be careful to complete the movement around in one instance.

The figure obtained is the required circle of radius 3.2 cm.

Given:

A common center point O.

Radii of the two circles: 4 cm and 2.5 cm.

To Draw:

Two concentric circles with the given radii.

Construction:

The steps of construction are as follows:

1. Mark a point 'O' on a piece of paper. This will be the common center.

2. Open the compass to a radius of 4 cm.

3. Place the pointed leg of the compass at O and draw a circle.

4. Now, open the compass to a radius of 2.5 cm.

5. Place the pointed leg of the compass again at O and draw another circle.

The figure shows the two required concentric circles.

Solution:

1. Draw a circle with center O and any radius.

2. Draw any two diameters, say AB and CD, passing through the center O.

3. Join the endpoints of the diameters to form a quadrilateral ACBD.

The figure obtained by joining the ends of two diameters is a rectangle. This is because the diagonals of the quadrilateral ACBD (which are the diameters AB and CD) are equal in length and bisect each other at the center O.

When the diameters are perpendicular to each other:

If we draw two diameters, say PQ and RS, such that they are perpendicular to each other at the center O, and then join their endpoints, the figure obtained is a square.

This is because the diagonals are equal, they bisect each other, and they are perpendicular to each other. These are the properties of a square.

How to check the answer:

We can check the answers by measuring the properties of the resulting figures.

For the first case (rectangle): Measure the four angles of the quadrilateral ACBD. Each angle will be $90^\circ$. Also, measure the opposite sides; they will be equal ($AC = DB$ and $AD = CB$).

For the second case (square): Measure all four sides of the quadrilateral PSQR. All sides will be equal. Also, measure the four angles; each angle will be $90^\circ$.

Solution:

1. Draw a circle with center O and any suitable radius.

2. Mark point A: Mark a point A anywhere on the boundary (circumference) of the circle. This point is 'on the circle'.

3. Mark point B: Mark a point B anywhere inside the boundary of the circle. This region is the 'interior of the circle'.

4. Mark point C: Mark a point C anywhere outside the boundary of the circle. This region is the 'exterior of the circle'.

In the figure above, A is on the circle, B is in the interior, and C is in the exterior of the circle.

Given:

Two circles with centers A and B, having equal radii.

Each circle passes through the center of the other.

The circles intersect at points C and D.

Construction:

1. Draw a line segment $\overline{AB}$ of any length.

2. With A as the center and radius equal to AB, draw a circle.

3. With B as the center and radius equal to BA, draw another circle.

4. Let the two circles intersect at points C and D.

5. Join the points C and D to form the line segment $\overline{CD}$.

Examination:

Let the line segments $\overline{AB}$ and $\overline{CD}$ intersect at point M.

Consider the quadrilateral ACBD. In $\triangle ABC$, we have:

So, $AC = BC = AB$. Thus, $\triangle ABC$ is an equilateral triangle.

Similarly, in $\triangle ABD$, we have:

So, $AD = BD = AB$. Thus, $\triangle ABD$ is also an equilateral triangle.

Since all four sides of the quadrilateral ACBD are equal to the radius ($AC = CB = BD = DA$), the quadrilateral ACBD is a rhombus.

One of the properties of a rhombus is that its diagonals bisect each other at right angles.

Here, $\overline{AB}$ and $\overline{CD}$ are the diagonals of the rhombus ACBD.

Therefore, $\overline{AB}$ and $\overline{CD}$ intersect at right angles.

Yes, $\overline{AB}$ and $\overline{CD}$ are at right angles.

This can also be verified by measuring the angle $\angle CMB$ or $\angle AMD$ with a protractor; the angle will be $90^\circ$.

Given:

Length of the line segment = 7.3 cm.

To Draw:

A line segment of length 7.3 cm using a ruler.

Construction:

The steps of construction are as follows:

1. Place a ruler on a sheet of paper.

2. Mark a point A at the 0 cm mark on the ruler.

3. Mark another point B at the 7.3 cm mark on the ruler.

4. Join the points A and B along the edge of the ruler.

The resulting line segment $\overline{AB}$ is the required segment of length 7.3 cm.

Given:

Length of the line segment = 5.6 cm.

