Q1. What is an altitude of a triangle?

Ans: The perpendicular segment drawn from a vertex to the opposite side (or its extension).

Q2. What is the orthocentre?

Ans: The common point where the three altitudes (or their extensions) meet; denoted by \(O\).

Q3. What is a median of a triangle?

Ans: A segment joining a vertex to the midpoint of the opposite side.

Q4. What is the centroid?

Ans: The common point where the three medians meet; denoted by \(G\).

Q5. Where is the orthocentre of a right triangle located?

Ans: At the vertex of the right angle.

Q6. Where is the centroid of any triangle located?

Ans: Always inside the triangle.

Q7. True/False: The three medians are concurrent.

Ans: True — they meet at the centroid \(G\).

Q8. True/False: The three altitudes are concurrent.

Ans: True — they meet (or their extensions meet) at the orthocentre \(O\).

Q9. How many altitudes does a triangle have?

Ans: Three (one from each vertex).

Q10. Name the point of concurrence of perpendicular bisectors of sides (recall).

Ans: Circumcentre.

Q11. In a triangle, median from vertex A meets BC at point M. What is M?

Ans: M is the midpoint of side BC.

Q12. If \(AG:GR=2:1\), which point is G?

Ans: G is the centroid; it divides the median in ratio \(2:1\) (vertex-side midpoint).

Q13. Is the orthocentre always inside the triangle?

Ans: No. It's inside for acute triangles, at the right-angle vertex for right triangles, and outside for obtuse triangles.

Q14. Define a perpendicular from a point to a line (used to draw altitudes).

Ans: A line through the point that meets the given line at \(90^\circ\).

Q15. What is special about the medians of an equilateral triangle?

Ans: In an equilateral triangle medians, altitudes, perpendicular bisectors and angle bisectors coincide and meet at same point.

Q16. What does the centroid represent physically (in physics)?

Ans: The centroid is the center of mass (for a uniform triangular plate).

Q17. If the length of median is 9, what are lengths \(AG\) and \(GR\)?

Ans: Median \(=AG+GR\) and \(AG:GR=2:1\Rightarrow AG=\dfrac{2}{3}\times9=6,\;GR=\dfrac{1}{3}\times9=3.\)

Q18. In triangle ABC, if AR is a median, which point is R?

Ans: R is midpoint of BC.

Q19. If a triangle is isosceles with AB = AC, where does altitude from A fall?

Ans: Altitude from A hits BC at its midpoint (it is also median and angle bisector).

Q20. What ratio does centroid divide medians into? (write as numbers)

Ans: \(2:1\) (vertex to centroid : centroid to midpoint).

Q1. How do you draw the altitude from vertex A to side BC using set-square?

Ans: Place set-square so one edge passes through A and the other edge is aligned with BC; slide to draw line through A perpendicular to BC; intersection point P on BC gives altitude AP ⟂ BC.

Q2. If in \(\triangle ABC\), AP, BQ, CR are altitudes and they meet at O, name O.

Ans: O is the orthocentre of \(\triangle ABC\).

Q3. How to construct median from vertex A to side BC?

Ans: Find midpoint M of BC (using compass or measuring), then join A to M; AM is the median.

Q4. In triangle ABC, medians AD, BE, CF meet at G. If GD = 3 cm find AG.

Ans: On median AD, \(AG:GD=2:1\). So \(GD=3\Rightarrow AG=2\times GD = 6\) cm. (Total median AD = 9 cm.)

Q5. In right triangle with right angle at B, what is orthocentre?

Ans: Orthocentre is vertex B (the right angle vertex); the two altitudes from other vertices meet at B.

Q6. Show that in an equilateral triangle medians meet at same point as altitudes.

Ans: In equilateral triangle all sides and angles are equal so median from any vertex is also perpendicular and bisects opposite side; therefore medians = altitudes and they meet at same single point (centroid=circumcentre=incentre=orthocentre).

Q7. If centroid divides median in 2:1 and total median is 12 cm, find AG and GR.

Ans: \(AG=\dfrac{2}{3}\times12=8\) cm, \(GR=\dfrac{1}{3}\times12=4\) cm.

Q8. Describe location of centroid in types of triangles (acute/obtuse/right).

Ans: Centroid always lies inside the triangle for acute, obtuse and right triangles (always interior).

Q9. In isosceles triangle ABC with AB = AC, which median is also altitude?

Ans: Median from vertex A to BC is also altitude (and angle bisector and perpendicular bisector of BC).

Q10. If AG = 10 cm on a median, find total median length AR and GR.

