## Triangles

Question 1 |

11 cm | |

6 cm | |

5 cm | |

9 cm |

**∠**EFG = ∠GCD **(Alternative angle)**

**∠**EFG = ∠CGD**(Vertically opposite angles)**

$\Delta$ GEF ~ $\Delta$ GCD

Thus,

GE/CG = EF/CD

or, 5/10 = EF/18

or, EF = 9 cm.

Question 2 |

5m, 7m, 10m | |

2m, 3m, 4m | |

3m, 4m, 8m | |

4m, 6m, 9m |

**So option b is not satisfying the condition 3 + 4 $\not\gt $ 8**

Question 3 |

Excentre | |

Orthocentre | |

Incentre | |

Centroid |

Question 4 |

^{0}, AL is drawn perpendicular to BC, Then ∠ BAL is equal to:

∠ALC | |

∠BAC | |

∠ACB | |

∠B - ∠BAL |

^{0}= 180

^{0}

or, ∠ BAL + ∠ B = 90^{0}

or, ∠ BAL = 90^{0} - ∠ B ----------- (1)

Now, in $\Delta$ ABC,

∠ ACB + ∠ B + ∠ A = 180^{0}

∠ ACB = 90^{0}

-∠ B ----- (2)

From, (1) and (2),

∠ BAL = ∠ ACB.

Question 5 |

90 ^{0} + 1/2 ∠Q | |

90 ^{0} - 1/2 ∠R | |

1/2 (∠ Q - ∠ R) | |

∠Q + ∠ R |

**[PS is bisector.] ------ (1)**

∠Q = 90^{0} - ∠1

∠R = 90^{0} -∠2 - ∠3

So,

∠Q - ∠R = (90^{0} - ∠1) - (90^{0} - ∠2 - ∠3)

∠Q - ∠R = ∠2 + ∠3 - ∠1

∠Q - ∠ R = ∠2 + (∠1 + ∠2) -∠1**[using equation 1]**

∠Q - ∠R = 2 ∠2

1/2 * (∠Q - ∠R) = ∠TPS.

Question 6 |

AB ^{2} - BD^{2} = AC^{2} - CD^{2} | |

AB ^{2} - AC^{2} = BD^{2} + CD^{2} | |

AB ^{2} + BD^{2} = AC^{2} - CD^{2} | |

AB ^{2} - BD^{2} = AC^{2} + CD^{2} |

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

In Right angled $\Delta$ ACD,

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

By (1) - (2)

AB^{2} - AC^{2} = BD^{2} - CD^{2}

AB^{2} - BD^{2} = AC^{2} - CD^{2}.

Question 7 |

I. Every equilateral triangle is necessarily an isosceles triangle.

II. Every right-angled triangle is necessarily an isosceles triangle.

III. A triangle in which one of the median is perpendicular to the side it meets, is necessarily an isosceles triangle.

The correct statements are:

I, II and III | |

I and III | |

II and III | |

I and II |

A triangle in which one of the median is perpendicular to the side it meets, is necessarily an isosceles triangle.

Question 8 |

5 : 9 | |

8 : 9 | |

6 :9 | |

7 : 9 |

So, We can say that $\Delta$ BCD and &delta ABC will be similar.

According to property of similarity,

AB/12 = 12/9

Hence,

AB = 16

AC/6 = 12/9

AC = 8

Hence, AD = 7 and AC = 8

Now,

Perimeter of Delta; ADC / Perimeter of &delta BDC,

= (6+7+8)/(9+6+12)

= 21/27 = 7/9.

Question 9 |

**∠**A = 60

^{0}, then length of AD is :

2 $\sqrt{3}$ | |

(6 $\sqrt{3}$) / 7 | |

(12 $\sqrt{3}$) / 7 | |

(15 $\sqrt{3}$) / 8 |

Let BC = x and Ad = y, then as per bisector theorem,

BD /DC = AB /AC = 4 /3

Hence, BD = 4x/7 and DC = 3x/7

Now, in $\Delta$ ABD using cosine rule,

cos 30^{0} = [{4^{2} +y^{2} - (16x^{2}/49)} /2*3*y]

Or, 4 $\sqrt{3y}$ =[{16 +y^{2} - (16x^{2}/49)}] --------- (i)

Similarly in $\Delta$ ADC,

cos 30^{0} = [{3^{2} +y^{2} - (9x^{2}/49)} /2*3*y]

Or, 3 $\sqrt{3y}$ =[{9 +y^{2} - (9x^{2}/49)}] --------- (ii)

From equation (i) and (ii), we get

y = 12 $\sqrt{3 }$/7.

Question 10 |

6 cm | |

8 cm | |

10 cm | |

7 cm |

QR +2 = 2PQ

QR = 2PQ - 2 ------- (2)

PR = PQ + 10 ----- (2)

Perimeter = 40 m

PQ + QR + Rp = 40

Putting the value of PQ and QR from equation (1) and (2),

Pq + 2PQ - 2 + PQ +10 = 40

4PQ = 32

PQ = 8 cm which is the smallest side of the triangle.