Mastery Guide: Practical Geometry - Triangle Constructions, Ambiguous Case, Angle Bisectors, Altitudes, Perp Bisectors & Centers
Mastery Guide: Practical Geometry
Single National Curriculum (SNC) • Triangle Constructions, Ambiguous Case (SSA), Concurrency of Angle Bisectors, Altitudes, Perpendicular Bisectors & Medians (Centers of Triangles)
📖 1. Unit Overview & Target Learning Outcomes
Practical Geometry provides the rigorous compass-and-straightedge construction techniques required to construct geometric figures accurately. Chapter 10 focuses on the systematic construction of triangles under various given conditions (including the classical Ambiguous Case), followed by the construction and verification of the Four Classical Points of Concurrency of a triangle: Incentre, Orthocentre, Circumcentre, and Centroid.
🎯 Target Learning Outcomes:
- Standard Triangle Constructions: Construct triangles given: (i) Two sides and included angle ($SAS$), (ii) One side and two angles ($ASA / AAS$), and (iii) Three sides ($SSS$).
- The Ambiguous Case ($SSA$): Mathematically analyze and construct triangles given two sides and a non-included angle opposite to one side. Determine when 0, 1, or 2 distinct triangles can be formed using altitude height $h = c \sin A$.
- Angle Bisectors & Incentre ($I$): Construct the internal angle bisectors of a triangle and verify they are always concurrent at an interior point called the Incentre (center of the inscribed circle).
- Altitudes & Orthocentre ($H$): Drop perpendiculars from vertices to opposite sides. Verify concurrency at the Orthocentre, and locate its position in acute (inside), right (at the right-angled vertex), and obtuse (outside) triangles.
- Perpendicular Bisectors & Circumcentre ($O$): Construct right bisectors of the three sides and verify concurrency at the Circumcentre (midpoint of hypotenuse in right triangles, inside in acute, outside in obtuse).
- Medians & Centroid ($G$): Construct line segments joining vertices to opposite midpoints. Verify concurrency at the Centroid (Centre of Gravity), and verify the fundamental $2:1$ division ratio.
- Coincidence in Equilateral Triangles: Prove that in an equilateral triangle, all four centers ($I, H, O, G$) coincide at the exact same point.
💡 2. Kid-Friendly Tips for Success & Memory Hooks
Angle Bisectors → Incentre (Inscribed circle)
Altitudes → Orthocentre (Height intersections)
Perpendicular Bisectors → Circumcentre (Circumcircle)
Medians → Centroid (Center of Gravity, 2:1 ratio)
In any right-angled triangle:
• Orthocentre is directly at the 90° Vertex!
• Circumcentre is exactly the Midpoint of the Hypotenuse!
The centroid is always twice as far from the vertex as it is from the midpoint of the opposite side. Distance from Vertex to $G = \frac{2}{3} \times \text{Median length}$; Distance from $G$ to Base $= \frac{1}{3} \times \text{Median length}$.
Given angle $A$ and adjacent side $c$, the critical altitude is $h = c \sin A$. If opposite side $a < h \implies \textbf{0 triangles}$; if $a = h \implies \textbf{1 right triangle}$; if $h < a < c \implies \textbf{2 triangles}$; if $a \ge c \implies \textbf{1 triangle}$.
🌍 3. Real-World Connections
🏗️ Structural Engineering & Trusses
Bridges, roof trusses, and cranes are constructed entirely out of triangles because triangles are the only rigid polygon that cannot deform without breaking a side.
⚖️ Center of Mass in Aerospace
The centroid is the physical balance point (center of gravity) of triangular aircraft wings and rocket fins. A fingertip placed under the centroid will support the entire wing in perfect balance.
📡 Emergency Services & Hospital Placement
Urban planners use the Circumcentre to locate a regional hospital or fire station equidistant from three surrounding towns, ensuring equal response times.
🛣️ Roundabout Design & Inscribed Roads
Highway engineers find the Incentre to design the largest possible circular roundabout or central park that touches all three connecting highways without overlapping.
