NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 6 **Triangles Class 10 NCERT Solutions Ex 6.6.**

- Triangles Class 10 Ex 6.1
- Triangles Class 10 Ex 6.2
- Triangles Class 10 Ex 6.3
- Triangles Class 10 Ex 6.4
- Triangles Class 10 Ex 6.5

Board | CBSE |

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 6 |

Chapter Name | Triangles |

Exercise | Ex 6.6 |

Number of Questions Solved | 10 |

Category | NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

NCERT Solutions for Class 10 Maths

**Question 1. **In the given figure, PS is the bisector of ∠QPR of ∆PQR. Prove that

**Solution:**

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

**Question 2.**In the given figure, D is a point on hypotenuse AC of ∆ABC, DM ⊥ BC and DN ⊥ AB, Prove that:

(i). DM^{2} = DN.MC

(ii). DN^{2} = DM.AN

**Solution:**

(i). Let us join DB.

DN || CB

DM || AB

So, DN = MB

DM = NB

The condition to be proved is the case when DNBM is a square or D is the midpoint of side AC.

Then ∠CDB = ∠ADB = 90°

∠2 + ∠3 = 90° (1)

In ∆CDM

∠ 1 + ∠ 2 + ∠ DMC = 180°

∠ 1 + ∠ 2 = 90° (2)

In ∆DMB

∠ 3 + ∠ DMB + ∠ 4 = 180°

∠ 3 + ∠ 4 = 90° (3)

From equation (1) and (2)

∠ 1 = ∠ 3

From equation (1) and (3)

∠ 2 = ∠ 4

∆BDM ~ ∆ DCM

(ii). Similarly in ∆DBN

∠ 4 + ∠ 3 = 90° (4)

In ∆DAN

∠ 5 + ∠ 6 = 90° (5)

In ∆DAB

∠4 + ∠ 5 = 90° (6)

From equation (4) and (6)

∠3 = ∠ 5

From equation (5) and (6)

∠4 = ∠ 6

∆DNA ~ ∆BND

**Question 3. **In the given figure, ABC is a triangle in which ∠ ABC> 90° and AD ⊥ CB produced.

Prove that AC^{2} = AB^{2} + BC^{2} + 2BC . BD.

**Solution:**

In ∆ADB applying Pythagoras theorem

AB

^{2}= AD

^{2}+ DB

^{2}(1)

In ∆ACD applying Pythagoras theorem

AC

^{2}= AD

^{2}+ DC

^{2}

AC

^{2}= AD

^{2}+ (DB + BC)

^{2}

AC

^{2}= AD

^{2}+ DB

^{2}+ BC

^{2}+ 2DB x BC

Now using equation (1)

AC

^{2}= AB

^{2}+ BC

^{2}+ 2BC . BD

**Question 4. **In the given figure, ABC is a triangle in which ∆ ABC < 90° and AD ⊥ BC. Prove that AC^{2} = AB^{2} + BC^{2} – 2BC.BD.

**Solution:**

In ∆ADB applying Pythagoras theorem

AD

^{2}+ DB

^{2}= AB

^{2}

AD

^{2}= AB

^{2}– DB

^{2}(1)

In ∆ ADC applying Pythagoras theorem

AD

^{2}+ DC

^{2}= AC

^{2}(2)

Now using equation (1)

AB

^{2}– BD

^{2}+ DC

^{2}= AC

^{2}

AB

^{2}– BD

^{2}+ (BC – BD)

^{2}= AC

^{2}

AC

^{2}= AB

^{2}– BD

^{2}+ BC

^{2}+ BD

^{2}– 2BC. BD

= AB

^{2}+ BC

^{2}– 2BC. BD

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

**Solution:**

(i). In ∆AMD

AM

^{2}+ MD

^{2}= AD

^{2 }(1)

In ∆AMC

AM

^{2}+ MC

^{2}= AC

^{2}(2)

AM

^{2}+ (MD + DC)

^{2}= AC

^{2}

(AM

^{2}+ MD

^{2}) + DC

^{2}+ 2MD.DC = AC

^{2}

Using equation (1) we may get

AD

^{2}+ DC

^{2}+ 2MD.DC = AC

^{2}

AB

^{2}= AM

^{2}+ MB

^{2}

= (AD

^{2}– DM

^{2}) + MB

^{2}

= (AD

^{2}– DM

^{2}) + (BD – MD)

^{2}

= AD

^{2}– DM

^{2}+ BD

^{2}+ MD

^{2 – }2BD.MD

= AD

^{2}+ BD

^{2 – }2BD.MD

(iii). In ∆AMB

AM

^{2}+ MB

^{2}= AB

^{2}(1)

In ∆ AMC

AM

^{2}+ MC

^{2}= AC

^{2}(2)

Adding equation (1) and (2)

2AM

^{2}+ MB

^{2}+ MC

^{2}= AB

^{2}+ AC

^{2}

2AM

^{2}+ (BD – DM)

^{2}+ (MD + DC)

^{2}= AB

^{2}+ AC

^{2}

2AM

^{2}+BD

^{2}+ DM

^{2}– 2BD.DM + MD

^{2}+ DC

^{2}+ 2MD.DC = AB

^{2}+ AC

^{2}

2AM

^{2}+ 2MD

^{2}+ BD

^{2}+ DC

^{2}+ 2MD (-BD + DC) = AB

^{2}+ AC

^{2}

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

**Solution:**

Let ABCD be a parallelogram

Let us draw perpendicular DE on extended side AB and AF on side DC.

