NCERT Class 10 Chapter 6: Triangles

Exercise 6.1

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

All circles are _ (congruent, similar)
All squares are _ (similar, congruent)
All triangles are similar _ (isosceles, equilateral)
Two polygons of the same number of sides are similar, if (a) their corresponding angles are _ and (b) their corresponding sides are _ (equal, proportional)

Solution

All circles are similar.
All squares are similar.
All equilateral triangles are similar.
Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.

2. Give two different examples of pair of

similar figures.
non-similar figures.

Solution

A pair of two squares with different length of its sides are similar.
A pair of two equilateral triangles with different lengths of sides are similar.
A pair of two isosceles triangles with different angles are non-similar.
A pair of square and a rhombus whose angles are not 90° are non-similar.

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

Solution

In the quadrilateral ABCD, all the angles are 90°.

However, in the quadrilateral PQRS, the angles are not equal to 90°.

Hence, the given quadrilaterals are not similar.

Exercise 6.2

1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution

(i) It is given that in ∆ABC, DE ‖ BC. Therefore, by using the basic proportionality theorem, we have:

(ii) It is given that in ∆ABC, DE ‖ BC. Therefore, by using the basic proportionality theorem, we have:

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 :

PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution

I. EF ‖ QR, if the basic proportionality theorem is satisfied, i.e.

Since, LHS ≠ RHS, the theorem isn't satisfied.

Hence, EF is not parallel to QR.

II. EF ‖ QR, if the basic proportionality theorem is satisfied, i.e.

Since, LHS = RHS, the theorem is satisfied.

Hence, EF is parallel to QR.

III.EF ‖ QR, if the basic proportionality theorem is satisfied, i.e.

Since, LHS = RHS, the theorem is satisfied.

Hence, EF is parallel to QR.

3. In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.

Solution

It is given that LM is parallel to CB in ∆ABC, therefore:

4. In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.

Solution

5. In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

Solution

6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution

7. Using Theorem 6.1, 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 you have proved it in Class IX).

For reference:-
(Theorem 6.1 : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.)

Solution

Let the triangle be ABC with DE as the line drawn through D as midpoint of side AB and E as a point on line AC.

To Prove : AE = EC

Proof :

8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

For reference:-
(Theorem 6.2 : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.)

Solution

Let the triangle be ABC, where DE is the line formed by joining the D which is the midpoint of AB and E which is the midpoint of AC.

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.

Solution

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.

Solution

Therefore, by using the converse basic proportionality theorem we can conclude that EO ‖ AB.

It is also known that EO ‖ CD, which implies that AB ‖ CD.

Since two of the sides of the given quadrilateral are parallel, so it satisfies the condition for being a trapezium.

Hence, ABCD is a trapezium.

Exercise 6.3

1. State which pairs of triangles in Fig. 6.34 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 :

Solution

(i) In ∆ABC and ∆PQR:

Hence, by AAA criterion ∆ABC ~ ∆PQR.

(ii) In ∆ABC and ∆PQR:

Hence, by SSS criterion ∆ABC ~ ∆PQR.

(iii) In ∆MPL and ∆DEF:

The ratios obtained are not same.

Hence, ∆MPL is not similar to ∆DEF.

(iv) In ∆LMN and ∆PQR:

Hence, by SAS criterion ∆LMN ~ ∆PQR.

(v) In ∆ABC and ∆DEF:

Hence, ∆ABC is not similar to ∆DEF.

(vi) By Angle Sum property of a triangle,

Hence, by AAA criterion ∆DEF ~ ∆PQR.

2. In Fig. 6.35, ∆ ODC ~ ∆ OBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

Solution

∠DOC and ∠BOC form a linear pair. Therefore

Hence ∠DOC = 55°, ∠DCO = 55° and ∠OAB = 55°.

3. 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.

Solution

In ∆AOB and ∆COD,

∠BAO = ∠OCD (Alternate interior angles)

∠ABO = ∠ODC (Alternate interior angles)

∠AOB = ∠COD (Vertically opposite angles)

Hence, by the AAA similarity criterion, ∆AOB ~ ∆COD.

Since, ∆AOB ~ ∆COD

Therefore, the corresponding sides will be proportional

Hence, proved that OB/OD = OA/OC.

4. In Fig. 6.36, QR/QS = QT/PR and ∠ 1 = ∠ 2. Show that ∆ PQS ~ ∆ TQR.

Solution

In ∆PQR,

∠PQR = ∠PRQ

Therefore, it is an isosceles triangle which implies that PQ = PR.

Using QR/QS = QT/PQ and ∠Q = ∠Q in ∆PQS and ∆TQR, we can conclude that the triangles are similar (by SAS similarity criterion).

Hence, proved that ∆PQS ~ ∆TQR.

