Maharashtra Board Class 10 Maths Chapter 2 Pythagoras Theorem Set 2.1 Solutions

Question 1. Identify, with reason, which of the following are Pythagorean triplets.
(i) (3,5,4)
(ii) (4,9,12)
(iii) (5,12,13)
(iv) (24,70,74)
(v) (10,24,27)
(vi) (11,60,61)
Answer:
i. Here, \(5^2 = 25\)
\(3^2 + 4^2 = 9 + 16 = 25\)
\(\therefore 5^2 = 3^2 + 4^2\)
The square of the largest number is equal to the sum of the squares of the other two numbers.
\(\therefore\) (3,5,4) is a Pythagorean triplet.

ii. Here, \(12^2 = 144\)
\(4^2 + 9^2 = 16 + 81 = 97\)
\(\therefore 12^2 \neq 4^2 + 9^2\)
The square of the largest number is not equal to the sum of the squares of the other two numbers.
\(\therefore\) (4,9,12) is not a Pythagorean triplet.

iii. Here, \(13^2 = 169\)
\(5^2 + 12^2 = 25 + 144 = 169\)
\(\therefore 13^2 = 5^2 + 12^2\)
The square of the largest number is equal to the sum of the squares of the other two numbers.
\(\therefore\) (5,12,13) is a Pythagorean triplet.

iv. Here, \(74^2 = 5476\)
\(24^2 + 70^2 = 576 + 4900 = 5476\)
\(\therefore 74^2 = 24^2 + 70^2\)
The square of the largest number is equal to the sum of the squares of the other two numbers.
\(\therefore\) (24, 70,74) is a Pythagorean triplet.

v. Here, \(27^2 = 729\)
\(10^2 + 24^2 = 100 + 576 = 676\)
\(\therefore 27^2 \neq 10^2 + 24^2\)
The square of the largest number is not equal to the sum of the squares of the other two numbers.
\(\therefore\) (10,24,27) is not a Pythagorean triplet.

vi. Here, \(61^2 = 3721\)
\(11^2 + 60^2 = 121 + 3600 = 3721\)
\(\therefore 61^2 = 11^2 + 60^2\)
The square of the largest number is equal to the sum of the squares of the other two numbers.
\(\therefore\) (11,60,61) is a Pythagorean triplet.
In simple words: A Pythagorean triplet consists of three positive integers a, b, and c, such that \(a^2 + b^2 = c^2\). To identify them, square the largest number and compare it to the sum of the squares of the other two numbers.

🎯 Exam Tip: When identifying Pythagorean triplets, always check if the square of the largest number exactly equals the sum of the squares of the other two numbers. Provide a clear reason based on this fundamental property of right-angled triangles.

Question 2. In the adjoining figure, \(\angle MNP = 90^\circ\), seg NQ \(\perp\) seg MP, MQ = 9, QP = 4, find NQ.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज MNP को दर्शाता है, जिसमें कोण N पर समकोण है। N से कर्ण MP पर NQ एक लंब है। इसमें MQ की लंबाई 9 इकाई और QP की लंबाई 4 इकाई है। यह चित्र ज्यामितीय माध्य प्रमेय को लागू करने के लिए आवश्यक संरचना दिखाता है।
Answer: Solution:
In \(\Delta MNP\), \(\angle MNP = 90^\circ\) [Given]
seg NQ \(\perp\) seg MP
\(NQ^2 = MQ \times QP\) [Theorem of geometric mean]
\(\therefore NQ = \sqrt{MQ \times QP}\) [Taking square root of both sides]
\( = \sqrt{9 \times 4}\)
\( = 3 \times 2\)
\(\therefore NQ = 6\) units
In simple words: Using the geometric mean theorem for a right-angled triangle with an altitude to the hypotenuse, the length of the altitude (NQ) is the square root of the product of the two segments it divides the hypotenuse into (MQ and QP).

🎯 Exam Tip: The theorem of geometric mean is crucial here. Ensure you correctly identify the altitude to the hypotenuse and the segments it creates. Squaring and then taking the square root correctly is important for final calculation.

