Class 8 Mathematics Ch 10: Practical Geometry Mastery Guide (FBISE)
Mastery Guide: Unit 10 • Practical Geometry (عملی جیومیٹری)
Comprehensive manual covering compass and straightedge construction of Triangles ($SSS, SAS, ASA, RHS$), Quadrilaterals (Square, Rectangle, Parallelogram, Rhombus, Trapezium, Kite), and Geometric Angle and Line Bisectors.
- Construct triangles using $SSS, SAS, ASA$, and $RHS$ criteria.
- Apply the Triangle Inequality Theorem ($a+b > c$) to assess construction feasibility.
- Construct all 6 major quadrilaterals with given sides, angles, and diagonals.
- Construct interior angle bisectors and perpendicular side bisectors to locate the Incenter and Circumcenter.
- Impossible Triangle Trap: If $a + b \le c$ (i.e. $a + b \le c$), compass arcs will never intersect!
- Parallelogram Diagonals: Diagonals bisect each other; remember to take half-diagonal ($d/2$) on each side of the center $O$.
- Kite vs Rhombus: A kite has 2 pairs of adjacent equal sides; diagonals are perpendicular but only one diagonal bisects the other.
📐 The 3-Step Practical Geometry Protocol
Draw a quick freehand diagram, label vertices in order ($ABCD$), mark given lengths, angles, and diagonals, and verify feasibility ($a+b > c$).
Draw the primary base with a ruler, construct given angles ($30^\circ, 45^\circ, 60^\circ, 90^\circ, 120^\circ$) using ruler and compass arcs.
Set compass to required radii, strike intersecting arcs to establish missing vertices, join with straight lines, and write formal Steps of Construction.
1. Triangle Construction Principles & Criteria
A unique triangle is determined when at least three independent measurements (with at least one side length) are provided:
- Side-Side-Side ($SSS$): Draw base $AB = c$. Draw arc of radius $b$ from $A$ and arc of radius $a$ from $B$. Their intersection gives vertex $C$. Condition: $a + b > c, b + c > a, c + a > b$.
- Side-Angle-Side ($SAS$): Draw base $AB$. Construct given angle $\angle A$ and draw ray $AX$. On ray $AX$, cut length $AC = b$. Join $C$ to $B$.
- Angle-Side-Angle ($ASA$): Draw base $AB$. Construct angle $\angle A$ at $A$ (ray $AX$) and angle $\angle B$ at $B$ (ray $BY$). The intersection of rays gives vertex $C$. If two angles and a non-included side are given ($AAS$), find the third angle via $\angle C = 180^\circ - (\angle A + \angle B)$.
- Hypotenuse-Side ($RHS / HS$): Draw one leg $AB$. Construct a right angle ($90^\circ$) at vertex $B$ to form ray $BX$. With centre $A$ and radius equal to the hypotenuse, draw an arc intersecting ray $BX$ at $C$. Join $A$ to $C$.
2. Quadrilateral Construction Protocols
Given 1 side: construct $90^\circ$ angles at ends. Given 1 diagonal $d$: draw $d$, construct its perpendicular bisector, and mark $d/2$ above and below the midpoint.
Given sides $a, b$: draw base $a$, erect $90^\circ$ perpendiculars of height $b$. Given diagonal $d$ and angle $\theta$: bisect $d$, draw line at angle $\theta$ through center, and mark $d/2$ on both sides.
Given sides $a, b$ and angle $\theta$: draw base $a$, draw ray at angle $\theta$ of length $b$. Arcs from opposite vertices of lengths $a$ and $b$ intersect at the 4th vertex.
Given side $a$ and diagonal $d$: draw $d$, strike arcs of radius $a$ from both ends on both sides of $d$. Intersection points form the opposite vertices.
Draw base $AB$. Draw a parallel line $l$ at given height $h$. Strike arcs from $A$ and $B$ of specified slant leg lengths to cut line $l$ at vertices $D$ and $C$.