5. S and T are points on sides PR and QR of ∆ PQR such that ∠ P = ∠ RTS. Show that ∆ RPQ ~ ∆ RTS.

Solution

In ∆ RPQ and ∆ RTS,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common Angle)

Hence, by the AA similarity criterion ∆ RPQ ~ ∆ RTS.

6. In Fig. 6.37, if ∆ ABE ≅ ∆ ACD, show that ∆ ADE ~ ∆ ABC.

Solution

In ∆ ABE and ∆ ACD

AB = AC (Using c.p.c.t) ... Equation (I)

AD = AE (Using c.p.c.t) ... Equation (II)

7. In Fig. 6.38, altitudes AD and CE of ∆ ABC intersect each other at the point P. Show that:

∆AEP ~ ∆ CDP
∆ABD ~ ∆ CBE
∆AEP ~ ∆ADB
∆ PDC ~ ∆ BEC

Solution

I. In ΔAEP and ΔCDP,

∠AEP = ∠CDP (Right Angles)

∠APE = ∠CPD (Vertically Opposite Angles)

Therefore, by AA similarity criterion ΔAEP ~ ΔCDP.

II. In ΔABD and ΔCBE,

∠ADB = ∠CEB (Right Angles)

∠ABD = ∠CBE& (Common Angle)

Therefore, by AA similarity criterion ΔABD ~ ΔCBE.

III. In ΔAEP and ΔABD,

∠AEP = ∠ADB (Right Angles)

∠PAE = ∠DAB (Common Angle)

Therefore, by AA similarity criterion ΔAEP ~ ΔABD.

IV. In ΔPDC and ΔBEC,

∠PDC = ∠BEC (Right Angles)

∠PCD = ∠BCE (Common Angle)

Therefore, by AA similarity criterion ΔPDC ~ ΔBEC.

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

Solution

In ∆ABE and ∆CFB,

∠BAE = ∠BCF (Opposite angles of a parallelogram)

∠ABE = ∠CFB (Alternate interior angles)

Therefore, by AA similarity criterion ΔABE ~ ΔCFB.

9. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

∆ ABC ~ ∆AMP
CA/PA = BC/MP

Solution

I. In ΔABC and ΔAMP,

∠ABC = ∠AMP (Right Angles)

∠BAC = ∠MAP (Common Angle)

Therefore, by AA similarity criterion ΔABC ~ ΔAMP.

II. Since, ∆ABC is similar to ∆AMP, their corresponding sides will have an equal ratio.

Hence,

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:

CD/GH = AC/FG
∆ DCB ~ ∆ HGE
∆ DCA ~ ∆ HGF

Solution

It is given that ∆ABC ~ ∆ FEG, therefore

∠A = ∠F

∠B = ∠E

∠ACB = ∠FGE

Divide both sides by 2

∠ACB/2 = ∠FGE/2

∠ACD = ∠FGH (Bisected Angles)

I. Now, in ∆ACD and ∆FGH:

∠A = ∠F (As shown above)

∠ACD = ∠FGH (As shown above)

Therefore, by the AA similarity criterion ∆ACD ~ ∆FGH.

Which implies that

II. Similarly,

∠DCB = ∠HGE

In ∆DCB and ∆HGE:

∠B = ∠E (As Shown above)

∠DCB = ∠HGE (As shown above)

Therefore, by the AA similarity criterion ∆DCB ~ ∆HGE.

III. In ∆DCA and ∆HGF

∠A = ∠F (As shown above)

∠DCA = ∠HGF (As shown above)

Therefore, by the AA similarity criterion ∆DCA ~ ∆HGF.

11. In Fig. 6.40, 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.

Solution

In an isosceles triangles, angles opposite to equal sides are also equal. Therefore,

∠ABD = ∠ACD = ∠ECF

In ∆ ABD and ∆ ECF:

∠ADB = ∠EFC (Right Angles)

∠ABD = ∠ECF (As Shown above)

Therefore, by the AA similarity criterion ∆ ABD ~ ∆ ECF.

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 Fig. 6.41). Show that ∆ ABC ~ ∆ PQR.

Solution

It is given that

Therefore, by the SSS similarity criterion, ∆ABD ~ ∆PQM.

This implies that

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

∠ABC = ∠PQR

In ∆ABC and ∆PQR

AB/PQ = BC/QR (Given)

∠ABC = ∠PQR (As shown above)

Hence, by the SAS similarity criterion ∆ABC ~ ∆PQR.

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

Solution

In ∆ADC and ∆BAC:

∠ADC = ∠BAC (Given)

∠ACD = ∠BCA (Common Angles)

Therefore, by the AA similarity criterion ∆ADC ~ ∆BAC.

This implies that

CA/CB = CD/CA (Corresponding sides of similar triangles)

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.

Solution

First, we will produce AD and PM lines to exterior points E and N respectively such that AD = DE and PM = MN, then join E and N with C and R respectively.

In ∆ABD and ∆CDE:

. ∠ADB = ∠ CDE (Vertically opposite angles)

AD = DE (By construction)

BD = DC (D is the mid-point of BC)

Therefore, by the SAS congruency criterion ∆ABD ≅ ∆CDE.