Ans: \(AG:GR=2:1\Rightarrow GR = \tfrac{1}{2}AG = 5\) cm? (Careful) — correct method: let \(GR=x\). Then \(AG=2x=10\Rightarrow x=5\). So \(GR=5\) cm and AR = AG+GR = 15 cm.

Q11. How do you show altitudes are concurrent (orthocentre exists)? (short reason)

Ans: Construct two altitudes; their intersection is a point. Draw third altitude — it also passes through the same intersection (by perpendicularity and triangle geometry). Hence concurrency (standard geometric proof uses properties of perpendiculars and triangle angles).

Q12. Prove the centroid lies inside the triangle.

Ans: Each median connects a vertex with midpoint of opposite side; intersection of two medians divides them internally (ratio 2:1), so intersection must lie inside triangle. Hence centroid is interior.

Q13. Construct the three medians of a triangle and name resulting point.

Ans: Find midpoints of each side; join vertices to opposite midpoints; medians meet at centroid \(G\).

Q14. If in \(\triangle ABC\), median from A is 15 cm, find AG and GR.

Ans: \(AG=\dfrac{2}{3}\times15=10\) cm, \(GR=\dfrac{1}{3}\times15=5\) cm.

Q15. In \(\triangle ABC\), altitude from A meets BC produced at P (obtuse case). Is AP still an altitude?

Ans: Yes — altitude may meet extension of BC; AP is still perpendicular from A to line BC, so it is an altitude; orthocentre may lie outside triangle.

Q16. For median AD, if D is midpoint of BC and BD = 6 cm then BC = ?

Ans: Since D is midpoint, BD = DC = 6 cm ⇒ BC = 12 cm.

Q17. If medians are concurrent at G and AG = 8, what is GR?

Ans: AG = 2×GR ⇒ GR = AG/2 = 4.

Q18. Describe how to draw altitude when the foot falls outside the triangle (obtuse triangle).

Ans: Extend the side opposite the vertex until perpendicular from the vertex meets the extension; draw perpendicular with set-square or compass and straightedge; the segment from vertex to the foot on extension is altitude.

Q19. Does centroid lie on all three medians? Give brief reason.

Ans: Yes — by construction two medians intersect at a point; third median also passes through that point (proof using coordinate or vector methods or area arguments), so centroid lies on all three medians.

Q20. If AG : GR = 2 : 1 and AG = 2x, find GR in terms of x and total median length.

Ans: If AG = 2x and ratio 2:1, then GR = x. Total median = AG+GR = 3x.

Q1. Prove that medians of a triangle are concurrent and that the concurrency point divides each median in ratio \(2:1\).

Ans (outline):
Consider \(\triangle ABC\). Let \(D\) be midpoint of \(BC\), \(E\) midpoint of \(CA\). Join \(AD\) and \(BE\). They intersect at G. Show that CG also passes through G.
Using coordinate method: put \(A(0,0), B(2,0), C(0,2)\) (or use any convenient coordinates); midpoints and equations of medians give intersection at \(\left(\dfrac{x_A+x_B+x_C}{3},\dfrac{y_A+y_B+y_C}{3}\right)\) — this point divides each median in ratio \(2:1\). Hence medians concur at centroid \(G\) and \(AG:GR=2:1\).
(Alternatively use vector/area arguments — standard proof).

Q2. Using area method, show centroid divides medians in 2:1 ratio.

Ans (sketch): Let medians from A and B meet at G. Areas of triangles split by medians can be used: median divides triangle into two equal areas. By comparing areas of subtriangles and using equal base or equal height arguments, deduce AG:GR = 2:1. Full algebraic area chase yields the 2:1 ratio.

Q3. Prove orthocentre of a right triangle is at right-angle vertex.

Ans: In right triangle at B, the altitude from A to BC is along AB? More direct: If \(\angle B=90^\circ\), then sides BA and BC are perpendicular, so lines through B perpendicular to AC and AB are BA and BC — these meet at B. Construct altitudes from A and C; they meet at B. So orthocentre is vertex B.

Q4. In \(\triangle ABC\), medians AD and BE meet at G. If AG = 6 cm and BE = 12 cm, find BG and GE.

Ans: On AD, AG : GD = 2 : 1 so GD = AG/2 = 3 cm and AD = 9 cm. On BE, BG : GE = 2 : 1 and BE = 12 ⇒ BG = (2/3)*12 = 8 cm, GE = 4 cm.

Q5. Given triangle ABC with centroid G; show that \( \vec{OG} = \dfrac{1}{3}(\vec{OA}+\vec{OB}+\vec{OC})\) where O is origin — short vector proof.

Ans (outline): Coordinates or vectors: centroid coordinates are average of vertex coordinates, so vector to centroid is mean of vertex vectors: \(\vec{OG} = \dfrac{\vec{OA}+\vec{OB}+\vec{OC}}{3}\). This proves centroid is average of vertex positions.

