10类RD Sharma解决方案–第4章三角形–练习4.7 |套装1

问题1.如果三角形的边长分别为3 cm，4 cm和6 cm，请确定该三角形是否为直角三角形。

解决方案：

According to the question

The sides of triangle are:

AB = 3 cm

BC = 4 cm

AC = 6 cm

According to Pythagoras Theorem:

 AB² = 3² = 9

BC² = 4² = 16

AC² = 6² = 36

Since, AB² + BC² ≠ AC²

Hence, by converse of Pythagoras theorem, triangle is not a right triangle

问题2。下面给出了某些三角形的边。确定其中哪些是直角三角形。

(i)a = 7厘米，b = 24厘米，c = 25厘米

(ii)a = 9厘米，b = 16厘米和c = 18厘米

(iii)a = 1.6厘米，b = 3.8厘米，c = 4厘米

(iv)a = 8厘米，b = 10厘米，c = 6厘米

解决方案：

(i) According to the question

The sides of triangle are: 

a = 7 cm, b = 24 cm and c = 25 cm

According to Pythagoras Theorem:

The hypotenuse or longest side of a right angle triangle is equal to the sum of 

squares of other two sides of the triangle

∴ a² = 49, b² = 576 and c² = 625

(Longest side)² = c² = 625

Sum of squares of shorter sides = (a)² + (b)²= 49 + 576 = 625

∴ 625 = 625

Hence, by converse of Pythagoras theorem, given triangle is a right triangle.

(ii) According to the question

The sides of triangle are: a = 9 cm, b = 16 cm and c = 18 cm

According to Pythagoras Theorem:

The hypotenuse or longest side of a right angle triangle is equal to the sum of 

squares of other two sides of the triangle

∴ a² = 81, b² = 256 and c² = 324

(Longest side)² = c² = 324

Sum of squares of shorter sides = (a)² + (b)² = 81 + 256 = 337

∴324 ≠ 337

Hence, by converse of Pythagoras theorem, given triangle is not a right triangle.

(iii) According to the question

The sides of triangle are: a = 1.6 cm, b = 3.8 cm and C = 4 cm

According to Pythagoras Theorem:

The hypotenuse or longest side of a right angle triangle is equal to the sum of 

squares of other two sides of the triangle

(Longest side)² = 16

Sum of squares of shorter sides = (1.6)² + (3.8)² = 2.56 + 14.44 = 17

∴ 16 ≠ 17

Hence, by converse of Pythagoras theorem, given triangle is not a light triangle.

(iv) According to the question

The sides of triangle are: a = 8 cm, b = 10 cm, c = 6 cm

According to Pythagoras Theorem:

The hypotenuse or longest side of a right angle triangle is equal to the sum of 

squares of other two sides of the triangle

(Longest side)² = (10)² = 100

Sum of squares of shorter sides = (8)² + (6)² = 64 + 36 = 100

∴ 100 = 100

Hence, by converse of Pythagoras theorem, given triangle is a light triangle.

问题3.一个人向西走15m，然后向北走8m。距离起点有多远?

解决方案：

Let the starting point of the man be O and final point be A.

So, In ∆ABO, 

By using Pythagoras theorem

 AO² = AB² + BO²         

⟹ AO2 = 8² + 15²

⟹ AO2 = 64 + 225 = 289

⟹ AO = √289 = 17m

Hence, the man is 17m far from the starting point.

问题4. 17 m长的梯子到达离地面15 m的建筑物的窗户。找到梯子脚与建筑物的距离。

解决方案：

In ∆ABC, 

By using Pythagoras theorem

AB² + BC² = AC²

⟹ 152 + BC² = 172

⟹ 225 + BC² = 172

⟹ BC² = 289 – 225

⟹ BC² = 64

⟹ BC = 8 m

Hence, the distance of the foot of the ladder from building = 8 m

问题5.两个高度分别为6 m和11 m的电线杆站立在平面地面上。如果两脚之间的距离为12 m，请找到两脚之间的距离。

解决方案：

Let us considered CD and AB be the poles of height 11 and 6 in.

Therefore, CP = 11 – 6 = 5 m

From the figure we may observe that AP = 12m

In ∆APC, 

By using Pythagoras theorem

AP² + PC² = AC²

12² + 5² = AC²

AC² = 144 + 25 = 169 m

AC = 13 m

Hence, the distance between their tops = 13 m.

问题6.在等腰三角形ABC中。 AB = AC = 25厘米，BC = 14厘米。从BC上的A计算海拔高度。

解决方案：

According to the question

AB = AC = 25 cm and BC = 14

In ∆ABD and ∆ACD

∠ADB = ∠ADC                   [Each = 90°]

AB = AC                              [Each = 25 cm]

AD = AD                              [Common]

So, by RHS condition

∆ABD ≅ ∆ACD   

Hence, by corresponding parts of congruent triangles

∴ BD = CD = 7 cm                

In ∆ADB, 

By using Pythagoras theorem

AD² + BD² = AB²             

⟹ AD² + 7² = 25²

⟹ AD² = 625 – 49 = 576

⟹ AD = √576 = 24 cm  

问题7.梯子的脚距墙6 m，梯子的顶部到达离地面8 m的窗户。如果梯子的移动方式使其脚距墙的距离为8 m，则其尖端可以达到什么高度?

