SOLUTION: A square piece of paper has side length 1. Esha cuts it in half along one of its diagonals. She labels the corners of one of the triangular halves A, B, C. Esha folds the triangl

Question 1193635: A square piece of paper has side length 1. Esha cuts it in half along one of its diagonals. She labels the corners of one of the triangular halves A, B, C.
Esha folds the triangle so that B is moved to any point B' on the side AC, making a triangle B'GH.
a) Area of ∆B'GH if B' is at C.
b) Area of ∆B'GH if B' is at A.
c) Find the length of AG if HB' is perpendicular to AC.
d) Find the area of ∆B'GH if HB' is perpendicular to AC.

Answer by parmen(42) (Show Source): You can put this solution on YOUR website!
**a) Area of ∆B'GH if B' is at C:**
* When B' is at C, triangle B'GH degenerates into a line segment.
* **Area of ∆B'GH = 0**
**b) Area of ∆B'GH if B' is at A:**
* When B' is at A, triangle B'GH coincides with the original right-angled triangle ABC.
* **Area of ∆B'GH = Area of ∆ABC**
* Since ABC is a right-angled isosceles triangle with sides 1, 1, and √2,
* **Area of ∆ABC = (1/2) * base * height = (1/2) * 1 * 1 = 1/2**
**c) Find the length of AG if HB' is perpendicular to AC:**
* Let AG = x
* Since HB' is perpendicular to AC, triangle AB'H is right-angled at B'.
* Using the Pythagorean theorem in triangle AB'H:
* AB'² + B'H² = AH²
* (1 - x)² + B'H² = x²
* 1 - 2x + x² + B'H² = x²
* B'H² = 2x - 1
* In triangle AB'H and triangle B'GH, ∠AB'H = ∠GB'H (angles of incidence and reflection are equal)
* Also, ∠AHB' = ∠B'GH = 90°
* Therefore, triangles AB'H and B'GH are similar.
* Using the property of similar triangles:
* AB' / B'H = B'H / GH
* (1 - x) / B'H = B'H / x
* B'H² = x(1 - x)
* Equating the two expressions for B'H²:
* 2x - 1 = x(1 - x)
* 2x - 1 = x - x²
* x² + x - 1 = 0
* Solving the quadratic equation for x using the quadratic formula:
* x = [-1 ± √(1² - 4 * 1 * (-1))] / (2 * 1)
* x = [-1 ± √5] / 2
* Since x represents a length, we take the positive value:
* x = (√5 - 1) / 2
* **Therefore, the length of AG is (√5 - 1) / 2**
**d) Find the area of ∆B'GH if HB' is perpendicular to AC:**
* Area of ∆B'GH = (1/2) * base * height
* Area of ∆B'GH = (1/2) * GH * B'H
* From the similar triangles property:
* GH / B'H = B'H / (1 - x)
* GH = B'H² / (1 - x)
* GH = x(1 - x) / (1 - x)
* GH = x
* Area of ∆B'GH = (1/2) * x * B'H
* Area of ∆B'GH = (1/2) * x * √(2x - 1)
* Substitute the value of x:
* Area of ∆B'GH = (1/2) * [(√5 - 1) / 2] * √[2 * (√5 - 1) / 2 - 1]
* Area of ∆B'GH = (1/4) * (√5 - 1) * √(√5 - 2)
**Therefore, the area of ∆B'GH if HB' is perpendicular to AC is (1/4) * (√5 - 1) * √(√5 - 2)**

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