The area of a right-angled isosceles triangle having hypotenuse

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Using Rule 1,

Let AB = BC = x;
AC = 16√2
∴ x² + x² = (16√2)²
⇒ 2x² = 16 × 16 × 2
⇒ x² = 16 × 16
⇒ x = 16

∴ Area of triangle =	1	× base × height
	2

=	1	× 16 × 16 = 128cm²
	2