Class 6 Mathematics - Ch 8: Mastery Guide: Practical Geometry, Line Bisectors, Compass Angle Constructions, Point & Triangle Angles (FBISE)

$60^\circ$ Angle:
Draw base ray. Strike an arc with radius $r$. From intersection on ray, strike arc of same radius $r$. Join to vertex!

$120^\circ$ Angle:
From $60^\circ$ arc mark, strike another arc of radius $r$ along main arc. $60^\circ + 60^\circ = 120^\circ$.

$90^\circ$ Angle:
Bisect the span between $60^\circ$ and $120^\circ$: $60^\circ + \frac{120^\circ - 60^\circ}{2} = 60^\circ + 30^\circ = 90^\circ$.

$30^\circ$ Angle:
Bisect the $60^\circ$ angle: $\frac{60^\circ}{2} = 30^\circ$.

$45^\circ$ Angle:
Bisect the $90^\circ$ right angle: $\frac{90^\circ}{2} = 45^\circ$. (Or bisect span between $30^\circ$ and $60^\circ$).

$75^\circ$ Angle:
Bisect the arc interval between $60^\circ$ and $90^\circ$: $60^\circ + \frac{90^\circ - 60^\circ}{2} = 75^\circ$.

$105^\circ$ Angle:
Bisect the arc interval between $90^\circ$ and $120^\circ$: $90^\circ + \frac{120^\circ - 90^\circ}{2} = 105^\circ$.

General Steps of Construction:

1. Place a straight edge (ruler) firmly on paper.
2. Mark an initial point $A$ at the $0\text{ cm}$ mark with a sharp pencil.
3. Look along the ruler and mark the terminal point $B$ at the exact specified measurement.
4. Join points $A$ and $B$ by drawing a straight pencil line along the ruler edge. $\overline{AB}$ is the required segment.

Detailed Part Conversions & Measurements:

• (a) $4\text{ cm}$ segment: Mark point $A$ at $0\text{ cm}$, point $B$ at $4\text{ cm}$. Length $AB = 4\text{ cm}$.
• (b) $3.5\text{ cm}$ segment: Mark point $A$ at $0\text{ cm}$, point $B$ midway between $3$ and $4\text{ cm}$ ($3.5\text{ cm}$). Length $AB = 3.5\text{ cm}$.
• (c) $6.2\text{ cm}$ segment: Mark point $A$ at $0$, point $B$ two small millimeter divisions past $6\text{ cm}$ ($6.2\text{ cm}$). Length $AB = 6.2\text{ cm}$.
• (d) $7\text{ cm } 4\text{ mm}$ segment: Since $1\text{ cm} = 10\text{ mm}$, $7\text{ cm } 4\text{ mm} = 7.4\text{ cm}$. Mark $A$ at $0$ and $B$ at $7.4\text{ cm}$. Length $AB = 7.4\text{ cm}$.
• (e) $47\text{ mm}$ segment: Convert to centimeters: $\frac{47}{10}\text{ cm} = 4.7\text{ cm}$. Mark $A$ at $0$ and $B$ at $4.7\text{ cm}$. Length $AB = 4.7\text{ cm}$.

• (a) $3AB$:
  Calculation: $3 \times AB = 3 \times 3\text{ cm} = 9\text{ cm}$.
  Construction: Draw a ray and use compass opened to $3\text{ cm}$ to cut off three consecutive segments from starting point $P$: $PQ_1 = 3\text{ cm}, Q_1Q_2 = 3\text{ cm}, Q_2Q_3 = 3\text{ cm}$. Total segment length = $9\text{ cm}$.
• (b) $AB + CD$:
  Calculation: $3\text{ cm} + 4\text{ cm} = 7\text{ cm}$.
  Construction: Cut off segment of $3\text{ cm}$ ($AB$), then from its endpoint cut off an adjoining segment of $4\text{ cm}$ ($CD$). Total length = $7\text{ cm}$.
• (c) $2CD$:
  Calculation: $2 \times CD = 2 \times 4\text{ cm} = 8\text{ cm}$.
  Construction: Cut off two consecutive segments of $4\text{ cm}$ each. Total length = $8\text{ cm}$.
• (d) $4AB - 2CD$:
  Calculation: $4(3\text{ cm}) - 2(4\text{ cm}) = 12\text{ cm} - 8\text{ cm} = 4\text{ cm}$.
  Construction: Draw a line segment of length $12\text{ cm}$ ($4AB$), and from its end mark backwards a length of $8\text{ cm}$ ($2CD$). The remaining segment measures $4\text{ cm}$.
• (e) $0.5CD$:
  Calculation: $0.5 \times 4\text{ cm} = \frac{1}{2} \times 4\text{ cm} = 2\text{ cm}$.
  Construction: Draw segment $CD = 4\text{ cm}$ and construct its right bisector, or measure $2\text{ cm}$ directly with a straight edge. Each half = $2\text{ cm}$.

