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WBB • कक्षा X • Mathematics • अध्याय 22

अनुमानित समय: 75 minutes

प्रगति: अध्ययनरत

Pythagoras Theorem

Welcome to the definitive master study guide for Chapter 22: Pythagoras Theorem (পিথাগোরাসের উপপাদ্য) under the West Bengal Board of Secondary Education (WBBSE) Class 10 Mathematics curriculum (Ganit Prakash). The theorem connecting the squares of the sides of a right-angled triangle is arguably the most famous and foundational result in all of mathematics. Historically known in ancient India through the Baudhayana Sulba Sutra (circa 800 BCE) for constructing sacrificial altars, the theorem was independently formalized in ancient Greece by Pythagoras of Samos and Euclid of Alexandria. In the WBBSE Class 10 syllabus, the core focus is Theorem 49: In any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides. Students master the elegant proof by similarity of triangles prescribed by the board. In right-angled triangle ABC with right angle at B, an altitude BD is drawn perpendicular to the hypotenuse AC. By the angle-angle similarity criterion, triangle ABD is similar to triangle ABC, yielding the ratio AD divided by AB equals AB divided by AC, which proves AB squared equals AD times AC. Similarly, triangle BCD is similar to triangle ABC, yielding CD divided by BC equals BC divided by AC, which proves BC squared equals CD times AC. Adding these two fundamental relations gives AB squared plus BC squared equals AC times the sum of AD and CD, which equals AC squared, completing the proof. The chapter also establishes Theorem 50, the Converse of Pythagoras Theorem: If in a triangle the area of the square on one side is equal to the sum of the areas of the squares on the other two sides, then the angle opposite to the first side is a right angle. Furthermore, the guide covers Pythagorean triplets (such as 3, 4, 5; 5, 12, 13; 7, 24, 25; 8, 15, 17) and their algebraic generation formula (2m, m squared minus 1, m squared plus 1), together with essential Madhyamik geometric riders, Apollonius' median theorem, and practical word problems.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें

प्रस्तावना अध्याय अवधारणाएँ सूत्र एवं सर्वसमिकाएँ हल उदाहरण सामान्य गलतियाँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

क्या आपने कभी सोचा है?

Over 2,800 years ago in ancient India, the sage-mathematician Baudhayana recorded in the Sulba Sutras: 'The rope stretched along the length of the diagonal of a rectangle produces an area which the vertical and horizontal sides make together.' Around the same era, Pythagoras of Samos and classical Greek geometers recognized this universal truth as the most fundamental relationship in spatial geometry. How does this simple equation a squared plus b squared equals c squared bridge planar geometry with trigonometry, Einstein's spacetime intervals, and modern 3D computer graphics? Let us explore Theorem 49 and its converse through rigorous Euclidean proofs and powerful Madhyamik riders.

यह अध्याय क्यों महत्वपूर्ण है

The Pythagoras Theorem is the supreme cornerstone of metric geometry, trigonometry, and coordinate analysis. Without this theorem, the concept of Euclidean distance between two points in Cartesian space would not exist, rendering modern computer graphics, GPS satellite triangulation, robotics kinematics, and architectural structural engineering impossible. In trigonometry, the cardinal identity sine squared theta plus cosine squared theta equals one is simply the Pythagoras Theorem in disguise on the unit circle. In Einstein's theory of relativity, the spacetime interval between events is computed via a four-dimensional pseudo-Pythagorean metric. In the WBBSE Madhyamik examination, Theorem 49 and Theorem 50 are primary candidates for the compulsory 5-mark geometric theorem question, while Pythagorean riders frequently appear in the 3-mark rider and 2-mark short answer sections. Mastery of this chapter guarantees students high scores and deep geometric intuition.

अध्ययन से पूर्व (आवश्यक ज्ञान)

Definition and properties of right-angled triangles (hypotenuse, perpendicular, base).
Conditions of similarity of triangles (AA / AAA criterion and proportional corresponding sides).
Conditions of congruence of triangles (SSS, SAS, RHS criteria).
Basic algebraic expansion of quadratic expressions and difference of squares.

