Question

A carpenter cuts a wooden cone into three parts A, B and C by
base as shown in the diagram. The heights of the three parts are equal
planes
Find the ratio of the volumes of parts A, B and C
(I) Find the ratio of the base areas of parts A, B and C
(2) If the volume of the original cone is 540 cu em,
find the volume of part B

if you can solve this then you can solve any question in class X board exams (pisa problems)
​

Answer 1

Answer:use similar triangle , as the three parts have same height then take the radius of the three parts as R,2R,3R

Explanation:

Answer 2

Recall the following formulae:

$\text {Volume of a cone } = \dfrac{1}{3} \pi r^2h$

$\text {Volume of a frustum } = \dfrac{1}{3} \pi h (R_2^2 + R_1^2 +R_1R_2)$

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Show the relationships among the 3 radii:

*See attached for a cross section of the cone.

△ABG is similar to △ACF

$\dfrac{R_{1} }{R_{2}} = \dfrac{h}{2h}$

$\dfrac{R_{1} }{R_{2}} = \dfrac{1}{2}$

$R_2 = 2R_1$

△ABG is similar to △ADE

$\dfrac{R_{1} }{R_{3}} = \dfrac{h}{3h}$

$\dfrac{R_{1} }{R_{3}} = \dfrac{1}{3}$

$R_3 = 3R_1$

Define r:

$\text{Let }R_1 = r$

$\text{Then }R_2 = 2r$

$\text{And }R_3 = 3r$

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Find the volume of Part A:

$\text {Volume of a cone } = \dfrac{1}{3} \pi r^2h$

$\text {Volume of A } = \dfrac{1}{3} \pi r^2h$

Find the volume of Part B:

$\text {Volume of a frustum } = \dfrac{1}{3} \pi h (R_2^2 + R_1^2 +R_1R_2)$

$\text {Volume of a frustum } = \dfrac{1}{3} \pi h ((2r)^2 + r^2 +(2r)(r))$

$\text {Volume of a frustum } = \dfrac{1}{3} \pi h (4r^2 + r^2 +2r^2)$

$\text {Volume of a frustum } = \dfrac{1}{3} \pi h (7r^2)$

$\text {Volume of a frustum } = \dfrac{7}{3} \pi hr^2$

Find the volume of Part C:

$\text {Volume of a frustum } = \dfrac{1}{3} \pi h (R_3^2 + R_2^2 +R_3R_2)$

$\text {Volume of a frustum } = \dfrac{1}{3} \pi h ((3r)^2 + (2r)^2 +(3r)(2r))$

$\text {Volume of a frustum } = \dfrac{1}{3} \pi h (9r^2 + 4r^2 +6r^2)$

$\text {Volume of a frustum } = \dfrac{1}{3} \pi h (19r^2)$

$\text {Volume of a frustum } = \dfrac{19}{3} \pi h r^2$

Find the ratio of the volumes:

$A : B : C = \dfrac{1}{3} \pi hr^2 : \dfrac{7}{3} \pi hr^2 : \dfrac{19}{3} \pi hr^2$

$A : B : C = \dfrac{1}{3} : \dfrac{7}{3} : \dfrac{19}{3}$

$A : B : C = 1 : 7 : 19$

Answer: The ratio of the volumes of A : B : C is 1 : 7 : 9

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Find the base area of A:

$\text{area } = \pi r^2$

Find the base area of B:

$\text{area } = \pi (2r)^2$

$\text{area } = 4 \pi r^2$

Find the base area of C:

$\text{area } = \pi (3r)^2$

$\text{area } =9 \pi r^2$

Find the ratio of A : B :C:

$A : B : C = \pi r^2 : 4 \pi r^2 : 9\pi r^2$

$A : B : C = 1 : 4 : 9$

Answer: The ratio of the base area of A : B : C is 1 : 4 : 9

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Find the volume of B:

$\text{Volume of the cone} = \dfrac{1}{3} \pi (3r)^2(3h)$

$\text{Volume of the cone} = 9\pi r^2h$

$9\pi r^2h = 540$

$\pi r^2h = 60$

$\text{Volume of B} = \dfrac{7}{3} \pi r^2h$

$\text{Volume of B} = \dfrac{7}{3} \bigg(\pi r^2h\bigg)$

$\text{Volume of B} = \dfrac{7}{3} \bigg(60\bigg)$

$\text{Volume of B} = 140 \text { cm}^3$

Answer: The volume of B is 140 cm³.