🌟 4. Section-by-Section Explanations & Visual Models
4.1 The Ambiguous Case (Side-Side-Angle / SSA)
When two sides ($c, a$) and a non-included angle ($\angle A$) opposite to side $a$ are given, the number of possible triangles depends on the perpendicular height $h = c \sin A$:
4.2 The Four Classical Centers of a Triangle
A line is concurrent if three or more lines pass through the exact same point in a plane. Every triangle possesses four fundamental centers of concurrency:
| Center | Formed By | Acute Δ | Right Δ | Obtuse Δ |
|---|---|---|---|---|
| Incentre (I) | Angle Bisectors | Inside | Inside | Inside |
| Orthocentre (H) | Altitudes | Inside | At 90° Vertex | Outside |
| Circumcentre (O) | Perpendicular Bisectors | Inside | Midpoint of Hypotenuse | Outside |
| Centroid (G) | Medians (2:1 ratio) | Inside | Inside | Inside |
🎯 5. Unit Synthesis Summary
Chapter 10 unites the practical mechanics of geometric construction with foundational Euclidean geometry theorems. The chapter rigorously establishes the conditions for unique and ambiguous triangle constructions (identifying 0, 1, or 2 triangles via the altitude relation $h = c \sin A$). It demonstrates that for any planar triangle, the three angle bisectors, three altitudes, three perpendicular bisectors of sides, and three medians each meet at a single concurrent point (the Incentre, Orthocentre, Circumcentre, and Centroid respectively). Furthermore, the spatial locations of these centers behave dynamically according to the triangle's angle classification: while the Incentre and Centroid remain strictly internal, the Orthocentre and Circumcentre transition from inside (acute) to the boundary (at the right-angle vertex and hypotenuse midpoint for right triangles) to outside the triangle (obtuse). In an equilateral triangle, geometric symmetry dictates that all four centers coincide at one unique point.
📝 Complete Solved Textbook Exercises & Examination Question Bank
Below is the exhaustive, step-by-step solution manual for every single textbook problem, example exercise, and review problem in Chapter 10, aligned strictly with FBISE scoring guidelines.
Exercise 10.1 • Step-by-Step Complete Solutions
Construct the following triangle: Triangle ABC when AB = 5.8 cm, AC = 4.2 cm, Angle A = 90°.
Step 2: At vertex $A$, construct an angle of $90^\circ$ using a compass ($ray AX \perp AB$).
Step 3: With center $A$ and radius $4.2\text{ cm}$, draw an arc intersecting ray $AX$ at point $C$.
Step 4: Join point $C$ to point $B$ with a straight line segment.
Final Answer: Triangle ABC is the required right-angled triangle.
Construct the following triangle: Triangle LMN when LM = 4 cm, MN = 4.7 cm, Angle M = 120°.
Step 2: At vertex $M$, construct an obtuse angle of $120^\circ$ using compass ($ray MY$).
Step 3: With center $M$ and radius $4.7\text{ cm}$, draw an arc cutting ray $MY$ at point $N$.
Step 4: Join point $N$ to point $L$.
Final Answer: Triangle LMN is the required obtuse-angled triangle.
Construct the following triangle: Triangle PQR when PQ = 7.2 cm, Angle P = 45°, Angle Q = 75°.
Step 2: At point $P$, construct an angle of $45^\circ$ ($ray PX$).
Step 3: At point $Q$, construct an angle of $75^\circ$ ($ray QY$).
Step 4: Let the two rays $PX$ and $QY$ intersect at point $R$. (Third angle $\angle R = 180^\circ - (45^\circ + 75^\circ) = 60^\circ$).
Final Answer: Triangle PQR is the required triangle.
Construct the following triangle: Triangle XYZ when XY = 5 cm, Angle X = 30°, Angle Z = 105°.
$$\angle Y = 180^\circ - (\angle X + \angle Z) = 180^\circ - (30^\circ + 105^\circ) = 180^\circ - 135^\circ = 45^\circ$$
Step 2: Draw base line segment $XY = 5\text{ cm}$.
Step 3: At vertex $X$, construct an angle of $30^\circ$ ($ray XA$).
Step 4: At vertex $Y$, construct an angle of $45^\circ$ ($ray YB$).
