NCERT Solutions Class 10 Maths Chapter 6 Triangles

Question

NCERT Solutions for Class 10 Maths Chapter 6 Triangles are provided here which is one of the most important study materials for the students preparing for the CBSE board examination. Our NCERT Solutions are based on the latest syllabus of the CBSE. Prepared and designed by subject matter experts these solutions are on to the point and explained in a point-wise method to make it easier for the students to learn, study, and last minute revision. We have also covered all the shortcuts, equations, rules, theorems, and axioms.

In the NCERT Solution Class 10 provided by us, we have explained all the topics in detail which will surely help students clear all the required concepts. Important topics discussed here are

• Pythagoras theorem – Pythagoras theorem states that “In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of squares of the other two sides.
• Similar figures – Two triangles are said to be similar when they have the same shape, irrespective of their size.
• Congruent figures – two triangles are said to be congruent if and only if they are an exact replica of each other, same size and same shape.
• Congruency and Similarity of triangles and their differences.
• Basic Proportionality Theorem (Thales Theorem) – It provides the relationship between the sides of two equiangular triangles.
• Angle-Angle-Angle (AAA) similarity – Two triangles that have all the angles equal to each other have AAA similarity.
• Comparison between AA and AAA, are they the same?
• Side-Side-Side (SSS) similarity – Two triangles that have all the sides equal to each other have SSS similarity.
• Side-Angle-Side (SAS) similarity – Two triangles that have two sides and the angle between them equal to each other are said to have SAS similarity.
• Ratios of areas of two similar triangles – In two similar triangles ratio of the area of two triangles is proportional to the square of the ratio of their corresponding sides.
• Perpendicular bisector of the right angle in a right angle triangle created who similar triangle in which smaller triangles are also similar triangles the larger triangle.

Our NCERT Solutions Class 10 Maths is the best resource to practice and learn about all the important concepts. Regular practice of our solutions will help students develop problem-solving skills related to the subject matter. We have solutions for all kinds of queries including exercise and in-text questions.

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Answer 1

NCERT Solutions Class 10 Maths Chapter 6 Triangles

1. Fill in the blanks using correct word given in the 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)

Answer: 

(i) Similar 

(ii) Similar 

(iii) Equilateral 

(iv) (a) Equal 

      (b) Proportional

2. Give two different examples of pair of 

(i) Similar figures 

(ii) Non-similar figures

Answer: 

(i) Two equilateral triangles with sides 1 cm and 2 cm

Two squares with sides 1 cm and 2 cm

(ii) Trapezium and square

Triangle and parallelogram

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

Answer: 

Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional, i.e. 1:2, but their corresponding angles are not equal.

4. In figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

(i)

(ii)

Answer: 

(i)

Let EC = x cm 

It is given that DE || BC. 

By using basic proportionality theorem, we obtain

(ii)

Let AD = x cm 

It is given that DE || BC. 

By using basic proportionality theorem, we obtain

5. 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 = 0.63 cm

Answer: 

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

Therefore, EF is not parallel to QR.

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm 

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm

Therefore, EF is not parallel to QR.

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

Therefore, EF is not parallel to QR.

6. In the following figure, if LM || CB and LN || CD, prove that  \(\frac{AM}{AB} =\frac{AN}{AD}\) .

Answer: 

In the given figure, LM || CB 

By using basic proportionality theorem, we obtain

7. In the following figure, DE || AC and DF || AE. Prove that \(\frac{BF}{FE} =\frac{BE}{EC}\) .

Answer: 

In ∆ABC, DE || AC

\(\therefore\frac{BD}{DA}=\frac{BE}{EC}\)    (Basic Proportionality Theorem)     (i)

In ∆ABC, DF || AE

\(\therefore\frac{BD}{DA}=\frac{BF}{FE}\)       (Basic Proportionality Theorem)     (ii)

From (i) and (ii), we obtain 

\(\frac{BE}{EC}=\frac{BF}{FE}\)

8. In the following figure, DE || OQ and DF || OR, show that EF || QR.

Answer: 

In ∆ POQ, DE || OQ

 \(\therefore\frac{PE}{EQ}=\frac{PD}{DO}\)  (Basic Proportionality Theorem)     (i)

In ∆ POQ, DF || OR

\(\therefore\frac{PF}{FR}=\frac{PD}{DO}\)   (Basic Proportionality Theorem)     (Ii)

From (i) and (ii), we obtain 

\(\frac{PE}{EQ}=\frac{PF}{FR}\)

\(\therefore EF\parallel QR\) (Converse of basic proportionality theorem)

Answer 2

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

Answer:

In ∆ POQ, AB || PQ

\(\therefore\frac{OA}{AP} = \frac{OB}{BQ}\)    (Basic Proportionality Theorem)      (i)

In ∆ POQ, AC || PR 

\(\therefore\frac{OA}{AP} = \frac{OC}{CR}\)     (Basic Proportionality Theorem)      (iI)

From (i) and (ii), we obtain

\(\frac{OB}{BQ}=\frac{OC}{CR}\)

\(\therefore BC\parallel QR\)  (By the converse of basic proportionality theorem)

10. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side.

