**NCERT Solutions for Class 10 Maths Chapter 6 Triangles**

Triangles Class 10 NCERT solutions have been prepared in simple language by our expert teachers. Triangles and their properties are defined in NCERT solutions for Class 10 Maths Chapter 6. NCERT Class 10 Maths Solutions Chapter 6 contains a detailed explanation for each question in the NCERT Class 10 Maths book’s exercises. Students will find solutions to problems based on the similarity of triangles in NCERT solutions for class 10 maths chapter 6. The solution to Class 10 Maths chapter 6 pdf download can be accessed offline.

## NCERT Solutions Class 10 Maths Chapter 6 – Sample Question

This following type of questions can be solved using the concept of similarity of triangles.

**Question: A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.**

**Answer:**

Given:- length of lamp post (AB) = 3.6m

Height of the girl = 90cm or 0.9m

Speed of the girl = 1.2m/sec

To find:- the length of the shadow DE.

Solution:- the girl walked BD distance in 4 seconds.

The distance traveled girl = length of BD = 4 x 1.2 => 4.8m

In Triangle ABE and Triangle CDE

Thus Triangle ABE and Triangle CDE are similar triangles.

From the theorem:- If two triangles are similar then the ratio of their sides is equal.

Hence the length of the shadow is 1.6m.

These types of examples are mentioned in these NCERT solutions for class 10 Maths chapter 6.. Here you will get NCERT solutions for Class 10 chapter wise also.

**Class 10 Maths Chapter 6 Triangles Excercise: 6.1**

Q1 (1) Fill in the blanks using the correct word given in brackets: All circles are ______ . (congruent, similar)

Answer:

All circles are similar.

Since all the circles have a similar shape. They may have different radius but the shape of all circles is the same.

Therefore, all circles are similar.

Q1 (2) Fill in the blanks using the correct word given in brackets: All squares are ______. (similar, congruent)

Answer:

All squares are similar.

Since all the squares have a similar shape. They may have a different side but the shape of all square is the same.

Therefore, all squares are similar.

Q1 (3) Fill in the blanks using the correct word given in brackets: All ______triangles are similar. (isosceles, equilateral)

### Answer:

All equilateral triangles are similar.

Since all the equilateral triangles have a similar shape. They may have different sides but the shape of all equilateral triangles is the same.

Therefore, all equilateral triangles are similar.

Q1 (4) Fill in the blanks using the correct word given in brackets : (iv) Two polygons of the same number of sides are similar if(a) there corresponding angles are _________and (b) their corresponding sides are________. (equal, proportional)

### Answer:

Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are proportional.

Thus, (a) equal

(b) proportional

Q2 (1) Give two different examples of a pair of similar figures.

### Answer:

The two different examples of a pair of similar figures are :

1. Two circles with different radii.

2. Two rectangles with different breadth and length.

Q2 (2) Give two different examples of a pair of non-similar figures.

### Answer:

The two different examples of a pair of non-similar figures are :

1.Rectangle and circle

2. A circle and a triangle.

Q3 State whether the following quadrilaterals are similar or not:

### Answer:

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

**Class 10 Maths Chapter 6 Triangles Excercise: 6.2**

Q1 In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

### Answer:

(i)

Let EC be x

Given: DE || BC

By using the proportionality theorem, we get

(ii)

Let AD be x

Given: DE || BC

By using the proportionality theorem, we get

Q2 (1) E and F are points on the sides PQ and PR respectively of a PQR. For each of the following cases, state whether EF || QR: PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

### Answer:

(i)

Given :

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

and

We have

Hence, EF is not parallel to QR.

Q2 (2) E and F are points on the sides PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF || QR : PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

### Answer:

(ii)

Given :

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

and

We have

Hence, EF is parallel to QR.

Q2 (3) E and F are points on the sides PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF || QR : PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

### Answer:

(iii)

Given :

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

and

We have

Hence, EF is parallel to QR.

