CBSE X Comprehensive Study material

Mathematics Kolabari Tutorial CBSE X Comprehensive Study material

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Ch6 TRIANGLES

Two triangles are similar if their corresponding angles are equal and corresponding sides are proportional.

1. (AAA similarity) If two triangles are equiangular, then the triangles are similar

2. Cor: (AA similarity): If two angles of one triangle are respectively equal to two angles of another triangle, then the two

triangles are similar.

3. (SSS similarity) If the corresponding sides of two triangles are proportional then they are similar

4. (SAS similarity) If in two triangles, one pair of corresponding sides are proportional and the included angles are equal, then the triangles are similar. Some Important Results And Theorems * Theorem no. 1: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides

in the same ratio. (BASIC PROPORTIONALITY THEOREM or BPT or THALES THEOREM). * proof may be asked.

* Theorem no 2: If a line is drawn intersecting the two sides of a triangle such that it divides the two sides in the same ratio,

then the line is parallel to the third side. (converse of BPT) * proof may be asked

Theorem no 3: The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides

containing the angle.(angle bisector theorem)

Theorem no 4: In a triangle ABC, if D is the point on BC such that BD/DC = AB/AC, prove that AD is the bisector of angle

A(converse)

Theorem no 5: The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides

containing the angle.

Theorem no 6: The line drawn from the mid point of one side of a triangle parallel to the other side bisects the third side.

Theorem no 7:The line joining the mid points of two sides of a triangle is parallel to the third side.

Theorem no 8:Prove that the diagonals of a trapezium divide each other proportionally.

Theorem no 9:If the diagonals of a quadrilateral divide each other proportionally , then it is a trapezium.

Theorem no 10:Any line parallel to the parallel sides of a trapezium divides the non parallel sides proportionally.

Theorem no 11:If two triangles are equiangular, prove that:

(i) Ratio of the corresponding sides is the same as the ratio of the corresponding medians.

(ii) Ratio of the corresponding sides is the same as the ratio of the corresponding angle bisector segments.

(iii) Ratio of the corresponding sides is the same as the ratio of the corresponding altitudes.

Theorem no 12: If one angle of a triangle is equal to one angle of another triangle and the bisectors of these equal angles divide

the opposite side in the same ratio, prove that the triangles are similar

Theorem no 13: If two sides and the median bisecting one of these sides of a triangle are respectively proportional to the two

sides and the corresponding median of another triangle, then the triangles are similar.

Theorem no 14: If two sides and a median bisecting the third side of a triangle are respectively proportional to the

corresponding sides and the median of another triangle, then the two triangles are similar.

* Theorem no 15: The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

* proof may be asked

Theorem no 16: The areas of two similar triangles are in the ratio of

(i) Squares of the corresponding altitudes



(ii) Squares of the corresponding medians

(iii) Squares of the corresponding angle bisectors.

Theorem no 17: if the areas of two similar triangles are equal then the triangles are congruent

* pythagoras theorem: in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the

other two sides. * proof may be asked

* Converse of pythagoras theorem: In a triangle, if the square of one side is equal to the sum of the squares of the other two

sides, then the angle opposite to the first side is a right angle. * proof may be asked.

Some important results on pythagoras theorem:

(i) ∆ ABC is an obtuse angled triangle, obtuse angled at B. If AD is perpendicular to CB, prove that, AC2 = AB2 + BC2 +2 BC. BD

(ii) ∆ ABC is an acute angled triangle, acute angled at B. If AD is perpendicular to CB, prove that, AC2 = AB2 + BC2 - 2 BC.

(iii) Prove that in any triangle, the sum of the squares of any two sides is equal to twice the square of half the third side together

with twice the square of the median which bisects the third side. (APPOLONIUS THEOREM)

(iv) Prove that three times the sum of the squares of the sides of the triangle is equal to four times the sum of the squares of the medians of the triangle. Proof of Theorems Basic Proportionality theorem (Or Thale’s Theorem):

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.

Proof: Given: A Δ ABC is a triangle in which DE||BC and DE intersects the other two sides AB and AC at D and E respectively.

