The diagram shows a semi-circle standing on a diameter [AC], and [BD] ⊥ [AC]. (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.

Photo AI

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]. (i) Prove that the triangles ABD and DBC are similar. (ii) If |AB| = x, |BC| = 1, and... 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.

96%

114 rated

Answer

To prove that triangles ABD and DBC are similar, we will utilize the Angle-Angle (AA) similarity criterion.

Identifying Angles:
- From the diagram, we know that angle |ABD| = |DBD| = 90° (right angles) because [BD] is perpendicular to [AC].
- Additionally, since D lies on the semicircle with AC as the diameter, angle |ADB| is an inscribed angle intercepting arc AB. By the inscribed angle theorem, this angle is equal to angle |BDC|. Thus:
  
  $|ABD| + |DBD| + |ADB| = 180°$ which implies that |ABD| + |DBC| + |BDC| = 180°.
Conclusion:
- Therefore, |ABD| = |DBC| = 90° and ( |ADB| = |BDC| ) implies that triangles ABD and DBC are similar by AA criterion, hence confirmed as equiangular.

Step 2

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

99%

104 rated

Answer

We will use the Pythagorean theorem in triangle ABD and triangle BDC.

For triangle ABD:

$|AD|^2 = |AB|^2 + |BD|^2$

Given |AB| = x, we have:

$|AD|^2 = x^2 + y^2 ag{1}$
For triangle BDC:

$|DC|^2 = |BC|^2 + |BD|^2$

Given |BC| = 1, we then find:

$|DC|^2 = 1^2 + y^2 = 1 + y^2 ag{2}$
Using the semicircle: Since |AC| is the diameter (and therefore |AD| + |DC| = |AC|), we know that:

$|AD|^2 + |DC|^2 = |AC|^2$

Therefore, substituting equations (1) and (2) into the semicircle equality, we have:

$x^2 + y^2 + 1 + y^2 = |AC|^2$
Simplifying: Thus:

$2y^2 + x^2 + 1 = |AC|^2$

Solving for y:

$2y^2 = |AC|^2 - (x^2 + 1)$
Expressing y: We can now express y in terms of x as:

$y = rac{ ext{sqrt}(|AC|^2 - (x^2 + 1))}{ ext{sqrt}(2)}$

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].

96%

101 rated

Answer

To construct a line segment that represents the length equal to ( \sqrt{|TU|} ) in centimeters, perform the following steps:

Measure Length:
- First, measure the length of the segment [TU]. Assume the measured length is L.
Setting Up Compass:
- Open the compass to a length of 1 cm.
Constructing the Square:
- From one endpoint of segment [TU], draw a perpendicular line of 1 cm.
- Continue constructing a right angle triangle to form a square based on the principle of the previous results. The length of the segments must be proportional to the sqrt values derived.
Finalizing the Length:
- Adjust accordingly to ensure the new line segment equals ( \sqrt{L} ) by equal partitioning based on the result from step (a)(ii).

This construction ensures you obtain a segment equivalent to ( \sqrt{|TU|} ).

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].

Join the Leaving Cert students using SimpleStudy...