Chapter Summary:

All congruent figures are similar, but it does not mean that all similar figures are congruent.

Two polygons of the same number of sides are similar, if:

- Their corresponding angles are equal.
- Their corresponding sides are in the same ratio.

Two triangles are similar, if:

- Their corresponding angles are equal.
- Their corresponding sides are in the same ratio.

According to Greek mathematician Thales, “The ratio of any two corresponding sides in two equiangular triangles is always the same.”

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.

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

According to the Indian mathematician Budhayan, “The diagonal of a rectangle produces by itself the same area as produced by its both sides (i.e., length and breadth).”

In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

If in a triangle, square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

THEOREM 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.

Construction: ABC is a triangle. DE || BC and DE intersects AB at D and AC at E.

Join B to E and C to D. Draw DN ⊥ AB and EM ⊥ AC.

To prove:

Proof:

Area of triangle

Similarly,

Hence,

Similarly,

Triangles BDE and DEC are on the same base, i.e. DE and between same parallels, i.e. DE and BC.

Hence,

From above equations, it is clear that;

THEOREM 2:

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Construction: ABC is a triangle in which line DE divides AB and AC in the same ratio. This means:

To prove: DE || BC

Let us assume that DE is not parallel to BC. Let us draw another line DE’ which is parallel to BC.

Proof:

If DE’ || BC, then we have;

According to the theorem;

Then according to the first theorem; E and E’ must be coincident.

This proves: DE || BC

THEOREM 3:

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar. This is also called AAA (Angle-Angle-Angle) criterion.

Construction: Two triangles ABC and DEF are drawn so that their corresponding angles are equal. This means:

∠ A =∠ D, ∠ B = ∠ E and ∠ C = ∠ F

To prove:

Draw a line PQ in the second triangle so that DP = AB and PQ = AC

Proof:

Because corresponding sides of these two triangles are equal

This means; ∠ B = ∠ P = ∠ E and PQ || EF

This means;

Hence;

Hence,

THEOREM 4:

If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. This is also called SSS (Side-Side-Side) criterion.

Construction: Two triangles ABC and DEF are drawn so that their corresponding sides are proportional. This means:

To prove: ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F

And hence; Δ ABC ~ Δ DEF

In triangle DEF, draw a line PQ so that DP = AB and DQ = AC

Proof:

Because corresponding sides of these two triangles are equal

This means;

This also means; ∠ P = ∠ E and ∠ Q = ∠ F

We have taken; ∠ A = ∠ D, ∠ B = ∠ P and ∠ C = ∠ Q

Hence; ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F

From AAA criterion;

Δ ABC ~ Δ DEF proved

THEOREM 5:

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. This is also called SAS (Side-Angle-Side) criterion.

Construction: Two triangles ABC and DEF are drawn so that one of the angles of one triangle is equal to one of the angles of another triangle. Moreover, two sides included in that angle of one triangle are proportional to two sides included in that angle of another triangle. This means;

∠ A = ∠ D and

To prove: Δ ABC ~ Δ DEF

Draw PQ in triangle DEF so that, AB = DP and AC = DF

Proof:

Because corresponding sides of these two triangles are equal

given
∠ A = ∠ D

Hence; from SSS criterion

Hence;

Hence; Δ ABC ~ Δ DEF proved

THEOREM 6:

The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Construction: Two triangles ABC and PQR are drawn so that, ΔABC ~ Δ PQR.

To prove:

Draw AD ⊥ BC and PM ⊥ PR

Proof:

Hence;

Now, in Δ ABD and Δ PQM;

∠ A = ∠ P, ∠ B = ∠ Q and ∠ D = ∠ M (because Δ ABC ~ Δ PQR)

Hence; Δ ABD ~ Δ PQM

Hence;

Since, Δ ABC ~ Δ PQR

So,

Hence;

Similarly, following can be proven:

THEOREM 7:

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

Construction: Triangle ABC is drawn which is right-angled at B. From vertex B, perpendicular BD is drawn on hypotenuse AC. To prove:

Δ ABC ~ Δ ADB ~ Δ BDC

Proof:

In Δ ABC and Δ ADB;

∠ ABC = ∠ ADB

∠ BAC = ∠ DAB

∠ ACB = ∠ DBA

From AAA criterion; Δ ABC ~ Δ ADB

In Δ ABC and Δ BDC;

∠ ABC = ∠ BDC

∠ BAC = ∠ DBC

∠ ACB = ∠ DBC

From AAA criterion; Δ ABC ~ Δ BDC

Hence; Δ ABC ~ Δ ADB ~ Δ BDC proved.