Question 3. In the adjoining figure, \(\angle QPR = 90^\circ\), seg PM \(\perp\) seg QR and Q - M - R, PM = 10, QM = 8, find QR.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज PQR को दर्शाता है, जिसमें कोण P पर समकोण है। P से कर्ण QR पर PM एक लंब है, और M, Q और R के बीच स्थित है। PM की लंबाई 10 इकाई और QM की लंबाई 8 इकाई है। यह चित्र ज्यामितीय माध्य प्रमेय का उपयोग करके QR की लंबाई ज्ञात करने के लिए जानकारी प्रदान करता है।
Answer: Solution:
In \(\Delta PQR\), \(\angle QPR = 90^\circ\) [Given]
seg PM \(\perp\) seg QR
\(\therefore PM^2 = QM \times MR\) [Theorem of geometric mean]
\(\therefore 10^2 = 8 \times MR\)
\(100 = 8 \times MR\)
\(\therefore MR = \frac{100}{8}\)
\( = 12.5\)
Now, QR = QM + MR [Q - M - R]
\( = 8 + 12.5\)
\(\therefore QR = 20.5\) units
In simple words: We first use the geometric mean theorem to find the length of MR. Once MR is known, the total length of the hypotenuse QR is found by adding the lengths of the segments QM and MR.

🎯 Exam Tip: Remember that the geometric mean theorem links the altitude to the hypotenuse with the segments of the hypotenuse. Calculate MR accurately and then sum the segments to find the total length of QR.

Question 4. See adjoining figure. Find RP and PS using the information given in \(\Delta PSR\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज PSR को दर्शाता है, जिसमें कोण S पर समकोण है। कोण P का मान 30 डिग्री है और भुजा RS की लंबाई 6 इकाई है। यह चित्र 30-60-90 त्रिभुज प्रमेय का उपयोग करके शेष भुजाओं RP और PS की लंबाई ज्ञात करने के लिए दिया गया है।
Answer: Solution:
In \(\Delta PSR\), \(\angle S = 90^\circ\), \(\angle P = 30^\circ\) [Given]
\(\therefore \angle R = 60^\circ\) [Remaining angle of a triangle]
\(\therefore \Delta PSR\) is a 30°-60°-90° triangle.
\(RS = \frac{1}{2} RP\) [Side opposite to 30°]
\(\therefore 6 = \frac{1}{2} RP\)
\(\therefore RP = 6 \times 2 = 12\) units
Also, \(PS = \frac{\sqrt{3}}{2} RP\) [Side opposite to 60°]
\( = \frac{\sqrt{3}}{2} \times 12\)
\( = 6\sqrt{3}\) units
\(\therefore RP = 12\) units, \(PS = 6\sqrt{3}\) units
In simple words: Given a 30-60-90 triangle, we use the properties that the side opposite the 30° angle is half the hypotenuse, and the side opposite the 60° angle is \(\frac{\sqrt{3}}{2}\) times the hypotenuse, to find the unknown side lengths.

🎯 Exam Tip: Accurately identifying the angles and applying the 30-60-90 triangle theorem is key. Remember the relationships: side opposite 30° = \(\frac{1}{2}\) Hypotenuse, and side opposite 60° = \(\frac{\sqrt{3}}{2}\) Hypotenuse.