Step 5: Let rays $XA$ and $YB$ intersect at point $Z$.
Final Answer: Triangle XYZ is the required triangle with Angle Z = 105°.
Construct the following triangle where possible: AB = 6.8 cm, BC = 8.1 cm, Angle A = 90°.
Step 2: In a right triangle, construction is possible if and only if hypotenuse $BC > AB$.
Since $8.1\text{ cm} > 6.8\text{ cm}$, exactly one unique right triangle can be constructed.
Step 3: Draw $AB = 6.8\text{ cm}$. At $A$, erect perpendicular $AX \perp AB$. With center $B$ and radius $8.1\text{ cm}$, draw an arc cutting $AX$ at $C$. Join $BC$.
Final Answer: Exactly 1 right triangle can be constructed.
Construct the following triangle where possible: DE = 5.2 cm, DF = 4 cm, Angle E = 45°.
Step 2: Calculate critical perpendicular altitude $h$ from $D$ to the ray from $E$:
$$h = DE \sin 45^\circ = 5.2 \times 0.7071 \approx 3.68\text{ cm}$$
Step 3: Compare sides: Since $h (3.68\text{ cm}) < DF (4\text{ cm}) < DE (5.2\text{ cm})$, an arc of radius $4\text{ cm}$ with center $D$ intersects the ray from $E$ in two distinct points $F_1$ and $F_2$.
Final Answer: Two distinct triangles (DE F1 and DE F2) can be constructed (Ambiguous Case).
Construct the following triangle where possible: QR = 7.0 cm, PQ = 5.6 cm, Angle R = 75°.
Step 2: Calculate perpendicular altitude $h$ from $Q$ to the base ray through $R$:
$$h = QR \sin 75^\circ = 7.0 \times 0.9659 \approx 6.76\text{ cm}$$
Step 3: Compare: Since opposite side $PQ (5.6\text{ cm}) < h (6.76\text{ cm})$, the arc of radius $5.6\text{ cm}$ drawn from $Q$ cannot reach the line through $R$.
Final Answer: No triangle can be constructed.
Construct the following triangle where possible: XY = 3.8 cm, XZ = 5 cm, Angle Y = 60°.
Step 2: Since the opposite side $XZ (5\text{ cm}) > XY (3.8\text{ cm})$, the arc drawn from $X$ intersects the ray from $Y$ in only one positive point $Z$.
Final Answer: Exactly 1 triangle can be constructed.
Construct the following triangle where possible: If two sides of lengths 5.7 cm and 7.5 cm are given and angle of 105° is opposite to the side of length 7.5 cm.
Step 2: In an obtuse triangle, the side opposite to the obtuse angle must be strictly greater than the adjacent side ($7.5\text{ cm} > 5.7\text{ cm}$).
Step 3: Since $7.5 > 5.7$, exactly one unique obtuse triangle can be constructed.
Final Answer: Exactly 1 obtuse triangle can be constructed.
Construct the following triangle where possible: If two sides of length 6.1 cm and 3.8 cm are given and angle of 30° is opposite to the side of length 3.8 cm.
Step 2: Calculate altitude $h = b \sin 30^\circ = 6.1 \times 0.5 = 3.05\text{ cm}$.
Step 3: Since $h (3.05\text{ cm}) < a (3.8\text{ cm}) < b (6.1\text{ cm})$, the arc intersects the base ray in two distinct points.
Final Answer: Two distinct triangles can be constructed (Ambiguous Case).
Exercise 10.2 • Step-by-Step Complete Solutions
Show that angle bisectors of the following triangle are concurrent: Triangle ABC when AB = 5.8 cm, BC = 4.9 cm, Angle B = 60°.
Step 2: Using compass, draw internal angle bisectors of $\angle A, \angle B,$ and $\angle C$.
Step 3: Observe that all three angle bisectors intersect at a single common interior point $I$ (Incentre).
Step 4: (Verification) Drop perpendicular from $I$ to $AB$ of radius $r$; a circle drawn with center $I$ touches all three sides internally.
Final Answer: The angle bisectors are concurrent at Incentre I.
