10类NCERT解决方案-第6章三角形-练习6.5 |套装1

问题1.三角形的边在下面给出。确定其中哪些是直角三角形?在直角三角形的情况下，写下其斜边的长度。

(i)7厘米，24厘米，25厘米

(ii)3厘米，8厘米，6厘米

(iii)50厘米，80厘米，100厘米

(iv)13厘米，12厘米，5厘米

解决方案：

(i) Given: 7cm, 24cm, 25cm

(25)²= 25 * 25 = 625

∴The are sides of a right ∆ and hypotenuse = 625  

(ii) Given: 3cm, 8cm, 6cm

(8)² = 8 * 8 = 64  

They are not side of right triangle.

(iii) Given: 50cm, 80cm, 100cm

(100)² = 100 * 100 = 10000

(50)² + (80)² = 2500 + 6400

Therefore, this is not right triangle.

(iv) Given: 13cm, 12cm, 5cm

(13)² = 13 * 13 = 169

(12)² + (5)² = 169

Therefore, this is the sides of right triangle and hypotenuse = 169

问题2。PQR是在P处成直角的三角形，M是QR上的点，使得PM⊥QR。证明PM2 = QM×MR。

解决方案：

Given: A right ∆PQR right-angled at P and PM⊥QR

To show: PM² = QM * MR

PM * PM = QM * MR

In ∆PMQ and ∆PMR

∠1 =∠2           -(each 90°)

∠1 +∠2 +∠3 = 180°

90° + ∠4 + ∠5 = 180°

∠4 +∠5 = 180° – 90°

∠3 + ∠4 = 180° – 90°

∠4 +∠5 = ∠3 +∠4

∴∠5 =∠3  

∴∆PMQ ~ ∆RMP  

$\frac{PM}{RM} = \frac{MQ}{MP} = \frac{QP}{PR}$  

$\frac{PM}{RM} = \frac{MQ}{MP}$

MP² = MQ.PM

问题3。在图中，ABD是与A和AC⊥BD成直角的三角形。显示

(i)AB ² = BC×BD

(ii)AC ² = BC×DC

(iii)广告² = BD×CD

解决方案：

Given: A right ∆ABD, right angled at A and AC⊥ BD

To show:  

(i)AB² = BC × BD

(ii) AC² = BC × DC

(iii) AD² = BD × CD

(i) AB² = BC × BD

AB * AB = BC.BD

In ∆ ABC and ∆ABD

∠B =∠B          -(common)

∠BCA = ∠A          -(each 90°)

∴ ∆ABC~∆DBA  

$\frac{AB}{DB} = \frac{BC}{BA} = \frac{CA}{AD}$

AB² = BC * DC

(ii) AC² = BC * DC

AC * AC = BC * DC

In ∆ABC and ∆ACD

∠ACB = ∠ACD          -(each 90°)

In ∆ ACB

∠1 + ∠2 + ∠3 = 180°

∠1 + ∠2 + 90° = 180°

∠1 + ∠2 = 90°           -(1)

∠1 + ∠3 = 90°          -(2)

∠1 + ∠2 =∠1 + ∠3

∴ ∠2 = ∠3  

∴ ∆ABC~∆DAC  

$\frac{AB}{DA} = \frac{BC}{AC} = \frac{CA}{CD}$

AC² = BC * CD

(iii) AD² = BD × CD

In ∆ABD and ∆ACD

∠A =∠ACD          -(each 90)

∠D =∠D          -(common)

∴ ∆ABD~∆CAD          -(AA similarity)  

$\frac{AB}{CA} = \frac{BD}{AD} = \frac{DA}{DC}$

AD² = BD * CD

问题4. ABC是一个等腰三角形，与C呈直角。证明AB ² = 2AC ² 。

解决方案：

Given: ∆ABC right angled at B

To prove: AB² = 2AC²

According to Pythagoras theorem,

(AB)² = (AC)² + (BC)²

(AB)² = (AC)² + (AC)²           -(BC = AC)

AB² = 2AC²

问题5. ABC是AC = BC的等腰三角形。如果AB ² = 2AC ² ，则证明ABC是直角三角形。

解决方案：

Given: ∆ABC is an isosceles triangle in which  

AC = BC

AB² = 2AC²

To prove: ABCD is right ∆  

AB² = 2AC²

AB² = AC² + AC²

∴ By the, converse of Pythagoras theorem ∆ACB is a right ∆ at C.  