Question 5. For finding AB and BC with the help of information given in the adjoining figure, complete the following activity.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज ABC को दर्शाता है, जिसमें कोण B पर समकोण है। भुजा AB और BC बराबर हैं, जो इसे एक समद्विबाहु समकोण त्रिभुज (45-45-90 त्रिभुज) बनाता है। कर्ण AC की लंबाई \(\sqrt{8}\) इकाई है। यह चित्र AB और BC की लंबाई ज्ञात करने के लिए जानकारी देता है।
Answer: Solution:
AB = BC [Given]
\(\therefore \angle BAC = \angle BCA\) [Isosceles triangle theorem]
Let \(\angle BAC = \angle BCA = x\) (i)
In \(\Delta ABC\), \(\angle A + \angle B + \angle C = 180^\circ\) [Sum of the measures of the angles of a triangle is \(180^\circ\)]
\(\therefore x + 90^\circ + x = 180^\circ\) [From (i)]
\(\therefore 2x = 90^\circ\)
\(\therefore x = \frac{90^\circ}{2}\) [From (i)]
\(\therefore x = 45^\circ\)
\(\therefore AB = BC = \frac{1}{\sqrt{2}} \times AC\) [Side opposite to 45°]
\( = \frac{1}{\sqrt{2}} \times \sqrt{8}\)
\( = \frac{1}{\sqrt{2}} \times 2\sqrt{2}\)
\(\therefore AB = BC = 2\) units
In simple words: Since angles A and C are equal and angle B is 90 degrees, it's a 45-45-90 triangle. In such a triangle, the sides opposite the 45-degree angles are \(1/\sqrt{2}\) times the hypotenuse.

🎯 Exam Tip: For a 45-45-90 triangle, the equal sides are \(1/\sqrt{2}\) times the hypotenuse. Simplify square roots carefully (e.g., \(\sqrt{8} = 2\sqrt{2}\)) to arrive at the final answer.

Question 6. Find the side and perimeter of a square whose diagonal is 10 cm.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वर्ग ABCD को दर्शाता है। वर्ग का विकर्ण AC 10 सेमी लंबा है। चित्र वर्ग की भुजा और परिमाप ज्ञात करने के लिए जानकारी देता है। विकर्ण वर्ग को दो समद्विबाहु समकोण त्रिभुजों में विभाजित करता है।
Answer: Solution:
Let ABCD be the given square.
l(diagonal AC) = 10 cm
Let the side of the square be 'x' cm.
In \(\Delta ABC\),
\(\angle B = 90^\circ\) [Angle of a square]
\(\therefore AC^2 = AB^2 + BC^2\) [Pythagoras theorem]
\(\therefore 10^2 = x^2 + x^2\)
\(\therefore 100 = 2x^2\)
\(\therefore x^2 = \frac{100}{2}\)
\(\therefore x^2 = 50\)
\(\therefore x = \sqrt{50}\) [Taking square root of both sides]
\( = \sqrt{25 \times 2} = 5\sqrt{2}\)
\(\therefore\) side of square is \(5\sqrt{2}\) cm.
Perimeter of square \( = 4 \times \text{side}\)
\( = 4 \times 5\sqrt{2}\)
\(\therefore\) Perimeter of square \( = 20\sqrt{2}\) cm
In simple words: For a square, the diagonal forms a right-angled isosceles triangle with two sides. Using Pythagoras theorem, we find the side length from the diagonal, and then calculate the perimeter by multiplying the side length by four.

🎯 Exam Tip: Remember that in a square, the diagonal forms a 45-45-90 triangle with two sides. Applying the Pythagoras theorem to find the side length (\(a^2 + a^2 = d^2\)) and then calculating the perimeter (\(4a\)) are the main steps.

Question 7. In the adjoining figure, \(\angle DFE = 90^\circ\), FG \(\perp\) ED. If GD = 8, FG = 12, find
(i) EG
(ii) FD, and
(iii) EF
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज DFE को दर्शाता है, जिसमें कोण F पर समकोण है। F से कर्ण ED पर FG एक लंब है। GD की लंबाई 8 इकाई और FG की लंबाई 12 इकाई है। यह चित्र EG, FD और EF की लंबाई ज्ञात करने के लिए जानकारी देता है।
Answer: Solution:
i. In \(\Delta DFE\), \(\angle DFE = 90^\circ\) and FG \(\perp\) ED [Given]
\(\therefore FG^2 = GD \times EG\) [Theorem of geometric mean]
\(\therefore 12^2 = 8 \times EG\).
\(144 = 8 \times EG\)
\(\therefore EG = \frac{144}{8}\)
\(\therefore EG = 18\) units