Show that angle bisectors of the following triangle are concurrent: Triangle XYZ when XY = 5.5 cm, Angle X = 45°, Angle Y = 75°.
Step 2: Draw internal angle bisectors of $\angle X, \angle Y,$ and $\angle Z$ using compass arcs.
Step 3: All three bisectors pass through a single point $I$ inside $\triangle XYZ$.
Final Answer: The angle bisectors are concurrent at Incentre I.
Show that altitudes of the following triangle are concurrent: Triangle ABC when AB = 4 cm, BC = 5 cm, AC = 6 cm (Acute triangle).
Step 2: From vertex $A$, draw perpendicular $AD \perp BC$; from vertex $B$, draw $BE \perp AC$; from vertex $C$, draw $CF \perp AB$.
Step 3: The three altitudes intersect at a single point $H$ (Orthocentre) lying inside $\triangle ABC$.
Final Answer: The altitudes are concurrent at Orthocentre H inside the triangle.
Show that altitudes of the following triangle are concurrent: Triangle PQR when PQ = 4.6 cm, QR = 6.5 cm, Angle P = 90° (Right triangle).
Step 2: The altitude from $Q$ to $PR$ is the side $QP$; the altitude from $R$ to $PQ$ is the side $RP$. Both meet at vertex $P$.
Step 3: The altitude from $P$ to hypotenuse $QR$ also originates from $P$. Hence, the Orthocentre $H$ is the right-angle vertex $P$ itself.
Final Answer: The altitudes are concurrent at vertex P (Orthocentre).
Show that altitudes of the following triangle are concurrent: Triangle LMN when LM = 4.2 cm, MN = 4 cm, Angle M = 105° (Obtuse triangle).
Step 2: Extend the sides $LM, MN, NL$ and draw perpendiculars from each vertex to the line containing the opposite side.
Step 3: Produce the three perpendiculars backwards; they intersect at a single point $H$ lying outside the triangle.
Final Answer: The altitudes are concurrent at Orthocentre H outside the triangle.
Show that right bisectors of sides of the following triangle are concurrent: Triangle XYZ when XY = 4.5 cm, YZ = 5 cm, ZX = 4.8 cm (Acute triangle).
Step 2: Using compass, construct the perpendicular bisectors of sides $XY, YZ,$ and $ZX$.
Step 3: The three right bisectors intersect at a single point $O$ (Circumcentre) inside $\triangle XYZ$.
Step 4: With center $O$ and radius $OX$, draw the circumscribed circle passing through all three vertices.
Final Answer: The right bisectors are concurrent at Circumcentre O inside the triangle.
Show that right bisectors of sides of the following triangle are concurrent: Triangle PQR when PQ = 4 cm, QR = 5.8 cm, Angle Q = 90° (Right triangle).
Step 2: Construct perpendicular bisectors of legs $PQ, QR$ and hypotenuse $PR$.
Step 3: All three perpendicular bisectors meet at point $O$, which is exactly the midpoint of hypotenuse $PR$.
Final Answer: The right bisectors are concurrent at the midpoint of hypotenuse PR.
Show that right bisectors of sides of the following triangle are concurrent: Triangle DEF when DE = 5 cm, EF = 4 cm, Angle E = 120° (Obtuse triangle).
Step 2: Construct the perpendicular bisectors of sides $DE, EF,$ and $FD$.
Step 3: The right bisectors intersect at a single point $O$ (Circumcentre) lying outside the triangle.
Final Answer: The right bisectors are concurrent at Circumcentre O outside the triangle.
Show that medians of the following triangle are concurrent: Triangle ABC when AB = 5.8 cm, BC = 5 cm, Angle B = 45°.
Step 2: Find midpoints $D, E, F$ of sides $BC, AC, AB$ by bisecting each side.
Step 3: Draw medians $AD, BE, CF$ by connecting vertices to opposite midpoints.
Step 4: The three medians intersect at a single point $G$ (Centroid) inside the triangle.
Final Answer: The medians are concurrent at Centroid G.
Show that medians of the following triangle are concurrent: Triangle DEF when DE = 6 cm, Angle D = 90°, Angle E = 30°.