This implies that

AB = CE (By c.p.c.t.)

In ∆PQM and ∆MNR:

. ∠PMQ = ∠NMR (Vertically opposite angles)

PM = MN (By construction)

QR = MR (M is the mid-point of QR)

Therefore, by the SAS congruency criterion ∆PMQ ≅ ∆NMR.

This implies that

PQ = RN (By c.p.c.t.)

It is given that

Therefore, by the SSS similarity criterion ∆ACE ~ ∆PRN.

This implies that

∠CAE = ∠RPN (Corresponding angles of similar triangles)

∠CAD = ∠RPM

Similarly, ∠BAD = ∠QPM.

Add both the above equations

∠CAD + ∠BAD = ∠RPM + ∠QPM

∠BAC = ∠QPR

In ∆ABC and ∆PQR:

∠BAC = ∠QPR (As shown above)

AB/PQ = AC/PR (Given)

Hence, by the SAS similarity criterion ∆ABC ~ ∆PQR.

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.

Solution

Let AB be the pole whose shadow is BC and let the tower be PQ with QR as the shadow.

If we connect the top of pole and tower to the end of their respective shadows, we can obtain triangles ABC and PQR.

Since, the shadows BC and QR are being cast at the same time, the sun's elevation will be the same.

Therefore,

∠ACB = ∠PRQ

In ∆ABC and ∆PQR:

∠ACB = ∠PRQ (As shown above)

∠ABC = ∠PQR (Right Angles)

Therefore, by the AA similarity criterion ∆ABC ~ ∆PQR.

This implies that

AB/PQ = BC/QR (Corresponding sides of similar triangles)

AB = 6 m, BC = 4 m

QR = 28 m

6/PQ = 4/28

6/PQ = 1/7

PQ = 42 m

Hence, the height of the tower is 42 m.

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

Solution

It is given that ∆ ABC ~ ∆ PQR, which implies that

∠ABC = ∠PQR (Corresponding Angles of similar triangles)

∠ABD = ∠PQM

In ∆ABD and ∆PQM:

AB/PQ = BD/QM (As shown above)

∠ABD = ∠PQM (As shown above)

Therefore, by the SAS similarity criterion ∆ABD ~ ∆PQM.

This implies that

AB/PQ = AD/PM (Corresponding sides of similar triangles)

Hence, proved that AB/PQ = AD/PM.

Exercise 6.4

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

Solution

Since ∆ ABC ~ ∆ DEF, therefore

Hence, BC = 11.2 cm².

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

SOLUTION

In ∆AOB and ∆COD:

∠AOB = ∠COD (Vertically opposite angles)

∠ABO = ∠CDO (Alternate interior angles)

Therefore, by the AA similarity criterion ∆AOB ~ ∆COD.

This implies that

Hence, the ratio of the areas of triangles AOB and COD is 4 : 1.

3. In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that ar (ABC)/ar (DBC) = AO/DO.

Solution

We need to construct two lines AP and DM where P and M are points on BC such that AP and DM are perpendicular to BC.

Area of triangle = ½ × Base × Altitude

Area of ∆ABC = ½ × BC × AP

Area of ∆DBC = ½ × BC × DM

Ratio of Areas of ∆ABC and ∆DBC = (½ × BC × AP)/ (½ × BC × DM)

ar (∆ABC)/ ar (∆DBC) = AP/DM ... Equation (I)

In ∆AOP and ∆DOM:

∠AOP = ∠DOM (Vertically opposite angles)

∠APO = ∠DMO& (Right Angles)

Therefore, by the AA similarity criterion ∆AOP ~ ∆DOM.

This implies that

AO/DO = PA/MD (Corresponding sides of similar triangles)

AO/DO = AP/DM ... Equation (II)

From Equation (I) and Equation (II),

ar (∆ABC)/ ar (∆DBC) = AO/DO

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

Solution

Let there be two similar triangles ABC and PQR where Area of ABC = Area of PQR.

AB/PQ = 1

AB = PQ

Similarly,

BC = QR

AC = PR

Therefore, by the SSS congruency criteria ∆ABC ≅ ∆PQR.

Hence, proved that two similar triangles with equal areas are congruent.

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

Solution

Since D and F are mid-points of sides AB and AC, therefore, by the mid-point theorem

DF ‖ BC.

In ∆ABC and ∆ADF:

∠BAC = ∠DAF (Common Angles)

∠ABC = ∠ADF (Corresponding Angles)

Therefore, by the AA similarity criterion ∆ABC ~ ∆ADF.

This implies that

From the above equations, we can conclude that

Ar (∆ADF) = �(Ar (∆ABC))

Ar (∆BED) = �(Ar (∆ABC))

Ar (∆CEF) = �(Ar (∆ABC))

Now,

Hence, the ratio of the areas of triangles DEF and ABC is 1 : 4.