Draw diagonal $AC$. From $A$, strike arcs of radius $a$ above and below. From $C$, strike arcs of radius $b$ above and below to intersect at $B$ and $D$.
3. Angle & Line Bisectors: Concurrency & Circle Inscription
Bisectors divide geometric entities into two strictly congruent halves:
- Angle Bisector: With vertex as center, draw an arc cutting both angle arms at $P$ and $Q$. With centers $P$ and $Q$, draw equal intersecting arcs at $R$. Ray through vertex and $R$ bisects the angle into two equal parts.
- Perpendicular (Right) Bisector of a Line Segment: With radius greater than half of segment $AB$, draw arcs above and below from $A$, and similarly from $B$. The line passing through arc intersections bisects $AB$ at $90^\circ$.
- Incenter ($I$): The intersection point of the 3 interior angle bisectors of a triangle. It is equidistant from all three sides and forms the center of the In-circle (incircle).
- Circumcenter ($O$): The intersection point of the 3 perpendicular side bisectors of a triangle. It is equidistant from all three vertices and forms the center of the Circumcircle.
📝 Solved Exercise 10.1 • Construction of Triangles
1. Construct the following triangles where possible:
(a) $\Delta ABC$ when $AB = 5.5\text{ cm}, BC = 4.3\text{ cm}, AC = 4.2\text{ cm}$
Steps of Construction:
1. Draw a line segment $AB = 5.5\text{ cm}$ with a ruler.
2. With point $A$ as center, draw an arc of radius $4.2\text{ cm}$ using a compass.
3. With point $B$ as center, draw another arc of radius $4.3\text{ cm}$ intersecting the first arc at point $C$.
4. Join point $C$ to point $A$ and point $B$.
Result: $\Delta ABC$ is the required triangle ($SSS$).
(b) $\Delta DEF$ when $EF = 6\text{ cm}, DE = 5.2\text{ cm}, DF = 5.8\text{ cm}$
Steps of Construction:
1. Draw base line segment $EF = 6\text{ cm}$.
2. With center $E$, draw an arc of radius $5.2\text{ cm}$.
3. With center $F$, draw an arc of radius $5.8\text{ cm}$ intersecting the previous arc at point $D$.
4. Join $D$ to $E$ and $F$.
Result: $\Delta DEF$ is the required triangle ($SSS$).
(c) $\Delta ABC$ when $AB = 5.8\text{ cm}, AC = 4.2\text{ cm}, \angle A = 90^\circ$
Steps of Construction:
1. Draw base segment $AB = 5.8\text{ cm}$.
2. At vertex $A$, construct an angle of $90^\circ$ using compass/protractor and draw ray $AX$.
3. With center $A$ and compass radius $4.2\text{ cm}$, cut an arc on ray $AX$ at point $C$.
4. Join point $C$ to $B$.
Result: $\Delta ABC$ is the required right-angled triangle ($SAS$).
(d) $\Delta LMN$ when $LM = 4\text{ cm}, MN = 4.7\text{ cm}, \angle M = 120^\circ$
Steps of Construction:
1. Draw line segment $LM = 4\text{ cm}$.
2. At point $M$, construct an obtuse angle of $120^\circ$ and draw ray $MX$.
3. From $M$, cut off $MN = 4.7\text{ cm}$ along ray $MX$ to locate point $N$.
4. Join point $N$ to $L$.
Result: $\Delta LMN$ is the required obtuse-angled triangle ($SAS$).
(e) $\Delta PQR$ when $PQ = 7.2\text{ cm}, \angle P = 45^\circ, \angle Q = 75^\circ$
Steps of Construction:
1. Draw base $PQ = 7.2\text{ cm}$.
2. At vertex $P$, construct an angle of $45^\circ$ and draw ray $PX$.
3. At vertex $Q$, construct an angle of $75^\circ$ and draw ray $QY$.
4. The two rays $PX$ and $QY$ intersect at point $R$.
Result: $\Delta PQR$ is the required triangle ($ASA$).
