Maharashtra Board Class 10 Maths Chapter 2 Pythagoras Theorem Set 2.2 Solutions

Question 1. In ΔPQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज PQR को दर्शाता है जहाँ बिंदु S भुजा QR का मध्यबिंदु है, जिससे PS एक माध्यिका बन जाती है। भुजा PQ की लंबाई 11 इकाई है, PR की लंबाई 17 इकाई है, और माध्यिका PS की लंबाई 13 इकाई है। इस ज्यामितीय आकृति का उपयोग अपोलोनियस प्रमेय को लागू करने के लिए किया जाएगा ताकि भुजा QR की लंबाई ज्ञात की जा सके।
Answer:
Solution:
In ΔPQR, point S is the midpoint of side QR. [Given]
.. seg PS is the median.
.. \(PQ^2 + PR^2 = 2 PS^2 + 2 SR^2\) [Apollonius theorem]
.. \(11^2 + 17^2 = 2 (13)^2 + 2 SR^2\)
.. \(121 + 289 = 2 (169)+ 2 SR^2\)
.. \(410 = 338+ 2 SR^2\)
.. \(2 SR^2 = 410 - 338\)
.. \(2 SR^2 = 72\)
.. \(SR^2 = \frac{72}{2} = 36\)
.. \(SR = \sqrt{36}\) [Taking square root of both sides]
= 6 units Now, QR = 2 SR [S is the midpoint of QR]
= \(2 \times 6\)
.. QR = 12 units
In simple words: We used Apollonius theorem which states that in a triangle, the sum of the squares of two sides is equal to twice the sum of the square of the median to the third side and the square of half of the third side. With the given values, we calculated SR, and since S is the midpoint of QR, QR is simply twice SR.

🎯 Exam Tip: Remember to clearly state the theorem (Apollonius theorem) used and show all calculation steps. Ensure correct application of square roots and multiplication for full marks.

Question 2. In ∆ABC, AB = 10, AC = 7, BC = 9, then find the length of the median drawn from point C to side AB.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जहाँ भुजाओं की लंबाई AB=10, AC=7 और BC=9 इकाई दी गई है। चित्र में एक माध्यिका CD को भी दर्शाया गया है जो शीर्ष C से भुजा AB पर खींची गई है, जिसमें D भुजा AB का मध्यबिंदु है। इस आरेख का उपयोग अपोलोनियस प्रमेय का उपयोग करके माध्यिका CD की लंबाई ज्ञात करने के लिए किया जाएगा।
Answer:
Solution:
Let CD be the median drawn from the vertex C to side AB.
\(BD = \frac{1}{2} AB\) [D is the midpoint of AB]
\( = \frac{1}{2} \times 10 = 5\) units
In ∆ABC, seg CD is the median. [Given]
.. \(AC^2 + BC^2 = 2 CD^2 + 2 BD^2\) [Apollonius theorem]
.. \(7^2 + 9^2 = 2 CD^2 + 2 (5)^2\)
.. \(49 + 81 = 2 CD^2 + 2 (25)\)
.. \(130 = 2 CD^2 + 50\)
.. \(2 CD^2 = 130 - 50\)
.. \(2 CD^2 = 80\)
.. \(CD^2 = \frac{80}{2} = 40\)
.. \(CD = \sqrt{40}\) [Taking square root of both sides]
= \(2 \sqrt{10}\) units
.. The length of the median drawn from point C to side AB is \(2 \sqrt{10}\) units.
In simple words: To find the length of the median CD, we first determine the length of BD, which is half of AB. Then, we apply Apollonius theorem to triangle ABC with median CD, substitute the given side lengths, and solve for CD.

🎯 Exam Tip: Carefully identify the median and the sides involved when applying Apollonius theorem. Double-check calculations, especially squares and square roots, to avoid errors.