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

Solution

Let there be two similar triangles ABC and PQR, where AD and PM are the respective medians of the triangles.

To Prove : ar (∆ABC)/ ar (∆PQR) = AD²/ PM²

Proof :

It is given that ∆ABC ~ ∆PQR, which implies that

∠ABC = ∠PQR (Corresponding angles of similar triangles)

∠ABD = ∠PQM

Now, in ∆ABD and ∆PQM:

∠ABD = ∠PQM (As shown above)

AB/PQ = BD/QM (As shown above)

Therefore, by the SAS similarity criterion ∆ABD ~ ∆PQM.

This implies that

Hence, proved that ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

7. 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.

Solution

Let there be a square ABCD with equilateral triangles ABD and ACE described on its one side and one diagonal.

To Prove : ar (∆ABD) = ½ (ar (∆ACE))

Proof :

By using the Pythagoras theorem in ABC, we get:

AC² = AB² + BC²

AB = BC as ABCD is a square

AC² = 2AB² ... Equation (I)

It is known that ABD and ACE are both equilateral triangles.

Therefore, by the AAA similarity criterion ∆ABD ~ ∆ACE.

This implies that

Hence, proved 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.

Tick the correct answer and justify:

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

(A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4

Solution

D is the mid-point of BC, therefore

BD = DC = ½(BC) ... Equation (I)

AB = BC (Sides of an equilateral triangle) ... Equation (II)

Since, ∆ABC and ∆BDE are both equilateral triangles, therefore

by the AAA similarity criterion ∆ABC ~ ∆BDE.

This implies that

Hence, the correct answer is (C) 4 : 1.

9. 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

Solution

Let there be two similar triangles ABC and DEF.

Then, according to question

Since, it is given that the triangles ABC and DEF are similar. Therefore,

Hence, the correct answer is (D) 16 : 81.

Exercise 6.5

1. 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.

7 cm, 24 cm, 25 cm
3 cm, 8 cm, 6 cm
50 cm, 80 cm, 100 cm
13 cm, 12 cm, 5 cm

Solution

I. Squares of the smaller two sides = 7² + 24² = 49 + 576 = 625 cm²

Square of the longest side = 25² = 625 cm²

Therefore, the triangle satisfies the Pythagoras theorem and is a right angled triangle.

The length of hypotenuse of this triangle is 25 cm.

II. Squares of the smaller two sides = 3² + 6² = 9 + 36 = 45 cm²

Square of the longest side = 8² = 64 cm²

Therefore, the triangle does not satisfy the Pythagoras theorem and is not a right angled triangle.

III. Squares of the smaller two sides = 50² + 80² = 2500 + 6400 = 8900 cm²

Square of the longest side = 100² = 10000 cm²

Therefore, the triangle does not satisfy the Pythagoras theorem and is not a right angled triangle.

IV. Squares of the smaller two sides = 5² + 12² = 25 + 144 = 169 cm²

Square of the longest side = 13² = 169 cm²

Therefore, the triangle satisfies the Pythagoras theorem and is a right angled triangle.

The length of hypotenuse of this triangle is 13 cm.

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

Solution

By using Pythagoras Theorem in ∆PQM, we get

PQ² = PM² + QM²

PM² = PQ² - QM^{2 ... Equation (I)}

By using Pythagoras Theorem in ∆PRM, we get

PR² = PM² + MR²

PM² = PR² - MR² ... Equation (II)

Add Equation (I) and Equation (II)

2PM² = PQ² - QM² + PR² - MR²

2PM² = (PQ² + PR²) - (QM² + MR²)

2PM² = (QR²) - (QM² + MR²) [PQ² + PR² = QR² by Pythagoras Theorem]

2PM² = QR² - QM² - MR²

2PM² = (QM + MR)² - QM² - MR² [QR = QM + MR]

2PM² = QM² + MR² + 2QM.MR - QM² - MR²

2PM² = 2QM.MR

PM² = QM.MR

Hence, proved that PM²

3. In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that

AB² = BC . BD
AC² = BC . DC
AD² = BD . CD

Solution

I. In ∆ABD and ∆CBA:

∠BAD = ∠BCA (Right Angles)

∠ABD = ∠ABC (Common Angles)

Therefore, by the AA similarity criterion ∆ABD ~ ∆CBA.

This implies that

AB/CB = BD/AB (Corresponding Sides of similar triangles)

AB² = BC.BD

II. Since, we have proven that ∆ABD ~ ∆CBA, therefore

∠CAB = ∠BDA (Corresponding angles of similar triangles)

∠CAB = ∠CDA

In ∆CBA and ∆CAD:

∠ACB = ∠ACD (Right Angles)

∠CAB = ∠CDA (As shown above)

Therefore, by the AA similarity criterion ∆CBA ~ ∆CAD.

