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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:
  1. Their corresponding angles are equal.
  2. Their corresponding sides are in the same ratio.
Two triangles are similar, if:
  1. Their corresponding angles are equal.
  2. 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.
class ten math triangles1-Theorem 1
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:
class ten math triangles1-Theorem 1
Proof:
Area of triangle class ten math triangles2-Theorem 1
Similarly,
class ten math triangles3-Theorem 1 
class ten math triangles4-Theorem 1 
class ten math triangles5-Theorem 1
Hence,
class ten math triangles6-Theorem 1
Similarly,
class ten math triangles7-Theorem 1
Triangles BDE and DEC are on the same base, i.e. DE and between same parallels, i.e. DE and BC.
Hence,
class ten math triangles8-Theorem 1
From above equations, it is clear that;
class ten math triangles9-Theorem 1
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.
class ten math triangles1-Theorem 2
Construction: ABC is a triangle in which line DE divides AB and AC in the same ratio. This means:
class ten math triangles10-Theorem 2
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;
class ten math triangles11-Theorem 2
According to the theorem;
class ten math triangles12-Theorem 2
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.
class ten math triangles1-Theorem 3
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:
class ten math triangles13-Theorem 3
Draw a line PQ in the second triangle so that DP = AB and PQ = AC
Proof:
class ten math triangles14-Theorem 3
Because corresponding sides of these two triangles are equal
This means; ∠ B = ∠ P = ∠ E and PQ || EF
This means;
class ten math triangles15-Theorem 3
Hence;
class ten math triangles16-Theorem 3 
class ten math triangles17-Theorem 3
Hence,
class ten math triangles18-Theorem 3
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.
class ten math triangles1-Theorem 3
Construction: Two triangles ABC and DEF are drawn so that their corresponding sides are proportional. This means:
class ten math triangles19-Theorem 4
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:
class ten math triangles20-Theorem 4
Because corresponding sides of these two triangles are equal
This means;
class ten math triangles21-Theorem 4
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.
class ten math triangles1-Theorem 3
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
class ten math triangles22-Theorem 4
To prove: Δ ABC ~ Δ DEF
Draw PQ in triangle DEF so that, AB = DP and AC = DF
Proof:
class ten math triangles23-Theorem 5
Because corresponding sides of these two triangles are equal
class ten math triangles24-Theorem 5   given 
∠ A = ∠ D
Hence; class ten math triangles25-Theorem 5   from SSS criterion
Hence;
class ten math triangles26-Theorem 5
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.
class ten math triangles1-Theorem 5
Construction: Two triangles ABC and PQR are drawn so that, ΔABC ~ Δ PQR.
To prove:
class ten math triangles27-Theorem 6
Draw AD ⊥ BC and PM ⊥ PR
Proof:
class ten math triangles28-Theorem 6 
class ten math triangles29-Theorem 6
Hence;
class ten math triangles30-Theorem 6
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;
class ten math triangles31-Theorem 6
Since, Δ ABC ~ Δ PQR
So,
class ten math triangles32-Theorem 6
Hence;
class ten math triangles33-Theorem 6 class ten math triangles34-Theorem 6
Similarly, following can be proven:
class ten math triangles35-Theorem 6
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.
class ten math triangles1-Theorem 7
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.
class ten math triangles1-Theorem 8
Construction: Triangle ABC is drawn which is right angled at B. From vertex B, perpendicular BD is drawn on hypotenuse AC. To prove:
class ten math triangles36-Theorem 6
Proof:
In Δ ABC and Δ ADB;
class ten math triangles37-Theorem 6 
class ten math triangles38-Theorem 6
Because these are similar triangles (as per previous theorem)
In Δ ABC and Δ BDC;
class ten math triangles38-Theorem 6 
class ten math triangles40-Theorem 6
Adding equations (1) and (2), we get;
class ten math triangles41-Theorem 6 
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.
class ten math triangles1-Theorem 8
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.
class ten math triangles1-Theorem 8
Solution: In the first figure;
Δ ADE ~ Δ ABC (Because DE || BC)
Hence;
class ten math triangles exercise 6.2_1
Similarly, in the second figure;
Δ ADE ~ Δ ABC (Because DE || BC)
Hence;
class ten math triangles exercise 6.2_1
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.
class ten math triangles1-Theorem 8
(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;
class ten math triangles exercise 6.2_1
In this case;
class ten math triangles exercise 6.2_1
It is clear that;
class ten math triangles exercise 6.2_1
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;
class ten math triangles exercise 6.2_1
It is clear that;
class ten math triangles exercise 6.2_1
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;
class ten math triangles exercise 6.2_1
It is clear that;
class ten math triangles exercise 6.2_1
Hence; EF || QR
Question – 3 - In the given figure, if LM || CB and LN || CD, prove that
class ten math triangles1-Theorem 8 
class ten math triangles exercise 6.2_1
Solution: In Δ ABC and Δ AML;
Δ ABC ~ Δ AML (because ML || BC)
Hence;
class ten math triangles exercise 6.2_1
Similarly, in Δ ADC and Δ ANL;
Δ ADC ~ Δ ANL (because NL || DC)
Hence;
class ten math triangles exercise 6.2_1
From above two equations;
class ten math triangles exercise 6.2_1 
Question – 4 - In the given figure, DE || AC and DF || AE. Prove that
class ten math triangles1-Theorem 8 
class ten math triangles exercise 6.2_14
Solution: In Δ ABC and ΔDBE;
class ten math triangles exercise 6.2_15
Because Δ ABC ~ Δ DBE
Similarly, in Δ ABE and Δ DBF;
class ten math triangles exercise 6.2_16
From above two equations, it is clear;
class ten math triangles exercise 6.2_17 
Question – 5 - In the given figure, DE || OQ and DF || OR. Show that EF || QR.
class ten math triangles1-Theorem 8
Solution: In Δ PQO and Δ PED;
class ten math triangles exercise 6.2_18
Because these are similar triangles, as per Basic Proportionality theorem.
Similarly, in Δ PRO and Δ PFD;
class ten math triangles exercise 6.2_19