Answer:

Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that \(PQ \parallel BC\)

By using basic proportionality theorem, we obtain

\(\frac{AQ}{QC}=\frac{AP}{PB}\)

\(\frac{AQ}{QC}=\frac{1}{1}\)        (P is the mid-point of AB. \(\therefore\) AP = PB)

⇒ AQ = QC 

Or, Q is the mid-point of AC.

11. Using Converse of basic proportionality theorem, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side.

Answer:

Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively. 

i.e., AP = PB and AQ = QC It can be observed that 

\(\frac{AP}{PB}= \frac{1}{1}\)

and \(\frac{AQ}{QC}= \frac{1}{1}\)

\(\therefore\frac{AP}{PB} = \frac{AQ}{QC}\)

Hence, by using basic proportionality theorem, we obtain \(PQ\parallel BC\) 

12.  ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(\frac{AO}{BO} = \frac{CO}{DO}\) .

Answer:

Draw a line EF through point O, such that \(EF \parallel CD\)

In ∆ADC,

 \(EO \parallel CD\)

By using basic proportionality theorem, we obtain

\(\frac{AE}{ED}=\frac{AO}{OC}\)      (1)

In ∆ABD,

 \(OE \parallel AB\)

So, by using basic proportionality theorem, we obtain

\(\frac{ED}{AE}=\frac{OD}{BO}\)

⇒ \(\frac{AE}{ED}=\frac{BO}{OD}\)       (2)

From equations (1) and (2), we obtain

\(\frac{AO}{OC} = \frac{BO}{OD}\)

⇒ \(\frac{AO}{BO} = \frac{OC}{OD}\)

13. The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac{AO}{BO}=\frac{CO}{DO}\). Show that ABCD is a trapezium.

Answer:

Draw a line OE || AB

In ∆ABD, 

OE || AB 

By using basic proportionality theorem, we obtain

\(\frac{AE}{ED}=\frac{BO}{OD}\)           (1)

However, it is given that

\(\frac{AO}{OC} = \frac{OB}{OD}\)              (2)

From equations (1) and (2), we obtain

\(\frac{AE}{ED}=\frac{AO}{OC}\)

⇒ EO || DC [By the converse of basic proportionality theorem] 

⇒ AB || OE || DC 

⇒ AB || CD 

∴ ABCD is a trapezium.

14. State which pairs of triangles in the following figure 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:

(i)

(ii) 

(iii)

(iv)

 

(v) 

(vi)

Answer:

(i) ∠A = ∠P = 60° 

∠B = ∠Q = 80° 

∠C = ∠R = 40° 

Therefore, ∆ABC ∼ ∆PQR [By AAA similarity criterion]

\(\frac{AB}{QR}=\frac{BC}{RP}= \frac{CA}{PQ}\)

(ii) \(\therefore \triangle ABC∼ \triangle QRP\) [By SSS similarity criterion]

(iii)The given triangles are not similar as the corresponding sides are not proportional.

(iv)The given triangles are not similar as the corresponding sides are not proportional. 

(v)The given triangles are not similar as the corresponding sides are not proportional.

(vi) In ∆DEF, ∠D +∠E +∠F = 180º (Sum of the measures of the angles of a triangle is 180º) 

70º + 80º +∠F = 180º 

∠F = 30º 

Similarly, in ∆PQR, 

∠P +∠Q +∠R = 180º 

(Sum of the measures of the angles of a triangle is 180º) 

∠P + 80º +30º = 180º 

∠P = 70º 

In ∆DEF and ∆PQR, 

∠D = ∠P (Each 70°) 

∠E = ∠Q (Each 80°) 

∠F = ∠R (Each 30°) 

∴ ∆DEF ∼ ∆PQR [By AAA similarity criterion]

Answer 3

15. In the following figure, ∆ODC ∼ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB

Answer:

DOB is a straight line. 

∴ ∠DOC + ∠COB = 180° 

⇒ ∠DOC = 180° − 125° = 55° 

In ∆DOC, 

∠DCO + ∠CDO + ∠DOC = 180° 

(Sum of the measures of the angles of a triangle is 180º.) 

⇒ ∠DCO + 70º + 55º = 180° 

⇒ ∠DCO = 55° 

It is given that ∆ODC ∼ ∆OBA. 

∴ ∠OAB = ∠ OCD [Corresponding angles are equal in similar triangles.] 

⇒ ∠OAB = 55°

16. 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 \(\frac{AO}{OC} = \frac{OB}{OD}\) .