Step 2: Locate midpoints of the three sides and draw the three medians.
Step 3: The medians meet at a single point $G$ inside $\triangle DEF$.
Final Answer: The medians are concurrent at Centroid G.
Construct a right triangle $ABC$ such that $\angle B = 90^\circ$. (a) Find its orthocentre, where does it lie? (b) Find circumcentre $M$ of the triangle. Is $M$ the mid-point of hypotenuse $AC$? Is $BM = CM$?
Step 2: (a) The altitude from $A$ to $BC$ is $AB$; the altitude from $C$ to $AB$ is $CB$. Both meet at vertex $B$. Thus, the Orthocentre lies directly at vertex B.
Step 3: (b) Construct the perpendicular bisectors of $AB, BC,$ and $AC$. They intersect at point $M$ on hypotenuse $AC$.
Step 4: Measure segments: $M$ is the exact midpoint of $AC$ ($AM = CM = \frac{1}{2}AC$). Since $M$ is the circumcentre, the distance to all three vertices is equal ($BM = CM = AM = r$).
Final Answer: (a) Orthocentre is at vertex B; (b) Yes, M is the midpoint of hypotenuse AC, and BM = CM.
Construct an obtuse angled triangle $DEF$. Find its (a) circumcentre (b) orthocentre. Check whether they lie inside or outside the triangle?
Step 2: (a) Draw perpendicular bisectors of sides $DE, EF, FD$. Their intersection point $O$ (Circumcentre) lies outside the triangle on the opposite side of the obtuse angle.
Step 3: (b) Draw altitudes from $D, E, F$ to the extended lines containing the opposite sides. Their intersection point $H$ (Orthocentre) lies outside the triangle behind the obtuse vertex.
Final Answer: Both the circumcentre and orthocentre lie outside the obtuse triangle.
Construct an isosceles triangle and find its (a) centroid (b) inscribed centre (incentre).
Step 2: (a) Draw medians to locate Centroid $G$ on the central line of symmetry.
Step 3: (b) Draw angle bisectors to locate Incentre $I$ on the same central line of symmetry.
Step 4: Observe that for an isosceles triangle, the centroid, incentre, orthocentre, and circumcentre are collinear (all lie on the altitude from vertex $A$).
Final Answer: Constructed; Centroid G and Incentre I are collinear along the main axis of symmetry.
Recognize altitudes, perpendicular bisectors and medians in the following triangle: In Triangle ABC: identify segment AD perpendicular to BC, line perpendicular through midpoint of BC, and segment BE connecting vertex to opposite midpoint.
Step 2: The line passing through midpoint of $BC$ making $90^\circ$ angle $\implies$ Perpendicular Bisector (Right Bisector).
Step 3: Segment $BE$ connects vertex $B$ to the midpoint $E$ of side $AC$ (indicated by equality hash marks) $\implies$ Median.
Final Answer: AD = Altitude, Midpoint perp line = Right Bisector, BE = Median.
Recognize altitudes, perpendicular bisectors and medians in the following triangle: In Triangle PRS: identify line segment RU connecting vertex to midpoint of side, line UQ perpendicular to base PS, and segment ST perpendicular to PR.
Step 2: Line $UQ$ passes through midpoint $U$ perpendicular to side $PS$ $\implies$ Perpendicular Bisector.
Step 3: Line segment $ST$ is drawn from vertex $S$ perpendicular to opposite side $PR$ $\implies$ Altitude.
Final Answer: RU = Median, UQ = Right Bisector, ST = Altitude.
Review Exercise 10 • Step-by-Step Complete Solutions
A line which bisects an angle into two equal parts is:
By definition, a ray/line that divides an angle into two equal halves is an angle bisector.
Final Answer: Option (B)
Which of the following is divided by a point called midpoint?
Only a line segment has finite measurable endpoints and can have a unique midpoint.
Final Answer: Option (C)
A line passing through mid point and perpendicular to a side of a triangle is called:
A line perpendicular to a line segment at its midpoint is called a right bisector (perpendicular bisector).
Final Answer: Option (A)
Two sides making arms of a right angle in right triangle are its:
In a right triangle, each perpendicular leg acts as an altitude to the other leg.