(f) $\Delta XYZ$ when $XZ = 5\text{ cm}, \angle X = 30^\circ, \angle Z = 105^\circ$
Steps of Construction:
1. Draw base $XZ = 5\text{ cm}$.
2. At $X$, construct $\angle X = 30^\circ$ and draw ray $XA$.
3. At $Z$, construct $\angle Z = 105^\circ$ and draw ray $ZB$.
4. Rays $XA$ and $ZB$ intersect at point $Y$.
Result: $\Delta XYZ$ is the required triangle ($ASA$).
(g) $\Delta XYZ$ when $XY = 6.3\text{ cm}, \angle X = 60^\circ, \angle Z = 90^\circ$
Calculations: $\angle Y = 180^\circ - (\angle X + \angle Z) = 180^\circ - (60^\circ + 90^\circ) = 30^\circ$.
Steps of Construction:
1. Draw base $XY = 6.3\text{ cm}$.
2. At $X$, construct an angle of $60^\circ$ and draw ray $XP$.
3. At $Y$, construct an angle of $30^\circ$ and draw ray $YQ$.
4. The rays $XP$ and $YQ$ intersect at point $Z$, forming $\angle Z = 90^\circ$.
Result: $\Delta XYZ$ is the required right triangle ($ASA$).
(h) $\Delta ABC$ when $AB = 6.8\text{ cm}, BC = 8.1\text{ cm}, \angle A = 90^\circ$
Steps of Construction:
1. Draw leg $AB = 6.8\text{ cm}$.
2. At vertex $A$, erect perpendicular ray $AX$ (angle $90^\circ$).
3. With center $B$ and radius equal to hypotenuse $BC = 8.1\text{ cm}$, draw an arc cutting ray $AX$ at point $C$.
4. Join $B$ to $C$.
Result: $\Delta ABC$ is the required right triangle ($RHS$).
(i) $\Delta PQR$ when $PQ = 5.1\text{ cm}, PR = 7.2\text{ cm}, \angle Q = 90^\circ$
Steps of Construction:
1. Draw leg $PQ = 5.1\text{ cm}$.
2. At vertex $Q$, erect perpendicular ray $QX$ ($90^\circ$).
3. With center $P$ and radius equal to hypotenuse $PR = 7.2\text{ cm}$, draw an arc intersecting ray $QX$ at point $R$.
4. Join $P$ to $R$.
Result: $\Delta PQR$ is the required right triangle ($RHS$).
(j) $\Delta DEF$ when $DE = 4.2\text{ cm}, EF = 3.3\text{ cm}, DF = 7.7\text{ cm}$
Triangle Inequality Test:
Sum of two smaller sides: $DE + EF = 4.2\text{ cm} + 3.3\text{ cm} = 7.5\text{ cm}$.
Third side: $DF = 7.7\text{ cm}$.
Since $4.2 + 3.3 = 7.5 < 7.7$, the sum of two sides is less than the third side.
Conclusion: Construction of this triangle is NOT POSSIBLE because the two compass arcs of radii $4.2\text{ cm}$ and $3.3\text{ cm}$ will never meet.
(k) $\Delta PQR$ when $PQ = 5\text{ cm}, QR = 4.5\text{ cm}, \angle P = 60^\circ$
Steps of Construction:
1. Draw base $PQ = 5\text{ cm}$.
2. At vertex $P$, construct an angle of $60^\circ$ and draw ray $PX$.
3. With center $Q$, draw an arc of radius $QR = 4.5\text{ cm}$ intersecting ray $PX$ at point $R$.
4. Join $Q$ to $R$.
Result: $\Delta PQR$ is the required triangle.
(l) $\Delta XYZ$ when $XY = 6.1\text{ cm}, YZ = 3\text{ cm}, \angle X = 75^\circ$
Check Feasibility: The perpendicular distance from $Y$ to the ray at $75^\circ$ is $6.1 \times \sin(75^\circ) \approx 6.1 \times 0.9659 = 5.89\text{ cm}$.
