Answer: Answer: Number 22, x = 10.58 units
Number 27, x = 15.26 units
Number 28, x = 2.12 units
Step-by-step explanation: In Number 22,let the triangle be labelled as ABC (from left, to bottom,to right). Let us label the perpendicular line from angle B as point D. Now we have triangles ABD and CBD, both right angled triangles. In triangle ABD,
AB^²= BD^² + DA^² (Pythagoras' theorem)
8^² = 6^² + DA^²
Subtract 6^² from both sides of the equation
8^²-6^² = DA^²
64 - 36 = DA^²
28 = DA^²
Add the square root sign to both sides of the equation
√28 = √DA^²
DA = 5.29
However, both triangles ABD and CBD are similar which means line DA equals line DC. Then line DA+ line DC = x
Therefore x = 5.29 + 5.29
x = 10.58 units
In Number 27, we shall label the triangle ABC, (from top, to left, to right). The perpendicular line from angle A shall be labelled as point D.
Hence in triangle ABD,
AB^² = AD^² + BD^²
10^² = AD^² + 6^²
100 = AD^² + 36
Subtract 36 from both sides of the equation
100 - 36 = AD^²
64 = AD^²
Add the square root sign to both sides of the equation
√64 = √AD^²
8 = AD
This now means in triangle ACD
AD^² + CD^² = AC^²
8^² + 13^² = x²
64 + 169 = x²
233 = x^²
Add the square root sign to both sides of the equation
√233 = √x^²
15.2643 = x
x = 15.26 units (approximately)
In Number 28, we shall label the triangle ABD (from the top, to left, to right). The line that touches line BC from angle A shall be called point D. Therefore, in triangle ABD, the hypotenuse AB measures 3 units. Using angle B (45°) as the reference angle, we shall apply trigonometrical ratios first
Cos B = adjacent/hypotenuse
Cos45° = AD/3
Multiply both sides of the equation by 3
3 Cos 45° = AD
3x 0.7071 = AD
2.12 = AD
At this point, we can now apply Pythagoras' theorem (we have two known sides)
AD^² + BD^² = AB^²
2.12^² + x^² = 3^²
4.49 + x^² = 9
Subtract 4.49 from both sides of the equation
x^² = 9 - 4.49
x^² = 4.51
Add the square root sign to both sides of the equation
√x^² = √4.51
x = 2.12