This implies that

AC/BC = DC/AC (Corresponding Sides of similar triangles)

AC² = BC.DC

III. In ∆DCA and ∆DAB:

∠DCA = ∠DAB (Right Angles)

∠CDA = ∠ADB (Common Angles)

Therefore, by the AA similarity criterion ∆DCA ~ ∆DAB.

This implies that

CD/AD = AD/BD (Corresponding Sides of similar triangles)

AD² = BD.CD

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

Solution

It is given that ∆ABC is an isosceles triangle. This implies that the sides opposite to the equal angles are equal.

Since, ∠ACB is given to be the right angle. Therefore, ∠ABC and ∠BAC are the equal angles with sides AC = BC.

By using Pythagoras Theorem in ABC, we get

AB² = AC² + BC²

AB² = AC² + AC²

AB² = 2AC²

Hence, proved that AB² = 2AC².

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

Solution

It is given that

AB² = 2AC²

AB² = AC² + AC²

AB² = AC² + BC² (Since AC = BC is given)

The above equation satisfies the Pythagoras Theorem.

Hence, ∆ABC is a right angles triangle.

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

Solution

Let us draw the altitude AD in the triangle ABC.

In ∆ADB and ∆ADC:

∠ADB = ∠ADC (Right Angles)

AD = AD (Common Side)

AB = AC (Sides of equilateral triangle ABC)

Therefore, by the RHS congruency criteria ∆ADB ≅ ∆ADC.

This implies that

BD = CD (By c.p.c.t.)

Therefore,

BC = 2BD

BD = BC/2

BD = a

By applying Pythagoras theorem in ∆ADB, we get

AB² = BD² + AD²

AD² = AB² - BD²

AD² = 4a² - a²

AD² = 3a²

AD = ?3a

Hence, the length of each of the altitude of the triangle ABC is ?3a.

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

Solution

Let there be a rhombus ABCD whose diagonals bisect each other at O.

To Prove : AB² + BC² + CD² + AD² = AC² + BD²

Proof :

We know that the diagonals of a rhombus bisect each other at right angles. Therefore, AOB, AOD, BOC, COD are all right angles triangles.

This also implies that AO = CO and BO = DO as O is the point of bisection.

By applying Pythagoras Theorem to ∆AOB, we get

AB² = AO² + BO²

Similarly,

BC² = BO² + CO²

CD² = CO² + DO²

AD² = AO² + DO²

Adding all of the above obtained equations, we get

AB² + BC² + CD² + AD² = AO² + BO² + BO² + CO² + CO² + DO² + AO² + DO²

AB² + BC² + CD² + AD² = 2(AO² + BO² + CO² + DO²)

AB² + BC² + CD² + AD² = 2(AO² + CO² + BO² + DO²)

Since, AO = CO and BO = DO. Therefore,

AB² + BC² + CD² + AD² = 2(2AO² + 2BO²)

AB² + BC² + CD² + AD² = 4AO² + 4BO²

AB² + BC² + CD² + AD² = (2AO)² + (2BO)²

O is the midpoint of AC and BD, so 2AO = AC and 2BO = BD. Therefore,

AB² + BC² + CD² + AD² = AC² + BD²

Hence, proved that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

8. In Fig. 6.54, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥AC and OF ⊥AB. Show that

OA² + OB² + OC² - OD² - OE² - OF² = AF² + BD² + CE²,
AF² + BD² + CE² = AE² + CD² + BF².

Solution

Let us join OB, OA and OC.

I. By applying Pythagoras Theorem in ∆AOF, we get

OA² = AF² + OF²

By applying Pythagoras Theorem in ∆BOD, we get

OB² = OD² + BD²

By applying Pythagoras Theorem in ∆COE, we get

OC² = OE² + EC²

Adding the above obtained equations, we get

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

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

Hence, proved.

II. OA² + OB² + OC² - OD² - OE² - OF²= AF² + BD²+ EC²

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

By applying Pythagoras Theorem in ∆AOE, we get

OA² = OE² + AE²

OA² - OE² = AE²

Similarly,

OC² - OD² = CD²

OB² - OF² = BF²

Hence,

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

AE²+ CD²+ BF²= AF² + BD²+ EC²

9. 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.

Solution

Let AB be the ladder which reaches the window at A with the base of the wall at C.

The wall will be perpendicular to the ground, therefore ABC will form a right triangle which is right angled at C.

AB = 10 m , AC = 8m

By applying Pythagoras Theorem in ABC, we get

AB² = AC² + BC²

10² = 8² + BC²

100 = 64 + BC²

36 = BC²

BC = 6 m

Hence, the distance between foot of the ladder and the base of the wall is 6 m.

10. 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?

Solution

Let the wire be AB and the pole be AC. BC is the distance between the stake and base of the pole.

∆ACB is a right triangle which is right angled at C.

By Applying Pythagoras Theorem in ABC, we get

AB² = AC² + BC²

24² = 18² + BC²

576 - 324 = BC²

252 = BC²

BC = 6?7 m

Hence, the stake needs to be driven 6?7 m away from the base of the pole to make the wire taut.

