Maharashtra Board Class 7 Maths Chapter 12 Set 48 Pythagoras Solutions

Question 1. In the figures below, find the value of 'x'.
ℹ️ चित्र व्याख्या (Diagram Explanation): पहले चित्र में एक समकोण त्रिभुज LMN है, जिसमें कोण M समकोण है। भुजा LM की लंबाई 7 है, भुजा MN की लंबाई 24 है और कर्ण LN की लंबाई x है। दूसरे चित्र में एक समकोण त्रिभुज PQR है, जिसमें कोण Q समकोण है। भुजा PQ की लंबाई 9 है, भुजा QR की लंबाई x है और कर्ण PR की लंबाई 41 है। तीसरे चित्र में एक समकोण त्रिभुज EDF है, जिसमें कोण D समकोण है। भुजा ED की लंबाई x है, भुजा DF की लंबाई 8 है और कर्ण EF की लंबाई 17 है।
Answer: i. In ∆LMN, ∠M = 90°. Hence, side LN is the hypotenuse. According to Pythagoras' theorem, \( \text{I(LN)}^2 = \text{I(LM)}^2 + \text{I(MN)}^2 \)
\( \implies x^2 = 7^2 + 24^2 \)
\( \implies x^2 = 49 + 576 \)
\( \implies x^2 = 625 \)
\( \implies x^2 = 25^2 \)
\( \implies x = 25 \text{ units} \) ii. In ∆PQR, ∠Q = 90°. Hence, side PR is the hypotenuse. According to Pythagoras' theorem, \( \text{I(PR)}^2 = \text{I(PQ)}^2 + \text{I(QR)}^2 \)
\( \implies 41^2 = 9^2 + x^2 \)
\( \implies 1681 = 81 + x^2 \)
\( \implies 1681 - 81 = x^2 \)
\( \implies 1600 = x^2 \)
\( \implies x^2 = 1600 \)
\( \implies x^2 = 40^2 \)
\( \implies x = 40 \text{ units} \) iii. In ∆EDF, ∠D = 90°. Hence, side EF is the hypotenuse. According to Pythagoras' theorem, \( \text{I(EF)}^2 = \text{I(ED)}^2 + \text{I(DF)}^2 \)
\( \implies 17^2 = x^2 + 8^2 \)
\( \implies 289 = x^2 + 64 \)
\( \implies 289 - 64 = x^2 \)
\( \implies 225 = x^2 \)
\( \implies x^2 = 225 \)
\( \implies x^2 = 15^2 \)
\( \implies x = 15 \text{ units} \) In simple words: The value of 'x' is found by applying the Pythagorean theorem (hypotenuse² = side1² + side2²) to each right-angled triangle, calculating the unknown side based on the given lengths.

🎯 Exam Tip: Always identify the hypotenuse (the side opposite the right angle) correctly before applying Pythagoras' theorem. Show all calculation steps clearly for full marks.

Question 2. In the right-angled ∆PQR, ∠P = 90°. If I(PQ) = 24 cm and I(PR) = 10 cm, find the length of seg QR.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज PQR दर्शाया गया है जहाँ कोण P समकोण है। भुजा PQ की लंबाई 24 सेमी है और भुजा PR की लंबाई 10 सेमी है। भुजा QR कर्ण है, जिसकी लंबाई ज्ञात करनी है।
Answer: In ∆PQR, ∠P = 90°. Hence, side QR is the hypotenuse. According to Pythagoras' theorem, \( \text{I(QR)}^2 = \text{I(PR)}^2 + \text{I(PQ)}^2 \)
\( \implies \text{I(QR)}^2 = 10^2 + 24^2 \)
\( \implies \text{I(QR)}^2 = 100 + 576 \)
\( \implies \text{I(QR)}^2 = 676 \)
\( \implies \text{I(QR)}^2 = 26^2 \)
\( \implies \text{I(QR)} = 26 \text{ cm} \) The length of seg QR is 26 cm. In simple words: For a right-angled triangle PQR with the right angle at P, the length of the hypotenuse QR is calculated using the Pythagorean theorem, where the sum of squares of the other two sides (PQ and PR) equals the square of QR.

🎯 Exam Tip: Remember that the hypotenuse is always the longest side in a right-angled triangle. Clearly state the given values and the formula used.