To Construct:

A line segment of length 5.6 cm using a ruler and compasses.

Construction:

The steps of construction are as follows:

1. Draw a line $l$ of any length. Mark a point A on this line.

2. Place the pointed leg of the compass at the zero mark of a ruler and open the pencil leg to the 5.6 cm mark.

3. Without changing the compass opening, place the pointed leg at point A and draw an arc to cut the line $l$ at a point B.

The segment $\overline{AB}$ is the required line segment of length 5.6 cm.

Given:

Length of line segment $\overline{AB}$ = 7.8 cm.

Length of line segment $\overline{AC}$ = 4.7 cm.

To Construct and Measure:

Construct $\overline{AB}$ and cut off $\overline{AC}$, then measure the length of $\overline{BC}$.

Construction:

1. Draw a line segment $\overline{AB}$ of length 7.8 cm.

2. With A as the center, and using a compass opened to a radius of 4.7 cm, draw an arc to cut the line segment $\overline{AB}$ at point C.

Measurement:

Now, we measure the length of the line segment $\overline{BC}$ using a ruler. The length of $\overline{BC}$ is found to be 3.1 cm.

Calculation for verification:

Since point C lies on $\overline{AB}$, we have $AB = AC + BC$.

$7.8 \text{ cm} = 4.7 \text{ cm} + BC$

$BC = 7.8 \text{ cm} - 4.7 \text{ cm}$

$\begin{array}{cc} & 7 & . & 8 \\ - & 4 & . & 7 \\ \hline & 3 & . & 1 \\ \hline \end{array}$

$BC = 3.1$ cm.

The measured value matches the calculated value.

Given:

A line segment $\overline{AB}$ of length 3.9 cm.

To Construct and Verify:

A line segment $\overline{PQ}$ such that its length is twice the length of $\overline{AB}$, and verify by measurement.

Construction:

1. Draw a line $l$ and mark a point P on it.

2. Open the compass to the length of $\overline{AB}$ (3.9 cm).

3. Place the compass pointer at P and draw an arc to cut the line at a point, say X. So, $PX = AB$.

4. Now, place the compass pointer at X and draw another arc to cut the line at point Q. So, $XQ = AB$.

The segment $\overline{PQ}$ is the required line segment. The length of $\overline{PQ}$ is $PX + XQ = AB + AB = 2 \times AB$.

Verification by Measurement:

The expected length of $\overline{PQ}$ is $2 \times 3.9 \text{ cm} = 7.8 \text{ cm}$.

Using a ruler, measure the length of the constructed line segment $\overline{PQ}$. The measurement will be approximately 7.8 cm, which verifies the construction.

Given:

A line segment $\overline{AB}$ of length 7.3 cm.

A line segment $\overline{CD}$ of length 3.4 cm.

To Construct and Verify:

A line segment $\overline{XY}$ such that its length is $AB - CD$, and verify by measurement.

Construction:

1. Draw a line $l$ and mark a point X on it.

2. Open the compass to the length of $\overline{AB}$ (7.3 cm).

3. Place the compass pointer at X and draw an arc to cut the line at a point, say Z. So, $XZ = AB = 7.3$ cm.

4. Now, open the compass to the length of $\overline{CD}$ (3.4 cm).

5. Place the compass pointer at Z and draw an arc that cuts the line segment $\overline{XZ}$ at a point Y.

The segment $\overline{XY}$ is the required line segment. The length of $\overline{XY}$ is $XZ - YZ = AB - CD$.

Verification by Measurement:

The expected length of $\overline{XY}$ is $7.3 \text{ cm} - 3.4 \text{ cm}$.

$\begin{array}{cc} & 7 & . & 3 \\ - & 3 & . & 4 \\ \hline & 3 & . & 9 \\ \hline \end{array}$

The expected length is 3.9 cm.

Using a ruler, measure the length of the constructed line segment $\overline{XY}$. The measurement will be approximately 3.9 cm, which verifies the construction.

Given:

A line segment $\overline{PQ}$ of any length.

To Construct:

A copy of the line segment $\overline{PQ}$ without measuring it.