Answer:

In ∆DOC and ∆BOA, 

∠CDO = ∠ABO [Alternate interior angles as AB || CD] 

∠DCO = ∠BAO [Alternate interior angles as AB || CD] 

∠DOC = ∠BOA [Vertically opposite angles]

∴ ∆DOC ∼ ∆BOA [AAA similarity criterion]

∴\(\frac{DO}{BO} = \frac{OC}{OA}\)           [Corresponding sides are proportional]

⇒ \(\frac{OA}{OC} = \frac{OB}{OD}\) 

17. In the following figure, \(\frac{QR}{QS}=\frac{QT}{PR}\) and \(\angle{1}=\angle{2}\)

Show that ∆PQS ∼ ∆TQR

Answer:

In ∆PQR, 

∠PQR = ∠PRQ 

∴ PQ = PR …………(i)

Given,

\(\frac{QR}{QS}=\frac{QT}{PR}\)

Using (i), we obtain

\(\frac{QR}{QS}=\frac{QT}{QP}\) ..............(ii)

In ∆PQS and ∆TQR

\(\frac{QR}{QS}=\frac{QT}{QP}\)     [Using (ii)]

\(\angle Q = \angle Q\)

∴ ∆PQS ∼ ∆TQR  [SAS Similarity criterion]

18. S and T are point on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ∼ ∆RTS.

Answer:

In ∆RPQ and ∆RST, 

∠RTS = ∠QPS (Given) 

∠R = ∠R (Common angle) 

∴ ∆RPQ ∼ ∆RTS (By AA similarity criterion)

19. In the following figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ∼ ∆ABC.

Answer:

It is given that ∆ABE ≅ ∆ACD. 

∴ AB = AC [By CPCT] ………………(1) 

And, AD = AE [By CPCT] …………….(2) 

In ∆ADE and ∆ABC, 

\(\frac{AD}{AB}= \frac{AE}{AC}\) [Dividing equation (2) by (1)] 

∠A = ∠A [Common angle] 

∴ ∆ADE ∼ ∆ABC [By SAS similarity criterion]

20. In the following 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

Answer:

(i) ∆AEP ∼ ∆CDP 

In ∆AEP and ∆CDP, 

∠AEP = ∠CDP (Each 90°) 

∠APE = ∠CPD (Vertically opposite angles) 

Hence, by using AA similarity criterion, 

∆AEP ∼ ∆CDP

(ii) ∆ABD ∼ ∆CBE 

In ∆ABD and ∆CBE, 

∠ADB = ∠CEB (Each 90°) 

∠ABD = ∠CBE (Common) 

Hence, by using AA similarity criterion, 

∆ABD ∼ ∆CBE

(iii) ∆AEP ∼ ∆ADB 

In ∆AEP and ∆ADB, 

∠AEP = ∠ADB (Each 90°) 

∠PAE = ∠DAB (Common) 

Hence, by using AA similarity criterion, 

∆AEP ∼ ∆ADB

(Iv) ∆PDC ∼ ∆BEC

In ∆PDC and ∆BEC, 

∠PDC = ∠BEC (Each 90°) 

∠PCD = ∠BCE (Common angle) 

Hence, by using AA similarity criterion, 

∆PDC ∼ ∆BEC

21. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ∼ ∆CFB

Answer:

In ∆ABE and ∆CFB, 

∠A = ∠C (Opposite angles of a parallelogram) 

∠AEB = ∠CBF (Alternate interior angles as AE || BC) 

∴ ∆ABE ∼ ∆CFB (By AA similarity criterion)

Answer 4

22. In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:

(i) ∆ABC ∼ ∆AMP 

(ii) \(\frac{CA}{PA} = \frac{BC}{MP}\)

Answer:

In ∆ABC and ∆AMP, 

∠ABC = ∠AMP (Each 90°) 

∠A = ∠A (Common) 

∴ ∆ABC ∼ ∆AMP (By AA similarity criterion)

⇒ \(\frac{CA}{PA} = \frac{BC}{MP}\)  (Corresponding sides of similar triangles are proportional)

23. 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:

(i) \(\frac{CD}{GH} = \frac{AC}{FG}\)

(ii) ∆DCB ∼ ∆HGE 

(iii) ∆DCA ∼ ∆HGF

Answer:

It is given that ∆ABC ∼ ∆FEG. 

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE 

∠ACB = ∠FGE 

∴ ∠ACD = ∠FGH (Angle bisector) 

And, ∠DCB = HGE (Angle bisector)

In ∆ACD and ∆FGH, 

∠A = ∠F (Proved above) 

∠ACD = ∠FGH (Proved above) 

∴ ∆ACD ∼ ∆FGH (By AA similarity criterion)

⇒ \(\frac{CD}{GH} = \frac{AC}{FG}\)

In ∆DCB and ∆HGE, 

∠DCB = ∠HGE (Proved above) 

∠B = ∠E (Proved above) 

∴ ∆DCB ∼ ∆HGE (By AA similarity criterion) 

In ∆DCA and ∆HGF, 

∠ACD = ∠FGH (Proved above) 

∠A = ∠F (Proved above) 

∴ ∆DCA ∼ ∆HGF (By AA similarity criterion)

24. In the following 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

Answer:

It is given that ABC is an isosceles triangle. 