Question 3. In the right-angled ∆LMN, ∠M = 90°. If I(LM) = 12 cm and I(LN) = 20 cm, find the length of seg MN.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज LMN दिखाया गया है जहाँ कोण M समकोण है। भुजा LM की लंबाई 12 सेमी है और कर्ण LN की लंबाई 20 सेमी है। भुजा MN की लंबाई ज्ञात करनी है।
Answer: In ∆LMN, ∠M = 90°. Hence, side LN is the hypotenuse. According to Pythagoras' theorem, \( \text{I(LN)}^2 = \text{I(LM)}^2 + \text{I(MN)}^2 \)
\( \implies 20^2 = 12^2 + \text{I(MN)}^2 \)
\( \implies \text{I(MN)}^2 = 20^2 - 12^2 \)
\( \implies \text{I(MN)}^2 = 400 - 144 \)
\( \implies \text{I(MN)}^2 = 256 \)
\( \implies \text{I(MN)}^2 = 16^2 \)
\( \implies \text{I(MN)} = 16 \text{ cm} \) The length of seg MN is 16 cm. In simple words: Using the Pythagorean theorem, we subtract the square of the given side (LM) from the square of the hypotenuse (LN) to find the square of the unknown side (MN), and then take the square root to get its length.

🎯 Exam Tip: When finding a shorter side, rearrange the Pythagorean theorem as \( \text{side}^2 = \text{hypotenuse}^2 - \text{other side}^2 \). Be careful with algebraic manipulation.

Question 4. The top of a ladder of length 15 m reaches a window 9 m above the ground. What is the distance between the base of the wall and that of the ladder?
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC दर्शाया गया है, जिसमें कोण B समकोण है। यह एक दीवार (AB), ज़मीन (BC), और एक सीढ़ी (AC) के बीच संबंध दिखाता है। सीढ़ी की लंबाई 15 मीटर है और यह दीवार पर 9 मीटर की ऊँचाई पर एक खिड़की तक पहुँचती है। दीवार के आधार (B) से सीढ़ी के आधार (C) तक की दूरी ज्ञात करनी है।
Answer: The wall and the ground are perpendicular to each other. Hence, the ladder leaning against the wall forms a right-angled triangle. In ∆ABC, ∠B = 90° According to Pythagoras' theorem, \( \text{I(AC)}^2 = \text{I(AB)}^2 + \text{I(BC)}^2 \)
\( \implies 15^2 = 9^2 + \text{I(BC)}^2 \)
\( \implies \text{I(BC)}^2 = 15^2 - 9^2 \)
\( \implies \text{I(BC)}^2 = 225 - 81 \)
\( \implies \text{I(BC)}^2 = 144 \)
\( \implies \text{I(BC)}^2 = 12^2 \)
\( \implies \text{I(BC)} = 12 \text{ m} \) The distance between the base of the wall and that of the ladder is 12 m. In simple words: This problem forms a right-angled triangle where the ladder is the hypotenuse, the wall's height is one side, and the distance from the wall's base to the ladder's base is the other side; we use the Pythagorean theorem to find the unknown distance.

🎯 Exam Tip: Visualize the problem as a right-angled triangle. Clearly label the hypotenuse and the two perpendicular sides. Show all steps for calculation, including the units.

Maharashtra Board Class 7 Maths Chapter 13 Pythagoras' Theorem Practice Set 48 Intext Questions and Activities

Question 1. Write the name of the hypotenuse of each of the right angled triangles shown below.
(i)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC दर्शाया गया है जहाँ कोण B समकोण है। त्रिभुज की भुजाएँ AB, BC, और AC हैं।
(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज LMN दर्शाया गया है जहाँ कोण M समकोण है। त्रिभुज की भुजाएँ LM, MN, और LN हैं।
(iii)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज XYZ दर्शाया गया है जहाँ कोण Y समकोण है। त्रिभुज की भुजाएँ XY, YZ, और XZ हैं।
Answer: i. AC ii. LN iii. XZ In simple words: The hypotenuse is always the side opposite the right angle in a right-angled triangle.

🎯 Exam Tip: The hypotenuse is the longest side of a right-angled triangle and is always opposite the 90-degree angle. Correct identification is crucial for applying Pythagoras' theorem.

Question 2. Draw right-angled triangles with the lengths of hypotenuse and one side as shown in the rough figures below. Measure the third side. Verify the Pythagoras' theorem. (Textbook pg. no. 87)
(i)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC दिखाया गया है, जहाँ कोण B समकोण है। कर्ण AC की लंबाई 5 है और भुजा BC की लंबाई 3 है। भुजा AB की लंबाई ज्ञात करनी है।
(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज PQR दिखाया गया है, जहाँ कोण Q समकोण है। कर्ण PR की लंबाई 10 है और भुजा QR की लंबाई 6 है। भुजा PQ की लंबाई ज्ञात करनी है।
(iii)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज XYZ दिखाया गया है, जहाँ कोण Y समकोण है। कर्ण XZ की लंबाई 17 है और भुजा XY की लंबाई 15 है। भुजा YZ की लंबाई ज्ञात करनी है।
Answer: i. From the figure, by measurement, I(AB) = 4 cm Now, in right-angled triangle ABC, \( \text{I(AB)}^2 + \text{I(BC)}^2 = (4)^2 + (3)^2 \) \( = 16 + 9 \) \( \implies \text{I(AB)}^2 + \text{I(BC)}^2 = 25 \quad ....(\text{i}) \) \( \text{I(AC)}^2 = (5)^2 = 25 \quad ....(\text{ii}) \) From (i) and (ii), \( \text{I(AC)}^2 = \text{I(AB)}^2 + \text{I(BC)}^2 \) Pythagoras' theorem is verified. (Students should draw the triangles PQR and XYZ and verify the Pythagoras' theorem) In simple words: By physically drawing the triangles and measuring the unknown side, then comparing the calculated value from Pythagoras' theorem, we can verify the theorem visually and numerically.

