Maharashtra State Board · Class 10 · Mathematics · Geometry Chapter 2

Pythagoras Theorem — Formula Sheet

Board Formulas

13 formulas

1.Pythagorean Triplet Formula
$(a^{2} + b^{2}, a^{2} - b^{2}, 2 ab), a > b, a, b \in N$
$a, b$ : Natural numbers with a > b
Gives a Pythagorean triplet for any natural numbers a > b. Example: a = 2, b = 1 gives (5, 3, 4). Multiplying a triplet by any natural number gives another triplet.
2.Similarity in a Right Triangle
$∠ A B C = 9 0^{\circ}, B D ⊥ A C \Rightarrow △ A D B \sim △ A B C \sim △ B D C$
$B D$ : Perpendicular from the right-angle vertex B to the hypotenuse AC
The altitude to the hypotenuse splits a right triangle into two triangles, each similar to the whole triangle and to each other (AA test: a common angle plus a right angle).
3.Theorem of Geometric Mean★
$B D^{2} = A D \times D C$
$B D$ : Altitude to the hypotenuse · $A D, D C$ : Segments of the hypotenuse AC
In a right triangle, the perpendicular from the right-angle vertex to the hypotenuse is the geometric mean of the two segments of the hypotenuse.
4.Pythagoras Theorem★
$A C^{2} = A B^{2} + B C^{2} (∠ A B C = 9 0^{\circ})$
$A C$ : Hypotenuse (side opposite the right angle) · $A B, B C$ : Perpendicular sides
In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
5.Converse of Pythagoras Theorem
$A C^{2} = A B^{2} + B C^{2} \Rightarrow ∠ A B C = 9 0^{\circ}$
If the square of one side equals the sum of the squares of the other two, the angle opposite that side is a right angle.
6.Theorem of 30°-60°-90° Triangle★
$side opposite 3 0^{\circ} = \frac{1}{2} hyp, side opposite 6 0^{\circ} = \frac{3}{2} hyp$
Sides are in the ratio 1 : √3 : 2 (opposite 30° : opposite 60° : hypotenuse).
7.Theorem of 45°-45°-90° Triangle
$each perpendicular side = \frac{1}{2} hyp$
An isosceles right triangle. Sides are in the ratio 1 : 1 : √2.
8.Diagonal of a Square
$d = 2 a$
$d$ : Diagonal (cm) · $a$ : Side of the square (cm)
The diagonal splits a square into two 45°-45°-90° triangles. Square of side 5 cm has diagonal 5√2 cm.
9.Height of an Equilateral Triangle
$h = \frac{3}{2} a$
$h$ : Height (altitude) (cm) · $a$ : Side of the equilateral triangle (cm)
The altitude splits an equilateral triangle into two 30°-60°-90° triangles; the altitude is the side opposite 60°.
10.Application in an Acute-angled Triangle
$A B^{2} = B C^{2} + A C^{2} - 2 B C \times D C (∠ C acute, A D ⊥ B C)$
$D$ : Foot of the perpendicular from A on BC
Opposite an acute angle, the square of the side is LESS than the sum of the squares of the other two.
11.Application in an Obtuse-angled Triangle
$A B^{2} = B C^{2} + A C^{2} + 2 B C \times C D (∠ A C B obtuse, A D ⊥ B C)$
Here D lies on BC produced beyond C. Opposite an obtuse angle, the square of the side is MORE than the sum of the squares of the other two.
12.Apollonius Theorem★
$A B^{2} + A C^{2} = 2 A M^{2} + 2 B M^{2} (M midpoint of B C)$
$A M$ : Median from A to side BC · $B M$ : Half of BC (BM = MC)
Relates the two sides, the median AM and half the third side. Since BM = BC/2, it can also be written AB² + AC² = 2AM² + BC²/2.
13.Sides and Diagonals of a Parallelogram
$A B^{2} + B C^{2} + C D^{2} + D A^{2} = A C^{2} + B D^{2}$
Sum of squares of the sides of a parallelogram = sum of squares of its diagonals. Follows from Apollonius theorem because the diagonals bisect each other.