Question 3. In the adjoining figure, seg PS is the median of ΔPQR and PT ⊥ QR. Prove that,
(i) \(PR^2 = PS^2 + QR \times ST + \left(\frac{QR}{2}\right)^2\)
(ii) \(PQ^2 = PS^2 – QR \times ST + \left(\frac{QR}{2}\right)^2\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज PQR को दर्शाता है जिसमें PS भुजा QR की माध्यिका है, जिसका अर्थ है कि S, QR का मध्यबिंदु है। PT भुजा QR पर लंब है, जिससे T, QR पर स्थित एक बिंदु है। यह आरेख अपोलोनियस प्रमेय और समकोण त्रिभुजों में पाइथागोरस प्रमेय के अनुप्रयोगों को सिद्ध करने में सहायता करेगा, विशेषकर त्रिभुज के कोणों के प्रकार (न्यूनकोण या अधिक कोण) के संबंध में।
Answer:
Solution:
(i) \(QS = SR = \frac{1}{2} QR\) (i) [S is the midpoint of side QR]
In ΔPSR, ∠PSR is an obtuse angle [Given]
and PT ⊥ SR [Given, Q-S-R]
.. \(PR^2 = SR^2 + PS^2 + 2 SR \times ST\) (ii) [Application of Pythagoras theorem]
.. \(PR^2 = \left(\frac{1}{2} QR\right)^2 + PS^2 + 2 \left(\frac{1}{2} QR\right) \times ST\) [From (i) and (ii)]
.. \(PR^2 = \left(\frac{QR}{2}\right)^2 + PS^2 + QR \times ST\)
.. \(PR^2 = PS^2 + QR \times ST + \left(\frac{QR}{2}\right)^2\)
ii. In ΔPQS, ∠PSQ is an acute angle and [Given]
PT ⊥ QS [Given, Q-S-R]
.. \(PQ^2 = QS^2 + PS^2 – 2 QS \times ST\) (iii) [Application of Pythagoras theorem]
.. \(PQ^2 = \left(\frac{1}{2} QR\right)^2 + PS^2 – 2 \left(\frac{1}{2} QR\right) \times ST\) [From (i) and (iii)]
.. \(PQ^2 = \left(\frac{QR}{2}\right)^2 + PS^2 – QR \times ST\)
.. \(PQ^2 = PS^2 – QR \times ST + \left(\frac{QR}{2}\right)^2\)
In simple words: This proof utilizes the extended Pythagoras theorem (Apollonius theorem variant for acute and obtuse angles) by considering two parts of the triangle: ΔPSR (obtuse angle at S) and ΔPQS (acute angle at S). We substitute SR and QS with half of QR, and then simplify the expressions to prove the given equations.

🎯 Exam Tip: When dealing with acute and obtuse angled triangles, remember the modified Pythagoras theorem (Apollonius theorem extension) which includes the projection term. Clearly label all parts and theorems used.

Question 4. In ∆ABC, point M is the midpoint of side BC. If \(AB^2 + AC^2 = 290\) cm, AM = 8 cm, find BC.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जहाँ बिंदु M भुजा BC का मध्यबिंदु है, जिससे AM त्रिभुज की माध्यिका है। प्रश्न में \(AB^2 + AC^2\) का मान 290 वर्ग सेमी और माध्यिका AM की लंबाई 8 सेमी दी गई है। यह आरेख अपोलोनियस प्रमेय का उपयोग करके भुजा BC की लंबाई ज्ञात करने के लिए सहायक है।
Answer:
Solution:
In ∆ABC, point M is the midpoint of side BC. [Given]
.. seg AM is the median.
.. \(AB^2 + AC^2 = 2 AM^2 + 2 MC^2\) [Apollonius theorem]
.. \(290 = 2 (8)^2 + 2 MC^2\)
.. \(145 = 64 + MC^2\) [Dividing both sides by 2]
.. \(MC^2 = 145 - 64\)
.. \(MC^2 = 81\)
.. \(MC = \sqrt{81}\) [Taking square root of both sides]
MC = 9 cm
Now, BC = 2 MC [M is the midpoint of BC]
= \(2 \times 9\)
.. BC = 18 cm
In simple words: We used Apollonius theorem, which relates the lengths of the sides of a triangle to the length of a median. By substituting the given values for \(AB^2 + AC^2\) and AM, we found MC. Since M is the midpoint of BC, BC is simply twice the length of MC.

🎯 Exam Tip: Always start by clearly stating the theorem being used (Apollonius theorem). Show each step of algebraic manipulation, including dividing by 2 and taking the square root, to ensure clarity and correctness.

