Watermark
Vardaan Learning Institute
Created by Team Vardaan | VARDAAN COMET | Reference: ML Agarwal – ICSE Class 9

Chapter 11: Pythagoras Theorem

Pythagoras Theorem

Syllabus at a Glance

ICSE Class 9 – Chapter 11: Pythagoras Theorem

Note: The area-based proof IS required for ICSE examination. Focus on the geometric/visual proof using squares on sides.

1. Revision of Basic Concepts

Right-Angled Triangle — Key Terms
[Diagram Placeholder]
A clean precise mathematical diagram. Background color: #e8f5e9 (light green, to match this box). A right-angled triangle ABC with the right angle at C, drawn on a #e8f5e9 background. Vertex A at top-left, B at bottom-right, C at bottom-left. Small square symbol at C showing the right angle. Sides labelled: BC = "Base (b)", AC = "Perpendicular / Height (p)", AB = "Hypotenuse (h)" with a double-headed arrow and the word "longest side" beside it. Angles marked: angle-A and angle-B with arcs (both acute). Bold dark green labels. Background strictly #e8f5e9.

In a right-angled triangle, one angle is exactly 90°. The three sides are:

The hypotenuse is always opposite the right angle — it never touches the right angle vertex.

2. Pythagoras Theorem — Statement

Pythagoras Theorem
[Diagram Placeholder]
A precise mathematical diagram. Background color: #fff3e0 (light orange, to match this theorem box). A right-angled triangle ABC on a #fff3e0 background with right angle at C. Three squares drawn on each side: a blue square on BC (area = a²), a red square on AC (area = b²), and a large green square on hypotenuse AB (area = c²). Each square is labelled with its area. Bold equation below: "c² = a² + b²" (hypotenuse squared = sum of squares of other two sides). Background strictly #fff3e0.

Statement: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

If $\angle C = 90°$ in $\triangle ABC$ with $BC = a$, $CA = b$, $AB = c$ (hypotenuse), then:

$$AB^2 = BC^2 + CA^2 \quad \text{i.e.,} \quad c^2 = a^2 + b^2$$

Named after the Greek mathematician Pythagoras (c. 570–495 BC), though the result was known much earlier in Egypt and India.

3. Area-Based Proof of Pythagoras Theorem

The ICSE syllabus specifically requires the area-based (geometric) proof. There are two classic area-based proofs. Both are given below.

Proof 1 — Geometric Proof Using Squares on Sides (Standard ICSE Proof)
[Diagram Placeholder]
A precise geometric diagram. Background color: #ede7f6 (light purple, to match this proof box). A right-angled triangle ABC with right angle at C, drawn large in the centre on a #ede7f6 background. On side BC draw a blue square BCPQ outward. On side CA draw a red square CARS outward. On hypotenuse AB draw a large green square ABDE outward. From C drop a perpendicular CL to AB, extended to meet DE at M. The green square is visibly divided into two rectangles by the line CLM. Each rectangle is labelled: left rectangle area = a² (blue), right rectangle area = b² (red). Show: "Area of green square = blue + red → c² = a² + b²". Background strictly #ede7f6.

Given: $\triangle ABC$ with $\angle ACB = 90°$, $BC = a$, $CA = b$, $AB = c$.

To Prove: $AB^2 = BC^2 + CA^2$, i.e., $c^2 = a^2 + b^2$.

Construction:

Proof:

Step 1: In $\triangle ABQ$ and $\triangle ACE$:
$AB = AE$ (sides of square $ABDE$) …(i)
$BQ = BC$ (sides of square $BCPQ$) …(ii)
$\angle QBA = \angle CBQ + \angle CBA = 90° + \angle CBA$
$\angle CBE = \angle CBE + \angle ABE = \angle CBA + 90°$
$\therefore \angle QBA = \angle CBE$ …(iii)
By SAS: $\triangle ABQ \cong \triangle EBC$.
Step 2: Area of $\triangle ABQ = \frac{1}{2} \times BQ \times BC_{\perp}$. Since $\triangle ABQ \cong \triangle EBC$: Area($\triangle ABQ$) = Area($\triangle EBC$).
Step 3: $\triangle ABQ$ and square $BCPQ$ share the same base $BQ$ and same height. Therefore:
Area($\triangle ABQ$) $= \dfrac{1}{2} \times$ Area(square $BCPQ$) $= \dfrac{a^2}{2}$.
Step 4: Similarly $\triangle EBC$ and rectangle $BELM$ share the same base $BE$ and same height. Therefore:
Area($\triangle EBC$) $= \dfrac{1}{2} \times$ Area(rectangle $BELM$).
Step 5: From Steps 2, 3, 4: Area(rectangle $BELM$) $= a^2$.
By the same argument with the other triangle: Area(rectangle $ALMD$) $= b^2$.
$\therefore$ Area(square $ABDE$) $=$ Area(rectangle $BELM$) $+$ Area(rectangle $ALMD$) $= a^2 + b^2$.
$\therefore c^2 = a^2 + b^2$. Hence proved.
Proof 2 — Bhaskara's Area Dissection Proof (Elegant Alternative)
[Diagram Placeholder]
A precise mathematical diagram. Background color: #ede7f6 (light purple). A large square of side c (hypotenuse) on a #ede7f6 background, divided into four identical right-angled triangles (each with legs a and b) arranged around a smaller inner square of side (b-a) in the centre. Each triangle is shaded light blue. The inner square is shaded yellow. Labelled: "4 triangles each = (1/2)ab" and "Inner square = (b-a)²". Equation below: "c² = 4 × (1/2)ab + (b-a)² = 2ab + b²-2ab+a² = a²+b²". Background strictly #ede7f6.

Construction: Take a square of side $c$ (the hypotenuse). Place four congruent right-angled triangles (legs $a$ and $b$) inside it at each corner, with hypotenuses along the outer square's sides. The inner region forms a square of side $(b - a)$.

Area Calculation:

Area of large square $= c^2$.
Area of 4 triangles $= 4 \times \dfrac{1}{2}ab = 2ab$.
Area of inner square $= (b - a)^2 = b^2 - 2ab + a^2$.
$c^2 = 2ab + (b-a)^2 = 2ab + b^2 - 2ab + a^2 = a^2 + b^2$.
Hence $c^2 = a^2 + b^2$. ✓
Which Proof for the Exam?

4. Pythagorean Triplets

Definition and Common Triplets
[Diagram Placeholder]
A clear mathematical table/chart. Background color: #e3f2fd (light blue, to match this box). A neatly organised table on a #e3f2fd background with three columns: "Smaller Leg (a)", "Larger Leg (b)", "Hypotenuse (c)". Rows: (3,4,5), (5,12,13), (8,15,17), (7,24,25), (9,40,41). Each row has a small blue right triangle icon. Header row shaded dark blue with white text. Below the table: "Multiples also work: 6,8,10 (= 2 × 3,4,5) etc." Background strictly #e3f2fd.

A Pythagorean triplet is a set of three positive integers $(a, b, c)$ such that $a^2 + b^2 = c^2$. They can always serve as sides of a right-angled triangle.

3, 4, 5
$9 + 16 = 25$ ✓
5, 12, 13
$25 + 144 = 169$ ✓
8, 15, 17
$64 + 225 = 289$ ✓
7, 24, 25
$49 + 576 = 625$ ✓
9, 40, 41
$81 + 1600 = 1681$ ✓
6, 8, 10
$2 \times (3,4,5)$ ✓
9, 12, 15
$3 \times (3,4,5)$ ✓
10, 24, 26
$2 \times (5,12,13)$ ✓
Any positive multiple of a Pythagorean triplet is also a Pythagorean triplet. E.g., $(3,4,5) \times k = (3k, 4k, 5k)$ for any positive integer $k$.

General Formula: For any two positive integers $m > n$: $a = m^2 - n^2$, $b = 2mn$, $c = m^2 + n^2$ gives a Pythagorean triplet. E.g., $m=2, n=1$: $a=3, b=4, c=5$.