Construction:

The steps of construction are as follows:

1. Draw the given line segment $\overline{PQ}$.

2. Draw a line $l$ and mark a point A on it. This will be the starting point of the copied segment.

3. Place the pointer of the compass on point P and open the pencil leg to touch point Q. The opening of the compass now equals the length of $\overline{PQ}$.

4. Without changing the compass opening, place the pointer on point A and draw an arc to cut the line $l$ at a point B.

The line segment $\overline{AB}$ is the required copy of the line segment $\overline{PQ}$.

Given:

A line segment $\overline{AB}$ of unknown length.

To Construct:

A line segment $\overline{PQ}$ such that its length is twice the length of $\overline{AB}$.

Construction:

The steps of construction are as follows:

1. Draw the given line segment $\overline{AB}$.

2. Draw a line $l$ and mark a point P on it.

3. Place the compass pointer on point A and open the pencil leg to touch point B. The compass opening is now equal to the length of $\overline{AB}$.

4. Without changing the compass opening, place the pointer on point P and draw an arc to cut the line $l$ at a point, let's call it X. So, the length of $\overline{PX}$ is equal to the length of $\overline{AB}$.

5. Again, without changing the compass opening, place the pointer on point X and draw another arc to cut the line $l$ at a point Q.

The resulting line segment $\overline{PQ}$ is twice the length of $\overline{AB}$, because $PQ = PX + XQ = AB + AB = 2 \times AB$.

Given:

A line segment $\overline{AB}$ and a point M on it.

To Construct:

A perpendicular to $\overline{AB}$ through M.

Construction:

The steps of construction are as follows:

1. Draw the line segment $\overline{AB}$ and mark a point M on it.

2. With M as the center and a convenient radius, draw an arc that intersects $\overline{AB}$ at two points, say C and D.

3. With C as the center and a radius greater than CM, draw an arc above $\overline{AB}$.

4. With D as the center and the same radius, draw another arc to intersect the previous arc at a point, say P.

5. Join the points P and M.

The line segment $\overline{PM}$ is the required perpendicular to $\overline{AB}$ at point M.

Given:

A line segment $\overline{PQ}$ and a point R not on it.

To Construct:

A perpendicular to $\overline{PQ}$ from the point R using a ruler and set-square.

Construction:

The steps of construction are as follows:

1. Draw the line segment $\overline{PQ}$ and mark a point R outside it.

2. Place a ruler along the line segment $\overline{PQ}$.

3. Place a set-square with one of its right-angle edges along the ruler.

4. Slide the set-square along the ruler until its other right-angle edge touches the point R.

5. Hold the set-square firmly and draw a line along this edge from point R to intersect $\overline{PQ}$ at a point, say S.

The line segment $\overline{RS}$ is the required perpendicular to $\overline{PQ}$ from point R.

Given:

A line $l$ and a point X on it.

To Construct:

1. A line segment $\overline{XY}$ perpendicular to $l$ at X.

2. A line perpendicular to $\overline{XY}$ at Y.

Construction:

Part 1: Perpendicular to $l$ at X

1. Draw a line $l$ and mark a point X on it.

2. With X as the center, draw an arc of a suitable radius that cuts line $l$ at two points, A and B.

3. With A and B as centers and a radius greater than AX, draw two arcs that intersect at a point, say Y.

4. Join X and Y. The line segment $\overline{XY}$ is perpendicular to the line $l$.

Part 2: Perpendicular to $\overline{XY}$ at Y

5. With Y as the center, draw an arc of a suitable radius that cuts the line containing $\overline{XY}$ at two points, P and Q.

6. With P and Q as centers and a radius greater than YP, draw two arcs that intersect at a point, say Z.

7. Join Y and Z. The line $\overline{YZ}$ is perpendicular to $\overline{XY}$ at Y.

The resulting line passing through Y and Z is parallel to the original line $l$.

Given:

A line segment $\overline{AB}$ of length 7.3 cm.

To Find:

The axis of symmetry of $\overline{AB}$.

Solution:

The axis of symmetry of a line segment is its perpendicular bisector. We need to construct the perpendicular bisector of $\overline{AB}$.

Construction:

1. Draw a line segment $\overline{AB}$ of length 7.3 cm.

2. With A as the center and a radius more than half of AB, draw arcs on both sides of $\overline{AB}$.

3. With B as the center and the same radius, draw arcs on both sides of $\overline{AB}$ to intersect the previous arcs at points P and Q.

