## NCERT10th MathematicsExercise 6.6Topic: 6 Triangles

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## Click here for ⇨ NCERT-10-6-Triangles - Ex- 6.5

### EXERCISE 6.6

**Q 1. In the following fig., PS is the bisector of**∠

**QPR of ∆ PQR.**

Prove that QS/SR = PQ/PR

### Solution:1) Let us draw a line parallel to PS, through R, which intersects QP at T 2) So, PS || RT.3) As PS is the bisector of ∠ QPR,

## ∠ QPS = ∠ SPR ------- equation 1

4) As PS || RT,

∠ SPR = ∠ PRT (Alertnate angles) ------- equation 2

5) From equations 1 and 2 we have,

∠ QPS = ∠ PRT ------- equation 3

6) As PS || RT,

∠ QPS = ∠ PTR (Corresponding angles) ------- equation 4

7) From equations 3 and 4 we have,

∠ PRT = ∠ PTR ------- equation 5

8) From equation 5 we have,

(PR) = (PT) ------- equation 6

9) In ∆ QTR,

(QS)/(SR) = (QP)/(PT) (By BPT) ------- equation 7

10) From equations 5 and 6 we have,

(QS)/(SR) = (QP)/(PR), hence proved.

**Q 2. In the following fig., D is a point on hypotenuse AC of**

**∆**

**ABC, such that**

BD ⟂ AC, DM ⟂ BC,and DN ⟂ AB. Prove that :

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

a) ∠ DCM + ∠ CDM = 90^{0}------ equation 1(sum of the remaining angles of right-angled ∆)b) ∠ CDM + ∠ BDM = 90^{0}(BD ⟂ AC) ------ equation 2

2) From equations 1 and 2 we have,

c) ∠ DCM + ∠ CDM = ∠ CDM + ∠ BDM

d) ∠ DCM = ∠ BDM ------ equation 3

### 3) In ∆ DMB,

3) In ∆ DMB,

a) ∠ DBM + ∠ BDM = 90^{0}------ equation 4

(sum of the remaining angles of right-angled ∆)

b) ∠ BDM + ∠ CDM = 90^{0}(BD ⟂ AC, as given in equation 2) equation 5

4) From equations 4 and 5 we have,

c) ∠ DBM + ∠ BDM = ∠ BDM + ∠ CDM

d) ∠ DBM = ∠ CDM ------ equation 6

### 5) From equations 3 and 6 and by AA similarity test,

5) From equations 3 and 6 and by AA similarity test,

a) ∆ DBM ~ ∆ CDM --------- equation 7

### 6) As corresponding sides of similar triangles are proportional, we have,a) (DM)/(MC) = (BM)/(DM)

b) (DM) x (DM) = (BM) x (MC)

c) (DM)^{2} = (BM) x (MC)

7) As □ DNBM is a rectangle, BM = DN, so, we have,a) (DM)^{2} = (DN) x (MC), hence proved.

a) (DM)/(MC) = (BM)/(DM)

^{2}= (BM) x (MC)

^{2}= (DN) x (MC), hence proved.

### 8) In ∆ AND,

8) In ∆ AND,

a) ∠ DAN + ∠ ADN = 90^{0}------ equation 8

(sum of the remaining angles of right-angled ∆)

### b) ∠ ADN + ∠ BDN = 90^{0} (BD ⟂ AC) ------ equation 9

b) ∠ ADN + ∠ BDN = 90

^{0}(BD ⟂ AC) ------ equation 9

9) From equations 8 and 9 we have,

### c) ∠ DAN + ∠ ADN = ∠ ADN + ∠ BDN

c) ∠ DAN + ∠ ADN = ∠ ADN + ∠ BDN

d) ∠ DAN = ∠ BDN ------ equation 10

### 10) In ∆ DNB,

10) In ∆ DNB,

a) ∠ DBN + ∠ BDN = 90^{0}------ equation 11

(sum of the remaining angles of right-angled ∆)

### b) ∠ BDN + ∠ ADN = 90^{0} (BD ⟂ AC, as given in equation 9) equation 12

b) ∠ BDN + ∠ ADN = 90

^{0}(BD ⟂ AC, as given in equation 9) equation 12### 11) From equations 11 and 12 we have,