Standard Steps of Construction for a Right Bisector:

1. Draw a line segment $AB$ of the given measure using a ruler.
2. Take compass with center $A$ and open radius to more than half of $AB$. Draw two arcs, one above and one below $AB$.
3. With center $B$ and the same radius, draw two arcs intersecting the first arcs at points $X$ (above) and $Y$ (below).
4. Draw a straight line through $X$ and $Y$. Line $\overleftrightarrow{XY}$ intersects $\overline{AB}$ at point $C$.
5. $\overleftrightarrow{XY}$ is the perpendicular (right) bisector of $\overline{AB}$, and $C$ is the midpoint ($AC = CB$).

Measurements for Each Part:

• (a) Segment $5.2\text{ cm}$ long:
  Each part $= \frac{5.2\text{ cm}}{2} = \mathbf{2.6\text{ cm}}$. Verification: $AC = 2.6\text{ cm}, CB = 2.6\text{ cm}$.
• (b) Segment $7\text{ cm}$ long:
  Each part $= \frac{7\text{ cm}}{2} = \mathbf{3.5\text{ cm}}$. Verification: $AC = 3.5\text{ cm}, CB = 3.5\text{ cm}$.
• (c) Segment $8\text{ cm}$ long:
  Each part $= \frac{8\text{ cm}}{2} = \mathbf{4\text{ cm}}$. Verification: $AC = 4\text{ cm}, CB = 4\text{ cm}$.

Steps of Construction:

1. Draw segment $AB = 6\text{ cm}$ with a straight edge.
2. With center $A$ and compass opened to $4\text{ cm}$ (more than half of $6\text{ cm}$), draw arcs above and below $\overline{AB}$.
3. With center $B$ and the same radius $4\text{ cm}$, draw arcs intersecting the previous arcs at points $P$ and $Q$.
4. Join $P$ and $Q$ by a line. Line $\overleftrightarrow{PQ}$ intersects $\overline{AB}$ at point $M$.
5. $\overleftrightarrow{PQ}$ is the required right bisector. Measuring gives $AM = MB = 3\text{ cm}$ and $\angle PMA = 90^\circ$.

Steps of Construction:

1. Draw line segment $CD = 8\text{ cm}$.
2. First Bisection: Draw the right bisector of $CD$ intersecting at midpoint $M$. Now $CM = MD = 4\text{ cm}$ (2 equal parts).
3. Second Bisections: Draw the right bisector of segment $CM$ to get point $P$, and the right bisector of segment $MD$ to get point $Q$.
4. The line segment $CD$ is now divided into 4 equal segments: $CP, PM, MQ, QD$.

Length of Each Part: $\frac{8\text{ cm}}{4} = \mathbf{2\text{ cm}}$ each ($CP = PM = MQ = QD = 2\text{ cm}$).

Steps of Construction (Perpendicular from Point ON Line):

1. Draw a horizontal line $\overleftrightarrow{PQ}$ and mark a point $X$ on it.
2. With center $X$ and any convenient radius, draw a semicircle intersecting line $PQ$ at points $D$ and $E$.
3. With center $D$ and radius greater than $DX$ (more than half of $DE$), draw an arc above the line.
4. With center $E$ and the same radius, draw another arc intersecting the previous arc at point $F$.
5. Draw a line passing through $X$ and $F$. Line $\overleftrightarrow{XF}$ is the required perpendicular on $\overleftrightarrow{PQ}$ at point $X$ ($\angle PXF = \angle QXF = 90^\circ$).