In ∆DEA

DE^{2} + EA^{2} = DA^{2} (i)

In ∆DEB

DE^{2} + EB^{2} = DB^{2}

DE^{2} + (EA + AB)^{2} = DB^{2}

(DE^{2} + EA^{2}) + AB^{2} + 2EA. AB = DB^{2}

DA^{2} + AB^{2} + 2EA.AB = DB^{2} (ii)

In ∆ADF

AD^{2} = AF^{2} + FD^{2}

In ∆ AFC

AC^{2} = AF^{2} + FC^{2}

= AF^{2} + (DC – FD)^{2}

= AF^{2} + DC^{2} + FD^{2} – 2DC – FD

= (AF^{2} + FD^{2}) + DC^{2} – 2DC . FD

AC^{2} = AD^{2} + DC^{2} – 2DC FD (iii)

Since ABCD is a parallelogram

AB = CD (iii)

And BC = AD (iv)

In ∆DEA and ∆ADF

∠ DEA = ∠ AFD

∠ EAD = ∠ FDA (EA || DF)

∠ EDA = ∠ FAD (AF || ED)

AD is common in both triangles.

Since respective angles are same and respective sides are same

∆DEA ≅ ∆AFD

So EA = DF (v)

Adding equation (ii) and (iii)

DA^{2} + AB^{2} + 2EA.AB + AD^{2} + DC^{2} – 2DC.FD = DB^{2} + AC^{2}

DA^{2} + AB^{2} + AD^{2 }+ DC^{2} + 2EA.AB – 2DC.FD = DB^{2} + AC^{2}

BC^{2} + AB^{2} + AD^{2} + DC^{2} + 2EA.AB-2AB.EA = DB^{2} + AC^{2}

AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}

**Question 7.** In the given figure, two chords AB and CD intersect each other at the point P.

prove that: (i) ∆APC ~ ∆DPB (ii) AP.PB = CP.DP

**Solution:**

Let us join CB

(i) In ∆APC and ∆DPB

∠APC = ∠DPB {Vertically opposite angles}

∠CAP = ∠BDP {Angles in same segment for chord CB}

∆APC ~ ∆DPB {BY AA similarly criterion}

(ii) We know that corresponding sides of similar triangles are proportional

**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

(i) ∆PAC ~ ∆PDB (ii) PA.PB = PC.PD

**Solution: **(i) In ∆PAC and ∆PDB

∠ P = ∠ P (common)

∠ PAC = ∠ PDB (exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

∠ PCA = ∠PBD

∆PAC ~ ∆PDB

(ii) We know that corresponding sides of similar triangles are proportional.

**Question 9. **In the given figure, D is a point on side BC of ∆ ABC such that \(\frac { BD }{ CD } =\frac { AB }{ AC }\) Prove that AD is the bisector of ∠ BAC.

**Solution:**

In ∆ DBA and ∆DCA

\(\frac { BD }{ CD } =\frac { AB }{ AC }\) (given)

AD = AD (common)

So, ∆DBA ~ ∆DCA (By SSS)

Now, corresponding angles of similar triangles will be equal.

∠ BAD = ∠ CAD

AD is angle 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 (see Fig. If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

**Solution:**

Let AB be the height of tip of fishing rod from water surface. Let BC be the horizontal distance of fly from the tip of fishing rod.

Then, AC is the length of string.

AC can be found by applying Pythagoras theorem in ∆ABC

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (1.8)^{2} + (2.4)^{2}

AC^{2} = 3.24 + 5.76

AC^{2} = 9.00

AC = √9 = 3

Thus, length of string out is 3 m.

Now, she pulls string at rate of 5 cm per second.

So, string pulled in 12 seconds = 12 x 5 = 60 cm = 0.6 m

Let after 12 second Fly be at point D.

Length of string out after 12 second is AD

AD = AC – string pulled by Nazima in 12 seconds

= 3.00 – 0.6

= 2.4

In ∆ ADB

AB^{2} + BD^{2} = AD^{2}

(1.8)^{2} + BD^{2} = (2.4)^{2}

BD^{2} = 5.76 – 3.24 = 2.52

BD = 1.587

Horizontal distance of fly = BD + 1.2

= 1.587 + 1.2

= 2.787

= 2.79 m.

We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6, drop a comment below and we will get back to you at the earliest.