Q3 In Fig. 6.18, if LM || CB and LN || CD, prove that

Answer:

Given : LM || CB and LN || CD

To prove :

Since , LM || CB so we have

Also, LN || CD

From equation 1 and 2, we have

Hence proved.

Q4 In Fig. 6.19, DE || AC and DF || AE. Prove that BF / FE = BE / EC

### Answer:

Given : DE || AC and DF || AE.

To prove :

Since , DE || AC so we have

Also,DF || AE

From equation 1 and 2, we have

Hence proved.

Q5 In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

### Answer:

Given : DE || OQ and DF || OR.

To prove EF || QR.

Since DE || OQ so we have

Also, DF || OR

From equation 1 and 2, we have

Thus, EF || QR. (converse of basic proportionality theorem)

Hence proved.

Q6 In Fig. 6.21, A, B, and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

### Answer:

Given : AB || PQ and AC || PR

To prove: BC || QR

Since, AB || PQ so we have

Also, AC || PR

From equation 1 and 2, we have

Therefore, BC || QR. (converse basic proportionality theorem)

Hence proved.

Q7 Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

### Answer:

Let PQ is a line passing through the midpoint of line AB and parallel to line BC intersecting line AC at point Q.

i.e. and .

Using basic proportionality theorem, we have

Since

Q is the midpoint of AC.

Q8 Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

### Answer:

Let P is the midpoint of line AB and Q is the midpoint of line AC.

PQ is the line joining midpoints P and Q of line AB and AC, respectively.

i.e. and .

we have,

From equation 1 and 2, we get

By basic proportionality theorem, we have

Q9 ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show

that

### Answer:

Draw a line EF passing through point O such that

To prove :

In , we have

So, by using basic proportionality theorem,

In , we have

So, by using basic proportionality theorem,

Using equation 1 and 2, we get

Hence proved.

Q10 The diagonals of a quadrilateral ABCD intersect each other at point O such that Show that ABCD is a trapezium.

### Answer:

Draw a line EF passing through point O such that

Given :

In , we have

So, by using basic proportionality theorem,

However, its is given that

Using equation 1 and 2 , we get

(By basic proportionality theorem)

Therefore, ABCD is a trapezium.

**Class 10 Maths Chapter 6 Triangles Excercise: 6.3**

Q1 State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

### Answer:

(i)

(By AAA)

So ,

(ii) As corresponding sides of both triangles are proportional.

(By SSS)

(iii) Given triangles are not similar because corresponding sides are not proportional.

(iv) by SAS similarity criteria.

(v) Given triangles are not similar because the corresponding angle is not contained by two corresponding sides

(vi) In , we know that

In , we know that

( By AAA)

### Answer:

Given : , and

(DOB is a straight line)

In

Since , , so

( Corresponding angles are equal in similar triangles).

Q3 Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that

### Answer:

In , we have

( Alternate interior angles as )

( Alternate interior angles as )

( Vertically opposite angles are equal)

( By AAA)

( corresponding sides are equal)

Hence proved.

Q4 In Fig. 6.36, and . Show that

### Answer:

Given : and

To prove :

In ,

(Given)

In ,

(Common)

( By SAS)

Q5 S and T are points on sides PR and QR of PQR such that P = RTS. Show that RPQ ~ RTS.

### Answer:

Given : P = RTS

To prove RPQ ~ RTS.

In RPQ and RTS,

P = RTS (Given )

R = R (common)

RPQ ~ RTS. ( By AA)

Q6 In Fig. 6.37, if ABE ACD, show that ADE ~ ABC.

Answer:

Given :

To prove ADE ~ ABC.