Question 5. In the adjoining figure, point T is in the interior of rectangle PQRS. Prove that, \(TS^2 + TQ^2 = TP^2 + TR^2\). (As shown in the figure, draw seg AB || side SR and A - T - B)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक आयत PQRS को दर्शाता है जिसके अंदर एक बिंदु T है। रचना के अनुसार, T से होकर एक रेखा AB खींची गई है जो भुजा SR के समानांतर है, जहाँ A भुजा PS पर और B भुजा QR पर स्थित है। यह आरेख पाइथागोरस प्रमेय के अनुप्रयोगों का उपयोग करके \(TS^2 + TQ^2 = TP^2 + TR^2\) संबंध को सिद्ध करने में सहायक होगा।
Answer:
Given: JPQRS is a rectangle.
Point T is in the interior of JPQRS.
To prove: \(TS^2 + TQ^2 = TP^2 + TR^2\)
Construction: Draw seg AB || side SR such that A - T - B.
Solution:
Proof:
JPQRS is a rectangle. [Given]
.. PS = QR (i) [Opposite sides of a rectangle]
In JASRB,
∠S = ∠R = 90° (ii) [Angles of rectangle PQRS]
side AB || side SR [Construction]
Also ∠A = ∠S = 90° [Interior angle theorem, from (ii)]
∠B = ∠R = 90°
.. ∠A = ∠B = ∠S = ∠R = 90° (iii)
.. JASRB is a rectangle.
.. AS = BR (iv) [Opposite sides of a rectangle]
In ΔPTS, ∠PST is an acute angle
and seg AT ⊥ side PS [From (iii)]
.. \(TP^2 = PS^2 + TS^2 – 2 PS.AS\) (v) [Application of Pythagoras theorem]
In ΔTQR, ∠TRQ is an acute angle
and seg BT ⊥ side QR [From (iii)]
.. \(TQ^2 = RQ^2 + TR^2 – 2 RQ.BR\) (vi) [Application of pythagoras theorem]
\(TP^2 – TQ^2 = PS^2 + TS^2 – 2PS.AS\)
- \(RQ^2 – TR^2 + 2RQ.BR\) [Subtracting (vi) from (v)]
.. \(TP^2 – TQ^2 = TS^2 – TR^2 + PS^2\)
- \(RQ^2 - 2 PS.AS + 2 RQ.BR\)
.. \(TP^2 – TQ^2 = TS^2 – TR^2 + PS^2\)
- \(PS^2 – 2 PS.BR + 2PS.BR\) [From (i) and (iv)]
.. \(TP^2 – TQ^2 = TS^2 – TR^2\)
.. \(TS^2 + TQ^2 = TP^2 + TR^2\)
In simple words: By drawing a line AB through T parallel to SR, we divide the rectangle into two smaller rectangles and use the property of opposite sides. We then apply the generalized Pythagoras theorem for acute angles in triangles ΔPTS and ΔTQR, and subtract the resulting equations. Through substitution using the properties of the rectangle, we simplify the equation to arrive at the desired proof.

🎯 Exam Tip: Construction is key in such proofs. Clearly state your construction and use it to establish intermediate rectangles (like JASRB). The application of the modified Pythagoras theorem for acute/obtuse angles requires careful identification of projection terms.

Question 1. In ∆ABC, ∠C is an acute angle, seg AD ⊥ seg BC. Prove that: \(AB^2 = BC^2 + AC^2 – 2 BC \times DC\). (Textbook pg. no. 44)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जहाँ कोण C एक न्यूनकोण है। AD भुजा BC पर लंब है, जिससे D भुजा BC पर स्थित है। चित्र में भुजाओं को a, b, c और p (AD) तथा x (DC) के रूप में लेबल किया गया है। यह आरेख त्रिभुज के न्यूनकोण होने पर पाइथागोरस प्रमेय के विस्तारित रूप को सिद्ध करने के लिए उपयोग किया जाएगा।
Answer:
Given: ∠C is an acute angle, seg AD ⊥ seg BC.
To prove: \(AB^2 = BC^2 + AC^2 – 2BC \times DC\)
Solution:
Proof:
.. LetAB = c, AC = b, AD = p,
BC = a, DC = x
BD + DC = BC [B – D – C]
.. BD = BC - DC
.. BD = a - x
In ∆ABD, ∠D = 90° [Given]
\(AB^2 = BD^2 + AD^2\) [Pythagoras theorem]
.. \(c^2 = (a – x)^2 + p^2\) (i)
.. \(c^2 = a^2 – 2ax + x^2 + p^2\)
In AADC, ∠D = 90° [Given]
\(AC^2 = AD^2 + CD^2\) [Pythagoras theorem]
.. \(b^2 = p^2 + x^2\)
.. \(p^2 = b^2 – x^2\) (ii)
.. \(c^2 = a^2 – 2ax + x^2 + b^2 – x^2\) [Substituting (ii) in (i)]
.. \(c^2 = a^2 + b^2 – 2ax\)
.. \(AB^2 = BC^2 + AC^2 – 2 BC \times DC\)
In simple words: To prove the relation for an acute-angled triangle, we use the Pythagorean theorem in the right-angled triangles formed by the altitude AD. We express \(AB^2\) in terms of BD and AD, and \(AC^2\) in terms of CD and AD. By substituting the expression for AD and BD (a-x) and simplifying, we derive the desired formula.

🎯 Exam Tip: This is a standard proof for the projection theorem in acute-angled triangles. Ensure you correctly identify the right-angled triangles formed by the altitude and apply Pythagoras theorem. Proper substitution and algebraic manipulation are crucial.

