**NCERT Solution For Class 10, Maths, Chapter 6 Triangles, Exercise 6.6 (Optional)** has some additional questions to discuss which are very important for exam. once, earlier exercises are well learnt then practice this ex.6.6, chapter 6 of class 10.

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Toggle## Class 10, Maths, Chapter 6, Exercise 6.6 (Optional) Solutions

**Q.1. In the given Figure, PS is the bisector of ****∠****QPR of Δ PQR. Prove that $\frac{QS}{SR}=\frac{PQ}{PR}$**

**Ans:**

**Q.2. In Fig. 6.57, D is a point on hypotenuse AC of Δ ABC, such that BD ****⊥**** AC, DM ****⊥**** BC and DN ****⊥**** AB. Prove that :**

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

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

** ****Ans:**

**Given**: We have, DM ⊥ BC and DN ⊥ AB

∴ DN||BC and DM||AB

Hence, quadrilateral BMDN is a rectangle.

DN = MB and DM = NB

**To prove: ****(i) DM ^{2} = DN . MC**

**Proof: **BD ⊥ AC [∠CDB = 90^{0}]

∠2 + ∠3 = 90^{0} ……… (i)

In ΔCDM, we have

∠1 + ∠2 + ∠M = 180^{0}

∠1 + ∠2 + 90^{0} = 180^{0}

∴ ∠1 + ∠2 = 90^{0} ……….. (ii)

In ΔDMB, we have

∠3 + ∠6 + ∠M = 180^{0}

∴ ∠3 + ∠6 = 90^{0 }……….. (iii)

From (i) and (ii)

∠1 = ∠3

From (i) and (iii)

∠2 = ∠6

In ΔDCM and ΔBDM, we have

∠1 = ∠3 and ∠2 = ∠6 [by AA]

∴ Δ DCM~ΔBDM

** $\frac{DM}{MB}=\frac{CM}{DM}$**

DM^{2} = MB × MC [∵MB = DN]

DM^{2} = DN × MC **(Hence proved)**

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

**Proof: **In ΔDBN, we have, ∠N = 90^{0}

∠4 + ∠7 + ∠N = 180^{0}

∠4 + ∠7 + 90^{0} = 180^{0}

∴ ∠4 + ∠7 = 90^{0} ………. (iv)

In ΔDAN, ∠N = 90^{0}

∠5 + ∠8 + ∠N = 180^{0}

∴ ∠5 + ∠8 = 90^{0 }………. (v)

In ΔADB, ∠D = 90^{0}

∠4 + ∠5 + ∠D = 180^{0}

∴ ∠4 + ∠5 = 90^{0 }………. (vi)

From (iv) and (vi)

∠7 = ∠5

From (v) and (vi)

∠4 = ∠8

In ΔDNA and ΔBND, we have

∠5 = ∠7 and ∠4 = ∠8 [by AA]

∴ Δ DNA~ΔBND

** $\frac{AN}{DN}=\frac{DN}{NB}$**

DN^{2} = AN × NB [∵NB = DM]

DN^{2} = AN × DM **(Hence proved)**

**Q.3. In Figure, ABC is a triangle in which ****∠****ABC > 90° and AD ****⊥**** CB produced. Prove that AC ^{2} = AB^{2} + BC^{2} + 2 BC . BD.**

** **

**Ans:**

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

** **

**Ans:**

**Q.5. In Figure, AD is a median of a triangle ABC and AM ****⊥**** BC. Prove that :**

** (i) AC ^{2} = AD^{2} + BC. DM + ${{\left( \frac{BC}{2} \right)}^{2}}$**

** (ii) AB ^{2} = AD^{2} – BC. DM + ${{\left( \frac{BC}{2} \right)}^{2}}$**

** (iii) AC ^{2} + AB^{2 }= 2 AD^{2} + $\frac{1}{2}B{{C}^{2}}$**

**Ans:**

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

**Ans:**

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

**(i) Δ APC ~ Δ DPB**

**(ii) AP . PB = CP . DP**

** **

** ****Ans:**

**Q.8. In Figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that**

**(i) Δ PAC ~ Δ PDB **

**(ii) PA . PB = PC . PD**

** **** **

**Ans:**

**Q.9. In Fig. 6.63, D is a point on side BC of Δ ABC such that $\frac{BD}{CD}=\frac{AB}{AC}$. Prove that AD is the bisector of ****∠**** BAC.**

**Ans:**

**Q.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?**

**Ans:**