第 10 类 RD Sharma 解决方案 - 第 4 章三角形 - 练习 4.6 |设置 1

问题 1. 三角形 ABC 和 DEF 相似。

(i) 若面积(△ABC) = 16cm ² ，面积(△DEF) = 25 cm ²且BC = 2.3 cm，求EF。

解决方案：

Since, △ABC ∼ △DEF,

So, 

ar.(△ACB)/ar.(△DEF) = (BC)²/(EF)²

16/25 = (2.3)²/EF²

(4)²/(5)² = (2.3)²/EF² 

4/5 = 2.3/EF

EF = (2.3 × 5)/4

EF = 11.5/4 = 2.875 cm

Hence, EF = 2.875 cm

(ii) 若面积(△ABC) = 9cm ² ，面积(△DEF) = 64 cm ²且DE = 5.1 cm，求AB。

解决方案：

Since, △ABC ∼ △DEF

So, 

ar.(△ACB)/ar.(△DEF) = (AB)²/(DE)²

9/5 = AB²/(5.1)²

(3)²/(8)² = AB²/(5.1)²

3/8 = AB/5.1

AB = (3 × 5.1)/8 = 15.3/8

Hence, AB = 1.9125 cm

(iii) 如果 AC = 19cm，DF = 8cm，求两个三角形的面积比。

解决方案：

Since, △ABC ∼ △DEF

So, 

ar.(△ACB)/ar.(△DEF) = (AB)²/(DE)²

= (19)/(8)² = 361/64

Hence, ar.(△ABC):ar.(△DEF) = 361:64

(iv) 若面积(△ABC) = 36cm ² ，面积(△DEF) = 64 cm ²且DE = 6.2 cm，求AB。

解决方案：

Since, △ABC ∼ △DEF

So, 

ar.(△ACB)/ar.(△DEF) = (AB)²/(DE)²

⇒ 36/64 = AB²/(6.2)²

⇒ 6/8 = AB/(6.2)

⇒ AB/6.2 = 6/8

⇒ AB = (6 × 6.2)/8 = 37.2/8  

AB = 4.65

Hence AB = 4.65 cm

(v) 若 AB = 1.2 cm，DE = 1.4 cm，求△ABC 与△DEF 的面积之比。

解决方案：

Since, △ABC ∼ △DEF

So, 

ar.(△ACB)/ar.(△DEF) = (AB)²/(DE)²

(1.2)²/(1.4)² = 1.44/1.96 = 144/196 = 36/49

Hence, ar.(△ABC):ar.(△DEF) = 36:49

问题 2。低于 ΔACB ~ ΔAPQ。如果 BC = 10 cm，PQ = 5 cm，BA = 6.5 cm 和 AP = 2.8 cm，求 CA 和 AQ。此外，求面积 (ΔACB)：面积 (ΔAPQ)。

解决方案：

According to the question

It is given that, BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm

Also, △ACB ∼ △APQ

So, BC/PQ = AB/AQ = AC/AP

10/5 = 6.5/AQ = AC/2.8

6.5/AQ = 10/5

⇒ AQ = (6.5 × 5)/10 = 3.25 

and AC/2.8 = 10/5

⇒ (2.8 × 10)/5 = 5.6

AC = 5.6m, AQ = 3.25cm

As we know that △ACB ∼ △APQ, 

So, ar.(△ACB)/ar.(△APQ) = (BC)²/(PQ)²

 ar.(△ACB)/ar.(△APQ) = (10)²/(5)² = 100/25 = 4/1

Hence, ar(△ACB):ar(△APQ) = 4:1

问题 3. 两个相似三角形的面积分别为 81 cm ²和 49 cm ² 。求它们对应的高度之比。它们对应的中位数的比例是多少？

解决方案：

Let us consider two similar triangles, ABC and PQR whose altitudes are AD and PO

It is given that the areas of two similar triangles are 81 cm² and 49 cm²

So, △ABC = 81 cm²  and △PQR = 49 cm²

So, ar.(△ABC)/ar.(△PQR) = (AB)²/(PQ)²

81/49 = (AB/PQ)²

9/7 = AB/PQ

Now, in △ABD and △PQO

∠B = ∠Q

∠ADB = ∠POQ = 90°

Hence, △ABD ~ △PQO

So, AB/PQ = AD/PO

Hence, AD/PO = 9/7 

Or

AD:PO = 9:7

As we know, that the ratio of the areas of two similar triangles are proportional to the square of their corresponding altitudes and also the squares of their corresponding medians.

So, the ratio in their medians = 9 : 7

问题 4. 两个相似三角形的面积分别为 169 cm ²和 121 cm ² 。如果较大三角形的最长边是 26 厘米，请找出较小三角形的最长边。

解决方案：

Given that both the triangles are similar 

So, the area of larger triangle(ABC) = 169 cm²

and area of the smaller triangle(PQR) = 121 cm²

The length of the longest sides of the larger triangles(AC) = 26 cm

Let us assume the length of longest side of the smaller triangle(PR) = x 

So, the ar.(△ABC)/ar.(△PQR) = (AC)²/(PR)²

 169/121 = (26)²/(x)²

 13/11 = 26/x

x = (13 × 26)/11

x = 22 

Hence, the length of the longest side of the smaller triangle is 22 cm.