Since the given side $YZ = 3\text{ cm} < 5.89\text{ cm}$, an arc of radius $3\text{ cm}$ drawn from $Y$ will not reach ray $X$.
Conclusion: Construction is NOT POSSIBLE (side $YZ$ is too short to intersect ray $X$).
📝 Solved Exercise 10.2 • Construction of Squares, Rectangles & Parallelograms
Q1. Construct the squares whose measures of sides are: (i) $4\text{ cm}$, (ii) $4.7\text{ cm}$, (iii) $5.5\text{ cm}$
Standard Construction Protocol (for side $s$):
1. Draw base segment $AB = s$.
2. Construct $90^\circ$ perpendicular rays at both $A$ (ray $AX$) and $B$ (ray $BY$).
3. With compass radius $s$, cut arc on $AX$ at point $D$ and on $BY$ at point $C$.
4. Join $D$ to $C$. $ABCD$ is the required square.
Applied Values:
- (i) $s = 4\text{ cm} \implies AB = BC = CD = DA = 4\text{ cm}, \angle = 90^\circ$.
- (ii) $s = 4.7\text{ cm} \implies AB = BC = CD = DA = 4.7\text{ cm}, \angle = 90^\circ$.
- (iii) $s = 5.5\text{ cm} \implies AB = BC = CD = DA = 5.5\text{ cm}, \angle = 90^\circ$.
Q2. Construct the squares whose diagonals have lengths: (i) $5.4\text{ cm}$, (ii) $6.2\text{ cm}$, (iii) $5.8\text{ cm}$
Standard Diagonal Protocol (for diagonal $d$):
1. Draw diagonal $AC = d$.
2. Construct the perpendicular bisector $XY$ of $AC$, intersecting $AC$ at midpoint $O$.
3. With center $O$ and compass radius $r = d/2$, strike arcs on the bisector line $XY$ above at point $B$ and below at point $D$.
4. Join $A$ to $B, B$ to $C, C$ to $D$, and $D$ to $A$. $ABCD$ is the required square.
Applied Radii:
- (i) $d = 5.4\text{ cm} \implies r = OA = OB = OC = OD = 2.7\text{ cm}$.
- (ii) $d = 6.2\text{ cm} \implies r = OA = OB = OC = OD = 3.1\text{ cm}$.
- (iii) $d = 5.8\text{ cm} \implies r = OA = OB = OC = OD = 2.9\text{ cm}$.
Q3. Construct the rectangles when the two sides are: (i) $5\text{ cm}$ and $3\text{ cm}$, (ii) $5.4\text{ cm}$ and $2.6\text{ cm}$, (iii) $6.6\text{ cm}$ and $4\text{ cm}$
Steps of Construction (for length $l$ and width $w$):
1. Draw base segment $AB = l$.
2. Construct perpendicular rays of $90^\circ$ at $A$ (ray $AX$) and $B$ (ray $BY$).
3. Cut off length $w$ on $AX$ at point $D$ and on $BY$ at point $C$.
4. Join $D$ to $C$. $ABCD$ is the required rectangle.
Applications: (i) $l = 5\text{ cm}, w = 3\text{ cm}$; (ii) $l = 5.4\text{ cm}, w = 2.6\text{ cm}$; (iii) $l = 6.6\text{ cm}, w = 4\text{ cm}$.
Q4. Construct a rectangle when: (i) $\text{diagonal} = 6.3\text{ cm}, \text{side} = 2.2\text{ cm}$, (ii) $\text{diagonal} = 5.7\text{ cm}, \text{side} = 3.1\text{ cm}$
Steps of Construction:
1. Draw given side $AB = s$.
2. Construct perpendicular ray $AX \perp AB$ at vertex $A$.
3. With center $B$ and radius equal to diagonal $d$, draw an arc cutting ray $AX$ at point $D$.
4. With center $D$, draw an arc of radius $AB = s$. With center $B$, draw an arc of radius $AD$. These arcs intersect at point $C$.