THEOREM 8:

Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Construction: Triangle ABC is drawn which is right angled at B. From vertex B, perpendicular BD is drawn on hypotenuse AC. To prove:

Proof:

In Δ ABC and Δ ADB;

Because these are similar triangles (as per previous theorem)

In Δ ABC and Δ BDC;

Adding equations (1) and (2), we get;

Proved.

## NCERT Solution of Exercise 6.1 (Math)

Question – 1 - Fill in the blanks using the correct word given in brackets:

(a) All circles are ………… (congruent, similar).

Answer: similar

Explanation: All circles are similar irrespective of difference in their radii. A smaller circle can be enlarged to the size of a larger circle and vice-versa is also true.

(b) All squares are ……………(similar, congruent).

Answer: Similar

Explanation: All squares are similar because all the angles in a square are right angles and all the sides are equal. Hence, a smaller square can be enlarged to the size of a larger square and vice-versa is also true.

(c) All …………..triangles are similar. (isosceles, equilateral).

Answer: equilateral

Explanation: All equilateral triangles are similar because all the angles in an equilateral triangle are 60°. Moreover, all the sides of an equilateral triangle are equal. Hence, a smaller equilateral triangle can be enlarged to the size of a larger equilateral triangle and vice-versa is also true.

(d) Two polygons of the same number of sides are similar, if their corresponding angles are …………and their corresponding sides are ……………(equal, proportional).

Answer: equal, proportional.

Explanation: This is similar to AAA criterion in case of triangles. If corresponding angles are same and the number of sides are similar in two polygons then the polygons are similar.

Question – 2 - Give two different examples of pair of

(a) Similar figures

Answer: Two equilateral triangles of different sides.

Explanation: Two equilateral triangles of different sides would be similar because all the angles are 60°. Hence, AAA criterion applies in this case.

(b) Non-similar figures

Answer: A rhombus and a rectangle

Explanation: In case of a rhombus, all the sides are equal and the angles can either be right angles or a combination of acute and obtuse angles. In case of rectangle; opposite sides are equal and all the angles are right angles. Hence, a rhombus and a rectangle are non-similar figures. There can be some exceptions; like in case of two squares of same sides. A square is a rhombus and a rectangle too.

Question – 3 - State whether the following quadrilaterals are similar or not.

Answer: These quadrilaterals are not similar.

Explanation: In the given quadrilaterals, the sides are in the same ration, i.e. all the sides are same in respective quadrilaterals. But angles in the smaller quadrilateral are not right angles but all the angles in the larger quadrilateral are right angle. Hence, these quadrilaterals are not similar.

## Exercise 6.2 (Math)

Question – 1 - In the given figures, DE || BC. Find EC in first figure and AD in second figure.

Solution: In the first figure;

Δ ADE ~ Δ ABC (Because DE || BC)

Hence;

Similarly, in the second figure;

Δ ADE ~ Δ ABC (Because DE || BC)

Hence;

Question – 2 - E and F are points on the sides PQ and PR respectively of a Δ PQR. For each of the following cases, state whether EF || QR.

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

Answer: For EF || QR, the figure should fulfill following criterion;

In this case;

It is clear that;

Hence; EF and QR are not parallel.

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

Solution: In this case;

It is clear that;

Hence; EF || QR

(c) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution: In this case;

It is clear that;

Hence; EF || QR

Question – 3 - In the given figure, if LM || CB and LN || CD, prove that

Solution: In Δ ABC and Δ AML;

Δ ABC ~ Δ AML (because ML || BC)

Hence;

Similarly, in Δ ADC and Δ ANL;

Δ ADC ~ Δ ANL (because NL || DC)

Hence;

From above two equations;

Question – 4 - In the given figure, DE || AC and DF || AE. Prove that

Solution: In Δ ABC and ΔDBE;

Because Δ ABC ~ Δ DBE

Similarly, in Δ ABE and Δ DBF;

From above two equations, it is clear;

Question – 5 - In the given figure, DE || OQ and DF || OR. Show that EF || QR.

Solution: In Δ PQO and Δ PED;

Because these are similar triangles, as per Basic Proportionality theorem.