4. Join the points P and Q. The line $\overline{PQ}$ intersects $\overline{AB}$ at a point M.

The line $\overline{PQ}$ is the required axis of symmetry of the line segment $\overline{AB}$.

Given:

A line segment of length 9.5 cm.

To Construct:

The perpendicular bisector of the line segment.

Construction:

1. Draw a line segment $\overline{PQ}$ of length 9.5 cm.

2. With P as the center and a radius more than half of PQ, draw arcs on both sides of $\overline{PQ}$.

3. With Q as the center and the same radius, draw arcs on both sides of $\overline{PQ}$ to intersect the previous arcs at points A and B.

4. Join the points A and B. The line $\overline{AB}$ intersects $\overline{PQ}$ at a point M.

The line $\overline{AB}$ is the required perpendicular bisector of the line segment $\overline{PQ}$.

Given:

A line segment $\overline{XY}$ of length 10.3 cm.

Construction of Perpendicular Bisector:

1. Draw a line segment $\overline{XY}$ of length 10.3 cm.

2. With X as the center and a radius more than half of XY, draw arcs on both sides of $\overline{XY}$.

3. With Y as the center and the same radius, draw arcs to intersect the previous arcs at points A and B.

4. Join A and B. This line $\overline{AB}$ is the perpendicular bisector of $\overline{XY}$.

(a) Examination of PX and PY:

Take any point P on the perpendicular bisector $\overline{AB}$. Join P to X and P to Y.

Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.

Therefore, by property, $PX = PY$.

To verify, we can measure the lengths of $\overline{PX}$ and $\overline{PY}$ with a ruler. We will find that they are equal.

Yes, PX = PY.

(b) Relation between MX and XY:

The perpendicular bisector $\overline{AB}$ intersects $\overline{XY}$ at its midpoint, M.

By the definition of a midpoint, M divides the line segment $\overline{XY}$ into two equal halves.

Therefore, the length of $\overline{MX}$ is half the length of $\overline{XY}$.

Given that $XY = 10.3$ cm, we have:

Given:

A line segment of length 12.8 cm.

To Construct and Verify:

Divide the line segment into four equal parts and verify by measurement.

Construction:

1. Draw a line segment $\overline{AB}$ of length 12.8 cm.

2. Construct the perpendicular bisector of $\overline{AB}$. Let it intersect $\overline{AB}$ at point M. M is the midpoint of $\overline{AB}$.

3. Now, construct the perpendicular bisector of the segment $\overline{AM}$. Let it intersect $\overline{AM}$ at point P. P is the midpoint of $\overline{AM}$.

4. Construct the perpendicular bisector of the segment $\overline{MB}$. Let it intersect $\overline{MB}$ at point Q. Q is the midpoint of $\overline{MB}$.

The points P, M, and Q divide the line segment $\overline{AB}$ into four equal parts: $\overline{AP}$, $\overline{PM}$, $\overline{MQ}$, and $\overline{QB}$.

Verification by Measurement:

The total length of $\overline{AB}$ is 12.8 cm.

The expected length of each part is $\frac{12.8}{4} = 3.2$ cm.

By measuring the lengths of $\overline{AP}$, $\overline{PM}$, $\overline{MQ}$, and $\overline{QB}$ with a ruler, we will find that each segment is approximately 3.2 cm long, thus verifying the construction.

Given:

A line segment $\overline{PQ}$ of length 6.1 cm is the diameter of a circle.

To Draw:

A circle with $\overline{PQ}$ as its diameter.

Construction:

1. Draw the line segment $\overline{PQ}$ of length 6.1 cm.

2. Construct the perpendicular bisector of $\overline{PQ}$. Let this bisector intersect $\overline{PQ}$ at point O. This point O is the midpoint of the diameter and is the center of the circle.

3. With O as the center and a radius equal to the length of $\overline{OP}$ (or $\overline{OQ}$), draw a circle using a compass.

The resulting circle has $\overline{PQ}$ as its diameter.

Construction:

1. Draw a circle with center C and radius 3.4 cm.

2. Draw any chord $\overline{AB}$ in the circle.

3. Construct the perpendicular bisector of the chord $\overline{AB}$. To do this, draw intersecting arcs from A and B with a radius more than half of AB. Join the intersection points of these arcs.