解决方案：

Let length of ladder be AD = BE = l m

In ∆ACD, 

By using Pythagoras theorem

AD² = AC² + CD²           

⟹ l² = 8² + 6² … (i)

In ∆BCE, 

By using Pythagoras theorem

BE² = BC² + CE²            

⟹ l² = BC² + 8² … (ii)

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

BC² + 8² = 8² + 6²

⟹ BC² + 6²

⟹ BC = 6 m

问题8.高度分别为9 m in和14 m的两个杆在平面地面上。如果两脚之间的距离为12 m英寸，请找到两脚之间的距离。

解决方案：

We have,

AC = 14 m. DC = 12m and ED = BC = 9 m

Construction: Draw EB ⊥ AC

AB = AC –  BC = 14 — 9 = 5 m

And, EB = DC = 12m

In ∆ABE, 

By using Pythagoras theorem

AE² = AB² + BE²                     

⟹ AE² = 5² + 12²

⟹ AE² = 25 + 144 = 169

⟹ AE= √169 = 13 m

Hence, the distance between their tops = 13 m

问题9.使用毕达哥拉斯(Pythagoras)定理，根据下图所示的b和c确定AD的长度。

解决方案：

We have,

In ∆BAC, 

By using Pythagoras theorem

BC² = AB² + AC²

⟹ BC² = c² + b²

$\to \space BC=\sqrt{{c^2} +{b^2}}$    ….(1)

In ∆ABD and ∆CBA

∠B = ∠B                                           [Common]

∠ADB = ∠BAC                                 [Each 90°]

So, by AA both the triangles are similar

∆ABD ͏~ ∆CBA             

Hence, by corresponding parts of similar ∆ are proportional

$\therefore \space\frac{AB}{CB}=\frac{AD}{CA}$       

$\to \frac{C}{\sqrt{C^2 + B^2}}=\frac{AD}{B}$

$\to AD= \frac{BC}{\sqrt{C^2 + B^2}}$

Hence, the length of AD in terms of b and c is$AD= \frac{BC}{\sqrt{C^2 + B^2}}$

问题10.三角形的边长为5厘米，12厘米和13厘米。找到从相反顶点到长度为13 cm的边的垂直线的小数点后一位的长度。

解决方案：

Let, AB = 5cm, BC = 12 cm and AC = 13 cm. Then, AC² = AB² + BC². 

This proves that ∆ABC is a fight triangle. right angles at B. 

Let BD be the length of perpendicular from B on AC.

Now, Area of ∆ABC = 1/2(BC x BA)

= 1/2(12 x 5)

= 30 cm

Also, Area of ∆ABC = 1/2(AC x BD) = 1/2(13 x BD) = 1/2(13 x 2)

BD = 60/13 cm

问题11. ABCD是一个正方形，F是AB的中点。 BE是BC的三分之一。如果∆FBE的面积= 108 cm ^2，则求出AC的长度。

解决方案：

Given,

ABCD is a square. 

 F is the mid-point of AB.

BE is one third of BC.

Area of ∆ FBE = 108cm²

To find: length of AC

Let’s the sides of the square to be y.

AB = BC = CD = DA = y cm

So, AF = FB = x/2 cm

and, BE = x/3 cm

Now, 

 Area of ∆ FBE = 1/2 x BE x FB

⇒ 108 = (1/2) x (x/3) x (x/2)

⇒ x² = 108 x 2 x 3 x 2 = 1296

⇒ x = √(1296)              

x = 36cm

 In ∆ ABC,

By using Pythagoras theorem

AC² = AB² + BC²                   

⇒ AC² = x² + x² = 2x²

⇒ AC² = 2 x (36)²

⇒ AC = 36√2 = 36 x 1.414 = 50.904 cm

Hence, the length of AC is 50.904 cm.

问题12.在等腰三角形ABC中，如果AB = AC = 13 cm，并且从BC到A的高度为5 cm，则求出BC。

解决方案：

Given that,

An isosceles triangle ABC, AB = AC = 13cm, AD = 5cm

 To find: BC

Now In ∆ ADB, 

By using Pythagoras theorem

AD² + BD² = 13²         

5² + BD² = 169

BD² = 169 – 25 = 144

BD = √144 = 12 cm

Similarly, In ∆ADC,

By using Pythagoras theorem

AC² = AD² + DC²          

13² = 5² + DC²

169 – 25 = DC²

DC = √144 = 12 cm

So, BC = BD + DC = 12 + 12 = 24 cm

Hence, in isosceles triangle ABC side BC is 24 cm

问题13：在∆ABC中，AB = BC = CA = 2a，AD = BC。证明

(i)AD = a√3

(ii)面积(∆ABC)=√3a ²

解决方案：

(i) In ∆ABD and ∆ACD, 

∠ADB = ∠ADC = 90°             [Given]

AB = AC                                [Given]

AD = AD                               [Common]

So, By RHS condition

∆ABD ≅ ∆ACD                  

Hence, by corresponding parts of congruent triangles

 ∴ BD = CD = a                

Now In ∆ABD, 

By using Pythagoras theorem

AD² + BD² = AB²                         

AD² + a² = 2a²

AD² = 4a² – a² = 3a²

AD = a√3

(ii) Area (∆ABC) = 1/2 x BC x AD

= 1/2 x (2a) x (a√3)

= √3 a²

Hence Proved

问题14.菱形的对角线长度分别是24厘米和10厘米。找到菱形的每一面。

解决方案：

To find: Each side of the rhombus. 

Let ABCD be a rhombus with diagonals AC and BD.

∴ AC = 24cm and BD = 10cm

We know that,

AO = OC = 12cm and BO = OD = 3cm   [Diagonals of a rhombus bisect each other at right angle]

Now In ∆AOB,

By using Pythagoras theorem

AB² = AO² + BO²                                

= 12² + 5²

= 144 + 25

= 169

∴ AB = √(169) = 13cm

∴ AB = BC = CD = AD = 13cm.    [The sides of rhombus are all equal.]

Hence, the sides of the rhombus are as follows AB = BC = CD = AD = 13cm.