Similarly, in Δ PRO and Δ PFD;

From above two equations, it is clear;

Hence,

Hence, EF || QR proved.

Question – 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.

Solution: In Δ OPQ and Δ OAB;

Because these are similar triangles as per BPT.

Similarly, in Δ OPR and Δ OAC;

From above two equations, it is clear;

Hence; BC || QR proved

Question – 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.

Solution: PQR is a triangle in which DE || QR. Line DE intersects PQ at D so that PD = DQ

To prove: PE = ER

In Δ PQR and Δ PDE;

Because as per BPT, these are similar triangles.

Hence; E is the midpoint of PR proved.

Question – 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.

Solution: The figure from the previous question can be used to solve this question.

ABC is a triangle in which D and E are mid points of PQ and PR respectively.

To prove: DE || QR

In Δ PQR and Δ PDE;

Hence, as per BPT these triangles are similar.

Hence; DE || QR proved.

Question – 9 - ABCD is a trapezium in which AB || CD and its diagonals intersect each other at the point O. show that;

Solution: Draw a line EF || CD which is passing through O.

In Δ ABC and Δ EOC;

These are similar triangles as per BPT.

Similarly, in Δ BOD and Δ FOD;

In Δ ABC and Δ BAD;

Because diagonals of a trapezium divide each other in same ratio

From above three equations, it is clear;

Hence, Δ ABC ~ Δ BAD

Using the third equation;

Question – 10 - The diagonals of a quadrilateral ABCD intersect each other at point O such that Show that ABCD is a trapezium.

Solution: This question can be proved by using the figure in previous question.

Since diagonals are dividing each other in the same ratio hence, ABCD is a trapezium; proved.

## NCERT Solution of Exercise 6.3 (Math)

Question – 1 - State which pairs of triangles in the given 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:

Answer: (i) Δ ABC ~ Δ PQR (AAA criterion)

(ii) Δ ABC ~ Δ QRP (SSS criterion)

Answer: (iii) Not similar

(iv) Δ LMN ~ Δ PQR (SAS criterion)

Answer: (v) Not similar

(vi) Δ DEF ~ Δ PQR (AAA criterion)

Question – 2 - In the given figure, Δ ODC ~ Δ OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution: ∠DOC + ∠COB = 180° (Linear pair of angles)

Or, ∠DOC + 120° = 180°

Hence, ∠DOC = 180° - 120° = 60°

In Δ DOC;

∠DCO + ∠CDO + ∠DOC = 180° (Angle sum of triangle)

Or, ∠DCO + 70° + 60° = 180°

Or, ∠DCO = 180° - 130° = 50°

∠OCD = ∠OAB = 50° (Because Δ ODC ~ Δ OBA; given)

Question – 3 - Diagonals AC and BD of a trapezium ABCD with AB || CD intersect each other at the point O. Using a similarity criterion for two triangles, show that

Solution:

Draw a line EF || CD which is passing through O.

In Δ ABC and Δ EOC;

These are similar triangles as per BPT.

Similarly, in Δ BOD and Δ FOD;

In Δ ABC and Δ BAD;

Because diagonals of a trapezium divide each other in same ratio

From above three equations, it is clear;

Hence, Δ ABC ~ Δ BAD

Using the third equation;

Or, Proved

Question – 4 - In the given figure, and ∠1 = ∠2. Show that Δ PQS ~ Δ TQR.

Solution: Proof:

∠1 = ∠2 (given)

Hence, PQ = PR (Sides opposite to equal angles are equal in isosceles triangle)

Hence,

PS || TR

And; Δ PQS ~ Δ TQR proved

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

Solution:

In Δ RPQ and Δ RTS;

∠RPQ = ∠RTS (given)

∠PRQ = ∠TRS (common)

Hence, Δ RPQ ~ Δ RTS (AAA Criterion)

Question – 6 - In the given figure, if Δ ABE ≈ Δ ACD, show that Δ ADE ~ Δ ABC.

Solution: Since Δ ABE ≈ Δ ACD

Hence, BE = CD

And ∠DBE = ∠ECD

In Δ DBE and Δ ECD

BE = CD (proved earlier)

∠DBE = ∠ECD (proved earlier)

DE = DE (common)

Hence, Δ DBE ≈ Δ ECD

This means; DB = EC

This also means;

Hence; DE || BC

Thus, Δ ADE ~ Δ ABC proved.