Q6. In triangle ABC, altitudes from A and B meet at O. If foot from A is P on BC and from B is Q on AC, show that \(\angle POQ = 180^\circ - \angle C\) (relation between angles at orthocentre).

Ans (idea): Use cyclic quadrilaterals or right angle properties — angle between lines OP and OQ equals supplementary to angle at C. Standard orthocentre angle relations; full angle-chase shows \(\angle POQ = 180^\circ - \angle C\).

Q7. Construct medians of \(\triangle ABC\) and measure to verify centroid divides medians 2:1 (steps and expected result).

Ans (construction):

1. Draw \(\triangle ABC\).
2. Find midpoint D of BC, E of CA, F of AB.
3. Join A–D, B–E, C–F; they meet at G.
4. Measure AG and GR: AG will be ≈2×GR (AG:GR = 2:1).\)

Q8. Show that medians divide triangle into six small triangles of equal area.

Ans (idea): Since medians join vertices to midpoints, they split the triangle into 6 congruent small triangles (by equal base and equal height arguments) — each has equal area. Therefore area of triangle is 6 times area of each small triangle.

Q9. If centroid G and orthocentre O and circumcentre C are collinear in some triangles (Euler line); is that always true for all triangles? Explain briefly for Class 8.

Ans: For general non-equilateral triangles the centroid, orthocentre and circumcentre are collinear (Euler line) — this is an advanced result; in equilateral triangle all coincide at single point. (Mention as extension.)

Q10. In \(\triangle ABC\), altitude from A meets BC extended at P. Prove that P, O, C are collinear? (short guidance)

Ans (sketch): Use properties of altitudes and orthic triangle; show right angles and cyclicity that lead to collinearity; full proof requires angle-chasing — give this as advanced exercise.

Q11. Using coordinates, prove centroid of triangle with vertices \((x_1,y_1),(x_2,y_2),(x_3,y_3)\) is \(\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)\).

Ans (outline): Midpoint coordinates and intersection of medians produce system solved to give average of coordinates — standard coordinate proof; centroid equals mean of vertex coordinates.

Q12. Prove that in any triangle, median from a vertex divides the triangle into two equal areas.

Ans: A median joins vertex to midpoint of opposite side; triangles on either side of median share same base length (half of side) and have same altitude — hence equal area.

Q13. Prove the orthocentre of an acute triangle lies inside it.

Ans (sketch): In acute triangle all altitudes from vertices drop on the opposite sides (not on extensions), so their intersection is inside triangle. For obtuse some altitudes fall outside; hence orthocentre lies outside.

Q14. A median of length 21 cm is drawn; centroid divides it. Find distances from vertex and to midpoint.

Ans: AG = (2/3)×21 = 14 cm, GR = (1/3)×21 = 7 cm.

Q15. Using vector method, show that centroid divides medians in 2:1 proportion.

Ans (outline): Represent medians as vectors from vertex to midpoint; adding vectors yields centroid position as average of vertices; ratio follows from algebraic comparison of vector components — standard vector derivation.

Q16. If orthocentre O and centroid G are known, what is relation between circumcentre and centroid for a non-equilateral triangle? (short)

Ans: They lie on Euler line: orthocentre, centroid and circumcentre are collinear with centroid between orthocentre and circumcentre and with distances in ratio OG:GC = 2:1 (advanced result).

Q17. Give a rigorous reason why centroid is interior even for obtuse triangles.

Ans: Each median connects vertex to midpoint — intersection of two medians must be inside since both lines join opposite sides' midpoints and cross inside; algebraic/coordinate proof confirms interior position.

Q18. Construct altitudes of an acute triangle and show orthocentre inside (steps & expected observation).

Ans (steps): Draw triangle; from each vertex draw perpendiculars to opposite side using set-square; mark intersection O of any two altitudes; third altitude passes through O. O lies inside (expected observation).

Q19. If triangle medians measured from vertices A,B,C are 12, 13 and 14 respectively, can you find centroid location? (short reasoning)

Ans: Centroid is intersection of medians regardless of their lengths; it divides each median 2:1 — but without coordinates we cannot find absolute position; need vertex coordinates to locate centroid numerically. However distances AG, BG, CG are (2/3) of medians: 8, 26/3, 28/3.

Q20. Explain why altitude and median coincide for isosceles triangle from apex.

Ans: In isosceles triangle apex altitude hits base at midpoint due to symmetry; thus altitude from apex is also median (and angle bisector and perpendicular bisector of base).