इस अध्याय के लक्ष्य

Explain the historical significance of the Baudhayana-Pythagoras theorem in ancient Indian and Greek mathematics.
State and rigorously prove Theorem 49 (Pythagoras Theorem) using the similarity of triangles formed by dropping an altitude from the right-angled vertex to the hypotenuse.
State and prove Theorem 50 (Converse of Pythagoras Theorem) using the method of auxiliary congruent right triangles.
Identify, verify, and generate Pythagorean triplets using the algebraic parametric formula (2m, m^2 - 1, m^2 + 1).
Apply Pythagoras Theorem to solve real-world problems involving leaning ladders, shadows, navigation vectors, and geometric heights.
Prove high-yield Madhyamik riders including Apollonius' Theorem, acute and obtuse angle geometric generalizations, and the British Flag Theorem in rectangles.

अध्याय रूपरेखा एवं प्रगति

1 Module 1: Historical Evolution & Th...

2 Module 2: Theorem 49 (Pythagoras Th...

3 Module 3: Theorem 50 (Converse of P...

4 Module 4: Pythagorean Triplets & Al...

5 Module 5: High-Yield Madhyamik Geom...

6 Module 6: Practical Word Problems &...

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

Module 1: Historical Evolution & The Sulba Sutra Heritage

1.1 The Ancient Indian Context: Baudhayana Sulba Sutra

Long before the Greek philosopher Pythagoras of Samos (circa 570–495 BCE) founded his school at Croton, ancient Indian Vedic mathematicians had discovered, formulated, and applied the theorem relating the sides of a right-angled triangle. In the Baudhayana Sulba Sutra (circa 800 BCE), the sage-mathematician Baudhayana composed the famous shloka:

"dīrghasyākṣaṇayā rajjuḥ pārśvamānī tiryaṅmānī ca yatpṛthagbhūte kurutastadubhayaṃ karoti."

Translation: The rope stretched along the length of the diagonal of a rectangle produces an area which the vertical side and the horizontal side produce separately.
In modern algebraic notation: $$ ext{Diagonal}^2 = ext{Length}^2 + ext{Breadth}^2 \implies d^2 = l^2 + b^2$$

This principle was essential for the construction of complex sacrificial brick altars (vedis) such as the falcon-shaped Śyenaciti, which required combining two squares into a single larger square of equal area.

1.2 Greek Geometric Formulation (Euclid Book I, Proposition 47)

In classical Greek geometry, Pythagoras and later Euclid framed the theorem in terms of geometric areas of physical squares constructed upon the sides of the triangle: the square erected on the hypotenuse has an area equal to the sum of the areas of the two squares erected on the perpendicular sides.

Module 2: Theorem 49 (Pythagoras Theorem) — Full Statement & Proof via Similarity

2.1 Formal Statement of Theorem 49

WBBSE Theorem 49 (Core Board Theorem):
In any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides.

2.2 Step-by-Step Proof by Triangle Similarity

Given: Let $ riangle ABC$ be a right-angled triangle in which $ngle ABC = 90^\circ$. The side opposite to the right angle is the hypotenuse $AC$.

To Prove:

$$AC^2 = AB^2 + BC^2$$

Construction: From the right-angled vertex $B$, draw altitude $BD$ perpendicular to the hypotenuse $AC$ ($BD \perp AC$, where $D$ lies on $AC$).

Proof:

Part 1: Similarity of $ riangle ABD$ and $ riangle ABC$

In $ riangle ABD$ and $ riangle ABC$:
1. $ngle BDA = ngle ABC = 90^\circ$ (by construction and given).
2. $ngle BAD = ngle BAC$ (common angle $ngle A$).
Therefore, by the Angle-Angle (AA) similarity criterion: $$ riangle ABD \sim riangle ABC$$ Since corresponding sides of similar triangles are proportional: $$rac{AD}{AB} = rac{AB}{AC} \implies AB^2 = AD \cdot AC \quad ext{--- (Equation 1)}$$