Question – 7 - In the given figure, altitudes AD and CE of Δ ABC intersect each other at point P. Show that:

(a) Δ AEP ~ Δ CDP

Solution: In Δ AEP and Δ CDP

∠AEP = ∠CDP (Right angle)

∠APE = ∠CPD (Opposite angles)

Hence; Δ AEP ~ Δ CDP proved (AAA criterion)

(b) Δ ABD ~ Δ CBE

Solution: In Δ ABD and Δ CBE

∠ADB = ∠CEB (Right angle)

∠DBA = ∠EBC (Common angle)

Hence; Δ ABD ~ Δ CBE proved (AAA criterion)

(c) Δ AEP ~ Δ ADB

Solution: In Δ AEP and Δ ADB

∠AEP = ∠ADB (Right angle)

∠EAP = ∠DAB (Common angle)

Hence; Δ AEP ~ Δ ADB proved (AAA criterion)

(d) Δ PDC ~ Δ BEC

Solution: In Δ PDC and Δ BEC

∠PDC = ∠BEC (Right angle)

∠PCD = ∠BCE (Common angle)

Hence; Δ PDC ~ Δ BEC proved (AAA criterion)

Question – 8 - E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. show that Δ ABE ~ Δ CFB.

Solution:

In Δ ABE and Δ CFB

∠ABE = ∠CFB (Alternate angles)

∠AEB = ∠CBF (Alternate angles)

Hence; Δ ABE ~ Δ CFB proved (AAA criterion)

Question – 9 - In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(a) Δ ABC ~ Δ AMP

Solution: In Δ ABC and Δ AMP

∠ABC = ∠AMP (Right angle)

∠CAB = ∠PAM (Common angle)

Hence; Δ ABC ~ Δ AMP proved (AAA criterion)

(b)

Solution: Since Δ ABC ~ Δ AMP;

Hence;

Proved

Question – 10 - CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If Δ ABC ~ Δ FEG, show that:

Solution: Δ ABC ~ Δ FEG (given)

Hence; ∠ABC = ∠FEG

∠ACB = ∠FGE

∠BAC = ∠GFE -------------(1)

Hence; ∠ACD = ∠FGH (Halves of ∠BCA and ∠FGE) -------------(2)

From equations (1) and (2);

ΔACD ~ Δ FGH

Hence; in these triangles,

(b) Δ DCB ~ Δ HGE

Solution: In Δ DCB and Δ HGE

∠DCB = ∠HGE (Halves of ∠BCA and ∠FGE)

∠DBC = ∠HEG (Because they are common to ∠ABC and ∠FEG)

Hence; Δ DCB ~ Δ HE proved

(c) Δ DCA ~ Δ HGF

Solution: In Δ DCA and Δ HGF

ΔACD ~ Δ FGH [proved in question (a)]

Hence; ΔACD ~ Δ FGH proved

(Note: points C and G are in middle and sequence of other points has been changed in naming these triangles. Hence; triangles in question (a) are similar in this question.)

Question – 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 Δ ABD ~ Δ ECF.

Solution: In Δ ABD and Δ ECF

∠ADB = ∠EFC (Right angle)

∠ABC = ∠ECF (Angles opposite to equal sides)

Hence; Δ ABD ~ Δ ECF (AAA criterion)

Question – 12 - Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Δ PQR. Show that Δ ABC ~ Δ PQR

Solution: In triangle Δ ABC and Δ PQR

(given) ---------(1)

Hence; ∠BAD = ∠QPM

∠DAC = ∠MPR

(These are angles made by a side and median of one triangle and corresponding side and median of another triangle)

Hence; ∠BAD + ∠DAC = ∠QPM + ∠MPR

Or, ∠BAC = ∠QPR -----------(2)

From equation (1) and (2);

Δ ABC ~ Δ PQR proved (SAS criterion)

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

Solution: In ΔBAC and ΔADC;

∠BAC = ∠ADC (given)

∠ACB = ∠DCA (Common angle)

Hence; ΔBAC ~ ΔADC

Hence;

Question – 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: Height of pole = AB = 6 m and its shadow = BC = 4 m

Height of tower = PQ = ? and its shadow = QR = 28 m

The angle of elevation of the sun will be same at a given time for both the triangles.

