Question

Please please help

17 A right-angled triangle has base 60 cm and height 80 cm. A rectangle is inscribed in the triangle, so that
one of its sides lies along the base of the triangle.
a Let the rectangle have length y cm and height x cm. By using similar triangles, show that
y = (80 - x).
I
b Hence find the dimensions of the rectangle of maximum area.
18 A coal chute is built in the shape of an upturned cone. The sum of the base radius r and the height h is
12 metres.
a Show that the volume V of the coal chute is given by V = 4r²r³. (Recall that the volume
of a cone is given by V = r²h.)
b
Find the radius of the cone that yields the maximum volume.

Answer 1

17.a) See below.

17.b) x = 34.3 cm, y = 34.3 cm

18.a) See below.

18.b) r = 8 m

Explanation:

Question 17

Given information (see attachment 1):

• A right-angled triangle has base 60 cm and height 80 cm.
• A rectangle is inscribed in the triangle, so that one of its sides lies along the base of the triangle.
• Let the rectangle have length y cm and height x cm.

Part (a)

According to the Similar Triangles Side Splitter Theorem, if a line parallel to one side of a triangle intersects the other two sides, then this line divides those two sides proportionally. Therefore, triangles ABC and ADE are similar (refer to attachment 1).

As corresponding sides of similar triangles are always in the same ratio, the ratio of the base to height of the smallest triangle is equal to the ratio of the base to height of the largest triangle:

`\implies \sf DE:AD=BC:AB`

`\implies y:80-x=60:80`

`\implies (y)/(80-x)=(60)/(80)`

`\implies 80y=60(80-x)`

`\implies y=(60(80-x))/(80)`

`\implies y=(3(80-x))/(4)`

Part (b)

The maximum area of a rectangle is when the length and height are equal:

`\implies x=y`

Therefore, to find the value of x when the area of the rectangle is at its maximum, substitute x = y into the equation from part (a) and solve for x:

`\implies x=(3(80-x))/(4)`

`\implies 4x=3(80-x)`

`\implies 4x=240-3x`

`\implies 7x=240`

`\implies x=(240)/(7)`

`\implies x=34.3\; \sf cm\;\;(nearest\;tenth)`

Therefore, the dimensions of the rectangle of maximum area to the nearest tenth are:

• x = 34.3 cm
• y = 34.3 cm

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Question 18

Given information:

• A coal chute is built in the shape of an upturned cone.
• The sum of the base radius r and the height h is 12 metres.

Part (a)

The formula for the volume of a cone is:

`V=(\pi r^2 h)/(3)`

If the sum of the base radius r and the height h is 12 metres then:

`\implies r + h = 12`

`\implies h=12-r`

Substitute the expression for h into the volume of a cone formula to create an equation for V in terms of r:

`\implies V=(\pi r^2 (12-r))/(3)`

`\implies V=(12\pi r^2-\pi r^3)/(3)`

`\implies V=(12\pi r^2)/(3)-(\pi r^3)/(3)`

`\implies V=4\pi r^2-(1)/(3)\pi r^3`

Part (b)

To find the radius of the cone that yields the maximum volume, differentiate V with respect to x:

`\implies \frac{\text{d}V}{\text{d}x}=2 \cdot 4\pi r^(2-1)-3 \cdot (1)/(3)\pi r^(3-1)`

`\implies \frac{\text{d}V}{\text{d}x}=8\pi r-\pi r^2`

`\implies \frac{\text{d}V}{\text{d}x}=\pi r(8-r)`

Set the differentiated equation equal to zero and solve for x:

`\implies \pi r(8-r)=0`

`\implies 8-r=0`

`\implies r=8`

(Note: r is also equal to zero, but we can discard that solution since the radius has to be more than zero).

Therefore, the radius of the cone that yields the maximum volume is 8 m.