Part 2: Similarity of $ riangle BCD$ and $ riangle ABC$

In $ riangle BCD$ and $ riangle ABC$:
1. $ngle BDC = ngle ABC = 90^\circ$ (by construction and given).
2. $ngle BCD = ngle BCA$ (common angle $ngle C$).
Therefore, by the Angle-Angle (AA) similarity criterion: $$ riangle BCD \sim riangle ABC$$ Since corresponding sides are proportional: $$rac{CD}{BC} = rac{BC}{AC} \implies BC^2 = CD \cdot AC \quad ext{--- (Equation 2)}$$

Part 3: Adding Equations 1 and 2

Adding Equation (1) and Equation (2): $$AB^2 + BC^2 = AD \cdot AC + CD \cdot AC$$ Factoring out the common term $AC$: $$AB^2 + BC^2 = AC \cdot (AD + CD)$$ Since point $D$ lies between $A$ and $C$ on line segment $AC$, we have $AD + CD = AC$. Therefore: $$AB^2 + BC^2 = AC \cdot AC = AC^2$$ $$AC^2 = AB^2 + BC^2$$ Hence Proved (Q.E.D.).

Module 3: Theorem 50 (Converse of Pythagoras Theorem)

3.1 Formal Statement of Theorem 50

WBBSE Theorem 50 (Converse of Pythagoras Theorem):
In a triangle, if the area of the square on one side is equal to the sum of the areas of the squares on the other two sides, then the angle opposite to the first side is a right angle.

3.2 Proof via Auxiliary Congruent Triangle

Given: In $ riangle ABC$, the relation between the three sides is: $$AC^2 = AB^2 + BC^2$$

To Prove: The angle opposite to side $AC$ is a right angle: $$ngle ABC = 90^\circ$$

Construction: Construct another triangle $ riangle DEF$ such that:
• $DE = AB$
• $EF = BC$
• $ngle DEF = 90^\circ$ (constructed as a right angle).

Proof:

Since $ riangle DEF$ is a right-angled triangle with $ngle DEF = 90^\circ$, applying Pythagoras Theorem (Theorem 49) to $ riangle DEF$ gives: $$DF^2 = DE^2 + EF^2$$ Substituting the constructed equalities $DE = AB$ and $EF = BC$: $$DF^2 = AB^2 + BC^2 \quad ext{--- (Equation 1)}$$ However, it is given that in $ riangle ABC$: $$AC^2 = AB^2 + BC^2 \quad ext{--- (Equation 2)}$$ Comparing Equation (1) and Equation (2): $$DF^2 = AC^2 \implies DF = AC$$ Now, compare $ riangle ABC$ and $ riangle DEF$:
1. $AB = DE$ (by construction).
2. $BC = EF$ (by construction).
3. $AC = DF$ (just proved).
Therefore, by the Side-Side-Side (SSS) congruence criterion: $$ riangle ABC \cong riangle DEF$$ Since corresponding angles of congruent triangles are equal (CPCT): $$ngle ABC = ngle DEF$$ Since $ngle DEF = 90^\circ$ by construction: $$ngle ABC = 90^\circ$$ Hence Proved (Q.E.D.).

Module 4: Pythagorean Triplets & Algebraic Generation

4.1 Definition of Pythagorean Triplets

A set of three positive integers $(a, b, c)$ is called a Pythagorean triplet if they satisfy the condition:

$$a^2 + b^2 = c^2$$

If $\gcd(a, b, c) = 1$, the triplet is called a primitive Pythagorean triplet. Any scalar multiple $(ka, kb, kc)$ for $k \in \mathbb{N}$ is also a valid Pythagorean triplet.

4.2 Algebraic Generation Formula

For any positive integer $m > 1$, the three numbers defined by:

$$a = 2m, \quad b = m^2 - 1, \quad c = m^2 + 1$$

always form a Pythagorean triplet. We verify this algebraically:

$$a^2 + b^2 = (2m)^2 + (m^2 - 1)^2 = 4m^2 + (m^4 - 2m^2 + 1) = m^4 + 2m^2 + 1 = (m^2 + 1)^2 = c^2$$