Hence; ΔABC ~ ΔPQR

This means;

Height of tower = 42 m

Question – 16 - If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR, prove that

Solution: Δ ABC ~ Δ PQR (Given)

Hence;

(A side and the median of one triangle are in same ratio as a corresponding side and median of another triangle)

Proved.

## NCERT Solution of Exercise 6.4 (Math)

Question – 1- Let Δ ABC ~ Δ DEF and their areas be, respectively, 64 sq cm and 121 sq m. If EF = 15.4 cm, find BC.

Solution: In Δ ABC ~ Δ DEF;

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

Solution:

In triangles AOB and COD;

∠ AOB = ∠ COD (Opposite angles)

∠ OAC = ∠ OCD (Alternate angles)

Hence; Δ AOB ~Δ COD

Hence;

Solution: In Δ ABC ~ Δ DEF;

Question – 3 - In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

Solution: Let us draw altitudes AM and DN on BC; respectively from A and D

In ΔAMO and ΔDNO;

∠ AMO = ∠ DNO (Right angle)

∠ AOM = ∠ DON (Opposite angles)

Hence; ΔAMO ~ ΔDNO

Hence;

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

Solution: Let us take two triangles ABC and PQR with equal areas.

Then, we have;

In this case;

Hence; the triangles are congruent.

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

Solution:

Since D, E and F are mid points of AB, BC and AC

Hence; ΔBAC ~ΔDFE

So,

So,

Question – 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: In case of two similar triangles ABC and PQR;

Let us assume AD and PM are the medians of these two triangles.

Then;

Question – 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 us take a square with side ‘a’

Then the diagonal of square will be

Area of equilateral triangle with side ‘a’

Area of equilateral triangle with side

Ratio of two areas can be given as follows:

Question – 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

- 2 : 1
- 1 : 2
- 4 : 1
- 1 : 4

Solution: (c) 4 : 1

For explanation; refer to question 5

Question – 9 - Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio

- 2 : 3
- 4 : 9
- 81 : 16
- 16 : 81

Solution: (d) 16 : 91

Note: Areas are in duplicate ratio of corresponding sides in case of similar figures.

## NCERT Solution of Exercise 6.5 (Math)

Question – 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.

(a) 7 cm, 24 cm, 25 cm

Solution: In a right triangle, the longest side is the hypotenuse. We also know that according to Pythagoras theorem:

Hypotenuse

^{2 }= Base^{2}+ Perpendicular^{2}
Let us check if the given three sides fulfill the criterion of Pythagoras theorem.

Here; LHS = RHS

Hence; this is a right triangle.

(b) 3 cm, 8 cm, 6 cm

Solution: Let us check if the given sides fulfill the criterion of Pythagoras theorem.

Here; LHS is not equal to RHS

Hence; this is not a right triangle.

(c) 50 cm, 80 cm, 100 cm

Solution: Let us check if the given sides fulfill the criterion of Pythagoras theorem.

Here; LHS is not equal to RHS

Hence; this is not a right triangle.

(d) 13 cm, 12 cm, 5 cm

Solution: Let us check if the given sides fulfill the criterion of Pythagoras theorem.

Here; LHS = RHS

Hence; this is a right triangle.

Question – 2 - PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM

^{2}= QM. MR.
Solution:

In triangles PMQ and RMP

∠ PMQ = ∠ RMP (Right angle)

∠ PQM = ∠ RPM (90 – MRP)

Hence; PMQ ~ RMP (AAA criterion)

Question – 3 - In the given figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that

(a) AB

^{2}= BC. BD
Solution: In triangles ACB and DAB

∠ ACB = ∠ DAB (Right angle)

∠ CBA = ∠ ABD (common angle)

Hence; ACB ~ DAB

(b) AC

^{2}= BC. DC
Solution: In triangles ACB and DCA

∠ ACB = ∠ DCA (right angle)

∠ CBA = ∠ CAD

Hence; ACB ~ DCA

(c) AD

^{2}= BD. CD
Solution: In triangles DAB and DCA

∠ DAB = ∠ DCA (right angle)

∠ ABD = ∠ CAD

Hence; DAB ~ DCA

Question – 4 - ABC is an isosceles triangle right angled at C. Prove that AB

^{2}= 2AC^{2}.
Solution: In this case; AB is hypotenuse and AC = BC are the other two sides

According to Pythagoras theorem:

Question – 5 - ABC is an isosceles triangle with AC = BC. If AB

^{2}= 2AC^{2}, prove that ABC is a right triangle.
Solution: This question will be sold in the same way as the earlier question.

