Chapter 6 : Triangles Class 10 Maths NCERT Solutions

Chapter 6: Triangles — Exercise 6.1

Triangles are the simplest polygons, but they hold the secrets to architecture and navigation! This exercise builds your foundation by explaining the difference between looking “the same” and being “the same size.”

Q1. Fill in the blanks using the correct word given in brackets:

i) All circles are __________. (congruent, similar)

The Logic: Think of a small coin and a large wheel. They are both perfectly round (same shape), but their sizes are different. In math, same shape but different size means “Similar.”
The Result: Similar

ii) All squares are __________. (similar, congruent)

The Logic: Every square has four 90-degree angles and equal sides. Whether it is a tiny square on your graph paper or a giant square courtyard, the shape remains identical.
The Result: Similar

iii) All __________ triangles are similar. (isosceles, equilateral)

The Logic: For triangles to be similar, their angles must be the same. Every equilateral triangle has three 60-degree angles, no matter how big it is. Isosceles triangles can have different angles, so they aren’t always similar.
The Result: Equilateral

iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

The Logic: This is the “Golden Rule” for similarity. To keep the shape the same, the corners (angles) must match exactly, and the sides must grow or shrink by the same ratio.
The Result: (a) Equal, (b) Proportional

Q2. Give two different examples of a pair of:

i) Similar figures

The Logic: Think of objects that are scaled versions of each other.
Example 1: Two equilateral triangles with sides 2 cm and 4 cm.
Example 2: Two circles with radii 1 cm and 5 cm.

ii) Non-similar figures

The Logic: Think of shapes that don’t share the same geometry.
Example 1: A triangle and a square.
Example 2: A right-angled triangle and an equilateral triangle.

Q3. State whether the following quadrilaterals are similar or not:

(Note: Refer to your textbook images of the rhombus PQRS and square ABCD)

The Logic: To be similar, a figure must pass two tests: 1. Are angles equal? 2. Are sides proportional?
The Steps:
- Step 1: Check sides. In the figures, the sides of the rhombus are 1.5 cm and the square are 3 cm. Ratio = 1.5/3 = 1/2. Sides are proportional.
- Step 2: Check angles. The square has 90-degree angles, but the rhombus does not.
Human Tip: Even if the sides look perfectly scaled, if the angles don’t match, the shape is “distorted” and cannot be similar!
The Result: No, they are not similar because their corresponding angles are not equal.

Chapter 6: Triangles — Exercise 6.2

This exercise is the foundation of modern geometry. By mastering the Basic Proportionality Theorem (BPT), you learn how parallel lines act as a “mirror,” creating perfectly balanced ratios on both sides of a triangle.

Q1. In figures (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

The Logic: Whenever you see parallel lines inside a triangle, think of BPT (Basic Proportionality Theorem). It tells us that if a line is parallel to one side, it divides the other two sides in the same ratio.
The Steps (i):
- Given: AD = 1.5 cm, DB = 3 cm, AE = 1 cm.
- Since DE || BC, by BPT: AD / DB = AE / EC
- 1.5 / 3 = 1 / EC
- EC = (3 x 1) / 1.5 = 2 cm
The Result (i): EC = 2 cm
The Steps (ii):
- Given: DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm.
- Since DE || BC, by BPT: AD / DB = AE / EC
- AD / 7.2 = 1.8 / 5.4
- AD / 7.2 = 1 / 3
- AD = 7.2 / 3 = 2.4 cm
The Result (ii): AD = 2.4 cm

Q2. E and F are points on the sides PQ and PR respectively of a triangle PQR. State whether EF || QR for the following cases:

The Logic: This is the Converse of BPT. If we calculate the ratios of the two sides and they turn out to be exactly equal, then the line EF must be parallel to the base.
The Steps (i): PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
- PE / EQ = 3.9 / 3 = 1.3
- PF / FR = 3.6 / 2.4 = 1.5
- Since 1.3 is not equal to 1.5, the ratios are not equal.
The Result (i): EF is not parallel to QR.
The Steps (ii): PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
- PE / QE = 4 / 4.5 = 8 / 9
- PF / RF = 8 / 9
- The ratios are identical!
The Result (ii): EF || QR.