ii. In \(\Delta FGD\), \(\angle FGD = 90^\circ\) [Given]
\(\therefore FD^2 = FG^2 + GD^2\) [Pythagoras theorem]
\( = 12^2 + 8^2 = 144 + 64\)
\( = 208\)
\(\therefore FD = \sqrt{208}\) [Taking square root of both sides]
\( = \sqrt{16 \times 13}\)
\(\therefore FD = 4\sqrt{13}\) units

iii. In \(\Delta EGF\), \(\angle EGF = 90^\circ\) [Given]
\(\therefore EF^2 = EG^2 + FG^2\) [Pythagoras theorem]
\( = 18^2 + 12^2 = 324 + 144\)
\( = 468\)
\(\therefore EF = \sqrt{468}\) [Taking square root of both sides]
\( = \sqrt{36 \times 13}\)
\(\therefore EF = 6\sqrt{13}\) units
In simple words: This problem uses the geometric mean theorem to find EG, then applies the Pythagorean theorem to two different right-angled triangles (\(\Delta FGD\) and \(\Delta EGF\)) to find FD and EF, respectively.

🎯 Exam Tip: This question combines the geometric mean theorem and Pythagoras theorem. Solve step-by-step: first use geometric mean for altitude, then apply Pythagoras in the smaller right triangles formed by the altitude.

Question 8. Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक आयत ABCD को दर्शाता है। इसकी चौड़ाई AB 12 सेमी है और लंबाई BC 35 सेमी है। आयत का विकर्ण AC दिखाया गया है, जिसकी लंबाई ज्ञात करनी है। विकर्ण आयत को दो समकोण त्रिभुजों में विभाजित करता है।
Answer: Solution:
Let ABCD be the given rectangle.
AB = 12 cm, BC = 35 cm
In \(\Delta ABC\), \(\angle B = 90^\circ\) [Angle of a rectangle]
\(\therefore AC^2 = AB^2 + BC^2\) [Pythagoras theorem]
\( = 12^2 + 35^2\)
\( = 144 + 1225\)
\( = 1369\)
\(\therefore AC = \sqrt{1369}\) [Taking square root of both sides]
\( = 37\) cm
\(\therefore\) The diagonal of the rectangle is 37 cm.
In simple words: A rectangle's diagonal forms the hypotenuse of a right-angled triangle, with the rectangle's length and breadth as the other two sides. We use the Pythagoras theorem to find the diagonal's length.

🎯 Exam Tip: Visualize the rectangle and its diagonal as a right-angled triangle. Applying Pythagoras theorem (\(diagonal^2 = length^2 + breadth^2\)) is the direct method to solve this. Be careful with calculations of squares and square roots.

Question 9. In the adjoining figure, M is the midpoint of QR. \(\angle PRQ = 90^\circ\). Prove that, \(PQ^2 = 4 PM^2 - 3 PR^2\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज PQR को दर्शाता है, जिसमें कोण R पर समकोण है (\(\angle PRQ = 90^\circ\))। भुजा QR का मध्यबिंदु M है। बिंदु P से M तक एक रेखा खींची गई है। यह चित्र दिए गए संबंध को सिद्ध करने के लिए जानकारी देता है।
Answer: Solution:
Proof:
In \(\Delta PQR\), \(\angle PRQ = 90^\circ\) [Given]
\(PQ^2 = PR^2 + QR^2\) (i) [Pythagoras theorem]
RM \( = \frac{1}{2}\) QR [M is the midpoint of QR]
\(\therefore 2RM = QR\) (ii)
\(\therefore PQ^2 = PR^2 + (2RM)^2\) [From (i) and (ii)]
\(\therefore PQ^2 = PR^2 + 4RM^2\) (iii)
Now, in \(\Delta PRM\), \(\angle PRM = 90^\circ\) [Given]
\(\therefore PM^2 = PR^2 + RM^2\) [Pythagoras theorem]
\(\therefore RM^2 = PM^2 - PR^2\) (iv)
\(\therefore PQ^2 = PR^2 + 4 (PM^2 - PR^2)\) [From (iii) and (iv)]
\(\therefore PQ^2 = PR^2 + 4 PM^2 - 4 PR^2\)
\(\therefore PQ^2 = 4 PM^2 - 3 PR^2\)
In simple words: This proof involves applying the Pythagoras theorem to two right-angled triangles, \(\Delta PQR\) and \(\Delta PRM\). We use the midpoint property to relate QR and RM, then substitute expressions to derive the required relationship.