11. 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*1/2 hours?

Solution

Let A be the current position of the plane flying to north and B be the current position of the plane flying to the west. Let O be the starting point for the two planes.

AOB will form a right triangle, right angled at O.

Speed of plane going north = 1000 km/h

= 1200 × 3/2 = 1800 km

By Applying the Pythagoras theorem in AOB, we get

AB² = AO² + BO²

AB² = 1500² + 1800²

AB² = 2250000 + 3240000

AB² = 5490000

AB = 300?61 km

12. 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.

SOLUTION

Let AB be the taller pole and CD be the smaller pole. The distance between their tops is AC and the distance between their feet is BD. We will draw an imaginary line from C to O which is a point on the pole AB such that OC ‖ BD.

The poles stand perpendicular to the ground, so ∠BDC is a right angle.

∠AOC = ∠BDC (Corresponding Angles)

Therefore, ∆AOC is a right triangle.

OCBD forms a rectangle as OC ‖ BD and AB ‖ CD.

This implies that

OC = BD = 12 m

OB = CD = 6 m

Now, AB = AO + OB

AO = AB - OB = 11 - 6 = 5 m

By Applying Pythagoras Theorem in AOC, we get

AC² = AO² + OC²

AC² = 5² + 12²

AC² = 25 + 144

AC² = 169

AC = 13 m

Hence, the distance between the top of the two poles is 13 m.

13. 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.

By Applying Pythagoras Theorem in ABC, we get

AB² = AC² + BC² ... Equation (I)

By Applying Pythagoras Theorem in AEC, we get

AE² = AC² + EC²

By Applying Pythagoras Theorem in BDC, we get

BD² = BC² + DC²

By Applying Pythagoras Theorem in DEC, we get

DE² = DC² + EC² ... Equation (II)

Therefore,

AE² + BD² = AC² + EC² + BC² + DC²

AE² + BD² = (AC² + BC²) + (EC² + DC²)

From Equation (I) and Equation (II), we get

AE² + BD² = AB² + DE²

Hence, proved.

14. The perpendicular from A on side BC of a ∆ ABC intersects BC at D such that DB = 3 CD (see Fig. 6.55). Prove that 2 AB² = 2 AC² + BC².

Solution

BC = BD + CD

BC = 3CD + CD

BC/4 = CD

By applying Pythagoras Theorem in ADB, we get

AB² = AD² + BD² ... Equation (I)

By applying Pythagoras Theorem in ADC, we get

AC² = AD² + CD² ... Equation (II)

Subtract Equation (II) from Equation (I):

AB² - AC² = BD² - CD²

AB² - AC² = (3CD)² - CD²

AB² - AC² = 9CD² - CD²

AB² - AC² = 8CD²

AB² - AC² = 8(BC/4)²

AB² - AC² = BC²/2

2(AB² - AC²) = BC²

2AB² - 2AC² = BC²

2AB² = 2AC² + BC²

Hence, proved.

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

Solution

Let us construct altitude AE in ∆ABC.

Since AE is the altitude of an equilateral triangle, therefore it will also act as a median.

BE = CE = BC/2

By applying Pythagoras Theorem in ABE, we get

AB² = BE² + AE²

AE² = AB² - BE²

AE² = AB² - BC²/4

By applying Pythagoras Theorem in ADE, we get

AD² = AE² + DE²

AD² = AE² + (BE - BD)²

AD² = AB² - BC²/4 + (BC/2 - BC/3)²

AD² = AB² - BC²/4 + (BC/6)²

AD² = AB² - BC²/4 + BC²/36

AD² = AB² - 8BC²/36

AD² = AB² - 2BC²/9

AD² = (9AB² - 2BC²)/9

9AD² = 9AB² - 2AB² (BC = AB as ABC is an equilateral triangle)

9AD² = 7AB²

Hence, proved.

16. 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.

Solution

Let there be an equilateral triangle ABC with altitude AD and length of each side a.

To prove : 3a² = 4AD²

Proof :

We know that the altitude of an equilateral triangle also acts as a median. Therefore,

BD = CD = BC/2

By applying Pythagoras Theorem in ABD, we get

AB² = AD² + BD²

AB² = AD² + BC²/4

a² = AD² + a²/4

3a²/4 = AD²

3a² = 4AD²

Hence, proved that three times the square of one side is equal to four times the square of one of its altitudes.

17. Tick the correct answer and justify : In ∆ABC, AB = 6?3 cm, AC = 12 cm and BC = 6 cm. The angle B is :

(A) 120° (B) 60° (C) 90° (D) 45°

Solution

AB² = (6?3)² = 108

AC² = 12² = 144

BC² = 6² = 36

It can be observed that AB² + BC² = 144 = AC².

This implies that the triangle satisfies the Pythagoras Theorem with AC as the hypotenuse. Therefore, angle B is 90°.

