10类NCERT解决方案-第6章三角形-练习6.4

问题1.令∆ ABC〜∆ DEF及其面积分别为64 cm ²和121 cm ² 。如果EF = 15.4 cm，则找到BC。

解决方案：

According to the theorem 1, we get

$\frac{ar(ΔABC)}{ar(ΔDEF)} = \frac{AB^2}{DE^2} = \frac{AC^2}{DF^2} = \frac{BC^2}{EF^2}$

$\frac{ar(ΔABC)}{ar(ΔDEF)} = \frac{BC^2}{EF^2}$

$\frac{64}{121} = \frac{BC^2}{15.4^2}$

$\frac{8^2}{11^2} = \frac{BC^2}{15.4^2}$

$\frac{8}{11} = \frac{BC}{15.4}$

BC = $\frac{8}{11}$ × 15.4

BC = 11.2 cm

问题2.梯形ABCD与AB ||的对角线DC在O点处相交。如果AB = 2 CD，则求出三角形AOB和COD的面积比。

解决方案：

Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.

In △AOB and △COD,

∠ AOB = ∠ COD (Opposite angles)

∠ 1 = ∠ 2 (Alternate angles of parallel lines)

△AOB ~ △COD by AA property.

According to the theorem 1, we get

$\frac{ar(ΔAOB)}{ar(ΔCOD)} = \frac{AB^2}{CD^2}$

As, AB = 2CD

= $\frac{(2CD)^2}{CD^2}$

= $\frac{4CD^2}{CD^2}$

= $\frac{4}{1}$

ar(AOB) : ar(COD) = 4 : 1

问题3.在图中，ABC和DBC是同一基础BC上的两个三角形。如果AD在O处与BC相交，则表明$\frac{ar(ΔABC)}{ar(ΔDBC)} = \frac{AO}{DO}$ 。

解决方案：

Let’s draw two perpendiculars AP and DM on line BC.

Area of triangle = ½ × Base × Height

$\frac{ar(ΔABC)}{ar(ΔDBC)} = \frac{½ × BC × AP}{½ × BC × DM}$

$\frac{ar(ΔABC)}{ar(ΔDBC)} = \frac{AP}{DM}$ ……………………………(1)

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each 90°)

∠AOP = ∠DOM (Vertically opposite angles)

ΔAPO ~ ΔDMO by AA similarity

$\frac{AP}{DM} = \frac{AO}{DO}$ ……………………………(2)

From (1) and (2), we can conclude that

$\mathbf{\frac{ar(ΔABC)}{ar(ΔDBC)} = \frac{AO}{DO}}$

问题4.如果两个相似的三角形的面积相等，请证明它们是全等的。

解决方案：

As it is given, ΔABC ~ ΔDEF

According to the theorem 1, we have

$\frac{Area of (ΔABC)}{Area of (ΔDEF)} = \frac{BC^2}{EF^2}$

$\frac{BC^2}{EF^2}$ =1 [Since, Area(ΔABC) = Area(ΔDEF)

BC² = EF²

BC = EF

Similarly, we can prove that

AB = DE and AC = DF

Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]

问题5. D，E和F分别是∆ ABC的AB，BC和CA边的中点。求出∆ DEF和∆ ABC的面积比。

解决方案：

As, it is given here

DF = ½ BC

DE = ½ AC

EF = ½ AB

So, $\frac{DF}{BC} = \frac{DE}{AC} = \frac{EF}{AB} = \frac{1}{2}$

Hence, ΔABC ~ ΔDEF

According to theorem 1,

$\frac{ar(ΔDEF)}{ar(ΔABC)} = \frac{DE^2}{AC^2} = \frac{EF^2}{AB^2} = \frac{DF^2}{AC^2} = \frac{1^2}{2^2}$

$\frac{ar(ΔDEF)}{ar(ΔABC)} = \frac{1}{4}$

ar(ΔDEF) : ar(ΔABC) = 1 : 4

问题6.证明两个相似三角形的面积之比等于其相应中值之比的平方。

解决方案：

Given: AM and DN are the medians of triangles ABC and DEF respectively.

ΔABC ~ ΔDEF

According to theorem 1,

$\frac{ar(ΔABC)}{ar(ΔDEF)} = \frac{AB^2}{DE^2} = \frac{BC^2}{EF^2} = \frac{AC^2}{DF^2}$

So, $\frac{AB}{DE} = \frac{BC}{EF} = \frac{2BP}{2EQ} = \frac{BP}{EQ}$

$\frac{AB}{DE} = \frac{BP}{EQ}$ ……………………….(1)

∠B = ∠E (because ΔABC ~ ΔDEF)

Hence, ΔABP ~ ΔDEQ [SAS similarity criterion]

$\frac{BP}{EQ} = \frac{AP}{DQ}$ ……………………….(2)

From (1) and (2), we conclude that

$\frac{ar(ΔABC)}{ar(ΔDEF)} = \frac{AP^2}{DQ^2}$

Hence, proved!

问题7.证明在正方形的一侧上描述的等边三角形的面积等于在其对角线之一上描述的等边三角形的面积的一半。

解决方案：

Let’s take side of square = a

Diagonal of square AC = a√2

As, ΔBCF and ΔACE are equilateral, so they are similar

ΔBCF ~ ΔACE

According to theorem 1,

$\frac{ar(ΔACE)}{ar(ΔBCF)} = \frac{AC^2}{BC^2}$

=$\frac{(a√2)^2}{a^2}$

$\frac{ar(ΔACE)}{ar(ΔBCF)}$ = 2

Hence, Area of (ΔBCF) = ½ Area of (ΔACE)

勾选正确答案并说明理由：

问题8. ABC和BDE是两个等边三角形，使得D是BC的中点。三角形ABC和BDE的面积比为

(A)2：1(B)1：2(C)4：1(D)1：4

解决方案：

Here,

AB = BC = AC = a

and, BE = BD = ED = ½a

ΔABC ~ ΔEBD (Equilateral triangle)

According to theorem 1,

$\frac{ar(ΔABC)}{ar(ΔEBD)} = \frac{AB^2}{EB^2} = \frac{a^2}{(½a)^2}$

$\frac{ar(ΔABC)}{ar(ΔEBD)} = \frac{4}{1}$

Area of (ΔABC) : Area of (ΔEBD) = 4 : 1

Hence, OPTION (C) is correct.

问题9.两个相似三角形的边之比为4：9。这两个三角形的面积之比为

(A)2：3(B)4：9(C)81:16(D)16:81

解决方案：

ΔABC ~ ΔDEF

$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{4}{9}$

According to theorem 1,

$\frac{ar(ΔABC)}{ar(ΔDEF)} = \frac{AB^2}{DE^2}$

$\frac{ar(ΔABC)}{ar(ΔDEF)} = \frac{4^2}{9^2} = \frac{16}{81}$

Area of (ΔABC) : Area of (ΔDEF) = 16 : 81

Hence, OPTION (D) is correct.