Steps of Construction (Perpendicular from External Point NOT ON Line):

1. Draw a straight line $\overleftrightarrow{XY}$ and mark an external point $P$ above or below the line.
2. With center $P$ and compass radius greater than the perpendicular distance to $\overleftrightarrow{XY}$, draw an arc cutting line $XY$ at two points, $A$ and $B$.
3. With center $A$ and radius greater than half of $AB$, draw an arc on the opposite side of the line from $P$.
4. With center $B$ and the same radius, draw another arc cutting the previous arc at point $Q$.
5. Draw a straight line connecting $P$ and $Q$. Line $\overleftrightarrow{PQ}$ intersects line $XY$ at point $M$.
6. $\overline{PM}$ is the required perpendicular. Placing a ruler between $P$ and $M$ gives the exact measured length (e.g., $3.2\text{ cm}$ depending on chosen point position).

• (a) $60^\circ$: Draw base ray $OA$. With center $O$ and any radius, draw an arc cutting $OA$ at $P$. With center $P$ and the same radius, draw an arc intersecting the first arc at $Q$. Draw ray $OB$ through $Q$. $\angle AOB = 60^\circ$.
• (b) $90^\circ$: Construct $60^\circ$ arc ($Q$) and $120^\circ$ arc ($R$) using the same radius. With centers $Q$ and $R$ and same radius, draw intersecting arcs above at point $S$. Draw ray $OC$ through $S$. $\angle AOC = 90^\circ$.
• (c) $45^\circ$: Construct $90^\circ$ angle ray $OC$. Let the base arc intersect ray $OA$ at $P$ and ray $OC$ at $S$. With centers $P$ and $S$ and radius greater than half of $PS$, draw two arcs intersecting at point $T$. Draw ray $OD$ through $T$. $\angle AOD = \frac{90^\circ}{2} = 45^\circ$.
• (d) $30^\circ$: Construct $60^\circ$ ray $OB$. The base arc cuts $OA$ at $P$ and ray $OB$ at $Q$. With centers $P$ and $Q$ and radius greater than half of $PQ$, draw intersecting arcs at point $K$. Draw ray through $K$. $\angle = \frac{60^\circ}{2} = 30^\circ$.
• (e) $120^\circ$: From the $60^\circ$ intersection point $Q$ on the main arc, draw another arc of the same radius cutting the main arc at $R$. Draw ray $OE$ through $R$. $\angle AOE = 60^\circ + 60^\circ = 120^\circ$.
• (f) $75^\circ$: Construct $60^\circ$ ray ($Q$) and $90^\circ$ ray ($S$). The arc distance between $60^\circ$ and $90^\circ$ is $30^\circ$. With centers $Q$ and $S$, draw intersecting arcs to bisect this angle interval: $60^\circ + \frac{30^\circ}{2} = 75^\circ$. Draw ray through intersection. $\angle = 75^\circ$.
• (g) $105^\circ$: Construct $90^\circ$ ray ($S$) and $120^\circ$ ray ($R$). The arc distance between $90^\circ$ and $120^\circ$ is $30^\circ$. With centers $S$ and $R$, draw intersecting arcs to bisect this interval: $90^\circ + \frac{30^\circ}{2} = 105^\circ$. Draw ray through intersection. $\angle = 105^\circ$.

Steps of Construction:

1. Draw segment $PQ = 7\text{ cm}$ using a straight edge.
2. At vertex $P$: With center $P$, draw a semicircle arc. Mark $60^\circ$ and $120^\circ$ marks. Bisect between them to get a perpendicular line. Draw ray $PX$ making $\angle XPQ = 90^\circ$.
3. At vertex $Q$: With center $Q$, draw an arc cutting $QP$. From the intersection, strike an arc of same radius ($60^\circ$). Bisect this $60^\circ$ angle to obtain $30^\circ$. Draw ray $QY$ making $\angle PQY = 30^\circ$.
4. Result: $\angle XPQ = 90^\circ$ and $\angle PQY = 30^\circ$ on base $PQ = 7\text{ cm}$.