∴ AB = AC 

⇒ ∠ABD = ∠ECF

In ∆ABD and ∆ECF, 

∠ADB = ∠EFC (Each 90°) 

∠BAD = ∠CEF (Proved above) 

∴ ∆ABD ∼ ∆ECF (By using AA similarity criterion)

25. 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 the given figure). Show that ∆ABC ∼ ∆PQR.

Answer:

Median divides the opposite side.

\(\therefore BD =\frac{BC}{2} \,and\,QM = \frac{QR}{2}\)

Given that,

In ∆ABD and ∆PQM,

\(\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}\)  (Proved above) 

∴ ∆ABD ∼ ∆PQM (By SSS similarity criterion) 

⇒ ∠ABD = ∠PQM (Corresponding angles of similar triangles) 

In ∆ABC and ∆PQR, 

∠ABD = ∠PQM (Proved above) 

\(\frac{AB}{PQ}=\frac{BC}{QR}\)

∴ ∆ABC ∼ ∆PQR (By SAS similarity criterion)

26. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD.

Answer:

In ∆ADC and ∆BAC, 

∠ADC = ∠BAC (Given) 

∠ACD = ∠BCA (Common angle) 

∴ ∆ADC ∼ ∆BAC (By AA similarity criterion) 

We know that corresponding sides of similar triangles are in proportion.

∴ \(\frac{CA}{CB}=\frac{CD}{CA}\)

⇒ CA² = CB x CD

Answer 5

27. 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

Answer:

Given that,

\(\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}\)

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. 

Then, join B to E, C to E, Q to L, and R to L.

We know that medians divide opposite sides. 

Therefore, BD = DC and QM = MR 

Also, AD = DE (By construction) 

And, PM = ML (By construction) 

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.

Therefore, quadrilateral ABEC is a parallelogram. 

∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal) 

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, 

PQ = LR 

It was given that

∴ ∆ABE ∼ ∆PQL (By SSS similarity criterion) 

We know that corresponding angles of similar triangles are equal. 

∴ ∠BAE = ∠QPL … (1) 

Similarly, it can be proved that ∆AEC ∼ ∆PLR and 

∠CAE = ∠RPL … (2) 

Adding equation (1) and (2), we obtain 

∠BAE + ∠CAE = ∠QPL + ∠RPL 

⇒ ∠CAB = ∠RPQ … (3)

In ∆ABC and ∆PQR,

\(\frac{AB}{PQ}=\frac{AC}{PR}\)  (Given)

∠CAB = ∠RPQ [Using equation (3)] 

∴ ∆ABC ∼ ∆PQR (By SAS similarity criterion)

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

Answer:

Let AB and CD be a tower and a pole respectively. 

Let the shadow of BE and DF be the shadow of AB and CD respectively. 

At the same time, the light rays from the sun will fall on the tower and the pole at the same angle. 

Therefore, ∠DCF = ∠BAE 

And, ∠DFC = ∠BEA 

∠CDF = ∠ABE (Tower and pole are vertical to the ground) 

∴ ∆ABE ∼ ∆CDF (AAA similarity criterion)

Therefore, the height of the tower will be 42 metres.

29. If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ∼ ∆PQR prove that \(\frac{AB}{PQ}=\frac{AD}{PM}\)

Answer:

It is given that ∆ABC ∼ ∆PQR 

We know that the corresponding sides of similar triangles are in proportion.

∴ \(\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}\) … (1)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

Since AD and PM are medians, they will divide their opposite sides.

∴ \(BD = \frac{BC}{2}\,and\,QM=\frac{QR}{2}\) … (3)

From equations (1) and (3), we obtain

\(\frac{AB}{PQ}=\frac{BD}{QM}\) … (4)

In ∆ABD and ∆PQM, 

∠B = ∠Q [Using equation (2)]

\(\frac{AB}{PQ}=\frac{BD}{QM}\) [Using equation (4)] 

∴ ∆ABD ∼ ∆PQM (By SAS similarity criterion)

\(\frac{AB}{PQ}=\frac{BD}{QM} =\frac{AD}{PM}\)

30. Let ∆ABC ∼ ∆DEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.

Answer:

It is given that  ∆ABC ∼ ∆DEF.

GIven that,

EF = 15.4 cm,

ar (∆ABC) = 64 cm²

ar (∆DEF) = 121 cm²

31. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Answer:

Since AB || CD, 

∴ ∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles) 

In ∆AOB and ∆COD, 

∠AOB = ∠COD (Vertically opposite angles) 

∠OAB = ∠OCD (Alternate interior angles) 

∠OBA = ∠ODC (Alternate interior angles) 

∴ ∆AOB ∼ ∆COD (By AAA similarity criterion)

32. In the following figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that 

\(\frac{area(\triangle ABC)}{area(\triangle DBC)}= \frac{AO}{DO}\)

Answer: 

Let us draw two perpendiculars AP and DM on line BC. 