In this case;

Square of the longest side = sum of squares of other two sides

Hence, this is a right triangle.

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

Answer: In case of an equilateral triangle, an altitude will divide the triangle into two congruent right triangles. In the right triangle thus formed, we have;

Hypotenuse = One of the sides of the equilateral triangle = 2a

Perpendicular = altitude of the equilateral triangle = p

Base = half of the side of the equilateral triangle = a

Using Pythagoras theorem, the perpendicular can be calculated as follows:

Question – 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: ABCD is a rhombus in which diagonals AC and BD intersect at point O.

Adding the above four equations, we get;

Now, let us take the sum of squares of diagonals;

From equations (1) and (2), it is clear;

Question – 8 - In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.

Show that

(a) OA

^{2}+ OB^{2}+ OC^{2}– OD^{2}– OE^{2}– OF^{2}= AF^{2}+ BD^{2}+ CE^{2}
(b) AF

^{2}+ BD^{2}+ CE^{2}= AE^{2}+ CD^{2}+ BF^{2}
Question – 9 - A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Solution: In this case, the ladder makes the hypotenuse, the height of top of the ladder is perpendicular and the distance of foot of the ladder from the base of the wall is the base.

So, h = 10 m, p = 8 m and b = ?

From Pythagoras theorem;

Question – 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: Here; h = 24 m, p = 18 m and b = ?

From Pythagoras theorem;

Question – 11 - An aeroplane leaves and airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1.5 hours?

Solution: Distance covered by the first plane in 1.5 hours = 1500 km

Distance covered by the second plane in 1.5 hours = 1800 km

The position of the two planes after 1.5 hour journey can be shown by a right triangle and we need to find the hypotenuse to know the aerial distance between them.

Here; h = ? p = 1800 km and b = 1500 km

From Pythagoras theorem;

Question – 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:

In this figure; AB = 6 m, CD = 11 m and BC = AE = 12 m

In right triangle DEA;

DE = CD – AB = 11 – 6 = 5 m, AE = 12 m and AD = ?

Distance between the tops of the two poles can be calculated by using Pythagoras theorem;

Question – 13- D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE

^{2}+ BD^{2}= AB^{2}+ DE^{2}.
Solution:

Question – 14- The perpendiculars from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD. Prove that 2AB

^{2}= 2AC^{2}+ BC^{2}.
Substituting the value of AD in equation (1), we get;

Substituting the value from above equation in equation (2), we get;

Question – 15 - In an equilateral triangle, ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD

^{2}= 7AB^{2}.
Solution:

Let us assume that each side of triangle ABC is ‘a’

Then, BD = a/3, MC = a/2 and

According to Pythagoras theorem, in triangle ADM;

Question – 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 us assume that each side of the triangle is ‘a’, then its altitude AM is as follows:

Four times of square of altitude can be calculates as follows:

Hence; three times of square of a side = four times of square of altitude proved

Question – 17 - Tick the correct answer and justify: In Δ ABC, AB = 6√ 3 cm, AC = 12 cm and BC = 6 cm. The angle B is:

- 120°
- 60°
- 90°
- 45°

Solution: Here; the longest side is 12 cm. Let us check if the sides of the given triangle fulfill the criterion of Pythagoras triplet.

Here; LHS = RHS and hence the given triangle is a right triangle.

Hence; angle B = 90°, i.e. option (c) is the correct answer.

## Exercise 6.6 (Optional)

Question – 1 - In the given figure, PS is the bisector of ∠ QPR or Δ PQR. Prove that

Solution: Draw a line RT || SP; which meets QP extended up to QT.

∠ QPS = ∠ SPQ (given)

∠ SPQ = ∠ PRT (Alternate angles)

∠ QPS = ∠ PTR (Corresponding angles)

From these three equations, we have;

∠ PRT = ∠ PTR

Hence, in triangle PRT;

PT = PR (Sides opposite to equal angles) -----------(1)

Now; in triangles SQP and RQT;

∠ QPS = ∠ QTR (Corresponding angles)

∠ QSP = ∠ QRT (Corresponding angles)

Hence; Δ SQP ~ ΔRQT (AAA criterion)

Question – 2- In the given figure, D is a point on hypotenuse AC of Δ ABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AN. Prove that:

(a) DM

^{2}= DN.MC
Solution: DN || BC and DM || AB

DNMB is a rectangle because all the four angles are right angles.

