10类NCERT解决方案-第6章三角形-练习6.5 |套装2

第6章三角形–练习6.5 |套装1

问题11：一架飞机离开机场，以每小时1000公里的速度向北飞行。同时，另一架飞机离开同一个机场，以每小时1200公里的速度向西飞行。之后的两架飞机相距多远$1\frac{1}{2}$小时?

解决方案：

Distance covered by place left towards north = 1000 * 1.5

Distance covered by place left towards west = 1200 * 1.5 = 1800km

In right ∆ABC by Pythagoras theorem

(AC)² = (AB)² + (BC)²

(AC)² = (1500)² + (1800)²

= 250000 + 3240000

= 5490000

= √(3 * 3 * 61 * 10 * 10 * 10 * 10)

= 3 * 10 * 10√61

AC = 300√61

Distance between the two poles 300√61km

问题12.两个高度分别为6 m和11 m的电线杆站立在平面地面上。如果两根脚之间的距离为12 m，请找到两根顶部之间的距离。

解决方案：

AB is pole of height = 11m

DC is another pole of height = 6m

BC = 12m

In fig.  DE = BC = 12m

AE = AB – EB

= 11 – 6

= 5m

In right ∆AED, by Pythagoras theorem

(AD)² = (AC)² + (DE)²

(AD)² = (5)² + (12)²

(AD)² = 25 + 144

AD = √169

AD = √(13 * 13)

AD = 13

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

问题13. D和E是分别在C处成直角的三角形ABC的CA和CB边上的点。证明AE ² + BD ² = AB ² + DE ² 。

解决方案：

To prove: (AE)² + (BD)² = (AB)² + (DE)²

Construction: Join AE, BD and DE

Proof:  

In ∆ACE, by Pythagoras theorem

(AE)² = (AC)² + (EC)²          -(1)

In ∆DCB, by Pythagoras theorem

(BD)² = (DC)² + (BC)²          -(2)

In ∆ACB, by Pythagoras theorem

(AB)² = (AC)² + (CB)²          -(3)

In ∆DCE, by Pythagoras theorem

(ED)² = (DC)² + (CE)²

Adding eq (1) and (2)

(AE)² + (BD)² = (AC)² + (EC)² + (DC)² + (DC)²

= (AC² + BC²) + (EC² + DC²)

= AB² + DE²          -(from 3 and 4)

问题14.ΔABC的BC边的A垂线与D处的BC相交，使得DB = 3CD(见图)。证明2AB ² = 2AC ² + BC ² 。

解决方案：

Given: DB = 3CD

To prove: 2AB² = 2AC² + BC²

Proof: DB = 3CD

BC = CD + DB

BC = CD + 3CD

BC = 4CD

BC/4 = CD           -(1)

DB = 3(BC/4)           -(2)

In right ∆ADB, by Pythagoras theorem

AB² = AD² + DB²           -(3)

In right ∆ADC, by Pythagoras theorem

AC² = AD² + CD²           -(4)

Subtract eq(3) from (4)

AB² – AC² = AD² + DB² – (AD² + CD²)

= AD² + DB² + AD² – CD²

= DB² – CD²

= (3/4BC)² – (BC/4)²

= 9/6BC² – BC/16

= 8BC²/16

= AB² – AC² = BC²/2

2AB² – 2AC² = BC²

2AB² = 2AC² + BC²      

问题15.在等边三角形ABC中，D是BC侧的一点，使得BD = 1 / 3BC。证明9AD ² = 7AB ² 。

解决方案：

Prove that 9AD² = 7AB²

Construction: Join AD and draw AE perpendicular BC

Let each side, AB = AC = BC = a

BD = 1/3BC = 1/3a

BC = 1/2BC = 1/2a

DE = BE – BD

= 1/2a – 1/3a

= 3a – 2a/6

= DE = a/6

In right ∆AED, by Pythagoras theorem

AD² = AE² + DE²

AD² = (√3a/2)² + (a/6)²

= 3a²/4+ a²/36

= 27a²/36 + a²/36

AD² = 28a²/36

AD² = 7a²/9

9AD² = 7a²

9AD² = 7AB²

问题16：在等边三角形中，证明一侧的正方形的三倍等于其高度之一的正方形的四倍。

解决方案：

To prove: 3AB² = 4AD²

 Proof: Let each side of one equilateral ∆ = a

BD = 1/2a          -[perpendicular bisects the side in on  an equilateral ∆]

In right ∆ADB, by Pythagoras theorem

(AB)²=(AD)² + (BD)²

(a)² = (AD)² + (1/2a)²

a² = AD² + (a/2)²

a² – a²/4 = AD²

3a²/4 = AD²

3a² = 4AD²

3AB² = 4AD²

问题17：勾选正确的答案并说明理由：在ΔABC中，AB =6√3cm，AC = 12 cm，BC = 6 cm。角度B为：

(A)120°(B)60°

(C)90°(D)45°

解决方案：

(AC)² = (12)² = 144

(AB)² = (6√3)² = 6 * 6√3 = 36 * 3 = 108

(BC)² = (6)² = 36

(AB)² + (BC)² = 108 + 36 = 144

∴ It is right ∆ thus ∠B = 90°