5. Join $D$ to $C$ and $B$ to $C$. $ABCD$ is the required rectangle.
Q5 & Q6. Construct Parallelograms from Two Adjacent Sides and Included Angle:
5(i) $ABCD$: $AB = 5.2\text{ cm}, BC = 3\text{ cm}, \angle ABC = 70^\circ$
1. Draw base $AB = 5.2\text{ cm}$. At $B$, draw ray at $70^\circ$ and cut $BC = 3\text{ cm}$.
2. With center $A$, draw arc of radius $3\text{ cm}$. With center $C$, draw arc of radius $5.2\text{ cm}$ to intersect at $D$. Join $AD$ and $CD$.
5(ii) $ABCD$: $AD = 2.7\text{ cm}, AB = 4.8\text{ cm}, \angle A = 110^\circ$
1. Draw base $AB = 4.8\text{ cm}$. Construct $\angle A = 110^\circ$ and cut $AD = 2.7\text{ cm}$.
2. From $D$, draw arc radius $4.8\text{ cm}$; from $B$, draw arc radius $2.7\text{ cm}$ intersecting at $C$. Join $DC$ and $BC$.
6(i) $EFGH$: $FG = 5\text{ cm}, GH = 3.5\text{ cm}, \angle G = 120^\circ$ (Base $FG=5$, angle $120^\circ$, cut $GH=3.5$, arcs give $E$).
6(ii) $EFGH$: $EF = 4\text{ cm}, FG = 3.6\text{ cm}, \angle F = 75^\circ$ (Base $EF=4$, angle $75^\circ$, cut $FG=3.6$, arcs give $H$).
Q7 & Q8. Construct Parallelograms from Two Diagonals and Angle between them:
Q7: Diagonals $d_1 = 6.3\text{ cm}, d_2 = 3.4\text{ cm}$, angle $\theta = 60^\circ$
1. Draw diagonal $AC = 6.3\text{ cm}$ and locate its midpoint $O$ ($OA = OC = 3.15\text{ cm}$).
2. At $O$, construct an angle of $60^\circ$ and extend line $XOY$.
3. From $O$, cut $OB = OD = d_2/2 = 3.4/2 = 1.7\text{ cm}$ along $XOY$.
4. Join $A$ to $B, B$ to $C, C$ to $D$, and $D$ to $A$. $ABCD$ is the required parallelogram.
Q8: Diagonals $d_1 = 8\text{ cm}, d_2 = 4\text{ cm}$, angle $\theta = 90^\circ$
1. Draw $AC = 8\text{ cm}$ and find midpoint $O$. At $O$, draw perpendicular line ($90^\circ$).
2. Cut $OB = OD = 4/2 = 2\text{ cm}$ on the perpendicular. Join vertices. (Note: Since diagonals are perpendicular, this forms a rhombus).
Q9 & Q10. Construct Rectangles from Diagonal and Angle between Diagonals:
Q9: Diagonal $d = 7\text{ cm}$, angle between diagonals $= 45^\circ$
1. In a rectangle, diagonals are equal ($d_1 = d_2 = 7\text{ cm}$) and bisect each other ($OA = OB = OC = OD = 3.5\text{ cm}$).
2. Draw diagonal $AC = 7\text{ cm}$, bisect at $O$. At $O$, construct angle $45^\circ$ and extend line $BD$.
3. Cut $OB = OD = 3.5\text{ cm}$ along the line. Join $AB, BC, CD, DA$.
Q10: Diagonal $d = 6.4\text{ cm}$, angle between diagonals $= 30^\circ$
1. Draw $AC = 6.4\text{ cm}$, bisect at $O$ ($OA = OC = 3.2\text{ cm}$).
2. At $O$, construct angle of $30^\circ$ and mark $OB = OD = 3.2\text{ cm}$. Join $A, B, C, D$.
📝 Solved Exercise 10.3 • Rhombus, Trapezium & Kite
Q1. Construct a Rhombus when side and base angle are given:
Protocol: Draw base $AB = s$. Construct base angle $\theta$ at $A$ and cut $AD = s$. With radius $s$, strike arcs from $D$ and $B$ intersecting at $C$. Join $DC$ and $BC$.