### c) ∠ DBN + ∠ BDN = ∠ BDN + ∠ ADN

c) ∠ DBN + ∠ BDN = ∠ BDN + ∠ ADN

d) ∠ DBN = ∠ ADN ------ equation 13

### 12) From equations 10 and 13 and by AA similarity test,

12) From equations 10 and 13 and by AA similarity test,

a) ∆ DBN ~ ∆ ADN --------- equation 14

### 13) As corresponding sides of similar triangles are proportional, we have,a) (DN)/(AN) = (BN)/(DN)

b) (DN) x (DN) = (BN) x (AN)

c) (DN)^{2} = (BN) x (AN)

14) As □ DNBM is a rectangle, BN = DM, so, we have,a) (DN)^{2} = (DM) x (AN), hence proved.

Q 3. In the following fig., ABC is a triangle in which ∠ ABC > 90° and AD **⟂** CB

a) (DN)/(AN) = (BN)/(DN)

^{2}= (BN) x (AN)

^{2}= (DM) x (AN), hence proved.

**⟂**CB

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

### Solution:

1) In ∆ ADB, by the theorem of Pythagoras, we have,(AB)^{2} = (AD)^{2} + (DB)^{2} ------- equation 1.

2) In ∆ ADC, by the theorem of Pythagoras, we have,(AC)^{2} = (AD)^{2} + (DC)^{2}

1) In ∆ ADB, by the theorem of Pythagoras, we have,

(AB)^{2}= (AD)^{2}+ (DB)^{2}------- equation 1.

2) In ∆ ADC, by the theorem of Pythagoras, we have,

(AC)^{2}= (AD)^{2}+ (DC)^{2}

(AC)^{2}= (AD)^{2}+ (DB + BC)^{2}

(AC)^{2}= (AD)^{2}+ (DB)^{2}+ (BC)^{2 }+ 2 . DB . BC ------- equation 2.

3) From equations 1 and 2, we have,

(AC)^{2}= [(AD)^{2}+ (DB)^{2}] + (BC)^{2 }+ 2 . DB . BC

(AC)^{2}= (AB)^{2 }+ (BC)^{2 }+ 2 . BC . BD, hence proved.

**Q 4. In the following fig., ABC is a triangle in which ∠ABC < 90° and AD ⟂ BC.**

Prove thatAC^{2}= AB^{2}+ BC^{2}– 2 BC . BD.

### Solution: 1) In ∆ ADB, by the theorem of Pythagoras, we have,(AB)^{2} = (AD)^{2} + (BD)^{2} ------- equation 1.

2) In ∆ ADC, by the theorem of Pythagoras, we have,(AC)^{2} = (AD)^{2} + (DC)^{2}

1) In ∆ ADB, by the theorem of Pythagoras, we have,

(AB)^{2}= (AD)^{2}+ (BD)^{2}------- equation 1.

2) In ∆ ADC, by the theorem of Pythagoras, we have,

(AC)^{2}= (AD)^{2}+ (DC)^{2}

(AC)^{2}= (AD)^{2}+ (BC - BD)^{2}

(AC)^{2}= (AD)^{2}+ (BC)^{2}+ (BD)^{2 }- 2 . BC . BD ------- equation 2.

3) From equations 1 and 2, we have,

## (AC)

^{2}= (AD)^{2}+ (BD)^{2}+ (BC)^{2 }- 2 . BC . BD## (AC)

^{2}= [(AD)^{2}+ (BD)^{2}]+ (BC)^{2 }- 2 . BC . BD## (AC)

^{2}= [(AB)^{2}]+ (BC)^{2 }- 2 . BC . BD## (AC)

^{2}= (AB)^{2}+ (BC)^{2 }- 2 . BC . BD, hence proved.

**Q 5. In the following fig., AD is a median of a triangle ABC and**

**AM ⟂ BC.**

Prove that :

(i) AC^{2}= AD^{2}+ BC . DM + (BC/2)^{2}^{(ii) AB2 = AD2 - BC . DM + (BC/2)2}^{(iii) (AC)2 + (AB)2 = 2 (AD)2 + (1/2)(BC)2 }

### Solution:**(i) AC**^{2} = AD^{2} + BC . DM + (BC/2)^{2}^{}1) In ∆ AMD, by the theorem of Pythagoras, we get,(AD)^{2} = (AM)^{2} + (MD)^{2} ------- equation 1.