Final Answer: Option (A)
Right bisectors of a right triangle meet at mid point of:
The circumcentre (intersection of right bisectors) of a right triangle is the midpoint of the hypotenuse.
Final Answer: Option (D)
Medians of a triangle are:
The three medians of any triangle always intersect at a single common point (Centroid), meaning they are concurrent.
Final Answer: Option (B)
In an equilateral triangle angle bisectors, medians, altitudes and right bisectors:
Due to complete rotational and reflectional symmetry, all four sets of lines coincide.
Final Answer: Option (C)
Right bisectors of equiangular triangle are its:
In an equiangular (equilateral) triangle, each right bisector is simultaneously an altitude, median, and angle bisector.
Final Answer: Option (D)
If three lines meet at a point, they are called:
Three or more lines passing through the same point are defined as concurrent lines.
Final Answer: Option (D)
Medians of a triangle intersect each other in the ratio:
The centroid divides each median in the ratio 2 : 1 from vertex to base (or 2 : 3 comparing the vertex segment to the total median length).
Final Answer: Option (D)
Construct the following triangle: Triangle PQR when PQ = 6 cm, Angle P = 30°, Angle R = 90°.
Step 2: Draw base line segment $PQ = 6\text{ cm}$.
Step 3: At vertex $P$, construct an angle of $30^\circ$ ($ray PX$).
Step 4: At vertex $Q$, construct an angle of $60^\circ$ ($ray QY$).
Step 5: The rays intersect at point $R$, creating $\angle R = 90^\circ$.
Final Answer: Triangle PQR is the required right-angled triangle.
Construct the following triangle: Triangle XYZ when XY = 5.6 cm, XZ = 5.2 cm, Angle Y = 60°.
Step 2: Since $h (4.85\text{ cm}) < XZ (5.2\text{ cm}) < XY (5.6\text{ cm})$, an arc of radius $5.2\text{ cm}$ from $X$ cuts the ray from $Y$ in two points $Z_1$ and $Z_2$.
Step 3: Draw $XY = 5.6\text{ cm}$, draw ray at $60^\circ$ from $Y$, cut arc from $X$ to locate $Z$.
Final Answer: Triangle XYZ is constructed (Ambiguous Case).
Construct the following triangle: Triangle ABC when BC = 7 cm, AC = 4.3 cm, Angle B = 45°.
Step 2: Compare: Since opposite side $AC (4.3\text{ cm}) < h (4.95\text{ cm})$, the arc drawn from $C$ cannot reach the line through $B$.
Step 3: Conclusion: No triangle can be constructed.
Final Answer: No triangle is possible (AC < h).
Construct a right isosceles triangle whose hypotenuse is $6\text{ cm}$.
Step 2: Construct the perpendicular bisector of $AB$ to locate its midpoint $M$ ($AM = MB = 3\text{ cm}$).
Step 3: With center $M$ and radius $3\text{ cm}$, draw a semicircle above $AB$.
Step 4: The perpendicular bisector intersects the semicircle at point $C$. Join $C$ to $A$ and $C$ to $B$.
Step 5: $\triangle ABC$ has $\angle C = 90^\circ$ (angle in semicircle) and $AC = BC = 3\sqrt{2} \approx 4.24\text{ cm}$.
Final Answer: Triangle ABC is the required right isosceles triangle.
Construct an equilateral triangle $ABC$. Find its: (a) incentre (b) circumcentre (c) orthocentre (d) centroid Do they coincide?
Step 2: (a) Draw internal angle bisectors; they intersect at Incentre $I$.
Step 3: (b) Draw perpendicular bisectors of sides; they intersect at Circumcentre $O$.
Step 4: (c) Draw altitudes from vertices; they intersect at Orthocentre $H$.
Step 5: (d) Draw medians connecting vertices to opposite midpoints; they intersect at Centroid $G$.
Step 6: Observe that all four points $I, O, H, G$ are identical ($I \equiv O \equiv H \equiv G$).
Final Answer: Yes, all four centers coincide at the exact same point.