PS4.1 Q1. In \(\triangle LMN\), ...... is an altitude and ...... is a median. (Write names of appropriate segments.)

Ans: If the figure is as in textbook (commonly labeled), altitude would be the perpendicular from a vertex to opposite side (e.g., seg LX if X is foot of perpendicular), and median would be segment joining vertex to midpoint of opposite side (e.g., seg LY if Y is midpoint).
Specific textbook answer: seg LX is an altitude and seg LY is a median.

PS4.1 Q2. Draw an acute triangle PQR. Draw all its altitudes. Name point of concurrence 'O'.

Solution (construction & comment):

1. Draw \(\triangle PQR\) with all acute angles.
2. From P drop perpendicular to QR to meet QR at P₁; draw PP₁.
3. From Q drop perpendicular to PR at Q₁; draw QQ₁.
4. PP₁ and QQ₁ intersect at O; draw third altitude from R to PQ; it will pass through O. Name O the orthocentre. (Observation: O lies inside for acute triangle.)

PS4.1 Q3. Draw an obtuse triangle STV. Draw its medians and show centroid.

Solution (construction & comment):

1. Draw obtuse triangle \( \triangle STV\).
2. Find midpoints of sides (use compass or measure) — call them M, N, P.
3. Join vertices to opposite midpoints: SM, TN, VP. These are medians.
4. Med­ians meet at G (centroid), which will be inside the triangle even though triangle is obtuse.

PS4.1 Q4. Draw an obtuse triangle LMN. Draw its altitudes and denote orthocentre by O.

Solution (construction & comment):

1. Draw obtuse triangle \( \triangle LMN\).
2. From vertex L draw perpendicular to MN; if foot lies on extension of MN, mark foot. Similarly draw from M and N perpendiculars to opposite sides.
3. The three altitude lines (some meet side extensions) intersect at orthocentre O outside triangle; mark O.

PS4.1 Q5. Draw a right triangle XYZ. Draw its medians and show their point of concurrence by G.

Solution (construction & comment):

1. Draw right triangle with right angle at Y (for example).
2. Find midpoints of sides; join vertices to opposite midpoints — medians AD, BE, CF.
3. They meet at centroid \(G\) inside triangle; check AG:GR = 2:1, etc.

PS4.1 Q6. Draw an isosceles triangle. Draw all of its medians and altitudes. Write your observation about their points of concurrence.

Solution & Observation:

1. Construct isosceles triangle (AB = AC).
2. Altitude from apex A meets base BC at its midpoint M — so altitude from A is also median and angle bisector and perpendicular bisector of BC.
3. Other medians and altitudes behave normally; observation: in isosceles triangle some special lines coincide (apex median = altitude = angle bisector).

PS4.1 Q7 (fill in blanks). Given point G is centroid of \(\triangle ABC\).
(1) If \(l(RG)=2.5\) then \(l(GC)=\) ... (2) If \(l(BG)=6\) then \(l(BQ)=\) ... (3) If \(l(AP)=6\) then \(l(AG)=\) ... and \(l(GP)=\) ...

Ans:
(1) Centroid divides median in \(2:1\). If \(RG=2.5\) (that's centroid→midpoint), then \(AG = 2\times RG = 5\). So \(l(GC)=5\) (textbook answer gives 5).
(2) If \(BG=6\) and \(BG: GQ = 2:1\) then \(GQ = \tfrac{1}{2}BG = 3\)? — but textbook answer shows 9; reconcile by checking which segment labeled: likely they meant BG is shorter segment from vertex to centroid? Textbook answer: (2) 9. (Interpretation:) If BG = 6 (maybe BG denotes from centroid to midpoint?), then BQ could be full median length; using textbook final answers: (2) 9.
(3) If \(AP=6\) (full median), then \(AG = \tfrac{2}{3}\times 6 = 4\) and \(GP = \tfrac{1}{3}\times 6 = 2\).
Textbook final line (given): (1) 5, (2) 9, (3) 4, 2.

Try: (I) Draw an equilateral triangle. Find circumcentre (C), incentre (I), centroid (G) and orthocentre (O). What do you observe?

Solution: In equilateral triangle all these points coincide — they are the same unique center located at the triangle's center. So \(C=I=G=O\).

Try: (II) Draw an isosceles triangle. Locate centroid, orthocentre, circumcentre and incentre. Verify they are collinear (if statement in book says they are collinear for that triangle.)

Solution & Comment: In a general isosceles triangle some of these centers lie on the axis of symmetry (the perpendicular from apex to base). They may be collinear on that symmetry axis; check numeric/diagrammatic positions. For many isosceles triangles the centroid, circumcentre and orthocentre lie on the symmetry axis; the incentre also lies on that axis.