Examination:

Upon completing the construction, we observe that the perpendicular bisector of the chord $\overline{AB}$ passes through the center C of the circle.

This is a fundamental property of circles: the perpendicular bisector of any chord of a circle always passes through the center of the circle.

Yes, the perpendicular bisector of $\overline{AB}$ passes through C.

Construction:

1. Draw a circle with center C and radius 3.4 cm.

2. Draw a diameter $\overline{AB}$. A diameter is a chord that passes through the center C.

3. Construct the perpendicular bisector of the diameter $\overline{AB}$.

Examination:

A diameter is a special type of chord. As established in the previous question, the perpendicular bisector of any chord passes through the center.

Therefore, the perpendicular bisector of the diameter $\overline{AB}$ must also pass through the center C.

Furthermore, the center C is the midpoint of the diameter $\overline{AB}$. The perpendicular bisector of a segment passes through its midpoint. So, it is certain that the perpendicular bisector of $\overline{AB}$ passes through C.

Yes, the perpendicular bisector of the diameter $\overline{AB}$ passes through C.

Construction:

1. Draw a circle with a radius of 4 cm. Let the center be O.

2. Draw any two non-parallel chords, say $\overline{AB}$ and $\overline{CD}$.

3. Construct the perpendicular bisector of chord $\overline{AB}$.

4. Construct the perpendicular bisector of chord $\overline{CD}$.

Observation:

The perpendicular bisector of any chord in a circle passes through the center of the circle.

Therefore, the perpendicular bisector of $\overline{AB}$ passes through the center O, and the perpendicular bisector of $\overline{CD}$ also passes through the center O.

When we observe the drawing, we will find that the two constructed perpendicular bisectors intersect at a single point, which is the center of the circle.

Construction:

1. Draw an angle with vertex O.

2. With O as the center and a suitable radius, draw an arc that cuts the arms of the angle at points A and B. By construction, $OA = OB$.

3. Construct the perpendicular bisector of the line segment $\overline{OA}$.

4. Construct the perpendicular bisector of the line segment $\overline{OB}$.

5. Let the two perpendicular bisectors meet at a point P.

6. Join P to A and P to B.

Examination:

We need to examine if $PA = PB$.

Let's use the property of a perpendicular bisector:

• Point P lies on the perpendicular bisector of $\overline{OA}$. Therefore, any point on this line is equidistant from O and A. So, $PO = PA$.
• Point P lies on the perpendicular bisector of $\overline{OB}$. Therefore, any point on this line is equidistant from O and B. So, $PO = PB$.

From these two statements, we have $PA = PO$ and $PB = PO$.

Therefore, by the transitive property of equality, we can conclude that $PA = PB$.

Yes, PA = PB.

This can be verified by measuring the lengths of the line segments $\overline{PA}$ and $\overline{PB}$ with a ruler.

Given:

An angle $\angle$POQ of measure $75^\circ$.

To Draw and Find:

Draw the angle $\angle$POQ and find its line of symmetry.

Construction:

The line of symmetry of an angle is its bisector. First, we will construct an angle of $75^\circ$, and then we will construct its bisector.

Steps to construct $\angle$POQ = $75^\circ$:

1. Draw a ray $\overline{OQ}$.

2. Construct an angle of $90^\circ$ at O. Let the ray for $90^\circ$ be $\overline{OR}$.

3. Construct an angle of $60^\circ$ at O. Let the ray for $60^\circ$ be $\overline{OS}$.

4. The angle $75^\circ$ is the bisector of the $30^\circ$ angle between the $60^\circ$ and $90^\circ$ rays ($\angle$SOR). Bisect $\angle$SOR. The bisecting ray, let's call it $\overline{OP}$, will form a $75^\circ$ angle with $\overline{OQ}$. So, $\angle$POQ = $75^\circ$.

Steps to find the line of symmetry (angle bisector of $\angle$POQ):

1. With O as the center and a suitable radius, draw an arc that intersects the arms $\overline{OP}$ and $\overline{OQ}$ at points A and B respectively.

