10类NCERT解决方案-第13章表面积和体积–练习13.5

问题1.将直径3 mm的铜线缠绕在长度为12 cm，直径为10 cm的圆柱体上，以覆盖圆柱体的曲面。假设铜的密度为8.88 g / cm ³ ，求出导线的长度和质量。

解决方案：

Given,

Radius of copper wire (r) = $\frac{3}{2}$ mm = $\frac{3}{20}$ cm

Height of cylinder (H) = 12 cm

Radius of Cylinder (R) = $\frac{10}{2}$ = 5 cm

Length of copper wire needed for one round will be circumference of cylindrical circle end = 2πR

=2π(5)

length of each round = 10π cm

Now, to completely cover the cylinder with wire no of rounds (n) of copper wire should be:

n = height of cylinder / diameter of copper wire

n = $\frac{12}{\frac{3}{10}}$

no. of rounds = 40

Now, Total length of copper wire will be:

h = no. of rounds × length of each round

h = 40×10π

h = 400×3.14                            (taking π = 3.14)

h = 1256 cm

Volume of copper wire = πr²h

=π×($\frac{3}{20}$)²×1256

=π×$\frac{3}{20} ×\frac{3}{20}$×1256

= 88.82 cm³                                (taking π = $\frac{22}{7}$)

Now, as it is given that,

For 1cm³ = 8.88 g of wire

so, for 88.82 cm³ = 8.88×88.82 g

Mass of copper wire = 788.7216 g

Hence, the length and mass of the wire are1256 cm and 788.216 grams respectively.

问题2。制作一个直角三角形，其侧面为3厘米和4厘米(斜边除外)围绕其斜边旋转。找到如此形成的双锥的体积和表面积。 (选择适当的π值。)

解决方案：

After revolving the triangle about its hypotenuse, we get two cones having slant height as 3 cm and 4 cm each.

Hypotenuse of triangle (CD) = √(3² + 4²) = 5 cm

l₁ = 3 cm

l₂ = 4 cm

ar(△ACD) = $\frac{1}{2}$×AC×AD = ½×AO×CD         (as AC⊥AD and AO⊥CD)

⇒ AC×AD = AO×CD

⇒ 4×3 = AO×5

AO = $\frac{12}{5}$ cm

Hence, radius of cone base is $\frac{12}{5}$ cm 

Now, Volume of double cone = Volume of cone 1 + Volume of cone 2

= ($\frac{1}{3}$)πr²×OD + ($\frac{1}{3}$)πr²×OC

= ($\frac{1}{3}$)πr²×(OD+OC)

= $\mathbf{\frac{1}{3}}$πr²×CD

= $\frac{1}{3}×\frac{22}{7}×\frac{12}{5}×\frac{12}{5}×5$

= 30.17 cm³

Now, Curve Surface area of double cone = CSA of cone 1 + CSA of cone 2

= πrl₁ + πrl₂

= πr(l₁+l₂)

= $\frac{22}{7}×\frac{12}{5}$×(3+4)

= 30.17 cm²

Hence, the volume and surface area of the double cone so formed is 30.17 cm³ and 30.17 cm² .

问题3.一个水箱，内部长150厘米×120厘米×110厘米，里面有129600厘米³的水。将多孔砖放入水中，直到水箱充满水。每块砖吸收其自身水量的十七分之一。在不溢出水的情况下可以放入多少块砖，每块砖为22.5厘米×7.5厘米×6.5厘米?

解决方案：

Given,

Length of Cistern (L) = 150 cm

Breadth of Cistern (W) = 120 cm

Height of Cistern (H) = 110 cm

Length of Brick (l) = 22.5 cm

Breadth of Brick (w) = 7.5 cm

Height of Brick (h) = 6.5 cm

Volume of Cistern = L×B×H

= 150×120×110

= 1980000 cm³

Volume of Water =  129600 cm³

Empty space left in Cistern = 1980000-129600

= 1850400 cm³

Volume of Brick = l×b×h

= 22.5×7.5×6.5

= 1096.88 cm³

Volume of n Bricks = 1096.88×n cm³

Volume absorbed by each brick = ($\frac{1}{17}$)^th (volume of brick)

= $\frac{1}{17}$×1096.88 cm³

= 64.522 cm³

Then, Volume absorbed by n bricks = 64.522×n

Volume of brick = Empty space left in Cistern + volume absorbed by bricks

1096.88×n = 1850400 + 64.522×n

n×(1096.88-64.522) = 1850400

n = $\frac{1850400}{1032.358}$

n = 1792.40

n ≈ 1792 

Hence, 1792 bricks can be put in without overflowing the water.

问题4：在给定月份的两个星期中，河谷降雨量为10厘米。如果山谷面积为7280 km ² ，则表明总降雨量大约等于三条河流的正常水量的补充，三条河流的长度分别为1072 km，宽75 m和深3 m。

解决方案：

Length of river (l) = 1072 km

Width of river (w) = $\frac{75}{1000}$ km = 0.075 km

Depth of river (h) = $\frac{3}{1000}$= 0.003 km

Depth of rainfall = $\frac{10}{(1000×100)}$ km

The total rainfall was approximately equivalent to the addition to the normal water of three rivers each

Which means,

(Volume of three rivers = ((Surface area of a valley)×Depth of rainfall) has to be equal.