Steps of Construction:

1. Draw a straight line $AB$ and mark a point $C$ on it.
2. At point $C$, construct an angle of $90^\circ$ and $120^\circ$.
3. Bisect the angle between $90^\circ$ and $120^\circ$ to get ray $CD$ such that $\angle BCD = 105^\circ$.

Calculation for Angle on Other Side ($\angle ACD$):
Since angles on a straight line are supplementary ($\sum = 180^\circ$): $$\angle ACD + \angle BCD = 180^\circ \implies \angle ACD + 105^\circ = 180^\circ$$ $$\angle ACD = 180^\circ - 105^\circ = \mathbf{75^\circ}$$ Answer: The angle on the other side measures $\mathbf{75^\circ}$.

• For $60^\circ$ Angle:
  Construct an angle of $60^\circ$ using compasses. Bisect it with an angle bisector ray.
  Measure of each part $= \frac{60^\circ}{2} = \mathbf{30^\circ}$.
• For $90^\circ$ Angle:
  Construct a right angle ($90^\circ$) using compasses. Bisect it with an angle bisector ray.
  Measure of each part $= \frac{90^\circ}{2} = \mathbf{45^\circ}$.

Steps of Construction:

1. Construct an angle of $120^\circ$ ($\angle AOB$) using compasses.
2. First Bisection: Bisect $\angle AOB$ to get ray $OC$. Now $\angle AOC = \angle COB = 60^\circ$ (2 equal halves).
3. Second Bisection: Bisect $\angle AOC$ to obtain ray $OD$ ($30^\circ$), and bisect $\angle COB$ to obtain ray $OE$ ($30^\circ$).
4. This yields four equal congruent angles: $\angle AOD, \angle DOC, \angle COE, \angle EOB$.

Measure of Each Part: $\frac{120^\circ}{4} = \mathbf{30^\circ}$.

Steps of Construction:

1. Draw base line segment $AB = 6\text{ cm}$.
2. At vertex $A$, construct an angle of $45^\circ$ using compasses and extend the arm $AX$.
3. At vertex $B$, construct an angle of $30^\circ$ directed towards the interior and extend arm $BY$.
4. The terminal rays $AX$ and $BY$ intersect at a unique point, let's call it $C$.

Observation ("What happens?"):
The two arms intersect to form a Triangle $\triangle ABC$!
Third Angle Check: $\angle C = 180^\circ - (45^\circ + 30^\circ) = 180^\circ - 75^\circ = 105^\circ$. Thus, a scalene obtuse-angled triangle is formed!

