Math Magic: Geometry Adventures

1 Journey East and North

A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point.

Step 1: Visualize the problem as a right-angled triangle where:

Eastward distance (adjacent side) = 18 m
Northward distance (opposite side) = 24 m

Step 2: Apply the Pythagorean theorem:

Hypotenuse² = Adjacent² + Opposite²
Distance² = 18² + 24²

Step 3: Calculate the squares:

18² = 324
24² = 576
Sum = 324 + 576 = 900

Step 4: Find the square root:

Distance = √900 = 30 m

Final Answer: The man is 30 meters away from his starting point.

2 Sarah's House to James' House

There are two paths from Sarah's house to James' house. One way is to take C street (direct path), and the other requires taking B street and then A street. How much shorter is the direct path along C street?

Step 1: Identify the distances:

B street = 50 m (vertical)
A street = 150 m (horizontal)

Step 2: Calculate the indirect path (B then A):

Total distance = B + A = 50 + 150 = 200 m

Step 3: Calculate the direct path (C street) using Pythagorean theorem:

C = √(50² + 150²) = √(2500 + 22500) = √25000 ≈ 158.11 m

Step 4: Find the difference:

Difference = 200 - 158.11 ≈ 41.89 m

Final Answer: The direct path along C street is approximately 41.9 meters shorter.

3 Avoiding the Pond

To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to walk through the pond?

Avoiding Path

Current Walk

Direct Path

Distance Saved

Step 1: Current path (avoiding pond):

South: 34 m
East: 41 m
Total distance = 34 + 41 = 75 m

Step 2: Direct path (through pond) - use Pythagorean theorem:

Distance = √(34² + 41²) = √(1156 + 1681) = √2837 ≈ 53.26 m

Step 3: Calculate the difference:

Meters saved = 75 - 53.26 ≈ 21.74 m

Step 4: Round to nearest meter:

≈ 22 meters

Final Answer: Approximately 22 meters would be saved by walking through the pond.

4 Rectangle Dimensions

In rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm. Calculate the length and breadth of the rectangle.

Step 1: Let length = l, breadth = b. From the problem:

XY + YZ = 17 cm ⇒ l + b = 17 ...(1)
XZ + YW = 26 cm ⇒ 2√(l² + b²) = 26 ⇒ √(l² + b²) = 13 ...(2)

Step 2: Square both sides of equation (2):

l² + b² = 169

Step 3: Square equation (1):

(l + b)² = 289 ⇒ l² + b² + 2lb = 289

Step 4: Substitute from step 2:

169 + 2lb = 289 ⇒ 2lb = 120 ⇒ lb = 60

Step 5: Now we have:

l + b = 17
l × b = 60

These are the sum and product of roots of a quadratic equation.

Step 6: Solve the quadratic equation:

x² - (l+b)x + lb = 0 ⇒ x² - 17x + 60 = 0

Step 7: Find roots:

x = [17 ± √(289 - 240)]/2 = [17 ± √49]/2 = [17 ± 7]/2
So x = 12 or x = 5

Final Answer: The rectangle has length = 12 cm and breadth = 5 cm.

5 Right Triangle Sides

The hypotenuse of a right triangle is 6 m more than twice the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.

Triangle Sides

Verification

Step 1: Let the shortest side be x meters.

Then hypotenuse = 2x + 6
Third side = hypotenuse - 2 = 2x + 4

Step 2: Apply Pythagorean theorem:

x² + (2x + 4)² = (2x + 6)²

Step 3: Expand the equation:

x² + 4x² + 16x + 16 = 4x² + 24x + 36

Step 4: Simplify:

x² + 4x² + 16x + 16 - 4x² - 24x - 36 = 0
x² - 8x - 20 = 0

Step 5: Solve the quadratic equation:

x = [8 ± √(64 + 80)]/2 = [8 ± √144]/2 = [8 ± 12]/2

Step 6: Take positive solution (since length can't be negative):

x = (8 + 12)/2 = 10 m

Step 7: Find other sides:

Hypotenuse = 2×10 + 6 = 26 m
Third side = 26 - 2 = 24 m

Final Answer: The sides of the triangle are 10 m, 24 m, and 26 m.

6 Sliding Ladder

A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.

Initial: 3m

Moved: 1.4m

Initial height: 4m

New height: 4.8m

↑ +0.8m

Step 1: Initial position - find distance from wall:

Using Pythagorean theorem:
Ladder² = Height² + Distance²
5² = 4² + Distance² ⇒ 25 = 16 + Distance² ⇒ Distance = 3 m

Step 2: After moving foot 1.6 m towards wall:

New distance from wall = 3 - 1.6 = 1.4 m

Step 3: Find new height using Pythagorean theorem:

Height² = 5² - 1.4² = 25 - 1.96 = 23.04
New height = √23.04 = 4.8 m

Step 4: Calculate upward slide:

Slide = New height - Original height = 4.8 - 4 = 0.8 m

Final Answer: The top of the ladder slides upwards by 0.8 meters.

7 Triangle Proof

The perpendicular PS on the base QR of a △PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ² = 2PR² + QR².

Step 1: Let SR = x, then QS = 3x (given QS = 3 SR), so QR = QS + SR = 4x.

Step 2: Let PS = h (the perpendicular height).

Step 3: Apply Pythagorean theorem to triangles PQS and PRS:

PQ² = PS² + QS² = h² + (3x)² = h² + 9x² ...(1)
PR² = PS² + SR² = h² + x² ...(2)

Step 4: We need to prove 2PQ² = 2PR² + QR².

Substitute from (1) and (2):
2(h² + 9x²) = 2(h² + x²) + (4x)²

Step 5: Expand both sides:

Left side: 2h² + 18x²
Right side: 2h² + 2x² + 16x² = 2h² + 18x²

Step 6: Both sides are equal (2h² + 18x²), so the statement is proved.

Final Answer: We have successfully proved that 2PQ² = 2PR² + QR².

8 Trisected Triangle Proof

In the adjacent figure, ABC is a right-angled triangle with right angle at B and points D, E trisect BC. Prove that 8AE² = 3AC² + 5AD².

Step 1: Let AB = a, BC = 3b (since D and E trisect BC).

So BD = b, DE = b, EC = b

Step 2: Coordinates approach (let's place B at origin):

Let B be (0,0), C be (3b,0), A be (0,a)
Then D is (b,0), E is (2b,0)

Step 3: Find expressions for the lengths:

AD² = (b-0)² + (0-a)² = b² + a²
AE² = (2b-0)² + (0-a)² = 4b² + a²
AC² = (3b-0)² + (0-a)² = 9b² + a²

Step 4: Compute right side of equation to prove:

3AC² + 5AD² = 3(9b² + a²) + 5(b² + a²) = 27b² + 3a² + 5b² + 5a² = 32b² + 8a²

Step 5: Compute left side:

8AE² = 8(4b² + a²) = 32b² + 8a²

Step 6: Both sides equal 32b² + 8a², so the equation holds.

Final Answer: We have successfully proved that 8AE² = 3AC² + 5AD².