Q3. In the figure, if LM || CB and LN || CD, prove that AM / AB = AN / AD.

The Logic: We apply BPT in two different triangles (ABC and ADC). Since both triangles share the same center line (AC), their side ratios become linked.
The Steps:
1. In triangle ABC, LM || CB. By BPT: AM / MB = AL / LC (Eq. 1)
2. In triangle ADC, LN || CD. By BPT: AN / ND = AL / LC (Eq. 2)
3. From Eq. 1 and Eq. 2, we see that AM / MB = AN / ND.
4. Pro-Tip: To reach the final form (AB and AD), we take the reciprocal and add 1 to both sides (a method called Componendo). This combines the segments into the full side of the triangle.
The Result: AM / AB = AN / AD. (Proved)

Q4. In the figure, DE || AC and DF || AE. Prove that BF / FE = BE / EC.

The Logic: Don’t get confused by the many lines! Focus on one parallel pair at a time. One pair works for the large triangle ABC, and the other pair works for the smaller triangle ABE.
The Steps:
1. Look at triangle ABE. Since DF || AE, by BPT: BF / FE = BD / DA (Eq. 1)
2. Look at triangle ABC. Since DE || AC, by BPT: BE / EC = BD / DA (Eq. 2)
3. Since both BF / FE and BE / EC are equal to the same ratio (BD / DA), they must be equal to each other.
The Result: BF / FE = BE / EC. (Proved)

Q5. In the figure, DE || OQ and DF || OR. Show that EF || QR.

The Logic: This problem is like a “Chain Reaction.” We use BPT twice to establish a common ratio, and then use the Converse of BPT to prove the final lines are parallel.
The Steps:
1. In triangle PQO, DE || OQ. By BPT: PE / EQ = PD / DO (Eq. 1)
2. In triangle POR, DF || OR. By BPT: PF / FR = PD / DO (Eq. 2)
3. From Eq. 1 and Eq. 2, we get PE / EQ = PF / FR.
4. Now, in triangle PQR, since the ratios of the sides are equal (PE/EQ = PF/FR), the line EF must be parallel to the third side.
The Result: EF || QR by Converse of BPT.

Q6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

The Logic: Think of this as a “Two-Step Bridge.” We use the parallel lines on the left and right sides to prove that the ratios on the bottom side must also be equal. Once the ratios are equal, the bottom lines must be parallel.
The Steps:
1. In triangle OPQ, AB || PQ. By BPT: OA / AP = OB / BQ (Eq. 1)
2. In triangle OPR, AC || PR. By BPT: OA / AP = OC / CR (Eq. 2)
3. Comparing both, we see both ratios equal OA/AP. Therefore: OB / BQ = OC / CR.
4. Now, look at triangle OQR. Since the segments on sides OQ and OR are in the same ratio, BC must be parallel to QR.
The Result: BC || QR (By Converse of BPT).

Q7. Using Basic Proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

The Logic: This is the “Mid-point Theorem” but proven using BPT. If you start exactly in the middle of one side and walk parallel to the base, you will land exactly in the middle of the other side.
The Steps:
1. Let ABC be a triangle where D is the mid-point of AB (so AD = DB, or AD / DB = 1).
2. A line DE is drawn parallel to BC (DE || BC).
3. By BPT: AD / DB = AE / EC.
4. Since AD / DB = 1, we get: 1 = AE / EC.
5. This means AE = EC.
The Result: E is the mid-point of AC. (Proved)

Q8. Using Converse of Basic Proportionality Theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

The Logic: This is the reverse of Q7. If we know we are starting at the mid-points of both sides, we calculate the ratios to prove the line is parallel.
The Steps:
1. Let D and E be mid-points of AB and AC.
2. Therefore, AD = DB (which means AD / DB = 1).
3. And AE = EC (which means AE / EC = 1).
4. Comparing them: AD / DB = AE / EC (Both are equal to 1).
The Result: DE || BC (By Converse of BPT).

Q9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO / BO = CO / DO.