We know that area of a triangle = \(\frac{1}{2}\) x Base x Height

In ∆APO and ∆DMO, 

∠APO = ∠DMO (Each = 90°) 

∠AOP = ∠DOM (Vertically opposite angles) 

∴ ∆APO ∼ ∆DMO (By AA similarity criterion)

Answer 6

33. If the areas of two similar triangles are equal, prove that they are congruent.

Answer: 

Let us assume two similar triangles as ∆ABC ∼ ∆PQR.

Given that, ar(∆ABC) = ar (∆PQR)

Putting this value in equation (!), we obatin

34. D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC. Find the ratio of the area of ∆DEF and ∆ABC.

Answer: 

D and E are the mid-points of ∆ABC.

∴ \(DE \parallel AC\, and \,DE = \frac{1}{2}AC\)

In ∆BED and ∆BCA,

35. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Answer: 

Let us assume two similar triangles as ∆ABC ∼ ∆PQR. 

Let AD and PS be the medians of these triangles. 

∵ ∆ABC ∼ ∆PQR

\(\therefore \frac{AB}{PQ }=\frac{BC}{QR}= \frac{AC}{PR} ....(1)\)

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ……… (2) 

Since AD and PS are medians,

∴ BD = DC = \(\frac{BC}{2}\)

And, QS = SR = \(\frac{QR}{2}\)

Equation (1) becomes

\(\frac{AB}{PQ }=\frac{BD}{QS}= \frac{AC}{PR} ....(3)\)

In ∆ABD and ∆PQS, 

∠B = ∠Q [Using equation (2)] 

And, \(\frac{AB}{PQ}= \frac{BD}{QS}\) [Using equation (3)] 

∴ ∆ABD ∼ ∆PQS (SAS similarity criterion) 

Therefore, it can be said that

\(\frac{AB}{PQ }=\frac{BD}{QS}= \frac{AD}{PS} ....(4)\)

From equations (1) and (4), we may find that

\(\frac{AB}{PQ }=\frac{BC}{QR}= \frac{AC}{PR} =\frac{AD}{PS}\)

And hence,

\(\frac{ar(\triangle ABC)}{ar(\triangle PQR)}= (\frac{AD}{PS})^2\)

36. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Answer:

Let ABCD be a square of side a. 

Therefore, its diagonal = \(\sqrt{2a}\)

Two desired equilateral triangles are formed as ∆ABE and ∆DBF. 

Side of an equilateral triangle, ∆ABE, described on one of its sides = a

Side of an equilateral triangle, ∆DBF, described on one of its diagonals = \(\sqrt{2a}\)

We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

37. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is 

(A) 2 : 1 

(B)1 : 2 

(C) 4 : 1 

(D) 1 : 4

Answer:

We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other.

Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

Let side of ∆ABC = x 

Therefore, side of \(\triangle BDE= \frac{x}{2}\)

Hence, the correct answer is (C).

38. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio 

(A) 2 : 3 

(B) 4 : 9 

(C) 81 : 16 

(D) 16 : 81

Answer:

If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles. 

It is given that the sides are in the ratio 4:9. 

Therefore, ratio between areas of these triangles = \((\frac{4}{9})^2 = \frac{16}{81}\)

Hence, the correct answer is (D).

39. Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

(iii) 50 cm, 80 cm, 100 cm

(iv) 13 cm, 12 cm, 5 cm

Answer:

(i) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of the sides of the, we will get 49, 576, and 625.

49 + 576 = 625

(7)² + (24)² = (25)²

Therefore, the above equation satisfies, Pythagoras theorem. Hence, it is right angled triangle.

Length of Hypotenuse = 25 cm

(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.

Squaring the lengths of these sides, we will get 9, 64, and 36.

Clearly, 9 + 36 ≠ 64

Or, 3² + 6² ≠ 8²

Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.

Hence, the given triangle does not satisfies Pythagoras theorem.

(iii) Given, sides of triangle’s are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will get 2500, 6400, and 10000.

However, 2500 + 6400 ≠ 10000

Or, 50² + 80² ≠ 100²

As you can see, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle does not satisfies Pythagoras theorem.

Hence, it is not a right triangle.

(iv) Given, sides are 13 cm, 12 cm, and 5 cm.

Squaring the lengths of these sides, we will get 169, 144, and 25.

Thus, 144 +25 = 169

Or, 12² + 5² = 13²

The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

Hence, length of the hypotenuse of this triangle is 13 cm.

Answer 7

40. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM × MR.