4.3 Core Triplet Inventory

Parameter $m$	Side $a = 2m$	Side $b = m^2 - 1$	Hypotenuse $c = m^2 + 1$	Triplet $(a, b, c)$	Verification
$m = 2$	$4$	$3$	$5$	$(3, 4, 5)$	$9 + 16 = 25$
$m = 3$	$6$	$8$	$10$	$(6, 8, 10)$	$36 + 64 = 100$
$m = 4$	$8$	$15$	$17$	$(8, 15, 17)$	$64 + 225 = 289$
$m = 5$	$10$	$24$	$26$	$(10, 24, 26)$	$100 + 576 = 676$

Module 5: High-Yield Madhyamik Geometric Riders

5.1 Rider 1: Apollonius' Theorem (Median Relation)

Statement: In any triangle $ riangle ABC$, if $AD$ is the median bisecting side $BC$ (so $BD = CD$), then:

$$AB^2 + AC^2 = 2(AD^2 + BD^2)$$

Proof Outline: Draw altitude $AE \perp BC$. In right-angled triangles $ riangle ABE$ and $ riangle ACE$, express $AB^2 = AE^2 + BE^2$ and $AC^2 = AE^2 + CE^2$. Write $BE = BD - ED$ and $CE = CD + ED = BD + ED$. Expanding both squares and summing cancels the cross-term $2BD \cdot ED$, yielding $2(AE^2 + ED^2 + BD^2) = 2(AD^2 + BD^2)$.

5.2 Rider 2: Perpendicular Altitude Property in Triangles

Statement: In $ riangle ABC$, if $AD \perp BC$, prove that:

$$AB^2 + CD^2 = AC^2 + BD^2$$

Proof: In right $ riangle ABD$, $AB^2 = AD^2 + BD^2 \implies AD^2 = AB^2 - BD^2$. In right $ riangle ACD$, $AC^2 = AD^2 + CD^2 \implies AD^2 = AC^2 - CD^2$. Equating both expressions: $AB^2 - BD^2 = AC^2 - CD^2 \implies AB^2 + CD^2 = AC^2 + BD^2$.

5.3 Rider 3: The British Flag Theorem in Rectangles

Statement: If $P$ is any point inside or outside a rectangle $ABCD$, prove that:

$$PA^2 + PC^2 = PB^2 + PD^2$$

Proof: Draw lines through $P$ parallel to the sides of the rectangle, and decompose the distance squares using Pythagoras theorem in the four corner right triangles.

Module 6: Practical Word Problems & Madhyamik Examination Strategies

6.1 The Classic Sliding Ladder Problem

A ladder of length $L$ leans against a vertical wall, with its foot on level ground at distance $x$ from the wall and its top reaching height $y$ on the wall. By Pythagoras theorem:

$$x^2 + y^2 = L^2$$

If the top of the ladder slides down by distance $\Delta y$ to height $y_2 = y - \Delta y$, the foot slides outward by distance $\Delta x$ to a new position $x_2 = x + \Delta x$. Because the physical length $L$ of the ladder remains constant:

$$x_2^2 + y_2^2 = L^2 \implies (x + \Delta x)^2 + (y - \Delta y)^2 = L^2$$

This relationship forms the basis of many 3-mark word problems in board examinations.

6.2 WBBSE Exam Writing Strategy for Theorem 49

Draw the diagram first: Draw a clean right-angled triangle with the right angle at $B$. Drop perpendicular $BD \perp AC$. Label all vertices clearly.
Structure the answer into 4 formal sections: 1. Special Enunciation (বিশেষ নির্বচন); 2. To Prove (প্রামাণ্য); 3. Construction (অঙ্কন); 4. Proof (প্রমাণ).
State the similarity criterion explicitly: Explicitly mention "by AA similarity criterion, $ riangle ABD \sim riangle ABC$". Omitting the criterion name may cause mark deductions.
State the reason for side sum: Note that "point $D$ lies on $AC$, so $AD + CD = AC$".