To prove:

=

Construction: Let us join BE and CD and then draw DM ⊥AC and EN⊥AB

Proof: Now, area of ΔADE =

× base × height =

× AD × EN

So, ar(ADE) =

× AD × EN

Similarly, ar(BDE) =

× DB × EN

=> (∆ )

(∆ ) =

=

…….(i)

Again ar(AED) =

× AE × DM

And ar(DEC) =

× EC × DM

Therefore, (∆ )

(∆ ) =

=

…….(ii)

Note that ΔBDE and ΔDEC are on the same base DE and between

the same parallels BC and DE. So, ar(BDE) = ar(DEC). ..(iii)

Therefore, from (i), (ii) and (iii), we have:

=

Corollary: Also we have (i)

=

(ii)

=

Proof: (i) By adding 1 both side of the above result we get

+ 1 =

+1 =>

=

=>

=

(ii) By inverting fraction of both side of thale’s result and adding 1 to each, we get,

+ 1 =

+1 =>

=

=>

=

=>

=

Converse of Thale’s theorem:

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

Given: A ΔABC and a line intersecting AB and AC at D and E respectively, such that

=

To prove: DE||BC

Proof: If possible, let DE is not parallel to BC. Then there must be another line through D, which is parallel to BC. Let DF||BC.

In the given figure, if DE||BC, then from

Thale’s theorem we can say:

(i)

=

(ii)

=

(iii)

=



Then, by Thale’s theorem, we have:

=

But we have given that

=

This is possible only when ,

=

=>

+ 1 =

+1 =>

=

=>

=

=>

=

=> FC = EC

Which is possible only when E and F coincide.

Hence, DE||BC

AA criterion of similarity or AAA criterion:

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

hence the two triangles are similar.

Given: ΔABC and ΔDEF are two triangles such that ∠A = ∠D, ∠B =∠ E, ∠C = ∠F.

To prove: ΔABC ~ ΔDEF

Construction: Let us cut DP = AB and DQ = AC and join PQ.

Proof: In ΔABC and ΔDPQ,

DP = AB and DQ = AC (as per construction)

Also, ∠A = ∠D (given)

So, ΔABC ≅ ΔDPQ (SAS criteria)

This gives ∠B = ∠P (CPCT)

=> ∠B = ∠P = ∠E (Since given that ∠B =∠ E)

So, since corresponding angles are equal => PQ || EF

Therefore,

=

(by Thale’s theorem) =>

=

(Since DP = AB and DQ = AC )

Similarly,

=

and so

=

=

Thus, the corresponding sides og the given triangles are in the same ratio (or proportion)

Hence, ΔABC ~ ΔDEF

SSS criterion of similarity

If in two triangles, the sides of one triangle are proportional to (i.e. in the same ratio of) the sides of the other triangle, then

their corresponding angles are equal and hence the two triangles are similar.

Given: ΔABC and ΔDEF are two triangles such that

=

=

(<1)


Construction: let us cut DP = AB and DQ = AC and join PQ.

Proof: Given that

=

And as per construction, DP = AB and DQ = AC, Thus

=

=> PQ || EF (by the converse of Thale’s theorem)

So, corresponding angles ∠P = ∠E and ∠Q = ∠F

Therefore, ΔDPQ ~ ΔDEF (by AA-similarity)

=>

=

=>

=

(since, DP = AB) But given that

=

Therefore,

=

=> BC = PQ

So, now we have AB = DP, AC = DQ and BC = PQ.

Hence, ΔABC ≅ ΔDPQ (by SSS – congruence)

=> ∠A = ∠D, ∠B =∠P = ∠ E, ∠C = ∠Q = ∠F.

Thus, the corresponding angles of the given triangles are equal and hence the two triangles are similar.


SAS criterion of similarity

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.

Given: Δ ABC and Δ DEF are two triangles such that

=

(<1) and ∠A = ∠D




Construction: let us cut DP = AB and DQ = AC and join PQ.

