Chapter 6 of NCERT Maths Class 10 delves into the world of triangles, building upon your previous knowledge. The key focus is on the similarity of triangles and its applications.
Key Concepts:
- Similarity of Triangles:
- Triangles can be similar based on the equality of corresponding angles (AAA similarity) or the proportionality of corresponding sides (SSS and SAS similarity).
- The concept of similarity is crucial for understanding the relationship between different triangles and their properties.
- Criteria for Similarity of Triangles:
- AAA similarity: If two triangles have corresponding angles equal, they are similar.
- SSS similarity: If the corresponding sides of two triangles are in proportion, they are similar.
- SAS similarity: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in proportion, the triangles are similar.
- Basic Proportionality Theorem (BPT):
- If a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides the two sides in the same ratio.
- Pythagoras Theorem:
- In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
- Areas of Similar Triangles:
- The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Applications:
These concepts are used in various real-life situations, including:
- Surveying
- Architecture
- Engineering
- Navigation
By understanding these concepts, you can solve a variety of problems related to triangles, including finding unknown side lengths, angles, and areas.
Exercise 6.1
1. Fill in the blanks by using the correct word given in brackets.
(i) All circles are ……………. . (congruent/similar)
(ii) All squares are …………… . (similar/congruent)
(iii) All …………….. triangles are similar. (isosceles/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)
Ans :
(i) All circles are similar .
(ii) All squares are similar .
(iii) All equilateral triangles are similar.
iv)
(a) their corresponding angles are equal and
(b) their corresponding sides are proportional
2. Give two different examples of pairs of
(i) similar figures.
(ii) non-similar figures.
Ans :
(i) Similar Figures
- Two equilateral triangles: All equilateral triangles have the same shape, with all angles equal to 60 degrees and side lengths in proportion.
- Two squares: All squares have four equal sides and four right angles, making them similar.
(ii) Non-Similar Figures
- A circle and a square: These shapes have entirely different properties and cannot be scaled to match each other.
- A triangle and a parallelogram: While both are polygons, their angles and side lengths do not correspond in a way that allows them to be scaled into each other.
3. State whether the following quadrilaterals are similar or not.
Ans :
The given quadrilaterals are not similar.
Reasoning:
For two quadrilaterals to be similar, their corresponding angles must be equal, and their corresponding sides must be in proportion.
Exercise 6.2
1. In the given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Ans :
For triangle (i):
- Applying BPT, we get:
- AD/DB = AE/EC
- Substituting the given values, we get: 1.5/3 = 1/EC
- Cross-multiplying, we get: 1.5 * EC = 3
- Therefore, EC = 3 / 1.5 = 2 cm
For triangle (ii):
- Applying BPT, we get:
- AD/DB = AE/EC
- Substituting the given values, we get: AD/7.2 = 1.8/5.4
- Cross-multiplying, we get: 5.4 * AD = 1.8 * 7.2
- Therefore, AD = (1.8 * 7.2) / 5.4 = 2.4 cm
2. E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Ans :
Case (i): PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
- PE/EQ = 3.9/3 = 1.3
- PF/FR = 3.6/2.4 = 1.5
Since PE/EQ ≠ PF/FR, EF is not parallel to QR.
Case (ii): PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
- PE/EQ = 4/4.5 = 8/9
- PF/RF = 8/9
Since PE/EQ = PF/RF, EF is parallel to QR.
Case (iii): First, we need to find EQ and FR:
- EQ = PQ – PE
- = 1.28 – 0.18 = 1.1 cm
- FR = PR – PF
- = 2.56 – 0.36 = 2.2 cm
- PE/EQ = 0.18/1.1 ≠ PF/FR = 0.36/2.2
Since PE/EQ ≠ PF/FR, EF is not parallel to QR.
3. In the given figure, if LM || CB and LN || CD.
Prove that AM/AB =AN/AD
∙
Ans :
In triangle ABC, LM || CB By Basic Proportionality Theorem (BPT), we have:
AM/MB = AN/ND …(1)
(AM + MB)/MB = (AN + ND)/ND
AB/MB = AD/ND
Inverting both sides, we get:
MB/AB = ND/AD …(2)
Multiplying equations (1) and (2), we get:
(AM/MB) * (MB/AB) = (AN/ND) * (ND/AD)
AM/AB = AN/AD
Hence proved.