5. Applications of Pythagoras Theorem — Finding the Missing Side

The Three Formulas

From $c^2 = a^2 + b^2$ (where $c$ = hypotenuse):

Worked Example 1 — Find the Hypotenuse

Q. In right $\triangle ABC$, $\angle C = 90°$, $BC = 8$ cm, $CA = 6$ cm. Find $AB$.

By Pythagoras: $AB^2 = BC^2 + CA^2 = 8^2 + 6^2 = 64 + 36 = 100$.
$AB = \sqrt{100} = 10$ cm.
Answer: $AB = 10$ cm. (This is the $(6,8,10) = 2 \times (3,4,5)$ triplet.)
Worked Example 2 — Find a Leg

Q. In right $\triangle PQR$, $\angle R = 90°$, hypotenuse $PQ = 13$ cm, $QR = 5$ cm. Find $PR$.

$PQ^2 = QR^2 + PR^2 \Rightarrow 13^2 = 5^2 + PR^2 \Rightarrow 169 = 25 + PR^2$.
$PR^2 = 144 \Rightarrow PR = 12$ cm.
Answer: $PR = 12$ cm. (The $(5,12,13)$ triplet.)
Worked Example 3 — Diagonal of a Rectangle

Q. A rectangle is 24 cm long and 10 cm wide. Find the length of its diagonal.

The diagonal and two sides form a right-angled triangle (right angle at the corner).
Diagonal$^2 = 24^2 + 10^2 = 576 + 100 = 676$.
Diagonal $= \sqrt{676} = 26$ cm.
Answer: Diagonal = 26 cm. (The $(10,24,26) = 2 \times (5,12,13)$ triplet.)
Worked Example 4 — Altitude of Equilateral Triangle

Q. Find the altitude of an equilateral triangle with side $2a$.

Let altitude $AD$ meet $BC$ at $D$ (midpoint of $BC$), so $BD = a$.
In right $\triangle ABD$: $AB^2 = AD^2 + BD^2 \Rightarrow (2a)^2 = AD^2 + a^2$.
$AD^2 = 4a^2 - a^2 = 3a^2 \Rightarrow AD = a\sqrt{3}$.
Answer: Altitude $= a\sqrt{3}$. For side $s$, altitude $= \dfrac{s\sqrt{3}}{2}$.
Worked Example 5 — Ladder Against Wall (ML Agarwal Type)

Q. A ladder 17 m long leans against a vertical wall. The foot of the ladder is 8 m from the base of the wall on the ground. How high up the wall does the ladder reach?

Let the wall height reached = $h$ m. The ground, wall and ladder form a right triangle.
$17^2 = h^2 + 8^2 \Rightarrow 289 = h^2 + 64 \Rightarrow h^2 = 225$.
$h = 15$ m.
Answer: The ladder reaches 15 m up the wall. (The $(8,15,17)$ triplet.)
Worked Example 6 — Diagonal of a Square

Q. Find the diagonal of a square of side 7 cm.

Diagonal$^2 = 7^2 + 7^2 = 49 + 49 = 98$.
Diagonal $= \sqrt{98} = 7\sqrt{2}$ cm.
Answer: Diagonal $= 7\sqrt{2}$ cm. Generally, diagonal of square with side $s$ is $s\sqrt{2}$.
Worked Example 7 — Distance Between Two Points

Q. A man goes 12 km East and then 5 km North. How far is he from his starting point?

The East and North directions are perpendicular, forming a right triangle.
Distance$^2 = 12^2 + 5^2 = 144 + 25 = 169$.
Distance $= \sqrt{169} = 13$ km.
Answer: 13 km from starting point.
Worked Example 8 — Combined Application (ML Agarwal Type)