महत्वपूर्ण सूत्र, सर्वसमिकाएँ एवं प्रमेय

Pythagoras Theorem

$AC^2 = AB^2 + BC^2$

Converse of Pythagoras Theorem

$a^2 + b^2 = c^2 \implies \angle C = 90^\circ$

Altitude to Hypotenuse Identity

$BD^2 = AD \cdot CD$

Leg and Hypotenuse Segment Relation

$AB^2 = AD \cdot AC, \quad BC^2 = CD \cdot AC$

Altitude Length via Area

$BD = \frac{AB \cdot BC}{AC}$

Pythagorean Triplet Generator

$(2m, \; m^2 - 1, \; m^2 + 1) \quad (m > 1)$

Apollonius' Theorem

$AB^2 + AC^2 = 2(AD^2 + BD^2)$

British Flag Theorem

$PA^2 + PC^2 = PB^2 + PD^2$

अवधारणात्मक हल उदाहरण एवं अनुप्रयोग (Solved Examples)

उदाहरण 1

In a right-angled triangle $ABC$, $\angle B = 90^\circ$. An altitude $BD$ is drawn to the hypotenuse $AC$. If $AD = 4\text{ cm}$ and $CD = 9\text{ cm}$, find the lengths of $BD$, $AB$, and $BC$.

विस्तृत समाधान / उत्तर:

Step 1: Determine Altitude $BD$
In right $\triangle ABC$ with $BD \perp AC$, by similarity of $\triangle ABD$ and $\triangle DBC$: $$\frac{AD}{BD} = \frac{BD}{CD} \implies BD^2 = AD \cdot CD$$ Substitute $AD = 4\text{ cm}$ and $CD = 9\text{ cm}$: $$BD^2 = 4 \times 9 = 36 \implies BD = \sqrt{36} = 6\text{ cm}$$ Step 2: Calculate Side $AB$
The hypotenuse $AC = AD + CD = 4 + 9 = 13\text{ cm}$.
By the leg-hypotenuse relation ($AB^2 = AD \cdot AC$): $$AB^2 = 4 \times 13 = 52 \implies AB = \sqrt{52} = 2\sqrt{13} \approx 7.21\text{ cm}$$ Step 3: Calculate Side $BC$
By the second leg relation ($BC^2 = CD \cdot AC$): $$BC^2 = 9 \times 13 = 117 \implies BC = \sqrt{117} = 3\sqrt{13} \approx 10.82\text{ cm}$$ Step 4: Verification via Pythagoras Theorem
$$AB^2 + BC^2 = 52 + 117 = 169 = 13^2 = AC^2$$ Matches perfectly. Final values: $BD = 6\text{ cm}$, $AB = 2\sqrt{13}\text{ cm}$, $BC = 3\sqrt{13}\text{ cm}$.

उदाहरण 2

A ladder $25\text{ m}$ long reaches a window of a building $20\text{ m}$ above the ground. Determine the distance of the foot of the ladder from the building. If the top of the ladder slides down by $4\text{ m}$, by how much does the foot of the ladder slide outwards?

विस्तृत समाधान / उत्तर:

Step 1: Initial Distance of Foot from Building
Let length of ladder $L = 25\text{ m}$ (hypotenuse).
Height of window $y_1 = 20\text{ m}$.
Let initial distance of foot from wall be $x_1$.
By Pythagoras Theorem in the right-angled triangle formed by wall, ground, and ladder: $$x_1^2 + y_1^2 = L^2 \implies x_1^2 + 20^2 = 25^2$$ $$x_1^2 + 400 = 625 \implies x_1^2 = 225 \implies x_1 = \sqrt{225} = 15\text{ m}$$ Step 2: New Position after Top Slides Down
The top of the ladder slides down by $4\text{ m}$, so the new height is: $$y_2 = 20 - 4 = 16\text{ m}$$ The length of the ladder remains constant at $L = 25\text{ m}$.
Let the new distance of the foot from the building be $x_2$: $$x_2^2 + y_2^2 = L^2 \implies x_2^2 + 16^2 = 25^2$$ $$x_2^2 + 256 = 625 \implies x_2^2 = 369 \implies x_2 = \sqrt{369} \approx 19.21\text{ m}$$ Step 3: Distance the Foot Slides Outwards
$$\text{Outward slide } \Delta x = x_2 - x_1 = \sqrt{369} - 15 \approx 19.21 - 15 = 4.21\text{ m}$$ Conclusion: The initial distance of the foot was $15\text{ m}$, and it slides outward by approximately $4.21\text{ m}$.