4. In the given figure, DE || AC and DF || AE.
Prove that BF/FE=BE/EC
Ans :
In triangle AEC, DF || AE
By Basic Proportionality Theorem (BPT), we have:
BF/FE = BD/DE …(1)
In triangle ABC, DE || AC
By Basic Proportionality Theorem (BPT), we have:
BD/DE = BE/EC …(2)
From equations (1) and (2), we get:BF/FE = BE/EC
Hence proved.
5. In the given figure, DE || OQ and DF || OR. Show that EF || QR.
Ans :
In triangle PQR, DE || OQ:
- Applying the Basic Proportionality Theorem (BPT), we get:
- PD/DO = PE/EQ …(1)
In triangle PQR, DF || OR:
- Applying the Basic Proportionality Theorem (BPT), we get:
- PD/DO = PF/FR …(2)
From equations (1) and (2), we get:
- PE/EQ = PF/FR
In triangle PQR, PE/EQ = PF/FR:
- Applying the converse of the Basic Proportionality Theorem, we get:
- EF || QR
Hence proved.
6. In the given figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Ans :
In triangle OPQ, AB || PQ:
- Applying the Basic Proportionality Theorem (BPT), we get:
- OA/AP = OB/BQ …(1)
In triangle OPR, AC || PR:
- Applying the Basic Proportionality Theorem (BPT), we get:
- OA/AP = OC/CR …(2)
From equations (1) and (2), we get:
- OB/BQ = OC/CR
In triangle OQR, OB/BQ = OC/CR:
- Applying the converse of the Basic Proportionality Theorem, we get:
- BC || QR
Hence proved.
7. Using B.P.T., prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that your have proved it in class IX)
Ans :
Given:
In triangle ABC,
D is the midpoint of AB such that AD = DB. DE is parallel to BC.
To prove:
AE = EC.
Proof:
Since D is the midpoint of AB,
AD = DB
Therefore, AD/DB = 1
Given that DE || BC,
By Basic Proportionality Theorem (BPT),
AD/DB = AE/EC
Since AD/DB = 1,
AE/EC = 1
Hence, AE = EC.
Therefore, E is the midpoint of AC.
8. Using converse of B.P.T., prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that your have done it in class IX)
Ans :
Given:
In triangle ABC, D and E are the mid-points of AB and AC respectively.
To prove:
DE || BC
Proof:
Since D and E are mid-points of AB and AC respectively, AD = DB and AE = EC
Therefore, AD/DB = 1 and AE/EC = 1
Hence, AD/DB = AE/EC
By the converse of Basic Proportionality Theorem (BPT),
DE || BC
9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO
Ans :
Given: Trapezium ABCD with AB || DC. Diagonals AC and BD intersect at point O. To Prove: AO/BO = CO/DO
Proof:
Construction: Draw EF parallel to AB and DC, passing through point O.
In triangle ABC, EF || AB:
- Applying the Basic Proportionality Theorem (BPT), we get:
- AO/BO = AE/EC …(1)
In triangle ADC, EF || DC:
- Applying the Basic Proportionality Theorem (BPT), we get:
- CO/DO = AE/EC …(2)
From equations (1) and (2), we get:
- AO/BO = CO/DO
Hence proved.
Therefore, in a trapezium ABCD with AB || DC, and diagonals intersecting at point O, AO/BO = CO/DO.
10. 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.
Ans :
∙
Exercise 6.3
1. State which pairs of triangles in the given figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
Ans :
2. In the given figure, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Ans :
We are given a figure with triangles ODC and OBA, where:
- ∠BOC = 125°
- ∠CDO = 70°
- ∆ODC ~ ∆OBA (similar triangles)
We need to find ∠DOC, ∠DCO, and ∠OAB.
Solution
1. Finding ∠DOC:
- ∠BOC + ∠DOC = 180° (linear pair)
- ∠DOC = 180° – 125° = 55°
2. Finding ∠DCO:
- In triangle ODC, ∠CDO + ∠DOC + ∠DCO = 180° (angle sum property)
- 70° + 55° + ∠DCO = 180°
- ∠DCO = 180° – 125° = 55°
3. Finding ∠OAB:
- Since ∆ODC ~ ∆OBA, corresponding angles are equal.