Since

( By CPCT)

(By CPCT)

In ADE and ABC,

( Common)

and

( and )

Therefore, ADE ~ ABC. ( By SAS criteria)

Q7 (1) In Fig. 6.38, altitudes AD and CE of intersect each other at the point P. Show that:

### Answer:

To prove :

In ,

( Both angles are right angle)

(Vertically opposite angles )

( By AA criterion)

Q7 (2) In Fig. 6.38, altitudes AD and CE of intersect each other at the point P. Show that:

### Answer:

To prove :

In ,

( Both angles are right angle)

(Common )

( By AA criterion)

Q7 (3) In Fig. 6.38, altitudes AD and CE of intersect each other at the point P. Show that:

### Answer:

To prove :

In ,

( Both angles are right angle)

(Common )

( By AA criterion)

Q7 (4) In Fig. 6.38, altitudes AD and CE of intersect each other at the point P. Show that:

### Answer:

To prove :

In ,

( Both angles are right angle)

(Common )

( By AA criterion)

Q8 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that

### Answer:

To prove :

In ,

( Opposite angles of a parallelogram are equal)

( Alternate angles of AE||BC)

( By AA criterion )

Q9 (1) In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

### Answer:

To prove :

In ,

( Each )

( common)

( By AA criterion )

Q9 In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that

### Answer:

To prove :

In ,

( Each )

( common)

( By AA criterion )

( corresponding parts of similar triangles )

Hence proved.

Q10 (1) CD and GH are respectively the bisectors of and such that D and H lie on sides AB and FE of respectively. If , show that:

### Answer:

To prove :

Given :

( CD and GH are bisectors of equal angles)

( CD and GH are bisectors of equal angles)

In

( proved above)

( proved above)

( By AA criterion)

Hence proved.

Q 10 (2) CD and GH are respectively the bisectors of such that D and H lie on sides AB and FE of respectively. If , show that:

### Answer:

To prove :

Given :

In ,

( CD and GH are bisectors of equal angles)

( )

( By AA criterion )

Q10 (3) CD and GH are respectively the bisectors of such that D and H lie on sides AB and FE of respectively. If , show that:

### Answer:

To prove :

Given :

In ,

( CD and GH are bisectors of equal angles)

( )

( By AA criterion )

Q11 In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If and , prove that

### Answer:

To prove :

Given: ABC is an isosceles triangle.

In ,

( )

( Each )

( By AA criterion)

Q12 Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of (see Fig. 6.41). Show that

### Answer:

AD and PM are medians of triangles. So,

Given :

In

(SSS similarity)

( Corresponding angles of similar triangles )

In

(proved above)

Therefore, . ( SAS similarity)

Q13 D is a point on the side BC of a triangle ABC such that . Show that

### Answer:

In,

( given )

(common )

( By AA rule)

( corresponding sides of similar triangles )

Q14 Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that

### Answer:

(given)

Produce AD and PM to E and L such that AD=DE and PM=DE. Now,

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

AD and PM are medians of a triangle, therefore

QM=MR and BD=DC

AD = DE (By construction)

PM=ML (By construction)

So, diagonals of ABEC bisecting each other at D,so ABEC is a parallelogram.

Similarly, PQLR is also a parallelogram.

Therefore, AC=BE ,AB=EC and PR=QL,PQ=LR

(Given )

(SSS similarity)

……………….1 (Corresponding angles of similar triangles)

Similarity,

……………………2

Adding equation 1 and 2,

……………………….3

In

( Given )

( From above equation 3)

( SAS similarity)

Q15 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.

### Answer:

CD = pole

AB = tower

Shadow of pole = DF

Shadow of tower = BE

In

( Each )

(Angle of sun at same place )

(AA similarity)

cm

Hence, the height of the tower is 42 cm.

Q16 If AD and PM are medians of triangles ABC and PQR, respectively where , prove that

Answer:

( Given )

…………… ….1( corresponding sides of similar triangles )

………………………………2

AD and PM are medians of triangle.So,

……………………………………3

From equation 1 and 3, we have

………………………………………………………….4

In

(From equation 2)

(From equation 4)

(SAS similarity)

**Class 10 Maths Chapter 6 Triangles Excercise:6.4**

Q1 Let and their areas be, respectively, 64 and 121 . If EF = 15.4 cm, find BC.

### Answer:

( Given )

ar(ABC) = 64 and ar(DEF)=121 .

EF = 15.4 cm (Given )

Q2 Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

### Answer:

Given: Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.