Hence, correct options is (C) 90°.

Exercise 6.6 (Optional)

1. In Fig. 6.56, PS is the bisector of ∠ QPR of ∆ PQR. Prove that QS/SR = PQ/PR.

Solution

Let us produce the line QP to an external point T such that TR is parallel to PS.

∠QPS = ∠SPR as PS is the angle bisector for ∠QPR.

∠QPS = ∠QTR (Corresponding Angles)

∠SPR = ∠PRT (Alternate Interior Angles)

Therefore, ∠PRT = ∠QTR.

We know that if two angles of a triangle are equal then the triangle is isosceles and sides opposite to the equal angles are also equal. Therefore,

PT = PR

By using the basic proportionality theorem in QRT, we get

QS/SR = PQ/PT

QS/SR = PQ/PR

Hence, proved.

2. In Fig. 6.57, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥AC, DM ⊥ BC and DN ⊥ AB. Prove that :

DM² = DN . MC
DN² = DM . AN

Solution

It can be concluded from the figure that DBMN is a rectangle as its opposite sides are parallel and the angles are 90° each.

This implies that

DN = BM and DM = BN.

(i). ∠CDM = ∠CDB - ∠BDM

∠CDM = 90° - ∠BDM

∠CDM + ∠BDM = 90°

Also, in ∆CDM:

∠CDM + ∠CMD + ∠MCD = 180°

∠CDM + 90° + ∠MCD = 180°

∠CDM + ∠MCD = 90°

Therefore, we can say that

∠CDM + ∠MCD = ∠CDM + ∠BDM (Both sides are equal to 90°)

∠MCD = ∠BDM

In ∆CDM and ∆MDB:

∠MCD = ∠BDM (As shown above)

∠CMD = ∠BMD (Right Angles)

Therefore, by the AA similarity criterion ∆CDM ~ ∆MDB.

This implies that

DM/BM = MC/DM

DM² = MC.BM

We have already shown that BM = DN.

DM² = DN.MC

(ii). ∠ADB = ∠ADN + ∠BDN

∠ADN + ∠BDN = 90°

Also, in ∆ADN:

∠AND + ∠ADN + ∠DAN = 180°

90° + ∠ADN + ∠DAN = 180°

∠ADN + ∠DAN = 90°

Therefore, we can say that

∠ADN + ∠DAN = ∠ADN + ∠BDN (Both sides equal to 90°)

∠DAN = ∠BDN

In ∆ADN and ∆NDB:

∠DAN = ∠BDN (As shown above)

∠CMD = ∠BMD (Right Angles)

Therefore, by the AA similarity criterion ∆ADN ~ ∆NDB.

This implies that

DN/BN = NA/DN

DN² = AN.BN

We have already shown that BN = DM.

DN² = DM.AN

Hence, proved.

3. In Fig. 6.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2 BC . BD.

Solution

By applying Pythagoras Theorem in ABD, we get

AB² = AD² + BD²

By applying Pythagoras Theorem in ACD, we get

AC² = AD² + CD²

AC² = AD² + (BD + BC)²

AC² = AD² + BD² + BC² + 2 BC.BD

We already know that AD² + BD² = AB², therefore

AC² = AB² + BC² + 2 BC.BD

Hence, proved.

4. In Fig. 6.59, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² - 2 BC . BD.

Solution

By applying Pythagoras Theorem in ABD, we get

AB² = AD² + BD²

By applying Pythagoras Theorem in ACD, we get

AC² = AD² + CD²

AC² = AD² + (BC - BD)²

AC² = AD² + BD² + BC² - 2 BC.BD

We already know that AD² + BD² = AB², therefore

AC² = AB² + BC² - 2 BC.BD

Hence, proved.

5. In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :

AC² = AD² + BC . DM + (BC/2)²
AB² = AD² - BC . DM + (BC/2)²
AC² + AB² = 2 AD² + BC²/2

Solution

It is given that AD is the median of triangle ABC, therefore

BD = CD = BC/2 ... Equation (I)

By applying Pythagoras Theorem in AMD, we get

AD² = AM² + DM²

(i). By applying Pythagoras Theorem in ACM, we get

AC² = AM² + CM²

AC² = AM² + (CD + DM)²

AC² = AM² + DM² + CD² + 2 CD.DM

By Using Equation (I)

AC² = AM² +DM² + (BC/2)² + 2(BC/2)DM

We already know that AM² + DM² = AD², therefore

AC² = AD² + (BC/2)² + 2(BC/2)DM

AC² = AD² + BC .DM + (BC/2)²

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

AB² = AM² + BM²

We know that AD² = AM² + DM²

AM² = AD² - DM²

Now,

AB² = AD² - DM² + BM²

AB² = AD² - DM² + (BD - DM)²

AB² = AD² - DM² + BD² + DM² - 2BD.DM

By Using Equation (I)