The Logic: A trapezium is just a shape with one pair of parallel lines. To use BPT, we need a triangle. We “create” a triangle by drawing a line through O that is parallel to the base.
The Steps:
1. Draw a line EO through O such that EO || AB (and since AB || DC, EO || DC too).
2. In triangle ABD, EO || AB. By BPT: AE / ED = BO / DO (Eq. 1)
3. In triangle ADC, EO || DC. By BPT: AE / ED = AO / CO (Eq. 2)
4. From Eq. 1 and Eq. 2: BO / DO = AO / CO.
5. Rearranging to match the question: AO / BO = CO / DO.
The Result: AO / BO = CO / DO. (Proved)

Q10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO / BO = CO / DO. Show that ABCD is a trapezium.

The Logic: This is the converse of Q9. If the diagonals split each other in ratio, we prove that at least one pair of sides is parallel to show it’s a trapezium.
The Steps:
1. Given: AO / BO = CO / DO, which can be written as AO / CO = BO / DO.
2. Draw a line EO || AB.
3. In triangle ABD, since EO || AB, by BPT: AE / ED = BO / DO.
4. But we know BO / DO = AO / CO.
5. Therefore, AE / ED = AO / CO.
6. In triangle ADC, since the ratios are equal, EO || DC (By Converse of BPT).
7. Now, EO || AB (we drew it that way) and EO || DC. This means AB || DC.
The Result: Since AB || DC, ABCD is a trapezium.

Chapter 6: Triangles — Exercise 6.3

Welcome to the ultimate guide on Similarity of Triangles! This exercise is where you learn to identify “twins” in geometry—triangles that look exactly the same but might be different in size, helping you solve complex patterns with ease.

Q1. State which pairs of triangles in the given figures are similar. Write the similarity criterion used by you and also write the pairs of similar triangles in the symbolic form:

The Logic: To prove two triangles are similar, we don’t need every single side and angle to match. We just need to satisfy one of the three “Similarity Tests”: AAA/AA (Angles match), SSS (Sides are in ratio), or SAS (Two sides in ratio and the angle between them matches).
The Steps (i): In triangle ABC and triangle PQR:
- Angle A = Angle P = 60 degrees
- Angle B = Angle Q = 80 degrees
- Angle C = Angle R = 40 degrees
- Since all corresponding angles are equal, the triangles are similar.
The Result (i): Triangle ABC ~ Triangle PQR (AAA Similarity Criterion)
The Steps (ii): In triangle ABC and triangle PQR:
- AB/QR = 2 / 4 = 1/2
- BC/RP = 2.5 / 5 = 1/2
- CA/PQ = 3 / 6 = 1/2
- Since the ratios of all corresponding sides are equal, they are similar.
The Result (ii): Triangle ABC ~ Triangle QRP (SSS Similarity Criterion)

Q2. In the figure, triangle ODC ~ triangle OBA, Angle BOC = 125 degrees and Angle CDO = 70 degrees. Find Angle DOC, Angle DCO and Angle OAB.

The Logic: We use basic line properties (Linear Pairs) and triangle properties (Angle Sum Property). Once we find the angles in one triangle, the similarity tells us the angles in the other!
The Steps:
1. Find Angle DOC: DOB is a straight line. So, Angle DOC + Angle BOC = 180 (Linear Pair).
  
  Angle DOC + 125 = 180 => Angle DOC = 55 degrees.
2. Find Angle DCO: In triangle ODC, sum of angles is 180.
  
  70 + 55 + Angle DCO = 180 => Angle DCO = 55 degrees.
3. Find Angle OAB: Since triangle ODC ~ triangle OBA, the corresponding angles are equal.
  
  Angle OAB = Angle OCD (or DCO).
The Result: Angle DOC = 55, Angle DCO = 55, Angle OAB = 55 degrees.

Q3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA / OC = OB / OD.