Let’s check, for each case,

so, 

Volume of three rivers = 3×(l×b×h)

= 3×(1072×0.075×0.003) km³

 = 0.7236 km³ ………………………..(1)

Volume of rainfall in valley = Area of valley×depth of rainfall

= 7280×$\frac{10}{100000}$

= 0.728 km³……………………………….(2)

From (1) and (2) we can see that, 

Hence, proved, Total rainfall was approximately equivalent to the addition to the normal water of three rivers each.

问题5。由锡板制成的油漏斗由一个10厘米长的圆柱形部分组成，该部分附着在圆锥台上。如果总高度为22厘米，圆柱部分的直径为8厘米，漏斗顶部的直径为18厘米，请找到制作漏斗所需的锡板面积(见图13.25)。

解决方案：

Oil funnel contains two shapes = frustum + Cylinder

Given values, 

Larger radius of frustum (R) = $\frac{18}{2}$ = 9 cm

Smaller radius of frustum (r) = $\frac{8}{2}$ = 4 cm

Height of frustum (H) = 22-10 = 12 cm

Radius of cylinder (r) = $\frac{8}{2}$ = 4 cm

Height of cylinder (h) = 10 cm

Area of the tin sheet required = CSA of frustum + CSA of cylinder

CSA of frustum = π(r+R)l

= π×(9+4)×√(12² + (9-4)²)                       (l =√(H² + (R-r)²))

= π×13×13

= 169π cm²

CSA of cylinder = 2πrh

= 2×π×4×10

= 80π cm²

Area of the tin sheet required = 169π + 80π

= 249π

= 782.571 cm²                                                    (Taking π=$\frac{22}{7}$)

Hence, the area of the tin sheet required to make the funnel is 782.571 cm²

问题6.使用解释的符号推导第13.5节中给出的圆锥体的截头圆锥体的曲面表面积和总表面积的公式。

解决方案：

Let ADE be a cone. From the cone the frustum BCDE is cut by a plane parallel to its base. Here, r and R are the radii of the frustum ends of the cone and h be the frustum height.

Here, as BC||DE

In △ADG and △ABF

So, ΔADG ∼ ΔABF (as BC||DE)

$\frac{AB}{AD} = \frac{AF}{AG} = \frac{BF}{GD}$

$\frac{L-l}{L} = \frac{H-h}{H} = \frac{r}{R}$…………..(1)

from (1), we get

1-($\frac{l}{L}$) = $\frac{r}{R}$

$\frac{l}{L}$ = 1 – $\frac{r}{R}$

$\frac{l}{L}$ = $\frac{(R-r)}{R}$

L(R-r) = Rl………………………(2)                 (By rearranging)

Total surface area of frustum = CSA of frustum + Area of upper circular end + Area of the lower circular end

• CSA of frustum = CSA of cone ADE – CSA of cone ABC

= πRL – πr(L-l)

= πL(R-r)+πrl             

= πRl+πrl                 (from (2) replacing L(R-r) = Rl )

= π(R+r)l

• Area of upper circular end = πR²
• Area of the lower circular end = πr²

Total surface area of frustum = π(R+r)l + πR² + πr²

问题7.使用解释的符号推导第13.5节中给您的圆锥台的截锥体体积的公式。

解决方案：

Let ADE be a cone. From the cone the frustum BCDE is cut by a plane parallel to its base. Here, r and R are the radii of the frustum ends of the cone and h be the frustum height.

Here, as BC||DE

In △ADG and △ABF

So, ΔADG ∼ ΔABF (as BC||DE)

$\frac{AB}{AD} = \frac{AF}{AG} = \frac{BF}{GD}$

$\frac{L-l}{L} = \frac{H-h}{H} = \frac{r}{R}$…………………(1)

from (1), we get

1-($\frac{h}{H}$) = $\frac{r}{R}$

$\frac{h}{H}$ = 1 – $\frac{r}{R}$

$\frac{h}{H}$ = $\frac{R-r}{R}$

H(R-r) = Rh………………………(2)                   (By rearranging)

Total volume of frustum of the cone will be = Volume of cone ADE – Volume of cone ABC

= $\mathbf{\frac{1}{3}}$πR²H – $\mathbf{\frac{1}{3}}$πr²(H-h)

= $\frac{1}{3}$πR²H – $\frac{1}{3}$πr²H + $\frac{1}{3}$πr²h

= $\frac{1}{3}$π(H(R²-r²)+r²h)

= $\frac{1}{3}$π(H(R-r)(R+r)+r²h)                          (Replacing (R²-r²) = (R-r)(R+r))

= $\frac{1}{3}$π(Rh(R+r)+r²h)                                (from (2) replacing H(R-r) = Rh )

= $\frac{1}{3}$π(R²h+Rrh+r²h)

= $\frac{1}{3}$πh(R²+Rr+r²)                      (Taking h common)

Hence, Total volume of frustum of the cone will be = $\frac{1}{3}$πh(R²+Rr+r²)