Hence; DN = MB and DM = NB

In triangles DMB and CMD;

∠ DMB = ∠ CMD (Right angle)

∠ DBM = ∠ CDM

DM = DM

Hence; Δ DMB ~ Δ CMD

(b) DN

^{2}= DM.AN
Solution: In triangles DNB and AND;

∠ DNB = ∠ AND (Right angles)

∠ NDB = ∠ NAD

DN = DN

Hence; Δ DNB ~ Δ AND

Question – 3 - In the given figure, ABC is a triangle in which ∠ ABC > 90° and AD ⊥ CB produced. Prove that AC

^{2}= AB^{2}+ BC^{2}+ 2 BC.BD
Solution: In triangle ADB;

In triangle ADC;

Substituting the value of AB2 from equation (1) into equation (2), we get;

Question – 4 - In the given figure, ABC is triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC

^{2}= AB^{2}+ BC^{2}- 2BC.BD.
Solution: In triangle ABD;

In triangle ADC;

Substituting the value of AB2 from equation (1) in equation (2), we get;

Question – 5 - In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(i)

Solution: In triangle AMD;

Substituting the value of AD2 from equation (1) in equation (2), we get;

(ii)

Solution: In triangle ABM;

Substituting the value of AD

^{2}from equation (1) in equation (2), we get;
(iii)

Solution: From question (a), we have;

Question – 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: ABCD is a parallelogram in which AB = CD and AD = BC

Perpendicular AN is drawn on DC and perpendicular DM is drawn on AB extended up to M.

Substituting the value of AM2 from equation (1) in equation (2), we get;

In triangle ADN;

In triangle ANC;

Substituting the value of AD

^{2}from equation (4) in equation (5), we get;
We also have; AM = DN and AB = CD.

Substituting these values in equation (6), we get;

Question – 7 - In the given figure, two chords AB and CD intersect each other at the point P. Prove that:

(a) Δ APC ~ Δ DPB

Solution: In Δ APC and Δ DPB;

∠ CAP = ∠ BDP (Angles on the same side of a chord are equal)

∠ APC = ∠ DPB (Opposite angles)

Hence; Δ APC ~ Δ DPB (AAA Criterion)

(b) AP.PB = CP. DP

Solution: Since the two triangles are similar, hence;

Question – 8 - In the given figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:

(a) Δ PAC ~ Δ PDB

Solution: In triangle Δ PAC ~ Δ PDB;

∠ PAC + ∠ CAB = 180° (Linear pair of angles)

∠ CAB + ∠ BDC = 180° (Opposite angles of cyclic quadrilateral are supplementary)

Hence; ∠ PAC = ∠ PDB

Similarly; ∠ PCA = ∠ PBD can be proven

Hence; Δ PAC ~ Δ PDB

(b) PA.PB = PC.PD

Solution: Since the two triangles are similar, so;

Question – 9 - In the given figure, D is a point on side BC of Δ ABC such that Prove that AD is the bisector of ∠ BAC.

Solution: Draw a line CM which meets BA extended up to AM so that AM = AC

This means; ∠ AMC = ∠ ACM (Angles opposite to equal sides)

Hence; Δ ABD ~ Δ MBC

Hence; AD || MC

This means;

∠ DAC = ∠ ACM (Alternate angles)

∠ BAD = ∠ AMC (Corresponding angles)

Since, ∠ AMC = ∠ ACM

Hence; ∠ DAC = ∠ BDA

Hence; AD is the bisector of ∠ BAC

Question – 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? 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: Here; AD = 1.8 m, BD = 2.4 m and CD = 1.2 m

Retracting speed of string = 5 cm per second

In triangle ABD; length of string, i.e. AB can be calculated as follows:

Let us assume that the string reaches at point M after 12 seconds

Length of retracted string in 12 seconds = 5 x 12 = 60 cm

Remaining length of string = 3 m – 0.6 m = 2.4 m

In triangle AMD, we can find MD by using Pythagoras Theorem.

Hence; horizontal distance between the girl and the fly = CD + MD = 1.2 + 1.58 = 2.78 m

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