2. With A as the center and a radius more than half of AB, draw an arc.

3. With B as the center and the same radius, draw another arc to intersect the previous arc at a point, say C.

4. Join O and C and extend it to form a ray $\overline{OC}$.

The ray $\overline{OC}$ is the line of symmetry of $\angle$POQ. It divides $\angle$POQ into two equal angles of $\frac{75^\circ}{2} = 37.5^\circ$ each.

Given:

An angle of measure $147^\circ$.

To Draw and Construct:

The angle of $147^\circ$ and its bisector.

Construction:

1. Draw a ray $\overline{OA}$.

2. Using a protractor, draw an angle of $147^\circ$. Place the center of the protractor at O and the baseline along $\overline{OA}$. Mark a point B at $147^\circ$ and draw the ray $\overline{OB}$. Thus, $\angle AOB = 147^\circ$.

3. To construct the bisector, with O as the center, draw an arc of a suitable radius that intersects the arms $\overline{OA}$ and $\overline{OB}$ at points P and Q respectively.

4. With P as the center and a radius more than half of PQ, draw an arc.

5. With Q as the center and the same radius, draw another arc to intersect the previous arc at a point, say R.

6. Join O and R and extend it to form a ray $\overline{OR}$.

The ray $\overline{OR}$ is the required bisector of $\angle AOB$.

Given:

A right angle ($90^\circ$).

To Draw and Construct:

A right angle and its bisector.

Construction:

1. Draw a ray $\overline{OA}$.

2. Construct a perpendicular to $\overline{OA}$ at point O to get a right angle. Let this be ray $\overline{OB}$. So, $\angle AOB = 90^\circ$.

3. To construct the bisector, with O as the center, draw an arc of a suitable radius that intersects the arms $\overline{OA}$ and $\overline{OB}$ at points P and Q respectively.

4. With P as the center and a radius more than half of PQ, draw an arc.

5. With Q as the center and the same radius, draw another arc to intersect the previous arc at a point, say C.

6. Join O and C and extend it to form a ray $\overline{OC}$.

The ray $\overline{OC}$ is the required bisector of the right angle $\angle AOB$. It divides the angle into two equal angles of $45^\circ$ each.

Given:

An angle of measure $153^\circ$.

To Draw and Divide:

Draw the angle of $153^\circ$ and divide it into four equal parts.

Construction:

1. Using a protractor, draw an angle $\angle AOB = 153^\circ$.

2. First, bisect $\angle AOB$. Let the bisector be the ray $\overline{OC}$. Now, $\angle AOC = \angle COB = \frac{153^\circ}{2} = 76.5^\circ$.

3. Next, bisect the angle $\angle AOC$. Let the bisector be the ray $\overline{OD}$. Now, $\angle AOD = \angle DOC = \frac{76.5^\circ}{2} = 38.25^\circ$.

4. Finally, bisect the angle $\angle COB$. Let the bisector be the ray $\overline{OE}$. Now, $\angle COE = \angle EOB = \frac{76.5^\circ}{2} = 38.25^\circ$.

The rays $\overline{OD}$, $\overline{OC}$, and $\overline{OE}$ are the required lines that divide the angle $\angle AOB = 153^\circ$ into four equal parts.

(a) Construction of $60^\circ$:

1. Draw a ray $\overline{OA}$.

2. With O as the center, draw an arc of any radius intersecting $\overline{OA}$ at P.

3. With P as the center and the same radius, draw an arc intersecting the first arc at Q.

4. Join O and Q and extend to form a ray $\overline{OB}$. $\angle AOB = 60^\circ$.

(b) Construction of $30^\circ$:

1. Construct a $60^\circ$ angle, $\angle AOB$.

2. Bisect $\angle AOB$. The resulting angle will be $30^\circ$.

(c) Construction of $90^\circ$:

1. Draw a ray $\overline{OA}$.

2. Construct a $60^\circ$ angle and a $120^\circ$ angle from O.

3. Bisect the angle between the $60^\circ$ and $120^\circ$ marks. The resulting ray will form a $90^\circ$ angle with $\overline{OA}$.