Proof: In ΔABC and ΔDPQ,

DP = AB and DQ = AC (as per construction)

Also, ∠A = ∠D (given)

So, ΔABC ≅ ΔDPQ (SAS criteria)

=> ∠A = ∠D, ∠B =∠P and ∠C = ∠Q (cpct)

Now,

=

(Given)

=>

=

(since DP = AB and DQ = AC )

=> PQ || EF (by the converse of Thale’s theorem)

So, corresponding angles ∠P = ∠E and ∠Q = ∠F

=> ∠A = ∠D, ∠B =∠P = ∠ E, ∠C = ∠Q = ∠F.

Thus, the corresponding angles of the given triangles are equal and hence the two triangles are similar.


Areas of Similar Triangles:

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

Given: ΔABC and ΔPQR are two triangles such that ΔABC ~ΔPQR

To prove: (∆ )

(∆ ) =

=

=

Construction: For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles. AM ⊥ BC and PN ⊥ QR.

Proof: Now, ar(ABC) =

× BC × AM

and ar(PQR) =

× QR × PN

So, (∆ )

(∆ ) =

=

… (1)

Now, in ΔABM and ΔPQN,

∠B = ∠Q (As ΔABC ~ ΔPQR)

and ∠M = ∠N (Each is 90°)

So, ΔABM ~ ΔPQN (AA similarity criterion)

Therefore,

=

… (2)

Also, ΔABC ~ ΔPQR (Given)

So,

=

=

… (3)

Therefore, (∆ )

(∆ ) =

×

[From (1) & (2)]

=

×

[From (3)]

=> (∆ )

(∆ ) =

Now using (3) we get : (∆ )

(∆ ) =

=

=

Pythagoras Theorem:

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

Given: ΔABC is a right triangle right angled at B.

To prove: AC2 = AB2 + BC2

Construction: Let us draw BD ⊥ AC.

Proof: We may note that in ΔADB and ΔABC

∠A = ∠A

and ∠ADB = ∠ABC (Right angle)

So, ΔADB ~ ΔABC

Similarly, ΔBDC ~ ΔABC

Now, ΔADB ~ ΔABC



So,

=

(Sides are proportional)

or AD × AC = AB2 … (1)

Also, ΔBDC ~ ΔABC

So,

=

(Sides are proportional)

or CD × AC = BC2 … (2)

Adding (1) and (2),

AD × AC + CD × AC = AB2 + BC2

Or AC (AD + CD) = AB2 + BC2

Or AC × AC = AB2 + BC2

Or AC2 = AB2 + BC2

//Note: We have proved ΔADB ~ ΔABC and ΔBDC ~ ΔABC, Thus (i) the triangles on both sides of the perpendicular BD are

similar to the whole triangle ABC. (ii) Since ΔADB~ΔABC and ΔBDC~ΔABC So, it is also proved that ΔBDC ~ ΔADB

i.e. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both

sides of the perpendicular are similar to the whole triangle and to each other. //

Converse of Pythagoras Theorem:

In a right triangle, if the 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.

Given: ΔABC is a triangle in which AC2 = AB2 + BC2

To prove: ∠B = 90°

Construction: Let us construct a ΔPQR right angled at Q such that PQ = AB and QR = BC

Proof: Now, from ΔPQR, we have:

PR2 = PQ2 + QR2 (Pythagoras Theorem, as ∠Q = 90°)

Or PR2 = AB2 + BC2 (By construction). (1)

But AC2 = AB2 + BC2 (Given)… (2)

So, AC = PR [From (1) and (2)] … (3)

Now, in ΔABC and ΔPQR,

AB = PQ (By construction)

BC = QR (By construction)

AC = PR [Proved in (3) above]

So,ΔABC ≅ ΔPQR (SSS congruence)

Therefore,∠B = ∠Q (CPCT)

But ∠Q = 90° (By construction)

So, ∠B = 90°

1 MARKS QUESTIONS

Fill in the blanks:

1. All ……………………triangles are similar. (A: Equilateral)

2. If a line divides two sides of a triangle in the same ratio, the line is ..........................to the third side. (A: Parallel)

3. Sides of two similar triangles are in the ratio 4:9.The ratio of areas of these triangles is ……………. (A: 16:81)

4. Two triangle are said to be similar if.............. (A: Corresponding sides are in same ratio and corresponding angles are equal)

5. Traingle have ........diagonals. (A: No)

6. The six elements of a triangle are its three angles and the ------------------- (A: three sides)

2 MARKS QUESTIONS

7. Find the length of x in the following cases:

Solution: InABC, BC2= AB2 + AC2 (by Pythagoras theorem) => x2 = 52 + 122 = 25 + 144= 169 => x = 1 = 13



8. A Ladder 25m long reaches a window of a building 20m above the ground. Determine the distance of the foot of the ladder

from the building.