- (i) Side $= 3.5\text{ cm}, \angle = 45^\circ \implies AB = BC = CD = DA = 3.5\text{ cm}, \angle A = 45^\circ$.
- (ii) Side $= 4\text{ cm}, \angle = 30^\circ \implies AB = BC = CD = DA = 4\text{ cm}, \angle A = 30^\circ$.
- (iii) Side $= 4.5\text{ cm}, \angle = 75^\circ \implies AB = BC = CD = DA = 4.5\text{ cm}, \angle A = 75^\circ$.
Q2. Construct a Rhombus when side and diagonal are given:
Protocol: Draw diagonal $AC = d$. With compass radius equal to side $s$, draw arcs from $A$ above and below $AC$, and similarly from $C$. Arc intersections establish vertices $B$ and $D$. Join $AB, BC, CD, DA$.
- (i) Side $= 3.8\text{ cm}, \text{diagonal} = 6.5\text{ cm}$ (Arcs of radius $3.8$ from ends of $6.5\text{ cm}$).
- (ii) Side $= 4.4\text{ cm}, \text{diagonal} = 7.1\text{ cm}$ (Arcs of radius $4.4$ from ends of $7.1\text{ cm}$).
- (iii) Side $= 5.2\text{ cm}, \text{diagonal} = 7.5\text{ cm}$ (Arcs of radius $5.2$ from ends of $7.5\text{ cm}$).
Q3 & Q4. Construct Trapeziums:
Q3: $ABCD$ with $AB = 4.5\text{ cm}, AD = 4\text{ cm}, BC = 3.5\text{ cm}$, distance between parallel sides $= 3.2\text{ cm}$
1. Draw base $AB = 4.5\text{ cm}$. Draw a parallel line $l$ at perpendicular distance $3.2\text{ cm}$ above $AB$.
2. With center $A$, draw an arc of radius $4\text{ cm}$ cutting line $l$ at $D$.
3. With center $B$, draw an arc of radius $3.5\text{ cm}$ cutting line $l$ at $C$.
4. Join $A$ to $D, B$ to $C$, and $D$ to $C$. $ABCD$ is the required trapezium.
Q4: $EFGH$ with $EF = 5.4\text{ cm}, FG = 4.3\text{ cm}, GH = 3.8\text{ cm}, EG = 5.8\text{ cm}$
1. Draw base $EF = 5.4\text{ cm}$. With center $E$, draw arc radius $5.8\text{ cm}$; with center $F$, draw arc radius $4.3\text{ cm}$ to intersect at vertex $G$.
2. Through $G$, draw line $l \parallel EF$. From $G$, cut $GH = 3.8\text{ cm}$ on line $l$. Join $E$ to $H$ and $F$ to $G$.
Q5 & Q6. Construct Kites:
Q5: Sides $5.4\text{ cm}$ and $3.2\text{ cm}$, diagonal $= 6.5\text{ cm}$
1. Draw diagonal $AC = 6.5\text{ cm}$.
2. With center $A$, draw arcs of radius $5.4\text{ cm}$ above and below $AC$.
3. With center $C$, draw arcs of radius $3.2\text{ cm}$ above and below $AC$ intersecting first arcs at $B$ and $D$.
4. Join $AB, BC, CD, DA$. $ABCD$ is the required kite.
Q6: Sides $6\text{ cm}$ and $4\text{ cm}$, diagonal $= 7.1\text{ cm}$
1. Draw diagonal $AC = 7.1\text{ cm}$. Arcs of radius $6\text{ cm}$ from $A$ and $4\text{ cm}$ from $C$ intersect at $B$ and $D$. Join vertices.