(d) Construction of $120^\circ$:

1. Draw a ray $\overline{OA}$.

2. With O as the center, draw an arc intersecting $\overline{OA}$ at P.

3. With P as the center and the same radius, cut the arc at Q ($60^\circ$).

4. With Q as the center and the same radius, cut the arc again at R ($120^\circ$).

5. Join O and R and extend to form a ray $\overline{OB}$. $\angle AOB = 120^\circ$.

(e) Construction of $45^\circ$:

1. Construct a $90^\circ$ angle, $\angle AOB$.

2. Bisect $\angle AOB$. The resulting angle will be $45^\circ$.

(f) Construction of $135^\circ$:

1. Draw a line and mark a point O on it.

2. Construct a $90^\circ$ angle at O. This divides the straight angle ($180^\circ$) into two $90^\circ$ angles.

3. Bisect the second $90^\circ$ angle to get a $45^\circ$ angle.

4. The angle $135^\circ$ is the sum of the first $90^\circ$ angle and this $45^\circ$ angle. ($90^\circ + 45^\circ = 135^\circ$).

To Draw and Bisect:

An angle of $45^\circ$ and its bisector.

Construction:

1. First, construct a $45^\circ$ angle. To do this, construct a $90^\circ$ angle and then bisect it. Let the resulting $45^\circ$ angle be $\angle AOB$.

2. Now, bisect the $45^\circ$ angle $\angle AOB$.

3. With O as the center, draw an arc cutting $\overline{OA}$ and $\overline{OB}$ at P and Q.

4. With P and Q as centers and a radius more than half of PQ, draw two arcs that intersect at a point, say C.

5. Join O and C. The ray $\overline{OC}$ is the bisector of the $45^\circ$ angle.

The ray $\overline{OC}$ divides the $45^\circ$ angle into two equal angles of $22.5^\circ$ each.

To Draw and Bisect:

An angle of $135^\circ$ and its bisector.

Construction:

1. First, construct a $135^\circ$ angle using a ruler and compasses. Let this angle be $\angle AOB$.

2. Now, bisect the angle $\angle AOB$.

3. With O as the center, draw an arc cutting the arms $\overline{OA}$ and $\overline{OB}$ at P and Q.

4. With P and Q as centers and a radius more than half of PQ, draw two arcs that intersect at a point, say C.

5. Join O and C. The ray $\overline{OC}$ is the bisector of the $135^\circ$ angle.

The ray $\overline{OC}$ divides the $135^\circ$ angle into two equal angles of $67.5^\circ$ each.

Given:

An angle of measure $70^\circ$.

To Draw and Copy:

Draw a $70^\circ$ angle and make a copy of it using a straightedge and compasses.

Construction:

1. Using a protractor, draw an angle $\angle AOB = 70^\circ$.

2. To make a copy, draw a ray $\overline{QR}$.

3. With O as the center, draw an arc cutting the arms of $\angle AOB$ at C and D.

4. With Q as the center and the same radius, draw an arc cutting the ray $\overline{QR}$ at S.

5. Set the compass to the length of the segment $\overline{CD}$.

6. With S as the center and the compass opening equal to CD, draw an arc that cuts the previous arc at P.

7. Join Q and P and extend to form a ray $\overline{QP}$.

The angle $\angle PQR$ is the required copy of $\angle AOB = 70^\circ$.

Given:

An angle of measure $40^\circ$.

To Draw and Copy:

Draw a $40^\circ$ angle, identify its supplementary angle, and then copy the supplementary angle.

Solution:

1. The supplementary angle of $40^\circ$ is $180^\circ - 40^\circ = 140^\circ$.

2. First, we need to draw the $140^\circ$ angle. Draw a straight line PQ and take a point O on it. Draw an angle of $40^\circ$ with ray $\overline{OQ}$ as one arm. Let the other arm be $\overline{OR}$. Then, the angle $\angle POR$ will be $140^\circ$.

3. Now, we need to copy the angle $\angle POR = 140^\circ$.

4. Follow the same procedure as in Question 8 to copy the $140^\circ$ angle. Draw a ray $\overline{XY}$. With O as the center, draw an arc cutting the arms of $\angle POR$. With X as the center and the same radius, draw another arc. Measure the chord length of the first arc and use it to mark the second arc. Join the points to form the copied angle.

The constructed angle is the copy of the supplementary angle of $40^\circ$.