Pythagoras Theorem — Maharashtra (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Pythagoras Theorem" — important questions with detailed answers, download PDF for board exam preparation.

Pythagoras Theorem — Maharashtra (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Pythagoras Theorem" — important questions with detailed answers, download PDF for board exam preparation.

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Pythagoras Theorem — Textbook Questions Answered (8)

Q1: In right triangle ABC with right angle at B, AB = 3 cm and BC = 4 cm. Find AC.

Step 1: Apply Pythagoras theorem. In right triangle, (Hypotenuse)^2 = (Base)^2 + (Height)^2. Step 2: AC is the hypotenuse (opposite to right angle). AC^2 = AB^2 + BC^2. Step 3: Substitute values. AC^2 = 3^2 + 4^2 = 9 + 16 = 25. Step 4: Take square root. AC = sqrt(25) = 5 cm. Step 5: Verify using Pythagorean triplet. (3, 4, 5) is a well-known Pythagorean triplet. Answer: AC = 5 cm.

Q2: A ladder of length 13 m leans against a wall. The base of the ladder is 5 m from the wall. How high does the ladder reach on the wall?

Step 1: Set up the problem as a right triangle. Ladder (hypotenuse) = 13 m, distance from wall (base) = 5 m, height on wall = h. Step 2: Apply Pythagoras theorem. (Hypotenuse)^2 = (Base)^2 + (Height)^2. Step 3: 13^2 = 5^2 + h^2. Step 4: 169 = 25 + h^2. Step 5: Solve for h^2. h^2 = 169 - 25 = 144. Step 6: Take square root. h = sqrt(144) = 12 m. Step 7: Verify. 5^2 + 12^2 = 25 + 144 = 169 = 13^2. Correct. Answer: The ladder reaches 12 m high on the wall.

Q3: Check if a triangle with sides 6 cm, 8 cm, and 10 cm is a right triangle.

Step 1: Identify the longest side (potential hypotenuse). Longest side = 10 cm. Step 2: Apply Pythagoras theorem check. If 6^2 + 8^2 = 10^2, then it's a right triangle. Step 3: Calculate. 6^2 + 8^2 = 36 + 64 = 100. Step 4: Calculate hypotenuse squared. 10^2 = 100. Step 5: Compare. 100 = 100, so the equation is satisfied. Step 6: Conclusion. Since the sum of squares of two sides equals the square of the third side, it is a right triangle. Answer: Yes, it is a right triangle. The right angle is opposite to the side of 10 cm.

Q4: In right triangle PQR with right angle at Q, PQ = 7 cm and QR = 24 cm. Find PR and verify.

Step 1: Identify the hypotenuse. PR is opposite the right angle at Q, so PR is the hypotenuse. Step 2: Apply Pythagoras theorem. PR^2 = PQ^2 + QR^2. Step 3: Substitute. PR^2 = 7^2 + 24^2 = 49 + 576 = 625. Step 4: Take square root. PR = sqrt(625) = 25 cm. Step 5: Verify the calculation. 7^2 + 24^2 = 49 + 576 = 625. sqrt(625) = 25. Correct. Step 6: Check if it's a Pythagorean triplet. (7, 24, 25) is a Pythagorean triplet. Answer: PR = 25 cm.

Q5: Two vertical poles of heights 4 m and 9 m stand on a level ground 12 m apart. Find the distance between their tops.

Step 1: Set up coordinate system. Place base of first pole at origin (0, 0) and base of second at (12, 0). Step 2: Position of top of first pole: (0, 4). Position of top of second pole: (12, 9). Step 3: Find distance using distance formula or by creating a right triangle. Step 4: Horizontal distance = 12 m. Vertical distance = 9 - 4 = 5 m. Step 5: Apply Pythagoras theorem. Distance^2 = 12^2 + 5^2 = 144 + 25 = 169. Step 6: Distance = sqrt(169) = 13 m. Answer: The distance between the tops is 13 m.

Q6: State and prove the Converse of Pythagoras Theorem.

Statement: If the sum of squares of two sides of a triangle equals the square of the third side, then the triangle is a right triangle. Proof: Let ABC be a triangle where a^2 + b^2 = c^2, where c is the longest side. Step 1: Construct a right triangle DEF with right angle at E such that DE = a and EF = b. Step 2: By Pythagoras theorem, DF^2 = DE^2 + EF^2 = a^2 + b^2. Step 3: Since a^2 + b^2 = c^2 (given), we have DF^2 = c^2, so DF = c. Step 4: Now triangles ABC and DEF have all three sides equal (a, b, c), so they are congruent (SSS criterion). Step 5: Since triangle DEF has a right angle at E, and ABC is congruent to DEF, triangle ABC also has a right angle. Step 6: The right angle in ABC is at the vertex opposite the side of length c. Conclusion: If a^2 + b^2 = c^2, then the triangle is a right triangle with the right angle opposite the side of length c.

Q7: A right triangle has hypotenuse 15 cm and one side 9 cm. Find the other side.

Step 1: Identify known values. Hypotenuse = 15 cm, one side = 9 cm, other side = x. Step 2: Apply Pythagoras theorem. (Hypotenuse)^2 = (Side1)^2 + (Side2)^2. Step 3: 15^2 = 9^2 + x^2. Step 4: 225 = 81 + x^2. Step 5: Solve for x^2. x^2 = 225 - 81 = 144. Step 6: Take square root. x = sqrt(144) = 12 cm. Step 7: Verify. 9^2 + 12^2 = 81 + 144 = 225 = 15^2. Correct. Answer: The other side is 12 cm.

Q8: In a rectangle ABCD, diagonal AC = 13 cm and side AB = 5 cm. Find side BC.

Step 1: Understand rectangle properties. In rectangle ABCD, the diagonal AC divides it into two right triangles. Step 2: Consider right triangle ABC with right angle at B. Step 3: In this triangle, AC is the hypotenuse (13 cm), AB is one side (5 cm), and BC is the other side. Step 4: Apply Pythagoras theorem. AC^2 = AB^2 + BC^2. Step 5: 13^2 = 5^2 + BC^2. Step 6: 169 = 25 + BC^2. Step 7: Solve. BC^2 = 169 - 25 = 144. Step 8: BC = sqrt(144) = 12 cm. Step 9: Verify. 5^2 + 12^2 = 25 + 144 = 169 = 13^2. Correct. Answer: BC = 12 cm.

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