2) In ∆ AMC, by the theorem of Pythagoras, we get,(AC)^{2} = (AM)^{2} + (MC)^{2}

**(i) AC**

^{2}= AD^{2}+ BC . DM + (BC/2)^{2}

^{}^{2}= (AM)

^{2}+ (MD)

^{2}------- equation 1.

^{2}= (AM)

^{2}+ (MC)

^{2}

(AC)^{2}= (AM)^{2}+ (MD + DC)^{2}

## (AC)

^{2}= (AM)^{2}+ (MD)^{2}+ (DC)^{2}^{ }+ 2 (MD) (DC) ------- equation 2.

3) From equations 1 and 2, we have,

## (AC)

^{2}= (AM)^{2}+ (MD)^{2}+ (DC)^{2}^{ }+ 2 (MD) (DC)## (AC)

^{2}= [(AD)^{2}] + (DC)^{2 }+ 2 (MD) (DC) ------- equation 3.

4) As (DC) = (1/2) (BC), so from equation 3 we have

## (AC)

^{2}= [(AD)^{2}] + (BC/2)^{2 }+ 2 (MD) (DC)

(AC)^{2}= [(AD)^{2}] + (BC/2)^{2 }+ [2 (DC)] (MD) [since 2 (DC) = (BC)],

(AC)^{2}= [(AD)^{2}] + (BC/2)^{2 }+ (BC) (MD)(AC)^{2}= (AD)^{2}+ (BC) (DM) + (BC/2)^{2 }------- equation 4.

5) (AC)

^{2}= (AD)^{2}+ (BC) (DM) + (BC/2)^{2}, hence proved.**(ii) AB**

^{2}= AD^{2}- BC . DM + (BC/2)^{2}### 6) In ∆ AMB, by the theorem of Pythagoras, we get,(AB)^{2} = (AM)^{2} + (BM)^{2}

^{2}= (AM)

^{2}+ (BM)

^{2}

## (AB)

^{2}= (AM)^{2}+ (BD - DM)^{2}

(AB)^{2}= (AM)^{2}+ (BD)^{2}+ (DM)^{2 }- 2 (BD) (DM)

(AB)^{2}= [(AM)^{2}+ (DM)^{2}] + (BD)^{2 }- 2 (BD) (DM)

(AB)^{2}= [(AD)^{2}] + (BC/2)^{2 }- 2 (BC/2) (DM)(AB)^{2}= (AD)^{2}+ (BC/2)^{2 }- (BC) (DM) ------- equation 5.

7) (AB)

^{2}= (AD)^{2}+ (BC/2)^{2 }- (BC) (DM), hence proved### (iii) (AC)^{2} + (AB)^{2} = 2 (AD)^{2} + (1/2)(BC)^{2}

### 8) Adding equations 4 and 5, we get,(AC)^{2} + (AB)^{2 }= (AD)^{2} + (BC) (DM) + (BC/2)^{2} + (AD)^{2} + (BC/2)^{2 }- (BC) (DM)

^{2}+ (AB)

^{2 }= (AD)

^{2}+ (BC) (DM) + (BC/2)

^{2}+ (AD)

^{2}+ (BC/2)

^{2 }- (BC) (DM)

## (AC)

^{2}+ (AB)^{2 }= 2 (AD)^{2}+ 2 (BC/2)^{2}

(AC)^{2}+ (AB)^{2 }= 2 (AD)^{2}+ 2 (BC)^{2}/4

## (AC)

^{2}+ (AB)^{2 }= 2 (AD)^{2}+ 1/2 (BC)^{2}

9) (AC)

^{2}+ (AB)^{2 }= 2 (AD)^{2}+ 1/2 (BC)^{2}, hence proved.**Q 6. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum**

**of the squares of its sides.**

### Solution:

### 1) In ∆ ACF, by the theorem of Pythagoras, we get,(AC)^{2} = (AF)^{2} + (CF)^{2 }------- equation 1.

^{2}= (AF)

^{2}+ (CF)

^{2 }------- equation 1.

### 2) In ∆ AFD, by the theorem of Pythagoras, we get,(AF)^{2} = (AD)^{2} - (DF)^{2 }------- equation 2.

3) According to the diagram, we have,

^{2}= (AD)

^{2}- (DF)

^{2 }------- equation 2.