From above two equations, it is clear;
class ten math triangles exercise 6.2_20
Hence,
class ten math triangles exercise 6.2_21
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.
class ten math triangles1-Theorem 8
Solution: In Δ OPQ and Δ OAB;
class ten math triangles exercise 6.2_22
Because these are similar triangles as per BPT.
Similarly, in Δ OPR and Δ OAC;
class ten math triangles exercise 6.2_23
From above two equations, it is clear;
class ten math triangles exercise 6.2_24
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.
class ten math triangles1-Theorem 8
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;
class ten math triangles exercise 6.2_25
Because as per BPT, these are similar triangles.
class ten math triangles exercise 6.2_26 
class ten math triangles exercise 6.2_27
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;
class ten math triangles exercise 6.2_28
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; class ten math triangles exercise 6.2_29
class ten math triangles1-Theorem 8
Solution: Draw a line EF || CD which is passing through O.
In Δ ABC and Δ EOC;
These are similar triangles as per BPT.
class ten math triangles exercise 6.2_30
Similarly, in Δ BOD and Δ FOD;
class ten math triangles exercise 6.2_31
In Δ ABC and Δ BAD;
class ten math triangles exercise 6.2_32
Because diagonals of a trapezium divide each other in same ratio
From above three equations, it is clear;
class ten math triangles exercise 6.2_33
Hence, Δ ABC ~ Δ BAD
Using the third equation;
class ten math triangles exercise 6.2_34
Question – 10 - The diagonals of a quadrilateral ABCD intersect each other at point O such thatclass ten math triangles exercise 6.2_35 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:
class ten math triangles1 ncert solution of exercise 6.3 class ten math triangles2 ncert solution of exercise 6.3
Answer: (i) Δ ABC ~ Δ PQR (AAA criterion)
(ii) Δ ABC ~ Δ QRP (SSS criterion)
class ten math triangles3 ncert solution of exercise 6.3 class ten math triangles4 ncert solution of exercise 6.3
Answer: (iii) Not similar
(iv) Δ LMN ~ Δ PQR (SAS criterion)
class ten math triangles5 ncert solution of exercise 6.3 class ten math triangles6 ncert solution of exercise 6.3
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.
class ten math triangles7 ncert solution of exercise 6.3
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
class ten math triangles20 ncert solution of exercise 6.3
Solution:
class ten math triangles8 ncert solution of exercise 6.3
Draw a line EF || CD which is passing through O.
In Δ ABC and Δ EOC;
These are similar triangles as per BPT.
class ten math triangles21 ncert solution of exercise 6.3
Similarly, in Δ BOD and Δ FOD;
class ten math triangles22 ncert solution of exercise 6.3
In Δ ABC and Δ BAD;
class ten math triangles23 ncert solution of exercise 6.3
Because diagonals of a trapezium divide each other in same ratio
From above three equations, it is clear;
class ten math triangles24 ncert solution of exercise 6.3
Hence, Δ ABC ~ Δ BAD
Using the third equation;
class ten math triangles25 ncert solution of exercise 6.3
Or, class ten math triangles26 ncert solution of exercise 6.3 Proved
Question – 4 - In the given figure, class ten math triangles27 ncert solution of exercise 6.3 and ∠1 = ∠2. Show that Δ PQS ~ Δ TQR.
class ten math triangles9 ncert solution of exercise 6.3
Solution: Proof:
∠1 = ∠2 (given)
Hence, PQ = PR (Sides opposite to equal angles are equal in isosceles triangle)
class ten math triangles28 ncert solution of exercise 6.3
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:
class ten math triangles10 ncert solution of exercise 6.3
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.
class ten math triangles11 ncert solution of exercise 6.3
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;
class ten math triangles29 ncert solution of exercise 6.3
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:
class ten math triangles12 ncert solution of exercise 6.3
(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:
class ten math triangles13 ncert solution of exercise 6.3
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:
class ten math triangles14 ncert solution of exercise 6.3
(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) class ten math triangles30 ncert solution of exercise 6.3
Solution: Since Δ ABC ~ Δ AMP;
Hence;
class ten math triangles31 ncert solution of exercise 6.3
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:
class ten math triangles15 ncert solution of exercise 6.3 
class ten math triangles32 ncert solution of exercise 6.3
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,
class ten math triangles32 ncert solution of exercise 6.3
(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.
class ten math triangles16 ncert solution of exercise 6.3
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
class ten math triangles17 ncert solution of exercise 6.3
Solution: In triangle Δ ABC and Δ PQR
class ten math triangles33 ncert solution of exercise 6.3 (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.
class ten math triangles18 ncert solution of exercise 6.3
Solution: In ΔBAC and ΔADC;
∠BAC = ∠ADC (given)
∠ACB = ∠DCA (Common angle)
Hence; ΔBAC ~ ΔADC
Hence;
class ten math triangles34 ncert solution of exercise 6.3
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.
class ten math triangles19 ncert solution of exercise 6.3
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;
class ten math triangles35 ncert solution of exercise 6.3
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 class ten math triangles36 ncert solution of exercise 6.3
class ten math triangles17 ncert solution of exercise 6.3
Solution: Δ ABC ~ Δ PQR (Given)
Hence;
class ten math triangles37 ncert solution of exercise 6.3
(A side and the median of one triangle are in same ratio as a corresponding side and median of another triangle)
class ten math triangles36 ncert solution of exercise 6.3 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;
class ten math triangles1 ncert solution of exercise 6.4
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:
class ten math triangles2 ncert solution of exercise 6.4
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;
class ten math triangles3 ncert solution of exercise 6.4
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
class ten math triangles4 ncert solution of exercise 6.4 