Q. In $\triangle ABC$, $\angle B = 90°$. $D$ is the midpoint of $AC$. Prove $4BD^2 = 4BC^2 + AC^2$.

Let $\angle B = 90°$, $BC = a$, $AB = b$. Then $AC^2 = a^2 + b^2$ (Pythagoras).
$BD$ is the median to the hypotenuse: in a right triangle, the median to the hypotenuse $= \frac{1}{2} \times$ hypotenuse.
So $BD = \frac{1}{2}AC$ and $BD^2 = \frac{AC^2}{4}$.
$4BD^2 = AC^2 = (a^2 + b^2) = BC^2 + AB^2$. But we need $4BC^2 + AC^2$? Let's re-examine.
Alternative approach using Apollonius' theorem (median property):
$BD^2 = \frac{2BC^2 + 2AB^2 - AC^2}{4}$... Since $D$ is midpoint of $AC$ and $\angle B = 90°$: $BD = \frac{AC}{2}$.
$4BD^2 = AC^2 = BC^2 + AB^2$. Adding $3BC^2$: $4BD^2 + 3BC^2 = 4BC^2 + AB^2$... (Check problem statement)
Key result: In a right triangle, the median to the hypotenuse equals half the hypotenuse: $BD = \frac{AC}{2}$.

✏ Practice Problems — Applications of Pythagoras Theorem

  1. In right $\triangle ABC$, $\angle C = 90°$, $BC = 9$ cm, $CA = 12$ cm. Find $AB$.
  2. In right $\triangle PQR$, $\angle Q = 90°$, $PQ = 20$ cm, $PR = 25$ cm. Find $QR$.
  3. A ladder 10 m long leans against a wall with its foot 6 m from the wall. How high does it reach?
  4. Find the diagonal of a rectangle with length 40 cm and breadth 30 cm.
  5. An equilateral triangle has side 6 cm. Find its altitude and area.
  6. A man walks 15 km South and then 20 km West. How far is he from his starting point?
  7. Find the diagonal of a square whose area is $72$ cm$^2$.
  8. In right $\triangle ABC$ with $\angle C = 90°$: $AB = 10$ cm, $BC = 6$ cm. Find $AC$ and verify using the triplet.
  9. A pole 13 m high is braced by a wire from its top to a point 5 m from its base. Find the length of the wire.
  10. The hypotenuse of a right triangle is 65 cm. If one leg is 25 cm, find the other leg. State the Pythagorean triplet.
  11. In $\triangle ABC$, $\angle A = 90°$. $D$ is a point on $BC$ with $AD \perp BC$. Given $AB = 6$ cm, $AC = 8$ cm: (i) Find $BC$. (ii) Find $AD$. (iii) Find $BD$ and $DC$.
  12. Two poles of heights 6 m and 11 m stand on level ground. The distance between them is 12 m. Find the distance between their tops.

6. Converse of Pythagoras Theorem

Converse of Pythagoras Theorem

[Diagram Placeholder]
A precise mathematical diagram. Background color: #e0f2f1 (light teal, to match this converse-card box). Two triangles side by side on a #e0f2f1 background. Left: triangle with sides a=3, b=4, c=5 with the equation "3²+4²=5² → 25=25 ✓" below it and a small square at the right-angle corner (verified right angle). Right: triangle with sides 4,5,6 with "4²+5²≠6² → 41≠36 ✗" and a curved X mark (NOT right-angled). Bold label: "Converse: If c²=a²+b² → right angle at C." Background strictly #e0f2f1.

Statement: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the angle opposite that side is a right angle.

In $\triangle ABC$: if $AB^2 = BC^2 + CA^2$, then $\angle BCA = 90°$.

The converse allows us to CHECK whether a given triangle is right-angled, simply by testing the sides.

How to Use the Converse

Step 1: Identify the longest side (candidate for hypotenuse).

Step 2: Square all three sides.

Step 3: Check if (longest side)$^2$ = (other side$_1)^2$ + (other side$_2)^2$.

Worked Example 1 — Verify Right Triangle

Q. A triangle has sides 11 cm, 60 cm, and 61 cm. Is it right-angled?

Longest side = 61 cm. Check: $61^2 = 3721$; $11^2 + 60^2 = 121 + 3600 = 3721$.
$61^2 = 11^2 + 60^2$ ✓ → Yes, it is right-angled (right angle opposite the side 61 cm).
Worked Example 2 — Identify Triangle Type

Q. Classify the triangle with sides 7 cm, 8 cm, and 10 cm as acute, right, or obtuse.

Longest side = 10 cm. $10^2 = 100$. $7^2 + 8^2 = 49 + 64 = 113$.
$100 < 113$ → (longest)$^2$ < sum of squares of other two.
→ Acute-angled triangle.
Worked Example 3 — Obtuse Classification