• (i) Three equal angles around a central point, labeled $2x, 2x, 2x$:
  Sum of all angles around a point is $360^\circ$: $$2x + 2x + 2x = 360^\circ \implies 6x = 360^\circ \implies x = \frac{360^\circ}{6} = 60^\circ$$ Each angle is $2x = 2(60^\circ) = \mathbf{120^\circ}$.
  Answer: $x = 60^\circ$, each angle $= \mathbf{120^\circ \text{ each}}$.
• (ii) Angles around a point: $25^\circ, a, 65^\circ, 140^\circ$:
  Sum of angles around a point $= 360^\circ$: $$25^\circ + a + 65^\circ + 140^\circ = 360^\circ$$ $$a + 230^\circ = 360^\circ \implies a = 360^\circ - 230^\circ = \mathbf{130^\circ}$$
  Answer: $a = \mathbf{130^\circ}$.
• (iii) Four angles meeting at a point with measures in terms of $x$: $4x, x, 2x, 3x$:
  Sum of angles around a point $= 360^\circ$: $$4x + x + 2x + 3x = 360^\circ \implies 10x = 360^\circ \implies x = \frac{360^\circ}{10} = 36^\circ$$ Calculating the individual angle values: $$\text{Angle 1: } x = 36^\circ$$ $$\text{Angle 2: } 2x = 2(36^\circ) = 72^\circ$$ $$\text{Angle 3: } 3x = 3(36^\circ) = 108^\circ$$ $$\text{Angle 4: } 4x = 4(36^\circ) = 144^\circ$$
  Answer: $\mathbf{36^\circ, 72^\circ, 108^\circ, 144^\circ}$.
• (iv) Intersecting lines with angles $130^\circ, c, a, b$:
  By vertically opposite angles: $$a = 130^\circ \quad (\text{vertically opposite to } 130^\circ)$$ Angles on a straight line add to $180^\circ$: $$b + 130^\circ = 180^\circ \implies b = 180^\circ - 130^\circ = 50^\circ$$ $$c = b = 50^\circ \quad (\text{vertically opposite})$$
  Answer: $b = c = \mathbf{50^\circ}, a = \mathbf{130^\circ}$.
• (v) Angles around a point: $(4x + 10)^\circ, (3x - 5)^\circ, (2x - 5)^\circ$:
  Sum of angles around a point $= 360^\circ$: $$(4x + 10) + (3x - 5) + (2x - 5) = 360^\circ$$ $$(4x + 3x + 2x) + (10 - 5 - 5) = 360^\circ \implies 9x + 0 = 360^\circ \implies 9x = 360^\circ \implies x = 40^\circ$$ Now evaluate each angle: $$\text{Angle 1: } 4x + 10 = 4(40^\circ) + 10 = 160^\circ + 10^\circ = \mathbf{170^\circ}$$ $$\text{Angle 2: } 3x - 5 = 3(40^\circ) - 5 = 120^\circ - 5^\circ = \mathbf{115^\circ}$$ $$\text{Angle 3: } 2x - 5 = 2(40^\circ) - 5 = 80^\circ - 5^\circ = \mathbf{75^\circ}$$
  Answer: $\mathbf{115^\circ, 75^\circ, 170^\circ}$ (Sum $= 115^\circ + 75^\circ + 170^\circ = 360^\circ$).
• (vi) Figure with right angle ($90^\circ$), $35^\circ$, $5x$, $z$, $y$:
  On the lower right quadrant, adjacent angles form a right angle: $5x + 35^\circ = 90^\circ \implies 5x = 55^\circ \implies a = 5x = 55^\circ$.
  Angles on straight line: $y = 55^\circ$, $z = 180^\circ - 55^\circ = 125^\circ$.
  Answer: $a = 55^\circ = y, z = \mathbf{125^\circ}$.

Analysis from Figure:
Rays intersect at a central point. We are given $d = 25^\circ$, $c = 40^\circ$, with angle $3c$: $$3c = 3(40^\circ) = 120^\circ$$ By vertically opposite angles: $$b = 3c = \mathbf{120^\circ}$$ Along the straight line containing $a, b, d$ or vertically opposite: $$a + b + d = 180^\circ \implies a + 120^\circ + \dots$$ Looking at the straight line containing $a, 3c, e$: $$a = 60^\circ$$ On the lower line: $$e + 3c + d = \dots \implies e + 120^\circ + 25^\circ = 180^\circ \implies e = 180^\circ - 145^\circ = \mathbf{35^\circ}$$
Answer: $e = \mathbf{35^\circ}, b = \mathbf{120^\circ}, a = \mathbf{60^\circ}$.

Step-by-step Solution:

1. Substitute $a = 3b$ into $a + b = 100^\circ$: $$3b + b = 100^\circ \implies 4b = 100^\circ \implies b = \frac{100^\circ}{4} = \mathbf{25^\circ}$$
2. Find $a$: $$a = 3b = 3(25^\circ) = \mathbf{75^\circ}$$
3. From the intersecting lines figure, angles $a$, $b$, and $c$ lie along the straight line: $$a + b + c = 180^\circ$$ Substitute $a + b = 100^\circ$: $$100^\circ + c = 180^\circ \implies c = 180^\circ - 100^\circ = \mathbf{80^\circ}$$

Answer: $a = \mathbf{75^\circ}, b = \mathbf{25^\circ}, c = \mathbf{80^\circ}$.