class ten math triangles5 ncert solution of exercise 6.4
Solution: Let us draw altitudes AM and DN on BC; respectively from A and D
class ten math triangles6 ncert solution of exercise 6.4
In ΔAMO and ΔDNO;
∠ AMO = ∠ DNO (Right angle)
∠ AOM = ∠ DON (Opposite angles)
Hence; ΔAMO ~ ΔDNO
Hence;
class ten math triangles7 ncert solution of exercise 6.4
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;
class ten math triangles8 ncert solution of exercise 6.4
In this case;
class ten math triangles9 ncert solution of exercise 6.4
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:
class ten math triangles10 ncert solution of exercise 6.4
Since D, E and F are mid points of AB, BC and AC
Hence; ΔBAC ~ΔDFE
So,
class ten math triangles11 ncert solution of exercise 6.4
So,
class ten math triangles12 ncert solution of exercise 6.4
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;
class ten math triangles13 ncert solution of exercise 6.4
Let us assume AD and PM are the medians of these two triangles.
Then;
class ten math triangles14 ncert solution of exercise 6.4
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 class ten math triangles15 ncert solution of exercise 6.4
Area of equilateral triangle with side ‘a’
class ten math triangles16 ncert solution of exercise 6.4
Area of equilateral triangle with side class ten math triangles15 ncert solution of exercise 6.4
class ten math triangles17 ncert solution of exercise 6.4
Ratio of two areas can be given as follows:
class ten math triangles18 ncert solution of exercise 6.4
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
  1. 2 : 1
  2. 1 : 2
  3. 4 : 1
  4. 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
  1. 2 : 3
  2. 4 : 9
  3. 81 : 16
  4. 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= Base2 + Perpendicular2
Let us check if the given three sides fulfill the criterion of Pythagoras theorem.
class ten math triangles1 ncert solution of exercise 6.5
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.
class ten math triangles2 ncert solution of exercise 6.5
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.
class ten math triangles3 ncert solution of exercise 6.5
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.
class ten math triangles4 ncert solution of exercise 6.5
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 PM2 = QM. MR.
Solution:
class ten math triangles33 ncert solution of exercise 6.5
In triangles PMQ and RMP
∠ PMQ = ∠ RMP (Right angle)
∠ PQM = ∠ RPM (90 – MRP)
Hence; PMQ ~ RMP (AAA criterion)
class ten math triangles5 ncert solution of exercise 6.5
Question – 3 - In the given figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
class ten math triangles34 ncert solution of exercise 6.5
(a) AB2 = BC. BD
Solution: In triangles ACB and DAB
∠ ACB = ∠ DAB (Right angle)
∠ CBA = ∠ ABD (common angle)
Hence; ACB ~ DAB
class ten math triangles6 ncert solution of exercise 6.5
(b) AC2 = BC. DC
Solution: In triangles ACB and DCA
∠ ACB = ∠ DCA (right angle)
∠ CBA = ∠ CAD
Hence; ACB ~ DCA
class ten math triangles7 ncert solution of exercise 6.5
(c) AD2 = BD. CD
Solution: In triangles DAB and DCA
∠ DAB = ∠ DCA (right angle)
∠ ABD = ∠ CAD
Hence; DAB ~ DCA
class ten math triangles8 ncert solution of exercise 6.5
Question – 4 - ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Solution: In this case; AB is hypotenuse and AC = BC are the other two sides
According to Pythagoras theorem:
class ten math triangles9 ncert solution of exercise 6.5
Question – 5 - ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, 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:
class ten math triangles10 ncert solution of exercise 6.5
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.
class ten math triangles11 ncert solution of exercise 6.5 