- Therefore, ∠OAB = ∠DCO = 55°
Hence, ∠DOC = 55°, ∠DCO = 55°, and ∠OAB = 55°.
3. Diagonals AC and BD of a trape∠ium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD
Ans :
Given: Trapezium ABCD with AB || DC. Diagonals AC and BD intersect at point O. To Prove: OA/OC = OB/OD
Proof:
Consider triangles OAB and OCD.
- ∠AOB = ∠COD (Vertically opposite angles)
- ∠OAB = ∠OCD (Alternate angles as AB || DC)
Since two corresponding angles of triangles OAB and OCD are equal, by the AA similarity criterion, triangles OAB and OCD are similar.
Therefore, corresponding sides are proportional:
- OA/OC = OB/OD
Hence proved.
4. In the given figure, QR/QS = QT/PR and ∠1 = ∠2. show that ∆PQR ~ ∆TQR.
Ans :
Given:
- QR/QS = QT/PR
- ∠1 = ∠2
To prove:
ΔPQS ~ ΔTQR
Proof:
We are given that:
- QR/QS = QT/PR …(1)
- ∠1 = ∠2 …(2)
Also, ∠PQR is common to both triangles PQS and TQR.
Therefore, in triangles PQS and TQR,
- ∠PQR = ∠PQR (common angle)
- QR/QS = QT/PR (given)
- ∠1 = ∠2 (given)
Hence, by SAS (Side-Angle-Side) similarity criterion,ΔPQS ~ ΔTQR
Therefore, the two triangles PQS and TQR are similar.
5. S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Ans :
Given: Triangle PQR with points S and T on sides PR and QR respectively such that ∠P = ∠RTS. To Prove: Triangle RPQ ~ Triangle RTS
Proof:
In triangles RPQ and RTS,
- ∠P = ∠RTS (Given)
- ∠R is common to both triangles.
Therefore, by the AA similarity criterion, Triangle RPQ ~ Triangle RTS.
Hence proved
6. In the given figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.
Ans :
Given:
∆ABE ≅ ∆ACD
To Prove:
∆ADE ~ ∆ABC
Proof:
Since ∆ABE ≅ ∆ACD, their corresponding parts are equal.
Therefore, AB = AC and AE = AD.
Now, consider triangles ADE and ABC:
- ∠A is common to both triangles.
- AD/AB = AE/AC (since AB = AC and AD = AE)
Hence, by the SAS (Side-Angle-Side) similarity criterion,∆ADE ~ ∆ABC
Therefore, triangles ADE and ABC are similar.
7. In the given figure, altitudes AD and CE of ∆ABC intersect each other at the point P. Show that:
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC
Ans :
i) ∆AEP ~ ∆CDP
- ∠AEP = ∠CDP = 90° (Given, AD and CE are altitudes)
- ∠APE = ∠CPD (Vertically opposite angles) Therefore, by AA similarity criterion, ∆AEP ~ ∆CDP.
ii) ∆ABD ~ ∆CBE
- ∠ADB = ∠CEB = 90° (Given, AD and CE are altitudes)
- ∠ABD = ∠CBE (Common angle) Therefore, by AA similarity criterion, ∆ABD ~ ∆CBE.
iii) ∆AEP ~ ∆ADB
- ∠AEP = ∠ADB = 90° (Given, AD and CE are altitudes)
- ∠PAE = ∠DAB (Common angle) Therefore, by AA similarity criterion, ∆AEP ~ ∆ADB.
iv) ∆PDC ~ ∆BEC
- ∠PDC = ∠BEC = 90° (Given, AD and CE are altitudes)
- ∠PCD = ∠BCE (Common angle) Therefore, by AA similarity criterion, ∆PDC ~ ∆BEC.
Hence proved.
All the given pairs of triangles are similar due to the AA similarity criterion.
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
Ans :
Given: Parallelogram ABCD with point E on AD produced and BE intersecting CD at F. To Prove: ∆ABE ~ ∆CFB
Proof:
In parallelogram ABCD:
- Opposite angles are equal.
- Therefore, ∠A = ∠C (opposite angles of a parallelogram)
In triangles ABE and CFB:
- ∠A = ∠C (proved above)
- ∠AEB = ∠CFB (alternate interior angles as AB || DC)
By the AA (Angle-Angle) similarity criterion,∆ABE ~ ∆CFB
Hence proved.