Solution: suppose that AB is a ladder, B is the window and CB is the building. Then, triangle ABC is right angle at C,

AB2 = AC2 + BC2

252 = AC2 + 202

AC2 = 625 – 400 = 225

AC = 225 = 15m

Hence, the foot of the ladder is at a distance 15 m from the building.

9. If ∆ ABC ~ ∆ DEF, such that AB = 1.2cm, and DE = 1.4cm. Find the ratio of areas of ∆ ABC and ∆ DEF.

Solution: Ratio of Areas of similar triangle is equal to ratio of square of corresponding sides.

Therefore ∆

∆ =

=

= .

.

=

=

10. 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; ∆

∆ =

Ratio of Areas of similar triangle is equal to ratio of square of corresponding sides. So in this case;

∆

∆ =

=

=

=>

=

=

=> AB = PQ and AC = PR

Also ∠A = ∠P (Since ∆ ABC ~ ∆ PQR) Hence; the triangles are congruent.(SAS)

11. In figure PQ||MN and

=

, KN = 20.4cm. Find KQ.

Solution: In ΔKMN

=

(PQ||MN given)

=>

=

=>

=

. => 4(20.4 – KQ) = 13 KQ => 81.6 – 4KQ = 13 KQ => KQ = 4.8cm

12. In figure, DE||BC. If AD = x, DB = x-2, AE = x+2 and EC = x – 1, find the value of ‘x’.

Solution: In ΔABC

=

(by Thale’s theorem) =>

=

=> x(x-1) = (x-2)(x+2) => x2 – x = x2 – 4 => x = 4

13. AD is the bisector of ∠A, if BD = 4 cm, DC = 3cm and AB = 6cm determine AC.



Solution: In Δ ABC , AD is the bisector of ∠A. Therefore,

=

=>

=

=> AC =

= 4.5cm

3 MARKS QUESTIONS

14. In ΔABC, AB = AC = x, BC = 5 cm and the area of the triangle ABC is 15 cm². Find x.

Solution: Construction AD BC. Since ΔABC is an isosceles triangle therefore AD bisects BC i:e BD = DC = 5/2

Area of ΔABC = ½ BC × AD = ½ ×5 × AD = ½ × 5 × x BD

=> 15 = ½ × 5 × x BD =>

= x

Squaring both sides we get, 36 = x2 -

=> x2 = 36 +

=> x = 6.5

Alternative method: Area of ΔABC = ½ ×BC × AD => 15 = ½ 5 AD => AD = 30/5= 6

In right ΔABD, x2 = 36 +

(by Pythagoras theorem)

x= 6.5

15. If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR, prove that

=

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

Solution: Δ ABC ~ Δ PQR (Given)

=>

=

=>

=

=

In Δ ABD & Δ PQM

=

and ∠ B = ∠ Q

=> Δ ABD ~ Δ PQM (SAS criteria)

Hence;

=

(corresponding sides are in same ratio)

=>

=

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

similar triangle) Proved.