AB = 2 CD ( Given )

In

(vertically opposite angles )

(Alternate angles)

(Alternate angles)

(AAA similarity)

Q3 In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

Answer:

Let DM and AP be perpendicular on BC.

In

(Each )

(Vertically opposite angles)

(AA similarity)

Since

Q4 If the areas of two similar triangles are equal, prove that they are congruent.

### Answer:

Let , therefore,

(Given )

(SSS )

Q5 D, E, and F are respectively the mid-points of sides AB, BC and CA of . Find the ratio of the areas of

### Answer:

D, E, and F are respectively the mid-points of sides AB, BC and CA of . ( Given )

and DE||AC

In ,

(corresponding angles )

(corresponding angles )

(By AA)

Let be x.

Similarly,

and

Q6 Proves that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

### Answer:

Let AD and PS be medians of both similar triangles.

Purring these value in 1,

In

(proved above)

(proved above)

(SAS )

Therefore,

From 1 and 4, we get

Q7 Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

### Answer:

Let ABCD be a square of side units.

Therefore, diagonal =

Triangles form on the side and diagonal are ABE and DEF, respectively.

Length of each side of triangle ABE = a units

Length of each side of triangle DEF = units

Both the triangles are equilateral triangles with each angle of .

( By AAA)

Using area theorem,

Q8 Tick the correct answer and justify : ABC and BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangles ABC and BDE is

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

### Answer:

Given: ABC and BDE are two equilateral triangles such that D is the mid-point of BC.

All angles of the triangle are .

ABC BDE (By AAA)

Let AB=BC=CA = x

then EB=BD=ED=

Option C is correct.

Q9 Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio

(A) 2 : 3 (B) 4: 9 (C) 81: 16 (D) 16: 81

### Answer:

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

Let triangles be ABC and DEF.

We know that

Option D is correct.

**Class 10 Maths Chapter 6 Triangles Excercise:**** 6.5**

Q1 (1) Sides of triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse. 7 cm, 24 cm, 25 cm

### Answer:

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 7 cm, 24 cm

By Pythagoras theorem,

= given third side.

Hence, it is the right triangle with h=25 cm.

Q1 (2) Sides of triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse. 3 cm, 8 cm, 6 cm

### Answer:

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 3 cm, 6 cm

By Pythagoras theorem,

Hence, it is not the right triangle.

Q1 (3) Sides of triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse. 50 cm, 80 cm, 100 cm

### Answer:

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 50 cm, 80 cm

By Pythagoras theorem,

Hence, it is not a right triangle.

Q1 (4) 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. 13 cm, 12 cm, 5 cm

### Answer:

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 5cm, 12 cm

By Pythagoras theorem,

= given third side.

Hence, it is a right triangle with h=13 cm.

Q2 PQR is a triangle right angled at P and M is a point on QR such that . Show that

### Answer:

Let be x

In ,

Similarly,

In ,

In

(By AAA)

Hence proved.

Q3 (1) In Fig. 6.53, ABD is a triangle right angled at A and AC BD. Show that

### Answer:

In

(common )

(By AA)

, hence prooved .

Q3 (2) In Fig. 6.53, ABD is a triangle right angled at A and AC BD. Show that

Answer:

Let be x

In ,

Similarly,

In ,

In

( Each right angle)

(By AAA)

Hence proved

Q3 (3) In Fig. 6.53, ABD is a triangle right angled at A and AC BD. Show that

### Answer:

In

(common )

(By AA)

Hence proved.

Q4 ABC is an isosceles triangle right angled at C. Prove that

### Answer:

Given: ABC is an isosceles triangle right angled at C.

Let AC=BC

In ABC,

By Pythagoras theorem

(AC=BC)

Hence proved.

Q5 ABC is an isosceles triangle with AC = BC. If , prove that ABC is a right triangle.

### Answer:

Given: ABC is an isosceles triangle with AC=BC.

In ABC,

(Given )

(AC=BC)

These sides satisfy Pythagoras theorem so ABC is a right-angled triangle.

