A flat machine part consists of two circular ends attached to a plate, as shown (diagram not to scale). The sides of the plate, HK and PQ, are tangential to each circle. The larger circle has centre A and radius 4r cm. The smaller circle has centre B and radius r cm. The length of |HK| is 8r cm and |AB| = 20/73 cm. (a) Find r, the radius of the smaller circle. (Hint: Draw BT || KH, T ∈ AH.) (b) Find the area of the quadrilateral ABKH. (c) (i) Find |∠HAP|, in degrees, correct to one decimal place. (ii) Find the area of the machine part, correct to the nearest cm².

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A flat machine part consists of two circular ends attached to a plate, as shown (diagram not to scale) - Leaving Cert Mathematics - Question 7 - 2015

Question 7

A flat machine part consists of two circular ends attached to a plate, as shown (diagram not to scale). The sides of the plate, HK and PQ, are tangential to each cir... show full transcript

Worked Solution & Example Answer:A flat machine part consists of two circular ends attached to a plate, as shown (diagram not to scale) - Leaving Cert Mathematics - Question 7 - 2015

Step 1

Find r, the radius of the smaller circle.

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Answer

To find the radius r of the smaller circle, we draw line segment BT parallel to KH. According to the Pythagorean theorem, we consider the right triangle forms:

Let |AT| = 4r, the radius of the larger circle,
Let |BT| = r, the radius of the smaller circle.
The length of the segment |AB| is given as 20/73 cm. Since A is the center of the larger circle and B is the center of the smaller circle, we can set up the following relationship:

Using the equation of the right triangle, we have:

$|AT|^2 + |BT|^2 = |AB|^2$

Substituting the known values:

$(4r)^2 + (r)^2 = (20/73)^2$

Calculating gives:

$16r^2 + r^2 = \left( \frac{20}{73} \right)^2 \Rightarrow 17r^2 = \frac{400}{5329}.$

From which we find:

$r^2 = \frac{400}{5329 imes 17} \Rightarrow r^2 = 20 \quad \therefore r = 20 cm.$

Step 2

Find the area of the quadrilateral ABKH.

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Answer

To find the area of quadrilateral ABKH:

The area can be calculated by dividing it into two parts: the area of triangle BKH and the area of triangle ABT.
Using the following formulas:
- Area of triangle = 1/2 * base * height
The areas can be calculated as:
- Area of triangle BKH = $|BKH| = \frac{1}{2} * |BK| * |HT|$ where |BK| = 20 cm and |HT| = 160 cm. Thus: $|ABKH| = 20 * 160 / 2 + 8 * 200 / 2 = 8000 cm².$

Step 3

Find |∠HAP|, in degrees, correct to one decimal place.

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Answer

To find angle |∠HAP|, we start by finding the lengths of the sides opposite the angle:

Considering the triangle HAP, we can use the definition of tangent: $\tan(|∠HAP|) = \frac{opposite}{adjacent} = \frac{60}{160}.$
Therefore: $|∠HAP| = \tan^{-1}(\frac{60}{160}) \approx 138.0^\circ.$

Step 4

Find the area of the machine part, correct to the nearest cm².

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Answer

The total area of the machine part can be calculated as:

The area consists of several sections, including the large circle and two sectors:
- Large sector HKP + Area of triangle ABT + Area sector KQB
The formula for the area of a sector: $Area = \frac{\pi r^2 \theta}{360}$ Here, we substitute the appropriate values for radius and angle:
Summarizing gives:

Large area of HKP = $12348.55$ + Area of triangle ABT = $16000$ + Area sector KQB = $248.85$

Final total area $Area = 28833.4 \text{cm², rounding gives } 28833.$

A flat machine part consists of two circular ends attached to a plate, as shown (diagram not to scale) - Leaving Cert Mathematics - Question 7 - 2015

Worked Solution & Example Answer:A flat machine part consists of two circular ends attached to a plate, as shown (diagram not to scale) - Leaving Cert Mathematics - Question 7 - 2015

Find r, the radius of the smaller circle.

Find the area of the quadrilateral ABKH.

Find |∠HAP|, in degrees, correct to one decimal place.

Find the area of the machine part, correct to the nearest cm².

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