The Logic: In a trapezium with parallel lines, the triangles formed by the diagonals are usually similar because of Alternate Interior Angles.
The Steps:
1. In triangle OAB and triangle OCD:
2. Angle OAB = Angle OCD (Alternate interior angles as AB || DC).
3. Angle OBA = Angle ODC (Alternate interior angles as AB || DC).
4. By AA Similarity, triangle OAB ~ triangle OCD.
5. When triangles are similar, their corresponding sides are in the same ratio.
The Result: OA / OC = OB / OD. (Proved)

Q4. In the figure, QR / QS = QT / PR and Angle 1 = Angle 2. Show that triangle PQS ~ triangle TQR.

The Logic: This one is tricky! We first use the given angles to prove a small triangle is isosceles. This lets us swap a side in the given ratio to satisfy the SAS Criterion.
The Steps:
1. In triangle PQR, Angle 1 = Angle 2 (Given).
2. Therefore, PQ = PR (Sides opposite to equal angles are equal).
3. The given ratio is: QR / QS = QT / PR.
4. Substitute PR with PQ: QR / QS = QT / PQ.
5. In triangle PQS and triangle TQR:
  - QR / QS = QT / PQ (From above)
  - Angle Q = Angle Q (Common angle)
The Result: Triangle PQS ~ Triangle TQR (By SAS Similarity).

Q5. S and T are points on sides PR and QR of triangle PQR such that Angle P = Angle RTS. Show that triangle RPQ ~ triangle RTS.

The Logic: This is the simplest type of proof. You just need to find two matching angles to satisfy the AA test.
The Steps:
1. In triangle RPQ and triangle RTS:
2. Angle P = Angle RTS (Given).
3. Angle R = Angle R (Common angle for both triangles).
The Result: Triangle RPQ ~ Triangle RTS (By AA Similarity).

Summary Table

Criterion	What to check?	Human Tip
AA (Angle-Angle)	Check if 2 angles match.	The easiest and most common test!
SSS (Side-Side-Side)	Check if all 3 side ratios are equal.	Use this when no angles are given.
SAS (Side-Angle-Side)	2 side ratios + the angle between them.	Make sure the angle is “sandwiched” by the sides.
Symbolic Form	The order of letters matters!	Match the equal angles (A goes with P, etc.).

Q6. In the figure, if triangle ABE ≅ triangle ACD, show that triangle ADE ~ triangle ABC.

- The Logic: This is a “Congruence to Similarity” bridge. Congruence (≅) means the triangles are identical in size, which gives us equal sides. We then use those equal sides to create the ratios needed for SAS similarity.
- The Steps:
  1. Given: triangle ABE ≅ triangle ACD.
  2. From CPCT (Corresponding Parts of Congruent Triangles): AB = AC and AE = AD.
  3. We can write this as: AD / AB = AE / AC.
  4. Now, in triangle ADE and triangle ABC:
    - AD / AB = AE / AC (From above)
    - Angle A = Angle A (Common angle)
- The Result: triangle ADE ~ triangle ABC (By SAS Similarity Criterion).

Getty Images

Q7. In the figure, altitudes AD and CE of triangle ABC intersect each other at the point P. Show that:

(i) triangle AEP ~ triangle CDP (ii) triangle ABD ~ triangle CBE (iii) triangle AEP ~ triangle ADB (iv) triangle PDC ~ triangle BEC

The Logic: Altitudes create 90-degree angles. In almost all these sub-parts, you just need to find one right angle and one common (or vertically opposite) angle to satisfy the AA criterion.
The Steps (i): In triangle AEP and triangle CDP:
- Angle AEP = Angle CDP = 90° (Altitudes)
- Angle APE = Angle CPD (Vertically opposite angles)
- Result: triangle AEP ~ triangle CDP (AA).
The Steps (ii): In triangle ABD and triangle CBE:
- Angle ADB = Angle CEB = 90°
- Angle B = Angle B (Common angle)
- Result: triangle ABD ~ triangle CBE (AA).
The Steps (iii): In triangle AEP and triangle ADB:
- Angle AEP = Angle ADB = 90°
- Angle A = Angle A (Common angle)
- Result: triangle AEP ~ triangle ADB (AA).
The Steps (iv): In triangle PDC and triangle BEC:
- Angle PDC = Angle BEC = 90°
- Angle C = Angle C (Common angle)
- Result: triangle PDC ~ triangle BEC (AA).