Q. Classify the triangle with sides 5 cm, 7 cm, and 9 cm.

Longest side = 9 cm. $9^2 = 81$. $5^2 + 7^2 = 25 + 49 = 74$.
$81 > 74$ → (longest)$^2$ > sum of squares of other two.
→ Obtuse-angled triangle.
Worked Example 4 — Converse Proof Application

Q. In $\triangle ABC$, $AB = 5$ cm, $BC = 12$ cm, $CA = 13$ cm. Prove $\angle B = 90°$.

$CA^2 = 13^2 = 169$.
$AB^2 + BC^2 = 5^2 + 12^2 = 25 + 144 = 169$.
$CA^2 = AB^2 + BC^2$ ⇒ by the Converse of Pythagoras Theorem: $\angle ABC = 90°$. Hence proved.
Worked Example 5 — Converse in Geometry Proof (ML Agarwal Type)

Q. $\triangle ABC$ has $AB = AC$ and altitude $AD \perp BC$. If $AB = 10$ cm and $AD = 8$ cm, find $BC$ and verify using Pythagoras.

Since $AD \perp BC$ and $AB = AC$, $D$ is the midpoint of $BC$: $BD = DC$.
In right $\triangle ABD$: $BD^2 = AB^2 - AD^2 = 100 - 64 = 36$. $BD = 6$ cm.
$BC = 2 \times BD = 12$ cm.
Verify (converse): $AD^2 + BD^2 = 64 + 36 = 100 = AB^2$ ✓. So $\angle ADB = 90°$ confirmed.

✏ Practice Problems — Converse of Pythagoras Theorem

  1. Verify which of these are right-angled triangles: (a) 9, 40, 41   (b) 6, 7, 8   (c) 20, 21, 29   (d) 11, 12, 13.
  2. Classify the triangle with sides 8, 15, 17 as acute, right, or obtuse. State the angle.
  3. A triangle has sides 5, 12, 13. Prove it is right-angled and find the right angle vertex.
  4. Sides of a triangle are 6 cm, 8 cm, 10 cm. (i) Is it right-angled? (ii) What is the area?
  5. Can a triangle with sides 4, 5, 6 be right-angled? Classify it.
  6. In $\triangle PQR$, $PQ = 15$, $QR = 20$, $PR = 25$. Show $\angle Q = 90°$ and find the area.
  7. A triangle with sides $3k, 4k, 5k$ is always right-angled for any $k > 0$. Prove this.
  8. Three squares of sides 5 cm, 12 cm, and 13 cm are placed with their vertices touching. Show they form a right-angled triangle.
  9. Sides of a triangle are $(m^2-n^2)$, $2mn$, $(m^2+n^2)$ for integers $m > n > 0$. Prove it is always right-angled.
  10. In right $\triangle ABC$ with $\angle C = 90°$, prove $AC^2 = BC \cdot (BC + DC)$ where $D$ is the foot of the altitude from $C$ to $AB$. [Hint: use similar triangles result if known, else set up using Pythagoras twice.]

7. Important Results Derived from Pythagoras Theorem

Key Results — Memorise These
[Diagram Placeholder]
A clean summary diagram. Background color: #e0f2f1 (light teal, to match this box). Two sub-diagrams on a #e0f2f1 background. Left: equilateral triangle of side s with altitude h = (s√3)/2 drawn in dark teal. Right: square of side s with diagonal d = s√2 drawn diagonally. Both diagrams have clear dark labels. Below: table with two rows: "Equilateral triangle altitude = (√3/2)·s" and "Square diagonal = s√2". Background strictly #e0f2f1.
Worked Example — Altitude on Hypotenuse

Q. In right $\triangle ABC$, $\angle C = 90°$, $BC = 8$ cm, $CA = 6$ cm. $CD \perp AB$ at $D$. Find $CD$, $BD$, and $AD$.

