Is it correct to say: Taking the longest given number as a perpendicular side and adding it to another side?

The hypotenuse is opposite the right angle and is always the longest side. Square it alone on one side of the equation.

Is it correct to say: In the geometric mean theorem, writing BD² = AB × BC?

The altitude is the geometric mean of the two SEGMENTS of the hypotenuse: BD² = AD × DC.

Is it correct to say: Swapping ½ and √3/2 in a 30°-60°-90° triangle?

The SHORTER side (opposite 30°) is half the hypotenuse. The side opposite 60° is the longer leg, (√3/2) × hypotenuse.

Is it correct to say: Writing Apollonius theorem as AB² + AC² = 2AM² + BC²?

It is 2AM² + 2BM², where BM is HALF of BC. Equivalent form: 2AM² + BC²/2.

Is it correct to say: Using the minus sign for an obtuse angle in the application results?

Acute angle at C ⇒ − 2BC × DC. Obtuse angle at C ⇒ + 2BC × CD (the foot D lies outside, on BC produced).

Is it correct to say: Leaving a surd answer as an unsimplified root such as √52?

Simplify surds: √52 = √(4 × 13) = 2√13.

📄 Formula sheet (PDF)

Pythagoras Theorem

Pythagorean triplets, similarity in right triangles, theorem of geometric mean, Pythagoras theorem and its converse, 30°-60°-90° and 45°-45°-90° triangles, acute/obtuse applications and Apollonius theorem — Maharashtra SSC Geometry Ch 2

📐 13 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 10✓ 2025–26 syllabus

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Board Exam Tips

→The proof of Pythagoras theorem using similarity (the altitude to the hypotenuse) is a standard textbook theorem. Learn it together with the theorem of geometric mean — they share the same figure.
→Before applying Pythagoras, say which angle is the right angle. The hypotenuse is always the side opposite the 90° angle.
→For 30°-60°-90° questions, first find the hypotenuse (or the side opposite 30°), then use ½ and √3/2. Leave answers in surd form like 7√3 unless a decimal is asked.
→Apollonius theorem needs M to be the MIDPOINT. Use 2BM², not BC², on the right-hand side.
→To test whether a triangle is right-angled, square all three sides and check whether the largest square equals the sum of the other two (converse of Pythagoras).

📐 Formulas(13)

Pythagorean Triplet Formula

(a^{2} + b^{2}, a^{2} - b^{2}, 2 ab), a > b, a, b \in N

Symbol	Meaning
$a, b$	Natural numbers with a > b

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

Symbol	Meaning
$B D$	Perpendicular from the right-angle vertex B to the hypotenuse AC

Theorem of Geometric Mean★ Board fav

B D^{2} = A D \times D C

Symbol	Meaning
$B D$	Altitude to the hypotenuse
$A D, D C$	Segments of the hypotenuse AC

Pythagoras Theorem★ Board fav

A C^{2} = A B^{2} + B C^{2} (∠ A B C = 9 0^{\circ})

Symbol	Meaning
$A C$	Hypotenuse (side opposite the right angle)
$A B, B C$	Perpendicular sides

Converse of Pythagoras Theorem

A C^{2} = A B^{2} + B C^{2} \Rightarrow ∠ A B C = 9 0^{\circ}

Theorem of 30°-60°-90° Triangle★ Board fav

side opposite 3 0^{\circ} = \frac{1}{2} hyp, side opposite 6 0^{\circ} = \frac{3}{2} hyp

Theorem of 45°-45°-90° Triangle

each perpendicular side = \frac{1}{2} hyp

Diagonal of a Square

d = 2 a

Symbol	Meaning
$d$	Diagonal (cm)
$a$	Side of the square (cm)

Height of an Equilateral Triangle

h = \frac{3}{2} a

Symbol	Meaning
$h$	Height (altitude) (cm)
$a$	Side of the equilateral triangle (cm)

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)

Symbol	Meaning
$D$	Foot of the perpendicular from A on BC

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)

Apollonius Theorem★ Board fav

A B^{2} + A C^{2} = 2 A M^{2} + 2 B M^{2} (M midpoint of B C)

Symbol	Meaning
$A M$	Median from A to side BC
$B M$	Half of BC (BM = MC)

Sides and Diagonals of a Parallelogram

A B^{2} + B C^{2} + C D^{2} + D A^{2} = A C^{2} + B D^{2}

✏️ Solved Examples

1Solved Exampleeasy2 steps

Using the formula for Pythagorean triplets with a = 5 and b = 2, write the triplet and verify it.

Substitute into (a² + b², a² − b², 2ab).