• i. $\angle A = 30^\circ, \angle B = 90^\circ, \angle C = ?$
  $\angle C = 180^\circ - (30^\circ + 90^\circ) = 180^\circ - 120^\circ = \mathbf{60^\circ}$.
• ii. $\angle A = ?, \angle B = 78^\circ, \angle C = 45^\circ$
  $\angle A = 180^\circ - (78^\circ + 45^\circ) = 180^\circ - 123^\circ = \mathbf{57^\circ}$.
• iii. $\angle A = 70.5^\circ, \angle B = ?, \angle C = 98.9^\circ$
  $\angle B = 180^\circ - (70.5^\circ + 98.9^\circ) = 180^\circ - 169.4^\circ = \mathbf{10.6^\circ}$.
• iv. $\angle A = 22\frac{1}{2}^\circ = 22.5^\circ, \angle B = 115\frac{1}{2}^\circ = 115.5^\circ, \angle C = ?$
  $\angle C = 180^\circ - (22.5^\circ + 115.5^\circ) = 180^\circ - 138^\circ = \mathbf{42^\circ}$.
• v. $\angle A = 100.5^\circ, \angle B = ?, \angle C = 40\frac{1}{2}^\circ = 40.5^\circ$
  $\angle B = 180^\circ - (100.5^\circ + 40.5^\circ) = 180^\circ - 141^\circ = \mathbf{39^\circ}$.

• (i) Triangle with angles $60^\circ$ and $60^\circ$:
  Third angle $= 180^\circ - (60^\circ + 60^\circ) = 180^\circ - 120^\circ = \mathbf{60^\circ}$ (Equilateral triangle).
• (ii) Right-angled triangle with one acute angle $50^\circ$:
  Third angle $= 180^\circ - (90^\circ + 50^\circ) = 180^\circ - 140^\circ = \mathbf{40^\circ}$.
• (iii) Triangle with angles $80^\circ$ and $50^\circ$:
  Third angle $= 180^\circ - (80^\circ + 50^\circ) = 180^\circ - 130^\circ = \mathbf{50^\circ}$ (Isosceles triangle).

Key Concept: In an isosceles triangle, the two angles opposite to equal sides are congruent ($a = b$). The third (vertex) angle $= 180^\circ - 2a$.

• (i) Equal angles $= 55^\circ$:
  Third angle $= 180^\circ - (55^\circ + 55^\circ) = 180^\circ - 110^\circ = \mathbf{70^\circ}$.
• (ii) Equal angles $= 70^\circ$:
  Third angle $= 180^\circ - (70^\circ + 70^\circ) = 180^\circ - 140^\circ = \mathbf{40^\circ}$.
• (iii) Equal angles $= 25^\circ$:
  Third angle $= 180^\circ - (25^\circ + 25^\circ) = 180^\circ - 50^\circ = \mathbf{130^\circ}$.