问题 5. 两个相似三角形的面积分别为 25 cm ²和 36 cm ² 。如果第一个三角形的高度是 2.4 厘米，求另一个三角形的对应高度。

解决方案：

Given that, the area of the first triangle = 25 cm²

and the area of second = 36 cm²

Altitude of the first triangle = 2.4 cm

Let us consider the altitude of the second triangle = x

It is given that both the triangles are similar, so

ar.(first triangle)/ar.(second triangle) = (Altitude of the first triangle)²/(Altitude of the second triangle)²

⇒ 25/36 = (2.4)²/x²

⇒ 2.4/x = 5/6

⇒ x = (2.4 × 6)/5 = 14.4/5 = 2.88

Hence, the altitude of the second triangle is 2.88cm

问题 6. 两个相似三角形对应的高分别是 6 厘米和 9 厘米。求它们的面积比。

解决方案：

Given that the length of the corresponding altitude of two triangles are 6 cm and 9 cm

Also, both the triangles are similar

So, 

ar.(first triangle)/ar.(second triangle) = (6)²/(9)²

= 36/81

= 4/9

Hence, the ratio of the areas of triangles is 4:9

问题 7. ABC 是一个三角形，其中∠A =90°，AN⊥ BC，BC = 12 cm，AC = 5cm。求 ΔANC 和 ΔABC 的面积比。

解决方案：

Given that, 

In ∆ABC, ∠A = 90°

AN ⊥ BC

BC = 12 cm, AC = 5 cm

So, in ∆ANC and ∆ ABC,

∠ANC = ∠BAC = 90°       

∠C =∠C              [Common]

So, by AA,

∆ANC ∼ ∆ ABC           

ar.(∆ANC)/ar.(∆ABC) = (AC)²/(BC)² = (5)²/(12)² = 25/144 

Hence, the ratio of the areas of ∆ANC and ∆ABC is 25:144

问题 8. 在图中，DE ||公元前

(i) 如果 DE = 4 cm, BC = 6 cm, 面积 (ΔADE) = 16 cm ² ，求 ΔABC 的面积。

(ii) 如果 DE = 4cm，BC = 8 cm，面积 (ΔADE) = 25 cm ² ，求 ΔABC 的面积。

(iii)如果 DE : BC = 3 : 5。计算 ΔADE 和梯形 BCED 的面积之比。

解决方案：

Given that, DE || BC  

So, In ∆ADE and ∆ABC

∠ADE = ∠B

∠BAC = ∠DAE

So, by AA 

∆ADE ~ ∆ABC

(i) Given that DE = 4 cm, BC = 6 cm and ar(∆ADE) = 16 cm²

As we know that ∆ADE ∼ ∆ABC

So, ar(∆ADE)/ar(∆ABC) = DE²/BC²

16/ar(∆ABC) = 4²/6² = 16/36

So,  16 × area ∆ABC = 16 × 36

⇒ ar.(∆ABC) = 36cm²

(ii) Given that DE = 4cm, BC = 8 cm and ar(∆ADE) = 25 cm²

As we know that ∆ADE ∼ ∆ABC

So, ar(∆ADE)/ar(∆ABC) = DE²/BC²

25/area(∆ABC) = (4)²/(8)² = 16/64

area(∆ABC) = (25 × 64)/16 = 100 cm² 

(iii) Given that, DE : BC = 3 : 5

As we know that ∆ADE ∼ ∆ABC

area(∆ADE)/area(∆ABC) = DE²/BC² = (3/5)² = 9/25 

25 (area(∆ADE)) = 9 (area ∆ABC)

25 (area(∆ADE)) = 9(area ∆ADE + area trapezium BCED)

area(∆ADE)/area of trapezium BCED = 9/16

Hence, the ratio of the areas of ∆ADE and the trapezium BCED is 9:16

问题 9. 在 ΔABC 中，D 和 E 分别是 AB 和 AC 的中点。求 ΔADE 和 ΔABC 的面积比。

解决方案：

Given that, in ∆ABC, D and E are the mid points of AB and AC

So, DE||BC and DE = 1/2BC

In ∆ADE and ∆ABC,

∠ADE = ∠B 

∠DAE = ∠BAC   [Common]

By AA

∆ADE ∼ ∆ABC 

So, ar(∆ADE)/ar(∆ABC) = (DE)²/(BC)² $=\frac{(\frac{1}{2}BC)^2}{BC^2}\\ =\frac{1×BC^2}{4BC^2}=\frac{1}{4}$ 

Hence, the ratio of the areas of ∆ADE and ∆ABC is 1:4

问题 10. 两个相似三角形的面积分别为 100 cm ²和 49 cm ² 。如果较大的三角形的高度为5厘米，则找到另一个相应的高度。

解决方案：

Let us consider ∆ABC and ∆DEF

It is given that, area ∆ABC = 100 cm²

and area ∆DEF = 49 cm²

AL perpendicular BC  and DM/EF

AL = 5cm,

Let DM = x cm

It is given that ∆ABC ~ ∆DEF    

So, ar(∆ABC)/ar(∆DEF) = AL²/DM²

100/49 = (5)²/(x)²

100/49 = 25/x²

x² = (25 × 49)/100 = 49/4

x = $\sqrt{\frac{49}{4}}=\frac{7}{2}=3.5$

Hence, the length of altitude of second triangle is 3.5cm