उदाहरण 3

In $\triangle ABC$, $AD \perp BC$. Prove that $AB^2 + CD^2 = AC^2 + BD^2$.

विस्तृत समाधान / उत्तर:

Step 1: Formulate Right Triangles
Since $AD \perp BC$, $D$ is the foot of the altitude on $BC$. This creates two right-angled triangles: $\triangle ABD$ and $\triangle ACD$, both sharing the common altitude $AD$. Step 2: Apply Pythagoras Theorem to $\triangle ABD$
In right $\triangle ABD$, $\angle ADB = 90^\circ$: $$AB^2 = AD^2 + BD^2 \implies AD^2 = AB^2 - BD^2 \quad \text{--- (Equation 1)}$$ Step 3: Apply Pythagoras Theorem to $\triangle ACD$
In right $\triangle ACD$, $\angle ADC = 90^\circ$: $$AC^2 = AD^2 + CD^2 \implies AD^2 = AC^2 - CD^2 \quad \text{--- (Equation 2)}$$ Step 4: Equating Both Expressions for $AD^2$
From Equation (1) and Equation (2): $$AB^2 - BD^2 = AC^2 - CD^2$$ Rearranging terms by transposing $-BD^2$ and $-CD^2$: $$AB^2 + CD^2 = AC^2 + BD^2$$ Hence Proved (Q.E.D.).

उदाहरण 4

In a right-angled triangle $ABC$ with $\angle C = 90^\circ$, points $D$ and $E$ lie on sides $CA$ and $CB$ respectively. Prove that $AE^2 + BD^2 = AB^2 + DE^2$.

विस्तृत समाधान / उत्तर:

Step 1: Identify the Right Triangles Involved
Since $\angle C = 90^\circ$, any triangle having $C$ as a vertex and two legs lying along $CA$ and $CB$ is a right-angled triangle. We identify four such triangles:
1. $\triangle ACE$ with hypotenuse $AE$.
2. $\triangle BCD$ with hypotenuse $BD$.
3. $\triangle ACB$ with hypotenuse $AB$.
4. $\triangle DCE$ with hypotenuse $DE$. Step 2: Express Hypotenuse Squares Using Pythagoras Theorem
In right $\triangle ACE$: $$AE^2 = AC^2 + CE^2 \quad \text{--- (1)}$$ In right $\triangle BCD$: $$BD^2 = CD^2 + BC^2 \quad \text{--- (2)}$$ Step 3: Sum the Left-Hand Side
Adding Equation (1) and Equation (2): $$AE^2 + BD^2 = (AC^2 + CE^2) + (CD^2 + BC^2)$$ Regrouping the terms on the right-hand side: $$AE^2 + BD^2 = (AC^2 + BC^2) + (CD^2 + CE^2) \quad \text{--- (3)}$$ Step 4: Substitute from the Remaining Right Triangles
In right $\triangle ACB$: $$AC^2 + BC^2 = AB^2$$ In right $\triangle DCE$: $$CD^2 + CE^2 = DE^2$$ Substituting these into Equation (3): $$AE^2 + BD^2 = AB^2 + DE^2$$ Hence Proved (Q.E.D.).

उदाहरण 5

In an isosceles triangle $ABC$, $AB = AC = 13\text{ cm}$ and base $BC = 10\text{ cm}$. Find the length of altitude $AD$ and the area of $\triangle ABC$.

विस्तृत समाधान / उत्तर:

Step 1: Properties of Isosceles Triangle Altitude
In an isosceles triangle with $AB = AC$, the altitude $AD \perp BC$ bisects the base $BC$.
Therefore: $$BD = CD = \frac{BC}{2} = \frac{10}{2} = 5\text{ cm}$$ Step 2: Apply Pythagoras Theorem to $\triangle ABD$
In right-angled triangle $\triangle ABD$, $\angle ADB = 90^\circ$: $$AB^2 = AD^2 + BD^2$$ Substitute $AB = 13\text{ cm}$ and $BD = 5\text{ cm}$: $$13^2 = AD^2 + 5^2 \implies 169 = AD^2 + 25$$ $$AD^2 = 169 - 25 = 144 \implies AD = \sqrt{144} = 12\text{ cm}$$ Step 3: Calculate Area of $\triangle ABC$
$$\text{Area}(\triangle ABC) = \frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2} \times BC \times AD$$ $$\text{Area} = \frac{1}{2} \times 10 \times 12 = 60\text{ cm}^2$$ Conclusion: Length of altitude $AD = 12\text{ cm}$ and $\text{Area} = 60\text{ cm}^2$.