• (i) Triangle with interior angles $x, 2x$, and $60^\circ$:
  $$x + 2x + 60^\circ = 180^\circ \implies 3x + 60^\circ = 180^\circ$$ $$3x = 180^\circ - 60^\circ = 120^\circ \implies x = \frac{120^\circ}{3} = \mathbf{40^\circ}$$ Then $2x = 2(40^\circ) = \mathbf{80^\circ}$.
  Answer: Unknown angles are $\mathbf{40^\circ, 80^\circ}$.
• (ii) Triangle with angles in ratio $y, 2y, 3y$:
  $$y + 2y + 3y = 180^\circ \implies 6y = 180^\circ \implies y = \frac{180^\circ}{6} = \mathbf{30^\circ}$$ The angles are: $y = \mathbf{30^\circ}, 2y = \mathbf{60^\circ}, 3y = \mathbf{90^\circ}$.
  Answer: $\mathbf{30^\circ, 60^\circ, 90^\circ}$.
• (iii) Triangle with angles $50^\circ, 70^\circ$, interior angle $y$, exterior angle $x$:
  Interior angle sum: $50^\circ + 70^\circ + y = 180^\circ \implies 120^\circ + y = 180^\circ \implies y = \mathbf{60^\circ}$.
  Exterior angle on straight line: $x + y = 180^\circ \implies x = 180^\circ - 60^\circ = \mathbf{120^\circ}$.
  (Or by exterior angle theorem: $x = 50^\circ + 70^\circ = 120^\circ$).
  Answer: $y = \mathbf{60^\circ}, x = \mathbf{120^\circ}$.
• (iv) Exterior angle is $150^\circ$, adjacent interior is $x$, other two angles are $y$ and $2y$:
  Adjacent interior on straight line: $x + 150^\circ = 180^\circ \implies x = 180^\circ - 150^\circ = \mathbf{30^\circ}$ (or labeled $y$ depending on figure lettering).
  By exterior angle theorem: $y + 2y = 150^\circ \implies 3y = 150^\circ \implies y = \mathbf{50^\circ}$.
  Then $2y = 2(50^\circ) = \mathbf{100^\circ}$.
  Answer: $\mathbf{30^\circ, 50^\circ, 100^\circ}$.
• (v) Triangle with angles $2x, 3x$, and $35^\circ$:
  $$2x + 3x + 35^\circ = 180^\circ \implies 5x + 35^\circ = 180^\circ$$ $$5x = 180^\circ - 35^\circ = 145^\circ \implies x = \frac{145^\circ}{5} = 29^\circ$$ Calculating angles: $2x = 2(29^\circ) = \mathbf{58^\circ}$, and $3x = 3(29^\circ) = \mathbf{87^\circ}$.
  Answer: $\mathbf{58^\circ, 87^\circ}$.
• (vi) Triangle with right angle ($90^\circ$), vertically opposite angle $30^\circ$, unknowns $y, z, x$:
  Vertically opposite: $y = \mathbf{30^\circ}$.
  In right-angled triangle: $y + 90^\circ + z = 180^\circ \implies 30^\circ + 90^\circ + z = 180^\circ \implies z = 180^\circ - 120^\circ = \mathbf{60^\circ}$.
  Exterior angle on straight line: $x + z = 180^\circ \implies x = 180^\circ - 60^\circ = \mathbf{120^\circ}$.
  Answer: $y = \mathbf{30^\circ}, z = \mathbf{60^\circ}, x = \mathbf{120^\circ}$.

• (i) The line segment $AB$ is denoted by:
  Options: (a) $AB$   (b) $\overline{AB}$   (c) $\overrightarrow{AB}$   (d) $\overline{AB}$ [standard textbook key marks option (d)].
  Answer: (d) $\overline{AB}$. (A line segment has a straight bar over the letters).
• (ii) Sum of measures of two sides of a triangle is ... the measure of third side.
  Options: (a) equal to   (b) greater than   (c) less than   (d) none of these
  Answer: (b) greater than (Triangle Inequality Theorem: $a + b > c$).
• (iii) To bisect a line segment means to divide it into ... equal parts.
  Options: (a) 2   (b) 3   (c) 4   (d) 5
  Answer: (a) 2 ("Bi" means two).
• (iv) The point where a line segment is bisected, is called:
  Options: (a) Initial point   (b) end point   (c) mid point   (d) any point
  Answer: (c) mid point.
• (v) A perpendicular line makes an angle of ... with given line.
  Options: (a) $0^\circ$   (b) $45^\circ$   (c) $90^\circ$   (d) $180^\circ$
  Answer: (c) $90^\circ$.
• (vi) To bisect an angle, we use an instrument called ...
  Options: (a) ruler   (b) compasses   (c) divider   (d) set square
  Answer: (b) compasses.
• (vii) Sum of measures of three angles of a triangle is ...
  Options: (a) $45^\circ$   (b) $160^\circ$   (c) $180^\circ$   (d) $360^\circ$
  Answer: (c) $180^\circ$.
• (viii) $100^\circ$ and $y$ are two angles on a straight line. The value of $y$ is:
  Options: (a) $150^\circ$   (b) $100^\circ$   (c) $50^\circ$   (d) $80^\circ$
  Answer: (d) $80^\circ$ ($180^\circ - 100^\circ = 80^\circ$).
• (ix) $x, 2x$ and $3x$ are angles of a triangle. The value of $x$ is:
  Options: (a) $30^\circ$   (b) $60^\circ$   (c) $90^\circ$   (d) $180^\circ$
  Answer: (a) $30^\circ$ ($x + 2x + 3x = 6x = 180^\circ \implies x = 30^\circ$).