📝 Solved Exercise 10.4 • Angle & Line Bisectors in Figures
Q1. Construct scalene $\Delta DEF$ with $DE = 5.1\text{ cm}, EF = 4.8\text{ cm}, DF = 4.5\text{ cm}$. Draw angle bisector at vertex $E$.
1. Draw $DE = 5.1\text{ cm}$. Arcs of radius $4.8\text{ cm}$ from $E$ and $4.5\text{ cm}$ from $D$ intersect at $F$. Join $DF$ and $EF$.
2. At vertex $E$, draw an arc cutting arms $ED$ and $EF$ at $P$ and $Q$.
3. From $P$ and $Q$, draw equal intersecting arcs inside $\angle E$ at point $R$. Draw ray $ER$. Ray $ER$ is the angle bisector of $\angle E$.
Q2. Construct equilateral $\Delta GHI$ with $GH = HI = GI = 6.2\text{ cm}$. Draw right bisector of side $HI$.
1. Draw base $HI = 6.2\text{ cm}$. Arcs of radius $6.2\text{ cm}$ from $H$ and $I$ intersect at vertex $G$. Join $GH$ and $GI$.
2. With radius $> 3.1\text{ cm}$, draw arcs above and below side $HI$ from $H$ and $I$. Line passing through arc intersections is the perpendicular right bisector of $HI$ (which also passes through opposite vertex $G$).
Q3. Construct kite $EFGH$ with $EF = GH = 5.7\text{ cm}, FG = EH = 4.6\text{ cm}, \angle E = 60^\circ$. Draw angle bisector at vertex $F$ and measure both parts.
1. Draw $EF = 5.7\text{ cm}$. Construct $\angle E = 60^\circ$ and cut $EH = 4.6\text{ cm}$. From $F$ arc radius $4.6$ and from $H$ arc radius $5.7$ intersect at $G$. Join vertices.
2. Bisect $\angle F$ using compass arcs. Measuring both bisected parts yields equal angles (each half $= \frac{1}{2}\angle F$).
Q4. Construct right $\Delta ABC$ with $AB = BC = 5.9\text{ cm}, \angle B = 90^\circ$. Draw bisector of right angle meeting $AC$ at $D$. Measure $AD$ and $CD$.
1. Draw $AB = 5.9\text{ cm}$. Construct $90^\circ$ at $B$ and cut $BC = 5.9\text{ cm}$. Join $A$ to $C$.
2. Bisect $\angle B = 90^\circ$ (each half is $45^\circ$) meeting hypotenuse $AC$ at point $D$.
3. Measuring shows: $AD = CD \approx 4.17\text{ cm}$ (In an isosceles right triangle, the angle bisector of the right angle is also the median and perpendicular bisector of the hypotenuse).
Q5. Construct square $CDEF$ with side $4.5\text{ cm}$. Draw right bisector of $CD$ extending to point $P$ on $EF$. Measure $PE$ and $PF$.
1. Construct square $CDEF$ of side $4.5\text{ cm}$.
2. Draw perpendicular bisector of side $CD$ and extend it across the square to intersect opposite side $EF$ at point $P$.