(25 + 4, 25 - 4, 2 \times 5 \times 2) = (29, 21, 20)

2Solved Exampleboard4 steps

In ΔPQR, ∠PQR = 90° and seg QS ⊥ hypotenuse PR. If PS = 4 cm and SR = 9 cm, find QS, PQ and QR.

By the theorem of geometric mean:

Q S^{2} = P S \times S R = 4 \times 9 = 36 \Rightarrow QS = 6 cm

3Solved Exampleboard2 steps

In ΔABC, ∠B = 90°, ∠A = 30° and AC = 18 cm. Find BC and AB.

∠C = 60°. By the 30°-60°-90° theorem, BC is opposite 30°.

B C = \frac{1}{2} \times A C = \frac{1}{2} \times 18 = 9 cm

4Solved ExampleHOTS3 steps

In ΔABC, AB = 13 cm, AC = 9 cm and BC = 10 cm. M is the midpoint of BC. Find the length of median AM.

M is the midpoint, so BM = BC/2 = 5 cm. Apply Apollonius theorem.

A B^{2} + A C^{2} = 2 A M^{2} + 2 B M^{2}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Taking the longest given number as a perpendicular side and adding it to another side
✓The hypotenuse is opposite the right angle and is always the longest side. Square it alone on one side of the equation.
2
In the geometric mean theorem, writing BD² = AB × BC
✓The altitude is the geometric mean of the two SEGMENTS of the hypotenuse: BD² = AD × DC.
3
Swapping ½ and √3/2 in a 30°-60°-90° triangle
✓The SHORTER side (opposite 30°) is half the hypotenuse. The side opposite 60° is the longer leg, (√3/2) × hypotenuse.
4
Writing Apollonius theorem as AB² + AC² = 2AM² + BC²
✓It is 2AM² + 2BM², where BM is HALF of BC. Equivalent form: 2AM² + BC²/2.
5
Using the minus sign for an obtuse angle in the application results
✓Acute angle at C ⇒ − 2BC × DC. Obtuse angle at C ⇒ + 2BC × CD (the foot D lies outside, on BC produced).
6
Leaving a surd answer as an unsimplified root such as √52
✓Simplify surds: √52 = √(4 × 13) = 2√13.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Is (20, 21, 29) a Pythagorean triplet?
Q2
Find the height of an equilateral triangle of side 10 cm.
Q3
In ΔABC, ∠ABC = 90° and BD ⊥ AC. If BD = 12 cm and AD = 8 cm, find DC.
Q4
The hypotenuse of an isosceles right triangle is 10 cm. Find each of the other sides.
Q5
In ΔABC, ∠ACB is obtuse and AD ⊥ BC produced. If BC = 5 cm, CD = 3 cm and AC = 5 cm, find AB.
Q6
The adjacent sides of a parallelogram are 5 cm and 7 cm, and one diagonal is 6 cm. Find the other diagonal.

📝 Notes

Pythagoras Theorem

This chapter builds directly on Similarity. Drop the altitude from the right angle to the hypotenuse and three similar triangles appear — the core results of the chapter come from that one figure.

One figure, three results

In $\triangle ABC$ with $\angle B = 90^\circ$ and $BD \perp AC$ :

$\triangle ADB \sim \triangle ABC \sim \triangle BDC$ (similarity in right triangles).
$BD^2 = AD \times DC$ (theorem of geometric mean).
$AB^2 = AC \times AD$ and $BC^2 = AC \times DC$ ; adding them gives $AB^2 + BC^2 = AC^2$ (Pythagoras theorem).

If you can draw this figure and write these three lines, you have the core proofs of the chapter.

Special right triangles

Triangle	Sides in ratio
30°-60°-90°	1 : √3 : 2
45°-45°-90°	1 : 1 : √2

These give the height of an equilateral triangle $\left(\frac{\sqrt{3}}{2}a\right)$ and the diagonal of a square $(\sqrt{2}\,a)$ without any new formula.

Beyond right triangles

When the angle at C is not 90°, drop a perpendicular from A to BC and apply Pythagoras twice:

Acute angle at C: $AB^2 = BC^2 + AC^2 - 2\,BC \times DC$
Obtuse angle at C: $AB^2 = BC^2 + AC^2 + 2\,BC \times CD$

Adding the two versions about the midpoint M of BC gives Apollonius theorem, $AB^2 + AC^2 = 2AM^2 + 2BM^2$ . Use it whenever a median appears.

Quick checks

The hypotenuse must be the longest side. If your hypotenuse is shorter than a leg, recheck the squares.
In $(a^2 + b^2, a^2 - b^2, 2ab)$ , the first number is always the largest.
Opposite an acute angle, side² < sum of the other two squares; opposite an obtuse angle, side² > that sum.