Q8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that triangle ABE ~ triangle CFB.

The Logic: Parallelograms have a special property—opposite angles are equal. Combined with alternate interior angles from parallel lines, we get a perfect AA similarity setup.
The Steps:
1. In triangle ABE and triangle CFB:
2. Angle A = Angle C (Opposite angles of a parallelogram ABCD).
3. Angle AEB = Angle CBF (Alternate interior angles because AD || BC).
The Result: triangle ABE ~ triangle CFB (By AA Similarity Criterion).

Q9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) triangle ABC ~ triangle AMP (ii) CA / PA = BC / MP

The Logic: Part (i) uses the standard AA test. Part (ii) is just the “reward”—once triangles are similar, their sides are automatically in ratio!
The Steps:
1. In triangle ABC and triangle AMP:
2. Angle ABC = Angle AMP = 90° (Given).
3. Angle A = Angle A (Common angle).
4. Therefore, triangle ABC ~ triangle AMP (AA Similarity).
5. Since the triangles are similar, their corresponding sides are proportional.
The Result: CA / PA = BC / MP. (Proved)

Q10. CD and GH are respectively the bisectors of Angle ACB and Angle EGF such that D and H lie on sides AB and FE of triangle ABC and triangle EFG respectively. If triangle ABC ~ triangle FEG, show that:

(i) CD / GH = AC / FG (ii) triangle DCB ~ triangle HGE (iii) triangle DCA ~ triangle HGF

The Logic: Since the big triangles are similar, their corresponding angles are equal. If you cut equal angles in half (angle bisectors), the resulting smaller angles are also equal!
The Steps:
1. Given triangle ABC ~ triangle FEG, so Angle C = Angle G.
2. Since CD and GH bisect these angles, Angle ACD = Angle FGH and Angle BCD = Angle EGH.
3. For (iii): In triangle DCA and triangle HGF:
  - Angle A = Angle F (Corresponding angles of similar big triangles).
  - Angle ACD = Angle FGH (Half of equal angles).
  - Result: triangle DCA ~ triangle HGF (AA).
4. For (i): Since triangle DCA ~ triangle HGF, their sides are in ratio: CD / GH = AC / FG.
5. For (ii): In triangle DCB and triangle HGE:
  - Angle B = Angle E.
  - Angle BCD = Angle EGH.
  - Result: triangle DCB ~ triangle HGE (AA).

Student Summary Table (Triangles Phase 2)

Concept	The “Aha!” Moment	Key Property
Altitude	Always look for the 90° angle.	Perpendicular lines
Parallelogram	Opposite angles are always equal.	Angle A = Angle C
Bisector	Cuts an angle into two equal halves.	Half of equal = Equal
AA Test	Find 2 matching angles to win the proof.	Sum of angles is 180°

We have reached the final stretch of Exercise 6.3! These questions are the “Grand Finale” because they combine everything you’ve learned—Similarity, Medians, and Ratios—into advanced proofs that frequently appear in the 5-mark section of board exams.

Q11. In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD is perpendicular to BC and EF is perpendicular to AC, prove that triangle ABD ~ triangle ECF.

The Logic: Since the triangle is isosceles (AB = AC), the angles opposite to these sides are equal. We then use those equal angles along with the 90-degree angles (from the perpendiculars) to prove similarity using the AA criterion.
The Steps:
1. In triangle ABC, AB = AC. Therefore, Angle B = Angle C (Angles opposite to equal sides).
2. Now, look at triangle ABD and triangle ECF:
  - Angle ADB = Angle EFC = 90° (Given).
  - Angle ABD = Angle ECF (From step 1, where Angle B = Angle C).
The Result: triangle ABD ~ triangle ECF (By AA Similarity Criterion).

Q12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of triangle PQR. Show that triangle ABC ~ triangle PQR.