9. In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
Ans :
Proof of ∆ABC ~ ∆AMP:
- ∠ABC = ∠AMP (Both are right angles)
- ∠BAC = ∠MAP (Common angle)
Since two angles of one triangle are equal to two corresponding angles of another triangle, by the AA (Angle-Angle) similarity criterion,
∆ABC ~ ∆AMP
Proof of CA/PA = BC/MP:
Since ∆ABC ~ ∆AMP, their corresponding sides are proportional.
Therefore,CA/PA = BC/MP
Hence proved.
10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that
Ans :
11. In the given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF.
Ans :
Given: ∆ABC is isosceles with AB = AC, AD ⊥ BC, and EF ⊥ AC.
To prove: ∆ABD ~ ∆ECF
Proof:
- ∠ADB = ∠EFC = 90° (Given, AD ⊥ BC and EF ⊥ AC)
- Since ∆ABC is isosceles with AB = AC, ∠B = ∠C.
- In ∆ABD and ∆ECF,
- ∠ADB = ∠EFC (Both are right angles)
- ∠B = ∠C (Proved above)
- Therefore, by the AA (Angle-Angle) similarity criterion,∆ABD ~ ∆ECF
Hence proved.
12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see in given figure). Show that ∆ABC ~ ∆bPQR.
Ans :
In triangles ABD and PQM,
- AB/PQ = AD/PM (given)
- ∠ADB = ∠PQM (both are right angles, as AD and PM are medians)
Therefore, by SAS (Side-Angle-Side) similarity criterion,
- Triangle ABD ~ Triangle PQM
Since similar triangles have corresponding sides proportional,
- BD/QM = AB/PQ
But we know, AB/PQ = BC/QR (given)
Therefore, BD/QM = BC/QR
Now, in triangles BCD and QMR,
- BD/QM = BC/QR (proved above)
- ∠BDC = ∠QMR (both are right angles, as AD and PM are medians)
Therefore, by SAS (Side-Angle-Side) similarity criterion,
- Triangle BCD ~ Triangle QMR
Since Triangle ABD ~ Triangle PQM and Triangle BCD ~ Triangle QMR, their corresponding angles are equal.
Hence, ∠ABD = ∠PQM and ∠CBD = ∠QMR
Adding these two equations, we get: ∠ABC = ∠PQR
Now, in triangles ABC and PQR,
- AB/PQ = BC/QR (given)
- ∠ABC = ∠PQR (proved above)
Therefore, by SAS (Side-Angle-Side) similarity criterion, Triangle ABC ~ Triangle PQR
Hence proved.
13. D is a point on the side BC of a triangle ABC, such that ∠ADC = ∠BAC. Show that CA² = CB.CD.
Ans :
- Construct the altitude from A to side BC, intersecting at point E.
- Since AE is the altitude, we know:
- ∠CAE = ∠BAE = 90° (right angles)
- In triangles BAE and CAD:
- ∠BAE = ∠CAD (both are right angles)
- ∠ABE = ∠ACD (given)
Therefore, by AA (Angle-Angle) similarity,Triangle BAE ~ Triangle CAD
- Since similar triangles have corresponding sides proportional:
- BE/AE = CD/CA …(1)
- Now, consider right triangles BAE and CAE:
- Pythagoras theorem can be applied to both triangles:
- BE² + AE² = BA² …(2)
- CA² + AE² = CE² …(3)
- Pythagoras theorem can be applied to both triangles:
- Subtracting (2) from (3), we get:
- CA² = CE² – BE² …(4)
- In triangle BCE (right-angled at E), Pythagoras theorem applies:
- CE² = BC² + BE² …(5)
- Substitute equation (5) into equation (4):
- CA² = (BC² + BE²) – BE²
- CA² = BC²
- Now, substitute equation (1) into the result of step 8:
- CA² = BC² = CB * (CD/AE) * AE
- CA² = CB * CD (since AE cancels out)
Therefore, we have proven that CA² = CB * CD.
14. 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 ∆ABC ~ ∆PQR.
Ans :
15. 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.
Ans :
16. If AD and PM are medians of triangles ABC and PQR respectively, where
∆ABC ~ ∆PQR. Prove that AB/PQ = AD/PM
∙
Ans :