Steps of Construction:

1. Draw a line segment $AB = 7.4\text{ cm}$ using a straight edge.
2. With center $A$ and radius $> 3.7\text{ cm}$ (e.g., $4.5\text{ cm}$), draw arcs above and below $AB$.
3. With center $B$ and the same radius, draw arcs cutting the first arcs at $X$ and $Y$.
4. Draw a line through $X$ and $Y$ intersecting $\overline{AB}$ at midpoint $M$.

Measurement: Each part $= \frac{7.4\text{ cm}}{2} = \mathbf{3.7\text{ cm}}$ ($AM = MB = 3.7\text{ cm}$).

Steps of Construction:

1. Draw a line $PQ$ and mark a point $R$ on it.
2. With center $R$ and convenient radius, draw a semicircle cutting $PQ$ at points $E$ and $F$.
3. With center $E$ and radius greater than $ER$, draw an arc above the line.
4. With center $F$ and the same radius, draw an arc intersecting the previous arc at $K$.
5. Draw line through $R$ and $K$. Line $RK$ is the required perpendicular to $PQ$ at $R$ ($\angle PRK = \angle QRK = 90^\circ$).

• (i) Figure with intersecting lines having angles $3x, 2x, x, y, z$:
  Angles on straight line sum to $180^\circ$: $$3x + 2x + x = 180^\circ \implies 6x = 180^\circ \implies x = \frac{180^\circ}{6} = \mathbf{30^\circ}$$ Then: $$y = 3x = 3(30^\circ) = \mathbf{90^\circ} \quad (\text{vertically opposite})$$ $$z = 2x + x = 3x = \mathbf{90^\circ} \quad (\text{vertically opposite})$$
  Answer: $x = \mathbf{30^\circ}, y = z = \mathbf{90^\circ}$.
• (ii) Figure with intersecting lines with angles $60^\circ, 30^\circ, x, y, 140^\circ, z$:
  Angles on top half of straight line sum to $180^\circ$: $$60^\circ + 30^\circ + x + y = 180^\circ \dots$$ From the textbook figure and vertical angles: $$x = \mathbf{50^\circ}, \quad y = z = \mathbf{40^\circ}$$
  Answer: $x = \mathbf{50^\circ}, y = z = \mathbf{40^\circ}$.

• (i) Figure with two intersecting lines and an isosceles triangle:
  Vertically opposite angle: $$b = 40^\circ$$ In the triangle, the two base angles are equal ($a$ and $a$): $$a + a + b = 180^\circ \implies 2a + 40^\circ = 180^\circ$$ $$2a = 180^\circ - 40^\circ = 140^\circ \implies a = \frac{140^\circ}{2} = \mathbf{70^\circ}$$
  Answer: $a = \mathbf{70^\circ}, b = \mathbf{40^\circ}$.
• (ii) Figure with reflex angle $280^\circ$, parallel lines, angle $40^\circ$, and unknowns $a, b, c, d$:
  At vertex with reflex angle: Full turn is $360^\circ$: $$b = 360^\circ - 280^\circ = \mathbf{80^\circ}$$ In the triangle: $$a + b + 40^\circ = 180^\circ \implies a + 80^\circ + 40^\circ = 180^\circ \implies a + 120^\circ = 180^\circ \implies a = \mathbf{60^\circ}$$ Supplementary/exterior angles: $$c = 180^\circ - 40^\circ = \mathbf{140^\circ}$$ $$d = 180^\circ - a = 180^\circ - 60^\circ = \mathbf{120^\circ}$$
  Answer: $a = \mathbf{60^\circ}, b = \mathbf{80^\circ}, c = \mathbf{140^\circ}, d = \mathbf{120^\circ}$.