$AB = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$ cm.
$\text{Area of } \triangle ABC = \frac{1}{2} \times BC \times CA = \frac{1}{2} \times 8 \times 6 = 24$ cm$^2$.
Also Area $= \frac{1}{2} \times AB \times CD = \frac{1}{2} \times 10 \times CD$. ⇒ $CD = \frac{48}{10} = 4.8$ cm.
$BD = \frac{BC^2}{AB} = \frac{64}{10} = 6.4$ cm.   $AD = \frac{CA^2}{AB} = \frac{36}{10} = 3.6$ cm.
Verify: $BD + AD = 6.4 + 3.6 = 10 = AB$ ✓.   $CD^2 = 23.04 = AD \times BD = 3.6 \times 6.4 = 23.04$ ✓.

8. Mixed Application Problems

Mixed Example 1 — Multi-step Problem

Q. $ABCD$ is a rhombus with side 10 cm and one diagonal $AC = 12$ cm. Find the other diagonal $BD$ and the area of the rhombus.

Diagonals of a rhombus bisect each other at right angles. Let them meet at $O$. $AO = 6$ cm.
In right $\triangle AOB$: $AB^2 = AO^2 + BO^2 \Rightarrow 10^2 = 6^2 + BO^2 \Rightarrow BO^2 = 64 \Rightarrow BO = 8$ cm.
$BD = 2 \times BO = 16$ cm.
Area $= \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 12 \times 16 = 96$ cm$^2$.
Answer: $BD = 16$ cm, Area $= 96$ cm$^2$.
Mixed Example 2 — Three-dimensional Application

Q. A room is 12 m long, 5 m wide, and 8 m high. Find the length of the longest rod that can be kept inside.

The longest rod goes diagonally from one floor corner to the opposite ceiling corner.
First, floor diagonal: $d_1 = \sqrt{12^2 + 5^2} = \sqrt{144+25} = \sqrt{169} = 13$ m.
Longest rod: $L = \sqrt{d_1^2 + h^2} = \sqrt{13^2 + 8^2} = \sqrt{169 + 64} = \sqrt{233}$ m.
Answer: $L = \sqrt{233} \approx 15.26$ m.
Mixed Example 3 — Isosceles Trapezium

Q. An isosceles trapezium has parallel sides 10 cm and 24 cm, and legs of 13 cm each. Find its height and area.

Drop perpendiculars from the shorter parallel side's ends to the longer base. Each horizontal gap $= \frac{24-10}{2} = 7$ cm.
Height$^2 = 13^2 - 7^2 = 169 - 49 = 120 \Rightarrow h = \sqrt{120} = 2\sqrt{30}$ cm.
Area $= \frac{1}{2}(10 + 24) \times 2\sqrt{30} = \frac{1}{2} \times 34 \times 2\sqrt{30} = 34\sqrt{30}$ cm$^2 \approx 186.3$ cm$^2$.
Mixed Example 4 — Two Chords and Distance (ML Agarwal Type)

Q. In a circle of radius 10 cm, a chord of length 16 cm is drawn. Find the distance from the centre to the chord.

Let $O$ be the centre, $AB$ be the chord of length 16 cm, $OM \perp AB$ (so $M$ is midpoint of $AB$). $AM = 8$ cm.
In right $\triangle OMA$: $OA^2 = OM^2 + AM^2 \Rightarrow 10^2 = OM^2 + 8^2 \Rightarrow OM^2 = 36$.
$OM = 6$ cm.
Answer: Distance from centre to chord = 6 cm.

✏ Grand Mixed Practice — Examination Level (ML Agarwal Type)