AB² = AD² + (BC/2)² - 2 (BC/2).DM

AB² = AD² + (BC/2)² - BC .DM

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

AB² = AM² + BM²

By applying Pythagoras Theorem in AMC, we get

AC² = AM² + MC²

Adding the above two equations, we get

AB² + AC² = 2AM² + BM² + MC²

AB² + AC² = 2AM² + (BD - DM)² + (DM + CD)²

AB² + AC² = 2AM² + BD² + DM² - 2 BD.DM + DM² + CD² + 2 DM.CD

AB² + AC² = 2AM² + 2DM² + BD² + CD² + 2DM(CD - BD)

AB² + AC² = 2(AM² + DM²) + BD² + CD² + 2DM(CD - BD)

We know that AM² + DM² = AD², therefore

AB² + AC² = 2AD² + BD² + CD² + 2DM(CD - BD)

By Using Equation (I)

AB² + AC² = 2AD² + (BC/2)² + (BC/2)² + 2DM(BC/2 - BC/2)

AB² + AC² = 2AD² + BC²/4 + BC²/4

AB² + AC² = 2AD² + BC²/2

Hence, proved.

6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Solution

Let ABCD be a parallelogram with AB ‖ CD and altitude AE. We will produce BA to a point F such that DF ? BF.

To Prove : AC² + BD² = AB² + BC² + CD² + AD²

Proof :

By applying Pythagoras Theorem in ADF, we get

AD² = AF² + DF²

By applying Pythagoras Theorem in BDF, we get

BD² = BF² + DF²

BD² = (AB + AF)² + DF²

BD² = AB² + AF² + 2 AB.AF + DF²

We know that AF² + DF² = AD², therefore

BD² = AB² + AD² + 2 AB.AF; ... Equation (I)

By applying Pythagoras Theorem in ADE, we get

AD² = AE² + DE²

By applying Pythagoras Theorem in ACE, we get

AC² = AE² + CE²

AC² = AE² + (CD - DE)²

AC² = AE² + CD² - 2 CD.DE + DE²

We know that AE² + DE² = AD², therefore

AC² = AD² + CD² - 2 CD.DE ... Equation (II)

Add Equation (I) and Equation (II)

AC² + BD² = AD² + CD² - 2 CD.DE + AB² + AD² + 2 AB.AF

Since ABCD is a parallelogram, therefore

AB = CD and BC = AD (Opposite sides of a parallelogram)

Also, it can be observed that AEDF forms a rectangle, which implies that

AF = DE; (Opposite sides of a rectangle)

Therefore,

AC² + BD² = AD² + CD² + AB² + BC²+ 2 AB.AF - 2 AB.AF

AC² + BD² = AD² + CD² + AB² + BC²

Hence, proved that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

7. In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that :

∆APC ~ ∆ DPB
AP . PB = CP . DP

Solution

AP = BP = CP = DP as they are radii of the given circle.

Therefore, AP/DP = 1 = CP/BP.

(i). In ∆APC and ∆DPB:

AP/DP = CP/BP (As shown above)

∠APC = ∠DPB (Vertically opposite angles)

Therefore, by the SAS similarity criterion ∆APC ~ ∆DPB.

(ii). We have already shown that AP/DP = CP/BP. This implies that

AP.BP = CP.DP

AP.PB = PC.DP

Hence, proved.

8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

∆ PAC ~ ∆ PDB
PA . PB = PC . PD

Solution

(i). Since ABCD is a quadrilateral which is contained in a circle, therefore, ABCD forms a cyclic quadrilateral. This implies that,

∠PAC = ∠BDC (Exterior angle of a cyclic quadrilateral is equal to the

opposite interior angle)

In ∆PAC and ∆PDB:

∠PAC = ∠BDC (As shown above)

∠APC = ∠BPD (Common Angle)

Therefore, by the AA similarity criterion ∆PAC ~ ∆PDB.

(ii). We have already shown that ∆PAC ~ ∆PDB.

This implies that

AP/DP = PC/PB (Corresponding sides of similar triangles)

AP.PB = PC.DP

Hence, proved.

9. In Fig. 6.63, D is a point on side BC of ∆ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠ BAC.

Solution

Let us produce BA to an external point P such that AP = AC.

BC/ CD = AB/ AC (Given)

BC/CD = AB/AP (AP = AC)

Therefore, by using the converse of basic proportionality theorem, we can say that AD ‖ CP.

∠BAD = ∠APC (Corresponding Angles)

∠CAD = ∠ACP (Alternate Interior Angles)

∆ACP is an isosceles triangle as AP = AC. This implies that

∠APC = ∠ACP (Angles opposite to equal sides in an isosceles triangle)

Therefore,

∠BAD = ∠CAD

AD divides the ∠BAC in two equal parts.

Hence, AD is the bisector of ∠BAC.

10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Solution

The situation can be represented in form of a triangle ABC, where A is the tip of fishing rod, C is the end of the string and BC is the surface of water.