A student wants to find incentre of an equilateral triangle but he does not know how to bisect the angles. Can he find incentre by using any other method? Explain your answer by taking an example.
Step 2: Alternative Method (Medians / Centroid):
- Find the midpoints $D, E, F$ of sides $BC, AC, AB$ using a ruler or compass.
- Draw straight lines from vertices to opposite midpoints ($AD, BE, CF$).
- Their point of intersection $G$ (Centroid) is identically the Incentre $I$.
Step 3: Example: For equilateral $\triangle ABC$ of side $6\text{ cm}$, the medians meet at $G$, which serves directly as the center for the inscribed circle of radius $r = \frac{6}{2\sqrt{3}} = \sqrt{3} \approx 1.73\text{ cm}$.
Final Answer: Yes, the student can find the incentre by finding the Centroid (medians) or Circumcentre (right bisectors).
Construct a triangle and find its centre of gravity. (Hint: Find centroid)
Step 2: Find the midpoints $D, E, F$ of sides $BC, AC, AB$ respectively by drawing perpendicular bisectors.
Step 3: Draw line segments (medians) $AD, BE,$ and $CF$.
Step 4: The point of intersection $G$ of the three medians is the Centroid, which is the physical Centre of Gravity of the triangular lamina.
Final Answer: The Centre of Gravity is located at the Centroid G.
Extra Exercise • Step-by-Step Complete Solutions
: The center of a circle that touches all three sides of a triangle internally is the:
Final Answer: Option (B)
: In a right-angled triangle, the orthocentre lies:
Final Answer: Option (C)
: The circumcentre of a right-angled triangle with hypotenuse of length $10\text{ cm}$ is at a distance of how many cm from each vertex?
Final Answer: Option (B)
: If the length of a median $AD$ in a triangle is $9\text{ cm}$, what is the distance from vertex $A$ to the centroid $G$?
Final Answer: Option (C)
: Given an acute angle $A$ and adjacent side $c$, no triangle can be constructed if opposite side $a$ satisfies:
Final Answer: Option (D)
: In an equilateral triangle, how many distinct points of concurrency exist among Incentre, Circumcentre, Centroid, and Orthocentre?
Final Answer: Option (A)
: State True or False: (i) The circumcentre of an obtuse angled triangle lies outside the triangle. (ii) The centroid of a triangle can lie outside the triangle.
Step 2: (ii) The centroid is the center of mass and ALWAYS lies strictly inside the triangle (False).
Final Answer: (i) True, (ii) False.
: Fill in the blanks: (i) The point of concurrency of altitudes of a triangle is called ______. (ii) The point of concurrency of perpendicular bisectors of sides is called ______. (iii) The point of concurrency of medians of a triangle is called ______. (iv) The point of concurrency of angle bisectors of a triangle is called ______.
Step 2: (ii) Perpendicular bisectors of sides meet at Circumcentre.
Step 3: (iii) Medians meet at Centroid.
Step 4: (iv) Angle bisectors meet at Incentre.
Final Answer: (i) Orthocentre, (ii) Circumcentre, (iii) Centroid, (iv) Incentre.
: Match each line of a triangle in Column A with its corresponding center of concurrency in Column B: Column A: 1. Angle Bisectors 2. Altitudes 3. Perpendicular Bisectors of Sides 4. Medians 5. Center of Gravity Column B: A. Orthocentre B. Incentre C. Centroid D. Circumcentre E. Centroid (Point of Balance)
Step 2: 2. Altitudes → Orthocentre (A)
Step 3: 3. Perpendicular Bisectors → Circumcentre (D)
Step 4: 4. Medians → Centroid (C)
Step 5: 5. Center of Gravity → Centroid (E)
Final Answer: 1-B, 2-A, 3-D, 4-C, 5-E.
More Chapter Notes for Class 9 (FBISE)
MathematicsTest Your Knowledge on Chapter 10: Mastery Guide: Practical Geometry - Triangle Constructions, Ambiguous Case, Angle Bisectors, Altitudes, Perp Bisectors & Centers
Practice textbook-aligned solved MCQs with instant answer feedback, step-by-step solutions, and timed test simulation.