10类NCERT解决方案-第13章表面积和体积–练习13.4

问题1.酒杯的形状为截锥，截锥的高度为14厘米。其两个圆形端的直径分别为4厘米和2厘米。找到玻璃的容量。

解决方案：

Given values:

Height of frustum (h) = 14 cm

Radius of larger circle end (R) = $\frac{4}{2}$ = 2 cm

Radius of smaller circle end (r)= $\frac{2}{2}$ = 1 cm

Capacity of frustum-shaped glass = Volume of Frustum

= $\frac{1}{3}$ πh (r² + R² + rR)

= $\frac{1}{3}$ × π × 14 ((1 × 1) + (2 × 2) × (2 × 1))

= $\frac{1}{3} × \frac{22}{7}$ × 14 × 7                     (taking π=$\frac{22}{7}$)

= $\frac{(22 × 14)}{3}$

= 102.67 cm³

Hence, the capacity of frustum-shaped glass = 102.67 cm³

问题2.圆锥台的斜面高度为4厘米，其圆锥形端的周长(周长)为18厘米和6厘米。找到平截头体的曲面区域。

解决方案：

Slant height of frustum (l) = 4 cm

Let radius of smaller circle end = r

Let radius of larger circle end = R

Circumference of circle = 2π × (radius of circle)

Circumference of larger circle = 2πR

18 cm² = 2πR

R = $\frac{18}{2π}$

R = $\frac{9}{π}$cm

Circumference of smaller circle = 2πr

6 cm² = 2πr

r = $\frac{6}{2π}$

r = $\frac{3}{π}$cm

Now, as curve surface area of frustum = π (r+R) l

= π × ($\frac{9}{π}+\frac{3}{π}$) × 4                  

= 12 × 4                         (Taking π common and canceling it)

= 48cm²

Hence, the curved surface area of the frustum = 48cm²

问题3.土耳其人使用的帽子菲斯帽的形状像圆锥体的截头圆锥体(见图)。如果其在敞开侧的半径为10厘米，在上底的半径为4厘米，其倾斜高度为15厘米，请找到用于制造它的材料区域。

解决方案：

Given values:

Slant height of frustum (l)= 15 cm

Let radius of smaller circle end (r) = 4 cm

Let radius of larger circle end (R) = 10 cm

Area of material used for making it = Curve surface area + area of upper base

= (π(r+R)l) + (πr²)

= π ((r+R)l + r²)                                              (Taking π common)

= π ((4+10) × 15 + (4 × 4))

= $\frac{22}{7}$× (226)                                         (Taking π = $\frac{22}{7}$)

= 710.286 cm²

Hence, the area of material used for making it = 710.286 cm² 

问题4：一个从顶部打开并由金属板制成的容器，其形状为圆锥台的截头锥体，圆锥台的高度为16 cm，其下端和上端的半径分别为8 cm和20 cm。找到可以完全装满容器的牛奶的成本，每升的价格为₹20。^{如果每100 cm 2}花费₹8，则还要查找用于制造容器的金属板的成本。 (取π= 3.14)

解决方案：

Given values:

Height of frustum (h)= 16 cm

Let radius of smaller circle end (r) = 8 cm

Let radius of larger circle end (R) = 20 cm

The amount of milk to fill the container = Volume of frustum

= $\frac{1}{3}$ πh (r² + R² + rR)

= $\frac{1}{3}$ × 3.14 × 16 (8×8 + 20×20 + 8×20)                (Taking π=3.14)

= $\frac{1}{3}$ × 3.14 × 16 × (624)

= 10449.92 cm³

Cost of 1 litre milk = ₹ 20

And as, 1 m³ = 1000 cm³ = 1 litre

10449.92 cm³ = ($\frac{1}{1000}$) ×10449.92 litres

cost of 10449.92 cm³ = ($\frac{10449.92}{1000}$) × 20

= ₹ 208.998

Now, metal sheet used to make the container = Curve surface area + area of lower base

= (π(r+R)l) + (πr²)                         

= π ((r+R)l + r²)                                              (Taking π common)

= π ((20+8) × (√(16²+(20-8)²)) + (8 × 8))                                (Slant height (l) = √(h²+(R-r)²))

= 3.14 × (28 × √400 + 64)                                       (Taking π = 3.14)

= 3.14 × (624)     

= 1959.36 cm²

Hence, the metal sheet used to make the container = 1959.36 cm² 

As, cost of 100 cm²= ₹ 8

1959.36 cm² = (8/100) × 1959.36

= ₹156.748

Hence, the cost of the milk which can completely fill the container = ₹ 208.998

and, the cost of metal sheet used to make the container = ₹156.748

问题5.一个20厘米高，垂直角度为60°的金属直角圆锥在其高度的中间被一个平行于其底部的平面切成两部分。如果将这样得到的截锥体拉成直径为1/16 cm的金属丝，请找到金属丝的长度。

解决方案：

As the angle is cut into two equal parts, the height gets half too.

Let radius of smaller circle end = r

Let radius of larger circle end = R

In ∆PFR and ∆PEB

tan ∝ = $\frac{(opposite \hspace{0.1cm}side)}{(adjacent \hspace{0.1cm}side)}$

tan 30° = $\frac{EB}{PE} = \frac{FR}{PF}$

$\frac{1}{√3} = \frac{r} {10} = \frac{R}{20}$

R = $\frac{20}{√3}$

r = $\frac{10}{√3}$

and as height of frustum = 10 cm

So according to the question,

Frustum is converted to cylindrical wire having diameter $\mathbf{\frac{1}{16}}$ cm

Volume of Frustum = Volume of Cylinder

Volume of Frustum = $\frac{1}{3}$ πh (r² + R² + rR)

= $\frac{1}{3} × π × 10 (\frac{10}{√3} × \frac{10}{√3} + \frac{20}{√3} × \frac{20}{√3} + \frac{20}{√3} × \frac{10}{√3})$  

= $\frac{1}{3} × π × 10 × (\frac{100}{3} + \frac{400}{3} + \frac{200}{3})$

= $\frac{1}{3} × π × 10 × \frac{700}{3}$

= $\frac{7000π}{ 9}$ cm³  ………………………..(1)

Volume of Cylinder = $\frac{1}{3}$ π(radius)²H

= $\frac{1}{3}$π($\frac{1}{16×2}$)²H …………………(2)

As (1) = (2) , then

7000π / 9 = 1/3 π(1/(16×2))²H

H = $\frac{7000π × 32 × 32}{9 × π}$                   (cancel π from both side)

H = 796444.443 cm

H = 7964.44 m

Hence, the length of the wire = 7964.44 m