(CF) = (DC - DF)^{ }------- equation 3.

4) From equations 1, 2, and 3, we have,

(AC)^{2}= (AF)^{2}+ (CF)^{2}

(AC)^{2}= [(AD)^{2}- (DF)^{2}] + (DC - DF)^{2}

## (AC)

^{2}= (AD)^{2}- (DF)^{2}+ (DC)^{2 }+ (DF)^{2 }- 2 (DC) (DF)

(AC)^{2}= (AD)^{2}+ (DC)^{2 }- 2 (DC) (DF) ------- equation 4.

### 5) In ∆ BDE, by the theorem of Pythagoras, we get,(BD)^{2} = (ED)^{2} + (BE)^{2 }------- equation 5.

^{2}= (ED)

^{2}+ (BE)

^{2 }------- equation 5.

### 6) In ∆ DEA, by the theorem of Pythagoras, we get,(ED)^{2} = (AD)^{2} - (AE)^{2 }------- equation 6.

7) According to the diagram, we have,

^{2}= (AD)

^{2}- (AE)

^{2 }------- equation 6.

(BE) = (BA + AE)^{ }------- equation 7.

8) From equations 5, 6, and 7, we have,

(BD)^{2}= (ED)^{2}+ (BE)^{2}

(BD)^{2}= [(AD)^{2}- (AE)^{2}] + (BA + AE)^{2}

## (BD)

^{2}= (AD)^{2}- (AE)^{2}+ (BA)^{2 }+ (AE)^{2 }+ 2 (BA) (AE)

(BD)^{2}= (AD)^{2}+ (BA)^{2 }+ 2 (BA) (AE) ------- equation 8.

### 9) Adding equations 4 and 8, we get,(AC)^{2} + (BD)^{2 }= [(AD)^{2} + (DC)^{2 }- 2 (DC) (DF)] + [(AD)^{2} + (BA)^{2 }+ 2 (BA) (AE)]

^{2}+ (BD)

^{2 }= [(AD)

^{2}+ (DC)

^{2 }- 2 (DC) (DF)] + [(AD)

^{2}+ (BA)

^{2 }+ 2 (BA) (AE)]

------- equation 9.

10) According to the diagram, we know that,

a) (AD) = (BC)

b) (BA) = (DC)11) So, we will change the red-colored terms of equation 9 as follows:

a) (AE) = (DF)

(AC)^{2}+ (BD)^{2 }= [(AD)^{2}+ (DC)^{2 }- 2 (DC) (DF)] + [(AD)^{2}+ (BA)^{2 }+ 2 (BA) (AE)]

(AC)^{2}+ (BD)^{2 }= [(AD)^{2}+ (DC)^{2 }- 2 (DC) (DF)] + [(BC)^{2}+ (BA)^{2 }+ 2 (DC) (DF)]

(AC)^{2}+ (BD)^{2 }= [(AD)^{2}+ (DC)^{2}] + [(BC)^{2}+ (BA)^{2}] ------- equation 10.

12) From equation 10, we can say that,

The sum of the squares of the diagonals of a parallelogram is equal to the sumof the squares of its sides.

**Q 7. In the following fig., two chords AB and CD intersect each other**

at point P.Prove that :

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

### Solution:**(i) ∆ APC ~ ∆ DPB**

1) In ∆ APC and ∆ DPB,a) < APC = < DPB (vertically opposite angles are equal)

b) < CAP = < BDP (angles inscribed in the same arc are equal)2) So, by AA similarity test, ∆ APC ~ ∆ DPB, hence proved.

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

1) We proved that,

**(i) ∆ APC ~ ∆ DPB**

1) In ∆ APC and ∆ DPB,

a) < APC = < DPB (vertically opposite angles are equal)

b) < CAP = < BDP (angles inscribed in the same arc are equal)

2) So, by AA similarity test, ∆ APC ~ ∆ DPB, hence proved.

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

1) We proved that,

∆ APC ~ ∆ DPB

### 2) So, as corresponding sides are proportional, we have,(AP)/(DP) = (CP)/(PB)

2) So, as corresponding sides are proportional, we have,

(AP)/(DP) = (CP)/(PB)

(AP) . (PB) = (CP) . (DP), hence proved.

