Photo AI
In the diagram, |BC| = |BD| and |∠ABD| = 118° - Leaving Cert Mathematics - Question 4 - 2010
Question 4
In the diagram, |BC| = |BD| and |∠ABD| = 118°. (i) Find x. (ii) Find y. (b) Prove that if three parallel lines make intercepts of equal length on a transversal, ... show full transcript
Worked Solution & Example Answer:In the diagram, |BC| = |BD| and |∠ABD| = 118° - Leaving Cert Mathematics - Question 4 - 2010
Step 1
(i) Find x.
Answer
To find the value of x, we need to consider the triangle ABC where we know that the sum of angles in a triangle is 180°.
Given that |∠ABD| = 118°, we can find |∠CDB| as follows:
-
[ |∠CDB| = 180° - |∠ABD| = 180° - 118° = 62° ]
-
In triangle BCD, since |BC| = |BD|, triangle BCD is isosceles. Therefore, the angles at B and C are equal. We can say: [ |∠BCD| = x \ |∠DCB| = x ]
-
Now, using the sum of angles in triangle BCD: [ x + x + 62° = 180° ] [ 2x + 62° = 180° ] [ 2x = 180° - 62° ] [ 2x = 118° ] [ x = 59° ]
Step 2
(ii) Find y.
Answer
To find y, we can use the triangle BCD again. We already established that: [ |∠CDB| = 118° ]
-
We also know that: [ |∠CDB| + |∠BCD| + |∠DCB| = 180° ]
-
Substituting for |∠BCD| and |∠DCB|: [ 118° + 59° + |∠DCB| = 180° ]
-
Now solving for |∠DCB|: [ |∠DCB| = 180° - 118° - 59° ] [ |∠DCB| = 3° ]
So, we can conclude: [ y = 59° ]
Step 3
Prove that if three parallel lines make intercepts of equal length on a transversal, then they will also make intercepts of equal length on any other transversal.
Answer
To prove this, let's denote three parallel lines as A, B, and C, and let D be a transversal intersecting them.
-
Let the segments intercepted by the transversal D on lines A, B, and C be equal, denoting the length of these segments as |DE| = |EF| = |FG|.
-
If we draw another transversal intersecting lines A, B, and C at points P, Q, and R, we can establish similar triangles.
-
By the properties of similar triangles, the intercept lengths on any transversal remain proportional to the segments on the transversal that originally had equal lengths.
-
Therefore, any other transversal will also yield equal intercept lengths between lines A, B, and C.
Hence, it is proven.
Step 4
Step 5
(ii) Draw the image of the square under the enlargement with centre O and scale factor 2.5.
Answer
To find the image of square OABC under the enlargement:
- Use the centre O as the focal point for the enlargement.
- Each vertex in square OABC will be moved away from O by a factor of 2.5.
- Thus, the new vertices will be:
- O remains at O (0,0).
- A moves to (2.5×4, 0) = (10, 0).
- B moves to (2.5×4, 2.5×4) = (10, 10).
- C moves to (0, 2.5×4) = (0, 10).
Step 6
(iii) Calculate the ratio area of image square : area of original square.
Answer
To calculate the area, we know:
- The area of the original square OABC, with side length 4 cm: [ ext{Area}_{original} = 4^2 = 16 ext{ cm}^2 ]
- The area of the enlarged square, with side length 4 cm × 2.5: [ ext{Area}_{image} = (4 imes 2.5)^2 = (10)^2 = 100 ext{ cm}^2 ]
- Therefore, the ratio: [ ext{Ratio} = \frac{Area_{image}}{Area_{original}} = \frac{100}{16} = 6.25 ]
Step 7
(iv) Calculate the scale factor of this enlargement.
Answer
Given the area of square OQPR is 324 cm²:
-
We set the area of OABC to be 16 cm² and use the formula for scaled area: [ (k^2) \text{Area}{original} = ext{Area}{image} ]
Where k is the scale factor:
-
Thus: [ k^2 imes 16 = 324 ]
-
Solving for k: [ k^2 = \frac{324}{16} = 20.25 ]
-
Therefore: [ k = \sqrt{20.25} = 4.5 ] We conclude the scale factor is 4.5.