🎯 Exam Tip: For proofs, clearly state the given information and use Pythagoras theorem on all relevant right-angled triangles. Substitution and algebraic manipulation are key to reaching the desired result.

Question 10. Walls of two buildings on either side of a street are parallel to each other. A ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street, its top touches the window of the other building at a height 4.2 m. Find the width of the street.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो समानांतर इमारतों के बीच एक सड़क को दर्शाता है। एक 5.8 मीटर लंबी सीढ़ी (AC और CE) सड़क (BD) पर रखी है। सीढ़ी का एक सिरा (C) सड़क पर है। पहले मामले में, सीढ़ी का शीर्ष (A) एक इमारत पर 4 मीटर की ऊँचाई पर (AB) एक खिड़की तक पहुँचता है। दूसरे मामले में, सीढ़ी को दूसरी तरफ पलटा जाता है, और उसका शीर्ष (E) दूसरी इमारत पर 4.2 मीटर की ऊँचाई पर (ED) एक खिड़की तक पहुँचता है। चित्र सड़क की चौड़ाई (BD) ज्ञात करने के लिए जानकारी देता है।
Answer: Solution:
Let AC and CE represent the ladder of length 5.8 m, and A and E represent windows of the buildings on the opposite sides of the street. BD is the width of the street.
AB = 4 m and ED = 4.2 m
In \(\Delta ABC\), \(\angle B = 90^\circ\) [Given]
\(AC^2 = AB^2 + BC^2\) [Pythagoras theorem]
\(\therefore 5.8^2 = 4^2 + BC^2\)
\(\therefore 5.8^2 - 4^2 = BC^2\)
\(\therefore (5.8 - 4) (5.8 + 4) = BC^2\)
\(\therefore 1.8 \times 9.8 = BC^2\)
\(\therefore \frac{18 \times 98}{100} = BC^2\)
\(\therefore \frac{9 \times 2 \times 49 \times 2}{100} = BC^2\)
\(\therefore \frac{9 \times 4 \times 49}{100} = BC^2\)
\(\therefore BC = \frac{3 \times 2 \times 7}{10}\) [Taking square root of both sides]
\(\therefore BC = \frac{42}{10} = 4.2\) cm (i)
In \(\Delta CDE\), \(\angle D = 90^\circ\) [Given]
\(CE^2 = CD^2 + DE^2\) [Pythagoras theorem]
\(\therefore 5.8^2 = CD^2 + 4.2^2\)
\(\therefore 5.8^2 - 4.2^2 = CD^2\)
\(\therefore (5.8 - 4.2) (5.8 + 4.2) = CD^2\)
\(\therefore 1.6 \times 10 = CD^2\)
\(\therefore CD^2 = 16\)
\(\therefore CD = 4\)m (ii) [Taking square root of both sides]
Now, BD = BC + CD [B - C - D]
\( = 4.2 + 4\) [From (i) and (ii)]
\( = 8.2\) m
\(\therefore\) The width of the street is 8.2 metres.
In simple words: This problem involves two right-angled triangles formed by the ladder, the building walls, and the street. We use the Pythagoras theorem twice to find the segment of the street corresponding to each building and then add them to get the total width.

🎯 Exam Tip: This is a multi-step problem. Break it down into two right-angled triangles. Apply Pythagoras theorem for each to find the partial widths of the street. Ensure correct algebraic manipulation, especially with decimals and square roots, and add the partial widths for the final answer.