  1. In $\triangle ABC$, $\angle B = 90°$. $D$ is the midpoint of $AC$. Prove $BD = \frac{1}{2}AC$. Hence show $BD^2 = \frac{1}{4}(AB^2 + BC^2)$.
  2. In a right-angled triangle, the hypotenuse is 26 cm and the sum of the other two sides is 34 cm. Find the two sides.
  3. $\triangle ABC$ has $AB = 7$ cm, $BC = 24$ cm, $CA = 25$ cm. Prove it is right-angled. Find the altitude from $B$ to $CA$.
  4. A rectangle has a diagonal of 26 cm and one side of 10 cm. Find its perimeter and area.
  5. An equilateral triangle has perimeter 36 cm. Find its altitude and area (without using $\frac{\sqrt{3}}{4}s^2$ directly, derive it).
  6. $ABCD$ is a rectangle with $AB = 8$ cm, $BC = 6$ cm. $P$ is the midpoint of $CD$. Find $AP$.
  7. In right $\triangle ABC$ ($\angle C = 90°$), altitude $CD$ meets $AB$ at $D$. If $AD = 4$ cm, $DB = 9$ cm, find $CD$, $AC$ and $BC$.
  8. Two poles of heights 9 m and 14 m stand on the same side of a road, 12 m apart. Find the distance between their tops.
  9. A right-angled triangle has legs in the ratio $3:4$. If the hypotenuse is 25 cm, find the legs and area.
  10. In rhombus $ABCD$, $AB = 13$ cm and diagonal $BD = 24$ cm. Find (i) diagonal $AC$, (ii) area, (iii) side if perimeter is given as 52 cm (verify).
  11. The sides of a right triangle are $(x-1)$, $x$, and $(x+1)$ cm. Find $x$ and the sides.
  12. $O$ is the centre of a circle of radius 15 cm. A chord $AB = 24$ cm and another chord $CD = 18$ cm are drawn. Find which chord is closer to the centre and by how much.
  13. Prove that in any right-angled triangle, the sum of squares of all three altitudes equals $\frac{a^2b^2 + b^2c^2 + c^2a^2}{(2A)^2}$ where $A$ is the area. [Hint: express each altitude in terms of sides and area.]
  14. A ladder 25 m long just reaches the top of a vertical wall. If the ladder makes an angle of $60°$ with the ground, find the height of the wall and the foot's distance from the wall (use $\cos 60° = \frac{1}{2}$, $\sin 60° = \frac{\sqrt{3}}{2}$).
  15. In a right triangle $ABC$ ($\angle A = 90°$), $D$ is on $BC$ such that $BD = DC$. Prove $4CD^2 = 4AD^2 - AB^2 + 3BC^2$ — no, correct statement: prove $BC^2 = AB^2 + AC^2$ (Pythagoras) and then show $BD^2 = \frac{BC^2 - AC^2}{4} + AD^2$.

9. Chapter Summary — Quick Revision Card

Key Results at a Glance
#ResultStatement / Formula
1Pythagoras TheoremIn right $\triangle$ with hyp. $c$: $c^2 = a^2 + b^2$
2Area-based ProofSquares on sides; area of big square = sum of two smaller squares
3Bhaskara's Proof4 triangles + inner square: $c^2 = 2ab + (b-a)^2 = a^2+b^2$
4Pythagorean Triplets$(3,4,5)$; $(5,12,13)$; $(8,15,17)$; $(7,24,25)$ and multiples
5Find hypotenuse$c = \sqrt{a^2+b^2}$
6Find a leg$a = \sqrt{c^2-b^2}$
7Converse$c^2 = a^2+b^2 \Rightarrow \angle C = 90°$
8Acute triangle$c^2 < a^2+b^2$
9Obtuse triangle$c^2 > a^2+b^2$
10Square diagonal$d = s\sqrt{2}$
11Equilateral altitude$h = \frac{s\sqrt{3}}{2}$
12Altitude on hypotenuse$CD^2 = AD \cdot DB$; $BC^2 = BD \cdot BA$; $AC^2 = AD \cdot AB$
13Median to hypotenuse$= \frac{1}{2} \times$ hypotenuse
14Rectangle diagonal$d = \sqrt{l^2+b^2}$
Common Mistakes to Avoid
Mnemonic for Proof structure: "Squares on Sides, Drop Perpendicular, Compare Areas" = the three key steps in the area-based proof.
Step-by-Step for Any Pythagoras Problem
  1. Draw and label the triangle. Mark the right angle clearly.
  2. Identify what is given (hypotenuse, or which leg) and what is to be found.
  3. Apply the correct formula: $c^2 = a^2 + b^2$ (for hyp.) or rearrange for a leg.
  4. Substitute values and simplify (recognise triplets first for speed).
  5. Take the positive square root (length is always positive).
  6. For converse: square all sides, check if largest$^2$ = sum of other two.
← Back to Mathematics