The diagram shows a semi-circle standing on a diameter [AC], and [BD] ⊥ [AC]. (a) (i) Prove that the triangles ABD and DBC are similar. (ii) If |AB| = x, |BC| = 1, and |BD| = y, write y in terms of x. (b) Use your result from part (a)(ii) to construct a line segment equal in length (in centimetres) to the square root of the length of the line segment [TU] which is drawn below.

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The diagram shows a semi-circle standing on a diameter [AC], and [BD] ⊥ [AC] - Leaving Cert Mathematics - Question 4 - 2016

Question 4

The diagram shows a semi-circle standing on a diameter [AC], and [BD] ⊥ [AC]. (a) (i) Prove that the triangles ABD and DBC are similar. (ii) If |AB| = x, |BC| = 1,... show full transcript

Worked Solution & Example Answer:The diagram shows a semi-circle standing on a diameter [AC], and [BD] ⊥ [AC] - Leaving Cert Mathematics - Question 4 - 2016

Step 1

Prove that the triangles ABD and DBC are similar.

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Answer

To prove that triangles ABD and DBC are similar, we need to show that they have the same angle measures.

Angle Analysis: Since [AC] is the diameter of the semicircle, by the Inscribed Angle Theorem, the angle ∠ABD is 90 degrees because any angle inscribed in a semicircle is a right angle. Thus, we have:

$|\angle ABD| = |\angle DBC| = 90^\circ$
Corresponding Angles: Next, we see that both triangles share angle ∠DAB. So, we have:

$|\angle DAB| = |\angle DBC|$
Conclusion: Since two angles in triangle ABD are equal to two angles in triangle DBC, by the AA (Angle-Angle) criterion for similarity, triangle ABD is similar to triangle DBC. Therefore, the triangles ABD and DBC are similar.

Step 2

If |AB| = x, |BC| = 1, and |BD| = y, write y in terms of x.

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Answer

To find y in terms of x, we apply the Pythagorean theorem to both triangles.

For triangle ABD: We can express the sides as follows:

$|AD|^2 = |AB|^2 + |BD|^2\ =>\ |AD|^2 = x^2 + y^2\ (1)$
For triangle DBC: We express the sides for triangle DBC:

$|DC|^2 = |BC|^2 + |BD|^2\ =>\ |DC|^2 = 1^2 + y^2\ =>\ |DC|^2 = 1 + y^2\ (2)$
Equating the two triangles: Since |AC| = |AD| + |DC|, we have:

$|AC|^2 = |AB|^2 + |BC|^2\ =>\ |AC|^2 = x^2 + 1\ =>\ |AC|^2 = (x^2 + 1 + y^2)$

From equation (1) and (2), we can solve for y:

Equating: $(x^2 + y^2) + (1 + y^2) = |AC|^2 \ => \ 2y^2 = x^2 + 1 - y^2\ => \ 3y^2 = x^2 + 1\ => \ y = \sqrt{\frac{x^2 + 1}{3}}$

Step 3

Use your result from part (a)(ii) to construct a line segment equal in length (in centimetres) to the square root of the length of the line segment [TU].

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Answer

To construct the line segment that equals the square root of the length of line segment [TU], you'll utilize the previously derived relationship:

Determine TU Length: Measure the length of segment [TU]. Let’s say the length is L cm.
Calculate y: From part (a)(ii), we set up the equation for y based on the value of x. If you find |TU|, let’s assume:

$y = \sqrt{L}$
Construct the segment: Use a compass to measure y and mark that length from a point, T, on your drawing. Ensure this measures to the previously calculated length derived from your result in part (a)(ii).

By following these steps, you'll successfully construct the necessary line segment.

The diagram shows a semi-circle standing on a diameter [AC], and [BD] ⊥ [AC] - Leaving Cert Mathematics - Question 4 - 2016

Worked Solution & Example Answer:The diagram shows a semi-circle standing on a diameter [AC], and [BD] ⊥ [AC] - Leaving Cert Mathematics - Question 4 - 2016

Prove that the triangles ABD and DBC are similar.

If |AB| = x, |BC| = 1, and |BD| = y, write y in terms of x.

Use your result from part (a)(ii) to construct a line segment equal in length (in centimetres) to the square root of the length of the line segment [TU].

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