Question 1. Verify that (3,4,5), (5,12,13), (8,15,17), (24,25,7) are Pythagorean triplets. (Textbook pg. no. 30)
Answer: Solution:
i. Here, \(5^2 = 25\)
\(3^2 + 4^2 = 9 + 16 = 25\)
\(\therefore 5^2 = 3^2 + 4^2\)
The square of the largest number is equal to the sum of the squares of the other two numbers.
\(\therefore\) 3,4,5 is a Pythagorean triplet.

ii. Here, \(13^2 = 169\)
\(5^2 + 12^2 = 25 + 144 = 169\)
\(\therefore 13^2 = 5^2 + 12^2\)
The square of the largest number is equal to the sum of the squares of the other two numbers.
\(\therefore\) 5,12,13 is a Pythagorean triplet.

iii. Here, \(17^2 = 289\)
\(8^2 + 15^2 = 64 + 225 = 289\)
\(\therefore 17^2 = 8^2 + 15^2\)
The square of the largest number is equal to the sum of the squares of the other two numbers.
\(\therefore\) 8,15,17 is a Pythagorean triplet.

iv. Here, \(25^2 = 625\)
\(7^2 + 24^2 = 49 + 576 = 625\)
\(\therefore 25^2 = 7^2 + 24^2\)
The square of the largest number is equal to the sum of the squares of the other two numbers.
\(\therefore\) 24,25, 7 is a Pythagorean triplet.
In simple words: To verify if a set of three numbers forms a Pythagorean triplet, square each number. If the square of the largest number is equal to the sum of the squares of the other two numbers, then it is a Pythagorean triplet.

🎯 Exam Tip: Systematically apply the Pythagorean theorem's converse for each given set of numbers. Always identify the largest number first, square it, then sum the squares of the other two. Compare the results for verification.

Question 2. Assign different values to a and b and obtain 5 Pythagorean triplets. (Textbook pg. no. 31)
Answer: Solution:
i. Let a = 2, b = 1
\(a^2 + b^2 = 2^2 + 1^2 = 4 + 1 = 5\)
\(a^2 - b^2 = 2^2 - 1^2 = 4 - 1 = 3\)
\(2ab = 2 \times 2 \times 1 = 4\)
\(\therefore\) (5, 3, 4) is a Pythagorean triplet.

ii. Let a = 4, b = 3
\(a^2 + b^2 = 4^2 + 3^2 = 16 + 9 = 25\)
\(a^2 - b^2 = 4^2 - 3^2 = 16 - 9 = 7\)
\(2ab = 2 \times 4 \times 3 = 24\)
\(\therefore\) (25, 7, 24) is a Pythagorean triplet.

iii. Let a = 5, b = 2
\(a^2 + b^2 = 5^2 + 2^2 = 25 + 4 = 29\)
\(a^2 - b^2 = 5^2 - 2^2 = 25 - 4 = 21\)
\(2ab = 2 \times 5 \times 2 = 20\)
\(\therefore\) (29, 21, 20) is a Pythagorean triplet.

iv. Let a = 4, b = 1
\(a^2 + b^2 = 4^2 + 1^2 = 16 + 1 = 17\)
\(a^2 - b^2 = 4^2 - 1^2 = 16 - 1 = 15\)
\(2ab = 2 \times 4 \times 1 = 8\)
\(\therefore\) (17, 15, 8) is a Pythagorean triplet.

v. Let a = 9, b = 7
\(a^2 + b^2 = 9^2 + 7^2 = 81 + 49 = 130\)
\(a^2 - b^2 = 9^2 - 7^2 = 81 - 49 = 32\)
\(2ab = 2 \times 9 \times 7 = 126\)
\(\therefore\) (130,32,126) is a Pythagorean triplet.
Note: Numbers in Pythagorean triplet can be written in any order.
In simple words: Pythagorean triplets can be generated using Euclid's formula where, for any two positive integers a and b (with a > b), the three numbers \(a^2 + b^2\), \(a^2 - b^2\), and \(2ab\) will form a Pythagorean triplet.

🎯 Exam Tip: To generate Pythagorean triplets, use the formula \((a^2 + b^2, a^2 - b^2, 2ab)\) with different integer values for 'a' and 'b' where \(a > b\). Ensure 'a' and 'b' are coprime and one of them is even for primitive triplets, though for general triplets, any \(a > b\) works.