Introduction: A fundamental theorem in geometry with countless applications. Beyond simple calculations, it's a key tool in 3D problems (like finding the length of a diagonal in a cuboid) and is essential for coordinate geometry and trigonometry.

The "S Pass" Foundation (නියත S එකකට)

• Prompt 1 (State): In a right-angled triangle with sides a, b, and hypotenuse c, state Pythagoras' theorem as an equation.

• Prompt 2 (Calculate): The two shorter sides of a right-angled triangle are 9 cm and 12 cm. Find the length of the hypotenuse.

• Prompt 3 (Calculate): The hypotenuse of a right-angled triangle is 17 cm and one side is 8 cm. Find the length of the other side.

• Prompt 4 (Identify): Is a triangle with sides 7 cm, 10 cm, and 12 cm a right-angled triangle? Show your calculation.

• Prompt 5 (Apply): A rectangle has a length of 24 cm and a width of 7 cm. Find the length of its diagonal.

Climbing to a "C" (C එකට පාර)

• Prompt 1 (Problem Solve): The diagonals of a rhombus are 10 cm and 24 cm. Find the perimeter of the rhombus.

• Prompt 2 (Problem Solve): An equilateral triangle has a side length of 10 cm. Find the length of its altitude (perpendicular height).

• Prompt 3 (Application): A ladder 13 m long rests against a vertical wall. The foot of the ladder is 5 m from the base of the wall. How high up the wall does the ladder reach?

• Prompt 4 (Proof): In an isosceles triangle ABC with AB = AC, AD is the altitude to BC. Prove that AB2=AD2+BD2.

• Prompt 5 (Problem Solve): Find the length of the longest rod that can be placed inside a rectangular box of dimensions 12 cm x 9 cm x 8 cm.

Aiming for a "B" (B ඉලක්කය)

• Prompt 1 (Rider): In triangle ABC, AD is the altitude to BC. Prove that AB2−AC2=BD2−CD2.

• Prompt 2 (Rider): P is a point inside a rectangle ABCD. Prove that AP2+CP2=BP2+DP2. (Hint: Draw a line through P parallel to the sides).

• Prompt 3 (Coordinate Geometry): The coordinates of points A and B are (2, 3) and (7, 15) respectively. Find the distance AB using Pythagoras' theorem.

• Prompt 4 (Rider): In an equilateral triangle ABC of side length 'a', prove that the altitude is .

• Prompt 5 (Problem Solve): A ship sails 15 km due East and then 20 km due North. It then sails 12 km due West. How far is it from its starting point?

Securing the "A" Distinction (A සාමාර්ථය තහවුරු කරගන්න)

• Prompt 1 (Challenge Rider): The medians BD and CE of a triangle ABC are perpendicular to each other. Prove that b2+c2=5a2, where a, b, and c are the lengths of the sides opposite to vertices A, B, and C respectively.

• Prompt 2 (Apollonius' Theorem): In triangle ABC, AD is the median to side BC. Prove that AB2+AC2=2(AD2+BD2).

• Prompt 3 (Synthesis): The sides of a right-angled triangle form a Pythagorean triple. If the two shorter sides are x2−y2 and 2xy, prove that the hypotenuse is x2+y2.

• Prompt 4 (Problem Solve): From the top of a vertical cliff of height 80 m, the distance to a boat at sea is 100 m. The boat sails 44 m directly towards the cliff. Calculate the new distance from the top of the cliff to the boat.

• Prompt 5 (3D Geometry): A right pyramid has a square base of side 10 cm and a slant height of 13 cm. Find the perpendicular height of the pyramid.