Hence proved.

Q6 ABC is an equilateral triangle of side 2a. Find each of its altitudes.

### Answer:

Given: ABC is an equilateral triangle of side 2a.

AB=BC=AC=2a

AD is perpendicular to BC.

We know that the altitude of an equilateral triangle bisects the opposite side.

So, BD=CD=a

In ADB,

By Pythagoras theorem,

The length of each altitude is .

Q7 Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

### Answer:

In AOB, by Pythagoras theorem,

In BOC, by Pythagoras theorem,

In COD, by Pythagoras theorem,

In AOD, by Pythagoras theorem,

Adding equation 1,2,3,4,we get

(AO=CO and BO=DO)

Hence proved .

Q8 (1) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD BC, OE AC and OF AB. Show that

### Answer:

Join AO, BO, CO

In AOF, by Pythagoras theorem,

In BOD, by Pythagoras theorem,

In COE, by Pythagoras theorem,

Adding equation 1,2,3,we get

Hence proved

Q8 (2) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD BC, OE AC and OF AB.

### Answer:

Join AO, BO, CO

In AOF, by Pythagoras theorem,

In BOD, by Pythagoras theorem,

In COE, by Pythagoras theorem,

Adding equation 1,2,3,we get

Q9 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.

### Answer:

OA is a wall and AB is a ladder.

In AOB, by Pythagoras theorem

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

Q10 A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

### Answer:

OB is a pole.

In AOB, by Pythagoras theorem

Hence, the distance of the stack from the base of the pole is m.

Q11 An aeroplane leaves an 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 1/2 hours?

### Answer:

Distance travelled by the first aeroplane due north in hours.

Distance travelled by second aeroplane due west in hours.

OA and OB are the distance travelled.

By Pythagoras theorem,

Thus, the distance between the two planes is .

Q12 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 top

### Answer:

Let AB and CD be poles of heights 6 m and 11 m respectively.

CP=11-6=5 m and AP= 12 m

In APC,

By Pythagoras theorem,

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

Q13 D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that

### Answer:

In ACE, by Pythagoras theorem,

In BCD, by Pythagoras theorem,

From 1 and 2, we get

In CDE, by Pythagoras theorem,

In ABC, by Pythagoras theorem,

From 3,4,5 we get

Q14 The perpendicular from A on side BC of a ABC intersects BC at D such that DB = 3 CD (see Fig. 6.55). Prove that

### Answer:

In ACD, by Pythagoras theorem,

In ABD, by Pythagoras theorem,

From 1 and 2, we get

Given : 3DC=DB, so

From 3 and 4, we get

Hence proved.

Q15 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that

### Answer:

Given: An equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC.

To prove :

Let AB=BC=CA=a

Draw an altitude AE on BC.

So,

In AEB, by Pythagoras theorem

Given : BD = 1/3 BC.

In ADE, by Pythagoras theorem,

Q16 In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

### Answer:

Given: An equilateral triangle ABC.

Let AB=BC=CA=a

Draw an altitude AE on BC.

So,

In AEB, by Pythagoras theorem

Q17 Tick the correct answer and justify : In AB = cm, AC = 12 cm and BC = 6 cm.

The angle B is :

(A) 120°

(B) 60°

(C) 90°

(D) 45°

### Answer:

In AB = cm, AC = 12 cm and BC = 6 cm.

It satisfies the Pythagoras theorem.

Hence, ABC is a right-angled triangle and right-angled at B.

Option C is correct.

## NCERT solutions for class 10 maths chapter 6 Triangles Excercise: 6.6

Q1 In Fig. 6.56, PS is the bisector of . Prove that

### Answer:

A line RT is drawn parallel to SP which intersect QP produced at T.

Given: PS is the bisector of .

By construction,

(as PS||TR)

(as PS||TR)

From the above equations, we get

By construction, PS||TR

In QTR, by Thales theorem,

Hence proved.