Answer:

Given, ΔPQR is right angled at P is a point on QR such that PM ⊥QR

We have to prove, PM² = QM × MR

In ΔPQM, by Pythagoras theorem

PQ² = PM² + QM²

Or, PM² = PQ² – QM² …………..(i)

In ΔPMR, by Pythagoras theorem

PR² = PM² + MR²

Or, PM² = PR² – MR² ……………..(ii)

Adding equation, (i) and (ii), we get,

2PM² = (PQ² + PM²) – (QM² + MR²)

= QR² – QM² – MR²        [∴ QR² = PQ² + PR²]

= (QM + MR)² – QM² – MR²

= 2QM × MR

∴ PM² = QM × MR

41. In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB² = BC × BD
(ii) AC² = BC × DC
(iii) AD² = BD × CD

Answer:

(i) In ΔADB and ΔCAB,

∠DAB = ∠ACB (Each 90°)

∠ABD = ∠CBA (Common angles)

∴ ΔADB ~ ΔCAB [AA similarity criterion]

⇒ AB/CB = BD/AB

⇒ AB² = CB × BD

(ii) Let ∠CAB = x

In ΔCBA,

∠CBA = 180° – 90° – x

∠CBA = 90° – x

Similarly, in ΔCAD

∠CAD = 90° – ∠CBA

= 90° – x

∠CDA = 180° – 90° – (90° – x)

∠CDA = x

In ΔCBA and ΔCAD, we have

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA (Each 90°)

∴ ΔCBA ~ ΔCAD [AAA similarity criterion]

⇒ AC/DC = BC/AC

⇒ AC² =  DC × BC

(iii) In ΔDCA and ΔDAB,

∠DCA = ∠DAB (Each 90°)

∠CDA = ∠ADB (common angles)

∴ ΔDCA ~ ΔDAB [AA similarity criterion]

⇒ DC/DA = DA/DA

⇒ AD² = BD × CD

42. ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC² .

Answer:

Given, ΔABC is an isosceles triangle right angled at C.

In ΔACB, ∠C = 90°

AC = BC (By isosceles triangle property)

AB² = AC² + BC² [By Pythagoras theorem]

= AC² + AC² [Since, AC = BC]

AB² = 2AC²

43. ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.

Answer:

Given, ΔABC is an isosceles triangle having AC = BC and AB² = 2AC²

In ΔACB,

AC = BC

AB² = 2AC²

AB² = AC² + AC²

= AC² + BC² [Since, AC = BC]

Hence, by Pythagoras theorem ΔABC is right angle triangle.

44. ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Answer:

Draw, AD ⊥ BC

In ΔADB and ΔADC,

AB = AC

AD = AD

∠ADB = ∠ADC [Both are 90°]

Therefore, ΔADB ≅ ΔADC by RHS congruence.

Hence, BD = DC [by CPCT]

In right angled ΔADB,

AB² = AD² + BD²

(2a)² = AD² + a² 

⇒ AD^{2 =} 4a² – a²

⇒ AD^{2 =} 3a²

⇒ AD ⁼ √3a

45. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.

Answer:

Given, ABCD is a rhombus whose diagonals AC and BD intersect at O.

We have to prove, as per the question,

AB² + BC² + CD² + AD² = AC² + BD²

Since, the diagonals of a rhombus bisect each other at right angles.

Therefore, AO = CO and BO = DO

In ΔAOB,

∠AOB = 90°

AB² = AO² + BO² ………….. (i) [By Pythagoras theorem]

Similarly,

AD² = AO² + DO² ………….. (ii)

DC² = DO² + CO² ………….. (iii)

BC² = CO² + BO² ………….. (iv)

Adding equations (i) + (ii) + (iii) + (iv), we get,

AB² + AD² + DC² + BC² = 2(AO² + BO² + DO² + CO²)

= 4AO² + 4BO² [Since, AO = CO and BO =DO]

= (2AO)² + (2BO)² = AC² + BD²

AB² + AD² + DC² + BC² = AC² + BD²

Hence, proved.

Answer 8

46. In Figure, O is a point in the interior of a triangle.

ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that:
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE² ,
(ii) AF² + BD² + CE² = AE² + CD² + BF².

Answer:

Given, in ΔABC, O is a point in the interior of a triangle.

And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.

Join OA, OB and OC

(i) By Pythagoras theorem in ΔAOF, we have

OA² = OF² + AF²

Similarly, in ΔBOD

OB² = OD² + BD²

Similarly, in ΔCOE

OC² = OE² + EC²

Adding these equations,

OA² + OB² + OC² = OF² + AF² + OD² + BD² + OE² + EC²

OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE².

(ii) AF² + BD² + EC² = (OA² – OE²) + (OC² – OD²) + (OB² – OF²)

∴ AF² + BD² + CE² = AE² + CD² + BF².

47. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Answer:

Given, a ladder 10 m long reaches a window 8 m above the ground.

Let BA be the wall and AC be the ladder,

Therefore, by Pythagoras theorem,

AC² = AB² + BC²

10² = 8² + BC²

BC² = 100 – 64

BC² = 36

BC = 6m

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

48. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Answer:

Given, a guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.

Let AB be the pole and AC be the wire.

By Pythagoras theorem,

AC² = AB² + BC²

24² = 18² + BC²

BC² = 576 – 324

BC² = 252

BC = 6√7m

Therefore, the distance from the base is 6√7m.

49. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after \(1\frac{1}{2}\) hours?