16. 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.

i.e. ∠C = ∠R and also ∠B = ∠Q (= 0⁰)

Hence; ΔABC ~ ΔPQR (AA)

So corresponding sides will be in same ratio ;

=

=>

=

=> PQ =

= 42m; Height of tower = 42 m



17. 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 (AA)

Hence;

=

(corresponding sides are in same ratio)

=> CA × CA = CB × CD => CA2 = CB × CD proved

4 MARKS QUESTIONS

18. 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

∆

∆ =

=

....(i)

In ΔAMO and ΔDNO;

∠ AMO = ∠ DNO (Right angle)

∠ AOM = ∠ DON (Opposite angles)

Therefore, ΔAMO ~ ΔDNO (AA)

Hence;

=

....(ii)

From equation (i) and (ii) we get, ∆

∆ =

19. 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, ∆

∆ =

=

20. 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 a 2

Area of equilateral triangle with side ‘a’ =

a

Area of equilateral triangle with side a 2=

a 2

= 2×

a

Thus the area of an equilateral triangle described on one side of a square =

× the area of the equilateral triangle described on

one of its diagonals

21. The perpendicular from A on side BC of ∆ABC intersect BC at D such that DB=3CD. Prove that 2AB2=2AC2+BC2.

Solution: AD BC, DB=3CD BC=DB+CD=3CD+CD=4CD

CD =

BC, DB =

BC

In rt ∆ABD; AB2 = AD2+BD2 = AD2+ 3

4 BC

= AD2+

BC2

In rt ∆ACD, AC2 = AD2+CD2 = AD2+ 1

4 BC

= AD2+

BC2

=> AB2-AC2=

BC2 => 2AB2-2AC2=BC2 => 2AB2=2AC2+BC2

22. ABC is a right-angled triangle, right angled at B. AD and CE are two medians drawn from A and C respectively. If AC = 5 cm, and AD = (3 5)/ 2 cm, find the length of CE.

Solution: In right angled triangle. Let BD = DC = x and AE = BE = y In right angled triangle ABC, AB2 + BC2 = AC2 => (2y)2 + (2x)2 = 25 => 4y2 + 4x2 = 25 ….(i) In right angled triangle ABD, AB2 + BD2 = AD2

=>4y2 + x2 =

=> 4y2 + x2=

=> 16y2 + 4x2= 45 ….(ii)

Subracting equation (i) from (ii) we get, 12y2= 20 => y2 =

On puting this value in equation (i) we get, 4×

+4x2 = 25 => 4x2 = 25 -

=> x2 =

now in right triangle BEC, CE2 = BE2 + BC2 = y2 + (2x)2 = y2 + 4x2 =

+4×

= 20

CE = 2 5

23. 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) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 (b) AF2 + BD2 + CE2 = AE2 + CD2 + BF2

Solution:

(a) In ∆ AFO: AF2 = OA2 – OF2 In ∆ BDO: BD2 = OB2 – OD2 In ∆ CEO: CE2 = OC2 – OE2 Adding the above three equations, we get, AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2 proved

(b) In ∆ AEO; AE2 = OA2 – OE2 In ∆ CDO; CD2 = OC2 – OD2 In ∆ BFO: BF2 = OB2 – OF2

Adding the above three equations, we get, AE2 + CD2 + BF2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2

From the previous solution, we also have, AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2

C B

A

E

D



Comparing the RHS of the above two equations, we get, AF2 + BD2 + CE2 = AE2 + CD2 + BF2 proved

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

Solution: 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, p2 = h2 – b2 = (2a)2 – a2 = 4a2 – a2 = 3a2 => p = a 3

25. 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.

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

Proof: In ∆ AOB; AB2 = AO2 + BO2 In ∆ BOC; BC2 = CO2 + BO2 In ∆ COD; CD2 = CO2 + DO2 In ∆ AOD; AD2 = DO2 + AO2

Adding the above four equations, we get;

AB2 + BC2 + CD2 + AD2

= AO2 + BO2 + CO2 + BO2 + CO2 + DO2 + DO2 + AO2

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

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

(Because AO = CO and BO = DO)

Or, AB2 + BC2 + CD2 + AD2 = 4(AO2 + BO2) ………(1)

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

AC2 + BD2 = (AO + CO)2 + (BO + DO)2 = (2AO)2 + (2BO)2 = 4AO2 + 4BO2 ……(2)

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

AB2 + BC2 + CD2 + AD2 = AC2 + BD2 proved

26. 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 = 1500 km and b = 1800 km

From Pythagoras theorem; h2 = p2 + b2 = 15002 + 18002 = 2250000 + 3240000 = 5490000

Or, h = 300 1 km

CBSE X Comprehensive Study material

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