Q2 In Fig. 6.57, D is a point on hypotenuse AC of triangle ABC, such that BD AC, DM BC and DN AB. Prove that :

### Answer:

Join BD

Given : D is a point on hypotenuse AC of D ABC, such that BD AC, DM BC and DN AB.Also DN || BC, DM||NB

In CDM,

In DMB,

From equation 1 and 2, we get

From equation 1 and 3, we get

In

(By AA)

(BM=DN)

Hence proved

Q2 (2) In Fig. 6.57, D is a point on hypotenuse AC of D ABC, such that BD AC, DM BC and DN AB. Prove that:

Answer:

In DBN,

In DAN,

BD AC,

From equation 1 and 3, we get

From equation 2 and 3, we get

In

(By AA)

(NB=DM)

Hence proved.

Q3 In Fig. 6.58, ABC is a triangle in which ABC > 90° and AD CB produced. Prove that

### Answer:

In ADB, by Pythagoras theorem

In ACD, by Pythagoras theorem

(From 1)

Q4 In Fig. 6.59, ABC is a triangle in which ABC < 90° and AD BC. Prove that

### Answer:

In ADB, by Pythagoras theorem

In ACD, by Pythagoras theorem

(From 1)

Q5 (1) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that :

### Answer:

Given: AD is a median of a triangle ABC and AM BC.

In AMD, by Pythagoras theorem

In AMC, by Pythagoras theorem

(From 1)

(BC=2 DC)

Q5 (2) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that :

### Answer:

In ABM, by Pythagoras theorem

(BC=2 BD)

Q5 (3) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that:

### Answer:

In ABM, by Pythagoras theorem

In AMC, by Pythagoras theorem

…………………………….2

Adding equation 1 and 2,

Q6 Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

### Answer:

In parallelogram ABCD, AF and DE are altitudes drawn on DC and produced BA.

In DEA, by Pythagoras theorem

In DEB, by Pythagoras theorem

………………………………2

In ADF, by Pythagoras theorem

In AFC, by Pythagoras theorem

Since ABCD is a parallelogram.

SO, AB=CD and BC=AD

In

(AE||DF)

AD=AD (common)

(ASA rule)

Adding 2 and, we get

(From 4 and 6)

\

Q7 (1) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that :

### Answer:

Join BC

In

( vertically opposite angle)

(Angles in the same segment)

(By AA)

Q7 (2) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that :

### Answer:

Join BC

In

( vertically opposite angle)

(Angles in the same segment)

(By AA)

(Corresponding sides of similar triangles are proportional)

Q8 (1) In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

### Answer:

In

(Common)

(Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

So, ( By AA rule)

Q8 (2) In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that PA. PB = PC. PD

### Answer:

In

(Common)

(Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

So, ( By AA rule)

24440 (Corresponding sides of similar triangles are proportional)

Q9 In Fig. 6.63, D is a point on side BC of D ABC such that Prove that AD is the bisector of BAC.

Answer:

Produce BA to P, such that AP=AC and join P to C.

(Given )

Using converse of Thales theorem,

AD||PC (Corresponding angles)

(Alternate angles)

By construction,

AP=AC

From equation 1,2,3, we get

Thus, AD bisects angle BAC.

Q10 Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

### Answer:

Let AB = 1.8 m

BC is a horizontal distance between fly to the tip of the rod.

Then, the length of the string is AC.

In ABC, using Pythagoras theorem

Hence, the length of the string which is out is 3m.

If she pulls in the string at the rate of 5cm/s, then the distance travelled by fly in 12 seconds.

=

Let D be the position of fly after 12 seconds.

Hence, AD is the length of the string that is out after 12 seconds.

Length of string pulled in by nazim=AD=AC-12

=3-0.6=2.4 m

In ADB,

Horizontal distance travelled by fly = BD+1.2 m

=1.587+1.2=2.787 m

= 2.79 m

## Topics of NCERT Class 10 Maths solutions chapter 6

- Similarity of triangles
- Theorems based on similar triangles
- Areas of similar triangles
- Theorems related to Trapezium
- Pythagoras theorem