Question 2. In ∆ABC, ∠ACB is an obtuse angle, seg AD ⊥ seg BC. Prove that: \(AB^2 = BC^2 + AC^2 + 2 BC \times CD\). (Textbook pg. no. 40 and 4.1)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जिसमें कोण ACB एक अधिक कोण है। AD भुजा BC पर लंब है, लेकिन D भुजा BC के बाहर स्थित है, जिससे BD, BC + CD के बराबर है। चित्र में भुजाओं को a, b, c और p (AD) तथा x (DC) के रूप में लेबल किया गया है। यह आरेख त्रिभुज के अधिक कोण होने पर पाइथागोरस प्रमेय के विस्तारित रूप को सिद्ध करने के लिए उपयोग किया जाएगा।
Answer:
Given: ∠ACB is an obtuse angle, seg AD ⊥ seg BC.
To prove: \(AB^2 = BC^2 + AC^2 + 2BC \times CD\)
Solution:
Proof:
Let AD = p, AC = b, AB = c,
BC = a, DC = x
BD = BC + DC [B – C – D]
.. BD = a + x
In ∆ADB, ∠D = 90° [Given]
\(AB^2 = BD^2 + AD^2\) [Pythagoras theorem]
.. \(c^2 = (a + x)^2 + p^2\) (i)
.. \(c^2 = a^2 + 2ax + x^2 + p^2\)
Also, in AADC, ∠D = 90° [Given]
\(AC^2 = CD^2 + AD^2\) [Pythagoras theorem]
.. \(b^2 = x^2 + p^2\)
.. \(p^2 = b^2 – x^2\) (ii)
.. \(c^2 = a^2 + 2ax + x^2 + b^2 – x^2\) [Substituting (ii) in (i)]
.. \(c^2 = a^2 + b^2 + 2ax\)
.. \(AB^2 = BC^2 + AC^2 + 2 BC \times CD\)
In simple words: For an obtuse-angled triangle, we extend the side and draw an altitude from the opposite vertex, forming two right-angled triangles. By applying the Pythagorean theorem to both triangles (ΔADB and ΔADC) and substituting the expressions for the sides, particularly for BD (a+x) and AD, we can derive the formula relating the sides to the projection term.

🎯 Exam Tip: This proof is for the projection theorem in obtuse-angled triangles. Pay close attention to how BD is expressed (as a sum) and ensure correct application of Pythagoras theorem. The substitution steps are critical for reaching the final proof.

Question 3. In ∆ABC, if M is the midpoint of side BC and seg AM ⊥ seg BC, then prove that \(AB^2 + AC^2 = 2 AM^2 + 2 BM^2\). (Textbook pg, no. 41)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जहाँ M भुजा BC का मध्यबिंदु है, और माध्यिका AM भुजा BC पर लंब है। यह स्थिति दो समकोण त्रिभुज ΔAMB और ΔAMC बनाती है। इस आरेख का उपयोग अपोलोनियस प्रमेय को सिद्ध करने के लिए किया जाएगा, जो माध्यिका की लंबाई को त्रिभुज की भुजाओं से संबंधित करता है।
Answer:
Given: In ΔABC, M is the midpoint of side BC and seg AM ⊥ seg BC.
To prove: \(AB^2 + AC^2 = 2 AM^2 + 2 BM^2\)
Solution:
Proof:
In ΔAMB, ∠M = 90° [segAM ⊥ segBC]
.. \(AB^2 = AM^2 + BM^2\) (i) [Pythagoras theorem]
Also, in ΔAMC, ∠M = 90° [seg AM ⊥ seg BC]
.. \(AC^2 = AM^2 + MC^2\) (ii) [Pythagoras theorem]
.. \(AB^2 + AC^2 = AM^2 + BM^2 + AM^2 + MC^2\) [Adding (i) and (ii)]
.. \(AB^2 + AC^2 = 2 AM^2 + BM^2 + BM^2\) [.: BM = MC (M is the midpoint of BC)]
.. \(AB^2 + AC^2 = 2 AM^2 + 2 BM^2\)
In simple words: Since AM is perpendicular to BC and M is the midpoint, we can apply the Pythagorean theorem to the two right-angled triangles ΔAMB and ΔAMC. Adding the resulting equations for \(AB^2\) and \(AC^2\) and substituting MC with BM (as M is the midpoint), we directly arrive at Apollonius theorem.

🎯 Exam Tip: This proof is a direct derivation of Apollonius theorem when the median is also an altitude. Highlight the formation of two right-angled triangles. Clearly state Pythagoras theorem application and the reason for substituting MC with BM.