问题6。ABC是边2a的等边三角形。找到它的每个高度。

解决方案：

Let ABC is an equilateral triangle of each 2a units

Construction: Draw AD ⊥ BC

In right ∆ ADB

(AB)² = (AD)² + (BD)²

(2a)² = (AD)² + (a)²

(4a)² = (AD)² + a²

4a² – a² = (AD)²

3a² = (AB)²

√3 a² = AD               ∴ Each of its altitude= √3 a unit

√3 a = AD  

问题7：证明菱形边的平方和等于对角线的平方和。

解决方案：

Given: ABCD is a rhombus

To prove: AB² + BC² + CD² + DA²

Poof: ABCD is rhombus and let diagonals AC and BD bisect each other at O.

∴ ∠AOB =∠BOC =∠COD =∠DOA = 90°  

In ∆AOB  

AB² = AC² + BO²

AB² = (1/2AC²) + (1/2BD²)

AB² = 1/4(AC² + BD²)

4AB² = AC² + BD²          -(1)

Similarly, in ∆BOC  

4BC² = AC² + BD²          -(2)

4CD² = AC² + BD²          -(3)

4AD² = AC² + BD²          -(4)

Adding 1, 2, 3, and 4

4AB² + 4BC² + 4CD² + 4DA² = 4(AC² + BD²)

4(AB² + BC² + CD² + DA²) = 4(AC² + BD²)

∴ AB² + BC² + CD² + DA² = AC² + BD²  

问题8。在图6.54中，O是三角形对角线内部的一个点。

(i)OA ² + OB ² + OC ² -OD ² -OE ² -OF ² = AF ² + BD ² + CE ²

(ii)AF ² + BD ² + CE ² = AE ² + CD ² + BF ²

解决方案：

(i) In right ∆AFO, by Pythagoras theorem  

OA² = AP² + OF²          -(1) 

In right ∆ODB, by Pythagoras theorem

OB² = BD² + OD²          -(2)          

In right ∆OEC, by Pythagoras theorem

OC² = CE² + OE²          -(3)   

Adding 1, 2, and 3

OA² + OB² + OC² = AF² + BD² + CF² + OD² + OE²

OA² + OB² + OC² – OE² – OD² – OE² = AF² + BD² + CE²

(ii) AF² + BD² + CE² = AE² + CD² + BF²

OB² = BD² + OD²          -(1) 

OC² = DC² + OD²          -(2) 

Subtracting (1) from (2) we get,

OB² – OC² = BD² + OD² – (DC² + OD²)

= BD² + OD² – DC² – OD²

= BD² – DC²          -(3)      

Similarly,  

OC² – AD² = CE² – EA²          -(4) 

AO² = BO² = AF² – FB²          -(5) 

Adding 3, 4, and 5, we get,

OB² – OC² + OC² – AO² + AO² – BO² = BD² – DC² + CE² – EA² + AF² – FB²

0 + DC² + EA² + FB² = BD² + CD² + AF²

问题9. 10 m长的梯子到达离地面8 m的窗户。找出梯子脚距墙底的距离。

解决方案：

AB is wall  = 8m

AC is ladder = 10m

BC = ?

In right ∆ABC, by Pythagoras theorem

(AC)² = (AB)² + (BC)²

(10)² = (8)² + (BC)²

100 = 64 + (BC)²

100 – 64 = (BC)²

36 = (BC)²

√36 = BC  

√(6 * 6) = BC

6 = BC  

Hence, the foot of the ladder is at a distance of 6 m from the base of the wall.

问题10.连接到高度为18 m的垂直杆的拉线长度为24 m，并在另一端连接有木桩。应该将桩子从杆的根部打多远，以便拉紧电线?

解决方案：

In fig.  

AB is pole = 18m

AC is wire = 24m

BC = ?

In right ∆ABC, by Pythagoras theorem

(AC)² = (AB)² + (BC)²

(24)² = (18)² + (BC)²

576 = 324 + (BC)²

576 – 324 = (BC)²

252 = (BC)²

√252 = BC  

√(2 * 2 * 3 * 3 * 7) = BC 

2 * 3√7 = BC

6√7 = BC

∴ Hence, the stake may be placed at distance of 6√7 from the base of the pole.

第6章三角形–练习6.5 |套装2