**Q 8. In the following fig., 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) ∆ PAC ~ ∆ PDB**

1) In ∆ PAC and ∆ PDB,a) < APC = < DPB (common angles)

b) < CAP = < BDP (Exterior angle of a cyclic quadrilateral is equal to the

**(i) ∆ PAC ~ ∆ PDB**

1) In ∆ PAC and ∆ PDB,

a) < APC = < DPB (common angles)

b) < CAP = < BDP (Exterior angle of a cyclic quadrilateral is equal to the

opposite interior)

### 2) So, by AA similarity test, ∆ PAC ~ ∆ PDB, hence proved.

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

1) We proved that,

2) So, by AA similarity test, ∆ PAC ~ ∆ PDB, hence proved.

**(ii)**

**PA . PB = PC . PD**

1) We proved that,

∆ PAC ~ ∆ PDB

### 2) So, as corresponding sides are proportional, we have,(PA)/(PD) = (PC)/(PB)

2) So, as corresponding sides are proportional, we have,

(PA)/(PD) = (PC)/(PB)

(PA) . (PB) = (PC) . (PD), hence proved.

**Q 9. In the following fig., D is a point on side BC of**

**∆**

**ABC**

such that BD/CD = AB/AC

Prove that AD is the bisector of∠BAC.

### Solution:1) It is given that,

a) (BD)/(CD) = (BA)/(AC)

2) So, using the converse of the basic proportionality theorem,

a) AD || EC

b) ∠ BAD = ∠ AEC (corresponding angles) ------ equation 1

c) ∠ CAD = ∠ ACE (alternet interior angles) ------ equation 2

3) As (AE) = (AC),

a) ∠ AEC = ∠ ACE ------ equation 3

4) From equations 1 and 3, we get,

a) ∠ BAD = ∠ ACE ------ equation 4

5) From equations 2 and 4, we get,

a) ∠ BAD = ∠ CAD ------ equation 5

6) From equation 5, we get,

a) AD is the bisector of ∠ BAC, hence proved.

**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**

**(see the following 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?**

**(Note: See the original figure in the text-book)**

### Solution:

### 1) Let AB be the distance between the tip of the fishing rod and the water surface.2) Let BC be the distance between the fly and the rod.3) In ∆ ABC, by the theorem of Pythagoras, we get,

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

^{2}= (AB)

^{2}+ (BC)

^{2}

## (AC)

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

(AC)^{2}= (3 x 0.6)^{2}+ (3 x 0.8)^{2}(AC)^{2}= [(3)^{2}x (0.6)^{2}] + [(3)^{2}x (0.8)^{2}]

(AC)^{2}= (3)^{2}x [(0.6)^{2}+ (0.8)^{2}]

(AC)^{2}= (3)^{2}x [(0.36) + (0.64)]

(AC)^{2}= 9 x 1

(AC) = 3 ----------- equation 1

4) So, the length of the string is 3 m.

5) According to the problem, Nazima pulls in the string at the rate of 5 cm per

second, so, for 12 seconds, the fly comes to point D, so the string pulled by Nazima can be calculated using the formula, distance = speed x time, so we have,

distance of the string pulled = speed x time

distance of the string pulled = 5 x 12

distance of the string pulled = 60 cm

distance of the string pulled = 0.6 m

6) According to the figure, we have,

AD = AC - distance of the string pulled

AD = 3.0 - 0.6

AD = 2.4 m

### 7) In ∆ ABD, by the theorem of Pythagoras, we get,(BD)^{2} = (AD)^{2} - (AB)^{2}

^{2}= (AD)

^{2}- (AB)

^{2}

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

(BD)^{2}= (3 x 0.8)^{2}- (3 x 0.6)^{2}(BD)^{2}= (3)^{2}x [(0.8)^{2}- (0.6)^{2}]

(BD)^{2}= (3)^{2}x [(0.64) - (0.36)]

(BD)^{2}= 9 x (0.28)

(BD) = 3 x √(0.28)

(BD) = 3 x 0.529

(BD) = 1.587

8) The horizontal distance of a fly from Nazima is DE, so

(DE) = (EB) + (BD)

(DE) = (1.2) + (1.587)

(DE) = 2.787

(DE) = 2.79 m

9) So, the horizontal distance of the fly from Nazima after 12 seconds is 2.79 m.

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