3. Measuring reveals: $PE = PF = 2.25\text{ cm}$ ($P$ is the exact midpoint of side $EF$).
🎯 Review Exercise 10 • Comprehensive Question Set
Q1. Multiple Choice Questions (Complete Solutions):
Adjacent angles are supplementary: $180^\circ - 60^\circ = 120^\circ$. Ans: (a) $120^\circ$
$360^\circ - (3 \times 75^\circ) = 360^\circ - 225^\circ = 135^\circ$. Ans: (b) $135^\circ$
Opposite sides and opposite angles are both equal. Ans: (d) both (a) & (c)
Square and rhombus both have 4 equal sides. Ans: (d) both a and b
Diagonals of a square and rhombus bisect angles. Ans: (a) square
By definition, this is a trapezium. Ans: (c) trapezium
In a rhombus, diagonals bisect at $90^\circ$ but are unequal. Ans: (d) rhombus
Consecutive interior angles sum to $180^\circ$. Ans: (a) supplementary
All sides equal with non-right angle is a rhombus. Ans: (d) rhombus
$x + 2x + 3x = 180^\circ \implies 6x = 180^\circ \implies x = 30^\circ$. Angles: $30^\circ, 60^\circ, 90^\circ$. Ans: (a) $30^\circ, 60^\circ, 90^\circ$
Diagonal bisects $90^\circ$ corner: $\frac{90^\circ}{2} = 45^\circ$. Ans: (c) $45^\circ$
Opposite angles are equal ($\angle P = \angle R$), so $\angle P - \angle R = 0^\circ$. Ans: (b) $0^\circ$
Descriptive Review Problems (Q2 to Q10):
Q2. Triangle Constructions:
(a) $AB = 4.3\text{ cm}, BC = 6\text{ cm}, AC = 3.8\text{ cm}$ ($SSS$: draw base $BC=6$, arcs $4.3$ and $3.8$ meet at $A$).
(b) $XY = 5.6\text{ cm}, YZ = 4\text{ cm}, \angle Y = 75^\circ$ ($SAS$: base $XY=5.6$, angle $75^\circ$ at $Y$, cut $YZ=4$, join $X$ to $Z$).
(c) $RQ = 5.2\text{ cm}, \angle R = 30^\circ, \angle Q = 120^\circ$ ($ASA$: base $RQ=5.2$, rays at $30^\circ$ and $120^\circ$ meet at $P$).
(d) Right $\Delta DEF$ with hypotenuse $5.7\text{ cm}$, leg $4.5\text{ cm}$ ($RHS$: draw leg $4.5$, construct $90^\circ$, arc radius $5.7$ cuts perpendicular).
Q3. Perpendicular from Vertex to Base: In $\Delta LMN$, from vertex $N$, draw an arc cutting base $LM$ at two points. From these points, draw equal intersecting arcs below $LM$. Line through $N$ and intersection is the perpendicular altitude.
Q4. Rhombus: (i) Side $3.5\text{ cm}, \angle = 75^\circ$; (ii) Side $5.4\text{ cm}$, diagonal $7.4\text{ cm}$.
Q5. Parallelogram: Diagonals $6.2\text{ cm}$ and $3.3\text{ cm}$ with angle $45^\circ$ (Bisect $6.2$ at $O$, draw line at $45^\circ$, mark $1.65\text{ cm}$ each side, join vertices).
Q6. Kite: Sides $5.2\text{ cm}$ and $3\text{ cm}$, diagonal $6.2\text{ cm}$. Other diagonal is perpendicular and bisected by the main diagonal.
Q7. Varignon Parallelogram Theorem: Joining midpoints of any quadrilateral $ABCD$ forms a parallelogram whose opposite sides are strictly equal and parallel.
Q8. Rectangle: Diagonal $6.3\text{ cm}$, side $2.2\text{ cm}$, angle between diagonals $45^\circ$.
Q9. Bisecting Length & Width of Rectangle: The line bisectors of the length and width of any rectangle always intersect at $90^\circ$ right angles.
Q10. Concurrency in $\Delta ABC$ ($4.5, 5, 6\text{ cm}$): Right bisectors of sides meet at the circumcenter, while angle bisectors meet at the incenter.
🎯 Unit Synthesis Summary
Practical Geometry translates theoretical geometric properties into exact compass-and-straightedge constructions. Triangle feasibility is strictly governed by the Triangle Inequality ($a+b > c$). Quadrilaterals require varying degrees of independent constraints depending on symmetry: squares require 1 measurement, rectangles and rhombuses require 2, parallelograms and kites require 3, and general trapeziums require 4. Interior angle bisectors concur at the incenter (center of incircle), while perpendicular side bisectors concur at the circumcenter (center of circumcircle).
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