Answer:

Given,

Speed of first aeroplane = 1000 km/hr

Distance covered by first aeroplane flying due north in \(1\frac{1}{2}\) hours (OA) = 1000 × 3/2 km = 1500 km

Speed of second aeroplane = 1200 km/hr

Distance covered by second aeroplane flying due west in \(1\frac{1}{2}\) hours (OB) = 1200 × 3/2 km = 1800 km

In right angle ΔAOB, by Pythagoras Theorem,

AB² = AO² + OB²

⇒ AB² = (1500)² + (1800)²

⇒ AB = √(2250000 + 3240000)

= √5490000

⇒ AB = 300√61 km

Hence, the distance between two aeroplanes will be 300√61 km.

50. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Answer:

Given, Two poles of heights 6 m and 11 m stand on a plane ground.

And distance between the feet of the poles is 12 m.

Let AB and CD be the poles of height 6m and 11m.

Therefore, CP = 11 – 6 = 5m

From the figure, it can be observed that AP = 12m

By Pythagoras theorem for ΔAPC, we get,

AP² = PC² + AC²

(12m)² + (5m)² = (AC)²

AC² = (144+25) m² = 169 m²

AC = 13m

Therefore, the distance between their tops is 13 m.

51. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².

Answer:

Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.

By Pythagoras theorem in ΔACE, we get

AC² + CE² = AE² …………….(i)

In ΔBCD, by Pythagoras theorem, we get

BC² + CD² = BD² ……………..(ii)

From equations (i) and (ii), we get,

AC² + CE² + BC² + CD² = AE² + BD² …………..(iii)

In ΔCDE, by Pythagoras theorem, we get

DE² = CD² + CE²

In ΔABC, by Pythagoras theorem, we get

AB² = AC² + CB²

Putting the above two values in equation (iii), we get

DE² + AB² = AE² + BD².

Answer 9

52. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB² = 2AC² + BC².

Answer:

Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;

DB = 3CD.

In Δ ABC,

AD ⊥BC and BD = 3CD

In right angle triangle, ADB and ADC, by Pythagoras theorem,

AB² = AD² + BD² …………….(i)

AC² = AD² + DC² ……………..(ii)

Subtracting equation (ii) from equation (i), we get

AB² – AC² = BD² – DC²

= 9CD² – CD² [Since, BD = 3CD]

= 8CD²

= 8(BC/4)² [Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB² – AC² = BC²/2

⇒ 2(AB² – AC²) = BC²

⇒ 2AB² – 2AC² = BC²

∴ 2AB² = 2AC² + BC².

53. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD² = 7AB².

Answer:

Given, ABC is an equilateral triangle.

And D is a point on side BC such that BD = 1/3BC

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

And, AE = a√3/2

Given, BD = 1/3BC

∴ BD = a/3

DE = BE – BD = a/2 – a/3 = a/6

In ΔADE, by Pythagoras theorem,

AD² = AE² + DE² 

Let the sides of the equilateral triangle be of length a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

In ΔABE, by Pythagoras Theorem, we get

AB² = AE² + BE²

4AE² = 3a²

⇒ 4 × (Square of altitude) = 3 × (Square of one side)

Hence, proved.

⇒ 9 AD² = 7 AB²

54. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Answer:

Given, an equilateral triangle say ABC,

Let the sides of the equilateral triangle be of length a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

In ΔABE, by Pythagoras Theorem, we get

AB² = AE² + BE²

4AE² = 3a²

⇒ 4 × (Square of altitude) = 3 × (Square of one side)

Hence, proved.

55. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:

(A) 120°

(B) 60°

(C) 90° 

(D) 45°

Answer:

Given, in ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.

We can observe that,

AB² = 108

AC² = 144

And, BC² = 36

AB² + BC² = AC²

The given triangle, ΔABC, is satisfying Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B.

∴ ∠B = 90°

Hence, the correct answer is (C).

56. In Figure, PS is the bisector of ∠ QPR of ∆ PQR. Prove that QS/PQ = SR/PR

Answer:

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

Given, PS is the angle bisector of ∠QPR. Therefore,

∠QPS = ∠SPR…………..(i)

As per the constructed figure,

∠SPR=∠PRT(Since, PS||TR)……………(ii)

∠QPS = ∠QRT(Since, PS||TR) …………..(iii)

From the above equations, we get,

∠PRT=∠QTR

Therefore,

PT=PR

In △QTR, by basic proportionality theorem,

QS/SR = QP/PT

Since, PT=TR

Therefore,

QS/SR = PQ/PR

Hence, proved.

Answer 10

57. In Fig, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥AC, DM ⊥ BC and DN ⊥ AB. Prove that: 

(i) DM² = DN . MC 

(ii) DN² = DM . AN.

Answer:

(i) Let us join Point D and B.

Given,

BD ⊥AC, DM ⊥ BC and DN ⊥ AB

Now from the figure we have,

DN || CB, DM || AB and ∠B = 90 °

Therefore, DMBN is a rectangle.

So, DN = MB and DM = NB

The given condition which we have to prove, is when D is the foot of the perpendicular drawn from B to AC.