सामान्य गलतियाँ एवं परीक्षक के जाल (Examiner Traps)

सामान्य भ्रम / गलत उत्तर

Applying Pythagoras theorem to non-right-angled triangles without dropping an altitude.

सही वैज्ञानिक तथ्य

Pythagoras theorem strictly applies only to right-angled triangles. For acute or obtuse triangles, drop an altitude to create right triangles first.

सामान्य भ्रम / गलत उत्तर

Assuming the longest side is always labeled 'c' or 'AC'.

सही वैज्ञानिक तथ्य

Always identify the hypotenuse as the side opposite the specific vertex where the 90-degree angle is located.

सामान्य भ्रम / गलत उत्तर

In proving Theorem 49, confusing corresponding vertices in triangle similarity.

सही वैज्ञानिक तथ्य

Write the similarity correspondence strictly as ΔABD ~ ΔABC and ΔBCD ~ ΔABC.

सामान्य भ्रम / गलत उत्तर

In proving the converse (Theorem 50), assuming the given triangle is right-angled.

सही वैज्ञानिक तथ्य

Construct an independent second triangle DEF with angle E = 90 deg, and prove it congruent to triangle ABC by SSS.

सामान्य भ्रम / गलत उत्तर

Believing (2m, m^2 - 1, m^2 + 1) generates every possible Pythagorean triplet.

सही वैज्ञानिक तथ्य

This formula generates all primitive triplets when m is an integer, but scalar multiples (e.g. 9, 12, 15) arise by multiplying primitive triplets by a constant k.

Pythagoras Theorem & Similarity Proof Architecture (WBBSE Class 10 Ganit Prakash)

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1

Baudhayana formulated the theorem in the Sulba Sutras (c. 800 BCE): the diagonal of a rectangle produces the combined area of both sides.

मुख्य बिंदु 2

Theorem 49 (Pythagoras Theorem): In any right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides (AC^2 = AB^2 + BC^2).

मुख्य बिंदु 3

Proof of Theorem 49 is conducted via triangle similarity: altitude BD perpendicular to hypotenuse AC creates ΔABD ~ ΔABC and ΔBCD ~ ΔABC.

मुख्य बिंदु 4

From similarity: AB^2 = AD * AC and BC^2 = CD * AC. Summing both gives AB^2 + BC^2 = AC * (AD + CD) = AC^2.

मुख्य बिंदु 5

Theorem 50 (Converse of Pythagoras Theorem): If a triangle satisfies AC^2 = AB^2 + BC^2, the angle opposite side AC is 90 degrees.

मुख्य बिंदु 6

Proof of Theorem 50 is established by constructing an auxiliary right triangle DEF and demonstrating SSS congruence with triangle ABC.

मुख्य बिंदु 7

Pythagorean triplets are integer solutions to a^2 + b^2 = c^2, generated by the parametric formula (2m, m^2 - 1, m^2 + 1) for m > 1.

मुख्य बिंदु 8

Apollonius' Theorem relates triangle sides to its median: AB^2 + AC^2 = 2(AD^2 + BD^2) where AD is the median to BC.

मुख्य बिंदु 9

In rectangles, the British Flag Theorem states that PA^2 + PC^2 = PB^2 + PD^2 for any arbitrary point P in the plane.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

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ऑनलाइन CBT टेस्ट देकर तैयारी का मूल्यांकन करें

झारखण्ड बोर्ड परीक्षा पैटर्न पर आधारित बहुविकल्पीय प्रश्नों का ऑनलाइन टेस्ट दें। तुरंत परिणाम, समय विश्लेषण और प्रत्येक प्रश्न का विस्तृत हल प्राप्त करें।

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