🎯 Exam Tip: For practical verification, ensure accurate drawing and measurement. The calculated value from the theorem should closely match the measured value for verification.

Question 3. Without using a protractor, can you verify that every angle of the vacant quadrilateral in the adjacent figure is a right angle? (Textbook pg. no. 89)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक बड़ा वर्ग ABCD दर्शाया गया है। इसके अंदर एक छोटा वर्ग LMNP (या PLMN) इस तरह से घुमाकर रखा गया है कि इसके कोने बड़े वर्ग की भुजाओं पर पड़ते हैं। इसके कारण, बड़े वर्ग के कोनों पर चार समकोण त्रिभुज (जैसे ∆ALB, ∆BPC, ∆CQD, ∆DRA) बनते हैं। केंद्रीय चतुर्भुज PLMN है।
Answer: In the square ABCD the shaded triangles are right-angled and are the same. In ∆LBM, \( m∠BLM + m∠BML + m∠LBM = 180° \) ....(Sum of the measures of the angles of a triangles is 180°)
\( \implies m∠BLM + m∠BML + 90° = 180° \)
\( \implies m∠BLM + m∠BML = 90° \quad ....(\text{i}) \) Now, ∆LBM and ∆LAP are same.
\( \implies m∠BML = m∠ALP \quad ....(\text{ii}) \)
\( \implies m∠BLM + m∠ALP = 90° \quad [\text{From (i) and (ii)}] \) Now, \( m∠ALP + m∠PLM + m∠BLM = 180° \) ....(The measure of a straight angle is 180°)
\( \implies m∠ALP + m∠BLM + m∠PLM = 180° \)
\( \implies 90° + m∠PLM = 180° \)
\( \implies m∠PLM = 180°- 90° = 90° \)
\( \implies m∠PLM \) is a right angle. Similarly, we can prove that the other angles of the vacant quadrilateral are right angles. In simple words: By understanding that the sum of angles on a straight line is 180 degrees and knowing that the corner triangles are right-angled and congruent, we can deduce that the angles of the inner quadrilateral are also 90 degrees without a protractor.

🎯 Exam Tip: This problem uses geometric properties like angles on a straight line and properties of congruent right-angled triangles. Clearly state the reasons for each step in your proof.

Question 4. On a sheet of card paper, draw a right-angled triangle of sides 3 cm, 4 cm and 5 cm. Construct a square on each of the sides. Find the area of each of the squares and verify Pythagoras' theorem. (Textbook pg. no. 89)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC दर्शाया गया है जिसमें कोण B समकोण है। भुजाएँ BC = 3 सेमी, AB = 4 सेमी और कर्ण AC = 5 सेमी हैं। प्रत्येक भुजा पर एक वर्ग बनाया गया है: भुजा AB पर वर्ग ABLM, भुजा BC पर वर्ग BCPN, और भुजा AC पर वर्ग ACQR। इन वर्गों के क्षेत्रफलों का उपयोग करके पाइथागोरस प्रमेय का सत्यापन करना है।
Answer: Area of square ABLM = I(AB)² = 4² = 16 sq.cm Area of square BCPN = I(BC)² = 3² = 9 sq.cm Area of square ACQR = I(AC)² = 5² = 25 sq.cm Now, 25 = 16 + 9 i.e. \( 5^2 = 4^2 + 3^2 \) \( \implies \text{I(AC)}^2 = \text{I(BC)}^2 + \text{I(AB)}^2 \) \( \implies (\text{hypotenuse})^2 = (\text{base})^2 + (\text{height})^2 \) In simple words: By drawing squares on each side of a 3-4-5 right-angled triangle and calculating their areas, we can visually and numerically confirm that the area of the square on the hypotenuse equals the sum of the areas of the squares on the other two sides, thus verifying Pythagoras' theorem.

🎯 Exam Tip: This activity provides a visual and practical verification of Pythagoras' theorem. Ensure accurate drawing of the squares and correct calculation of their areas to demonstrate the theorem effectively.