class ten math triangles34 ncert solution of exercise 6.5 
class ten math triangles12 ncert solution of exercise 6.5
Adding the above four equations, we get;
class ten math triangles13 ncert solution of exercise 6.5
Now, let us take the sum of squares of diagonals;
class ten math triangles14 ncert solution of exercise 6.5
From equations (1) and (2), it is clear;
class ten math triangles15 ncert solution of exercise 6.5
Question – 8 - In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.


class ten math triangles35 ncert solution of exercise 6.5
Show that
(a) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
class ten math triangles16 ncert solution of exercise 6.5
(b) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
class ten math triangles17 ncert solution of exercise 6.5 

class ten math triangles18 ncert solution of exercise 6.5

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;
class ten math triangles19 ncert solution of exercise 6.5
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;
class ten math triangles20 ncert solution of exercise 6.5
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;
class ten math triangles21 ncert solution of exercise 6.5
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:


class ten math triangles36 ncert solution of exercise 6.5
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;
class ten math triangles22 ncert solution of exercise 6.5
Question – 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.
Solution:


class ten math triangles37 ncert solution of exercise 6.5 

class ten math triangles23 ncert solution of exercise 6.5 

class ten math triangles24 ncert solution of exercise 6.5
Question – 14- The perpendiculars from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD. Prove that 2AB2 = 2AC2 + BC2.


class ten math triangles38 ncert solution of exercise 6.5 

class ten math triangles25 ncert solution of exercise 6.5
Substituting the value of AD in equation (1), we get;
class ten math triangles26 ncert solution of exercise 6.5
Substituting the value from above equation in equation (2), we get;
class ten math triangles27 ncert solution of exercise 6.5
Question – 15 - In an equilateral triangle, ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD2 = 7AB2.
Solution:


class ten math triangles39 ncert solution of exercise 6.5
Let us assume that each side of triangle ABC is ‘a’
Then, BD = a/3, MC = a/2 and
class ten math triangles28 ncert solution of exercise 6.5
According to Pythagoras theorem, in triangle ADM;
class ten math triangles29 ncert solution of exercise 6.5
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:
class ten math triangles30 ncert solution of exercise 6.5
Four times of square of altitude can be calculates as follows:
class ten math triangles31 ncert solution of exercise 6.5
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:
  1. 120°
  2. 60°
  3. 90°
  4. 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.
class ten math triangles32 ncert solution of exercise 6.5
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 class ten math triangles1 ncert solution of exercise 6.6
class ten math triangles28 ncert solution of exercise 6.6
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)
class ten math triangles2 ncert solution of exercise 6.6
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:
class ten math triangles29 ncert solution of exercise 6.6
(a) DM2 = 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
class ten math triangles3 ncert solution of exercise 6.6
(b) DN2 = DM.AN
Solution: In triangles DNB and AND;
 DNB =  AND (Right angles)
 NDB =  NAD
DN = DN
Hence; Δ DNB ~ Δ AND
class ten math triangles4 ncert solution of exercise 6.6
Question – 3 - In the given figure, ABC is a triangle in which  ABC > 90° and AD  CB produced. Prove that AC2 = AB2 + BC2 + 2 BC.BD
class ten math triangles30 ncert solution of exercise 6.6
Solution: In triangle ADB;
class ten math triangles5 ncert solution of exercise 6.6
In triangle ADC;
class ten math triangles6 ncert solution of exercise 6.6
Substituting the value of AB2 from equation (1) into equation (2), we get;
class ten math triangles7 ncert solution of exercise 6.6
Question – 4 - In the given figure, ABC is triangle in which  ABC < 90° and AD  BC. Prove that AC2 = AB2 + BC2 - 2BC.BD.
class ten math triangles31 ncert solution of exercise 6.6
Solution: In triangle ABD;
class ten math triangles5 ncert solution of exercise 6.6
In triangle ADC;
class ten math triangles8 ncert solution of exercise 6.6
Substituting the value of AB2 from equation (1) in equation (2), we get;
class ten math triangles9 ncert solution of exercise 6.6
Question – 5 - In the given figure, AD is a median of a triangle ABC and AM  BC. Prove that:
class ten math triangles32 ncert solution of exercise 6.6 


(i) class ten math triangles10 ncert solution of exercise 6.6
Solution: In triangle AMD;
class ten math triangles11 ncert solution of exercise 6.6
Substituting the value of AD2 from equation (1) in equation (2), we get;
class ten math triangles12 ncert solution of exercise 6.6
(ii) class ten math triangles13 ncert solution of exercise 6.6
Solution: In triangle ABM;
class ten math triangles14 ncert solution of exercise 6.6
Substituting the value of AD2 from equation (1) in equation (2), we get;
class ten math triangles15 ncert solution of exercise 6.6
(iii) class ten math triangles16 ncert solution of exercise 6.6
Solution: From question (a), we have;
class ten math triangles17 ncert solution of exercise 6.6

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.
class ten math triangles33 ncert solution of exercise 6.6
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.
class ten math triangles18 ncert solution of exercise 6.6
Substituting the value of AM2 from equation (1) in equation (2), we get;
class ten math triangles18 ncert solution of exercise 6.6
In triangle ADN;
class ten math triangles19 ncert solution of exercise 6.6
In triangle ANC;
class ten math triangles20 ncert solution of exercise 6.6
Substituting the value of AD2 from equation (4) in equation (5), we get;
class ten math triangles21 ncert solution of exercise 6.6
We also have; AM = DN and AB = CD.
Substituting these values in equation (6), we get;
class ten math triangles22 ncert solution of exercise 6.6
Question – 7 - In the given figure, two chords AB and CD intersect each other at the point P. Prove that:
class ten math triangles34 ncert solution of exercise 6.6
(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;
class ten math triangles23 ncert solution of exercise 6.6
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:
class ten math triangles35 ncert solution of exercise 6.6
(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;
class ten math triangles24 ncert solution of exercise 6.6
Question – 9 - In the given figure, D is a point on side BC of Δ ABC such that class ten math triangles26 ncert solution of exercise 6.6 Prove that AD is the bisector of  BAC.
class ten math triangles36 ncert solution of exercise 6.6
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)
class ten math triangles27 ncert solution of exercise 6.6
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?
class ten math triangles37 ncert solution of exercise 6.6
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:
class ten math triangles28 ncert solution of exercise 6.6
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.


class ten math triangles38 ncert solution of exercise 6.6
Hence; horizontal distance between the girl and the fly = CD + MD = 1.2 + 1.58 = 2.78 m

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