The Logic: This is a classic “Median Property” question. We use the fact that a median bisects the side to show that the smaller triangles (containing the medians) are similar first. This then gives us the angle we need to prove the big triangles are similar.
The Steps:
1. Given: AB / PQ = BC / QR = AD / PM.
2. Since AD and PM are medians, BC = 2BD and QR = 2QM.
3. Substituting these: AB / PQ = 2BD / 2QM = AD / PM => AB / PQ = BD / QM = AD / PM.
4. By SSS similarity, triangle ABD ~ triangle PQM.
5. This means Angle B = Angle Q (Corresponding angles of similar triangles).
6. Now, in triangle ABC and triangle PQR:
  - AB / PQ = BC / QR (Given)
  - Angle B = Angle Q (From above)
The Result: triangle ABC ~ triangle PQR (By SAS Similarity Criterion).

Q13. D is a point on the side BC of a triangle ABC such that Angle ADC = Angle BAC. Show that CA^2 = CB.CD.

The Logic: Don’t let the “Squared” term scare you! $CA^2 = CB cdot CD$ is just another way of writing the ratio CA / CD = CB / CA. To get this ratio, we just need to prove that triangle ABC and triangle DAC are similar.
The Steps:
1. In triangle ABC and triangle DAC:
  - Angle BAC = Angle ADC (Given).
  - Angle C = Angle C (Common angle).
2. By AA Similarity, triangle ABC ~ triangle DAC.
3. Therefore, their sides are in ratio: CB / CA = CA / CD.
4. Cross-multiply the terms.
The Result: CA^2 = CB.CD (Proved)

Q14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that triangle ABC ~ triangle PQR.

The Logic: This is the harder version of Q12! Here, the proportional sides are AB and AC (instead of BC). We have to use a “Construction” to extend the medians and create parallelograms to solve this.
The Steps:
1. Extend AD to E such that AD = DE and PM to L such that PM = ML. Join BE, CE, QL, and RL.
2. By doing this, ABEC and PQLR become parallelograms (diagonals bisect each other).
3. This lets us show triangle ACE ~ triangle PRL using SSS similarity.
4. This eventually proves that Angle A = Angle P.
5. With Angle A = Angle P and the given side ratios AB/PQ = AC/PR…
The Result: triangle ABC ~ triangle PQR (By SAS Similarity Criterion).

Q15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

The Logic: This is a real-world application of similarity! Since the sun hits the pole and the tower at the same angle, the triangles formed by the objects and their shadows are similar.
The Steps:
1. Let height of tower be ‘h’.
2. Ratio of (Height of Pole / Shadow of Pole) = (Height of Tower / Shadow of Tower).
3. 6 / 4 = h / 28
4. h = (6 x 28) / 4
5. h = 6 x 7 = 42.
The Result: The height of the tower is 42 m.

Q16. If AD and PM are medians of triangles ABC and PQR, respectively, where triangle ABC ~ triangle PQR, prove that AB / PQ = AD / PM.

The Logic: This is the reverse of Q12. We start with the big triangles being similar and use that to prove the smaller “median-cut” triangles are similar.
The Steps:
1. Given: triangle ABC ~ triangle PQR. So, AB / PQ = BC / QR and Angle B = Angle Q.
2. Since BC = 2BD and QR = 2QM (Medians), the ratio becomes: AB / PQ = BD / QM.
3. In triangle ABD and triangle PQM:
  - AB / PQ = BD / QM
  - Angle B = Angle Q
4. By SAS Similarity, triangle ABD ~ triangle PQM.
The Result: AB / PQ = AD / PM (Proved).

Student Summary Table (Triangles Finale)

Tip Type	The “Mentor” Advice	Formula / Logic
Shadow Problems	Always assume triangles are similar.	H1 / S1 = H2 / S2
Square Terms	It’s just cross-multiplied ratios!	$x^2 = y cdot z$ -> $x/y = z/x$
Median Secret	It usually splits the base into 1:1.	$BD = BC / 2$
HOTS Hint	Q14 is the hardest; practice the construction twice!	Extension of median

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Chapter 6 : Triangles Class 10 Maths NCERT Solutions

Contents

Class 10 Maths NCERT Solutions - Chapter wise

Chapter 1: Real Numbers

Chapter 2: Polynomials

Chapter 3 : Pair of Linear Equations in Two Variables

Chapter 4 : Quadratic Equations

Chapter 5 : Arithmetic Progressions