∴ ∠CDB = 90° 

⇒ ∠2 + ∠3 = 90° ……………. (i)

In ∆CDM, 

∠1 + ∠2 + ∠DMC = 180°

⇒ ∠1 + ∠2 = 90° ……………….. (ii)

In ∆DMB, 

∠3 + ∠DMB + ∠4 = 180°

⇒ ∠3 + ∠4 = 90° …………….. (iii)

From equation (i) and (ii), we get

∠1 = ∠3

From equation (i) and (iii), we get

∠2 = ∠4

In ∆DCM and ∆BDM,

∠1 = ∠3 (Already Proved)

∠2 = ∠4 (Already Proved)

∴ ∆DCM ∼ ∆BDM (AA similarity criterion)

BM/DM = DM/MC

DN/DM = DM/MC (BM = DN)

⇒ DM² = DN × MC

Hence, proved.

(ii) In right triangle DBN,

∠5 + ∠7 = 90° ………….. (iv)

In right triangle DAN,

∠6 + ∠8 = 90° …………… (v)

D is the point in triangle, which is foot of the perpendicular drawn from B to AC.

∴ ∠ADB = 90° ⇒ ∠5 + ∠6 = 90° ………….. (vi)

From equation (iv) and (vi), we get,

∠6 = ∠7

From equation (v) and (vi), we get,

∠8 = ∠5

In ∆DNA and ∆BND,

∠6 = ∠7 (Already proved)

∠8 = ∠5 (Already proved)

∴ ∆DNA ∼ ∆BND (AA similarity criterion)

AN/DN = DN/NB

⇒ DN² = AN × NB

⇒ DN² = AN × DM (Since, NB = DM)

Hence, proved.

58. In Figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that

AC² = AB² + BC² + 2 BC.BD.

Answer:

By applying Pythagoras Theorem in ∆ADB, we get,

AB² = AD² + DB² ……………… (i)

Again, by applying Pythagoras Theorem in ∆ACD, we get,

AC² = AD² + DC²

AC² = AD² + (DB + BC) ²

AC² = AD² + DB² + BC² + 2DB × BC

From equation (i), we can write,

AC² = AB² + BC² + 2DB × BC

Hence, proved.

59. In Figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that

AC²= AB²+ BC² – 2 BC.BD.

Answer:

By applying Pythagoras Theorem in ∆ADB, we get,

AB² = AD² + DB²

We can write it as;

⇒ AD² = AB² − DB² ……………….. (i)

By applying Pythagoras Theorem in ∆ADC, we get,

AD² + DC² = AC²

From equation (i),

AB² − BD² + DC² = AC²

AB² − BD² + (BC − BD) ² = AC²

AC² = AB² − BD² + BC² + BD² −2BC × BD

AC² = AB² + BC² − 2BC × BD

Hence, proved.

60. In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :

(i) AC² = AD² + BC.DM + 2 (BC/2) ²

(ii) AB² = AD² – BC.DM + 2 (BC/2) ²

(iii) AC² + AB² = 2 AD² + 1/2 BC²

Answer:

(i) By applying Pythagoras Theorem in ∆AMD, we get,

AM² + MD² = AD² ……………. (i)

Again, by applying Pythagoras Theorem in ∆AMC, we get,

AM² + MC² = AC²

AM² + (MD + DC) ² = AC²

(AM² + MD² ) + DC² + 2MD.DC = AC²

From equation(i), we get,

AD² + DC² + 2MD.DC = AC²

Since, DC=BC/2, thus, we get,

AD² + (BC/2) ² + 2MD.(BC/2) ² = AC²

AD² + (BC/2) ² + 2MD × BC = AC²

Hence, proved.

(ii) By applying Pythagoras Theorem in ∆ABM, we get;

AB² = AM² + MB²

= (AD² − DM²) + MB²

= (AD² − DM²) + (BD − MD) ²

= AD² − DM² + BD² + MD² − 2BD × MD

= AD² + BD² − 2BD × MD

= AD² + (BC/2)² – 2(BC/2) MD

= AD² + (BC/2)² – BC MD

Hence, proved.

(iii) By applying Pythagoras Theorem in ∆ABM, we get,

AM² + MB² = AB² …………… (i)

By applying Pythagoras Theorem in ∆AMC, we get,

AM² + MC² = AC² ……………… (ii)

Adding both the equations (i) and (ii), we get,

2AM² + MB² + MC² = AB² + AC²

2AM² + (BD − DM) ² + (MD + DC) ² = AB² + AC²

2AM²+BD² + DM² − 2BD.DM + MD² + DC² + 2MD.DC = AB² + AC²

2AM² + 2MD² + BD² + DC² + 2MD (− BD + DC) = AB² + AC²

2(AM²+ MD²) + (BC/2) ² + (BC/2) ² + 2MD (-BC/2 + BC/2) ² = AB² + AC²

2AD² + BC²/2 = AB² + AC²