Samacheer Kalvi Class 10 Maths Solutions Chapter 4 Geometry Exercise 4.3

Question 1. A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
Answer:
Let the initial position of the man be 'O' and his final position be 'B'. The path taken forms a right-angled triangle OAB, where 'A' is the point 18m east from 'O'. The journey east is OA = 18 m, and the journey north is AB = 24 m. We need to find the distance OB, which is the hypotenuse of the triangle. A right-angled triangle helps us visualize the movement and apply the Pythagorean theorem.In the right \( \Delta OAB \), By Pythagoras theorem, \( OB^2 = OA^2 + AB^2 \) \( OB^2 = 18^2 + 24^2 \) \( OB^2 = 324 + 576 \) \( OB^2 = 900 \) \( OB = \sqrt{900} \) \( OB = 30 \) Therefore, the distance of his current position from the starting point is 30 m.
In simple words: The man walked east and then north, forming a right-angled path. We use the Pythagoras theorem to find the straight-line distance from his start to end point, which turns out to be 30 meters.

🎯 Exam Tip: Always draw a simple diagram for navigation problems to clearly visualize the directions and distances, which helps in applying the correct geometric theorem like Pythagoras.

Question 2. There are two paths that one can choose to go from Sarah's house to James house. One way is to take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C street? (Using figure).
Answer:
Let's analyze the paths from Sarah's house to James' house based on the figure. We can see two routes: 1. **Indirect path:** Taking A Street and then B Street. 2. **Direct path:** Taking C Street. The streets form a right-angled triangle, where A Street and B Street are the legs, and C Street is the hypotenuse. We can use the distances given in the diagram to find the lengths of these paths. Understanding which path is a hypotenuse helps us find the shortest distance.Let A, B, C be the points forming the right triangle. AB = 2 miles (A Street) BC = 1.5 miles (B Street) AC = C Street (direct path) Using Pythagoras theorem for \( \Delta ABC \): \( AC^2 = AB^2 + BC^2 \) \( AC^2 = 2^2 + 1.5^2 \) \( AC^2 = 4 + 2.25 \) \( AC^2 = 6.25 \) \( AC = \sqrt{6.25} \) \( AC = 2.5 \) miles Distance covered by using "A Street" and "B Street" (indirect path): \( = (2 + 1.5) \) miles \( = 3.5 \) miles Difference in distance (how much shorter the direct path is): \( = 3.5 \) miles \( - 2.5 \) miles \( = 1 \) mile The direct path along C Street is 1 mile shorter.
In simple words: The direct path (C Street) is like the shortcut in a right-angled triangle. By using Pythagoras, we find the shortcut is 2.5 miles long. The other path is 3.5 miles. So, taking the shortcut saves 1 mile.

🎯 Exam Tip: When comparing path lengths in geometry, always consider if the paths form a right-angled triangle. The hypotenuse will always represent the shortest straight-line distance between two points compared to going along the two perpendicular sides.

Question 3. To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?
Answer:
Imagine the starting point as 'A'. The walk forms a right-angled path: 34 m south and then 41 m east. This creates two sides of a right triangle. If we could walk directly through the pond, that path would be the hypotenuse. We need to find the length of this direct path using Pythagoras and then compare it to the total walking distance. Calculating the direct distance helps us find the saving.Let the points be A (start), B (after walking south), and C (final position after walking east). The walk from A to B is 34 m (south). The walk from B to C is 41 m (east). The direct path through the pond would be the hypotenuse AC of the right-angled triangle \( \Delta ABC \). By Pythagoras theorem: \( AC^2 = AB^2 + BC^2 \) \( AC^2 = 34^2 + 41^2 \) \( AC^2 = 1156 + 1681 \) \( AC^2 = 2837 \) \( AC = \sqrt{2837} \) \( AC \approx 53.26 \) m Total distance walked through the indirect path: \( = 34 \text{ m} + 41 \text{ m} = 75 \text{ m} \) The direct distance (if possible through the pond) is approximately 53.26 m. Difference in distance (meters saved): \( = 75 \text{ m} - 53.26 \text{ m} \) \( = 21.74 \text{ m} \) Approximately 21.74 meters would be saved if a direct path through the pond were possible.
In simple words: Walking around the pond makes you travel 75 meters. If you could go straight through, it would only be about 53.26 meters. So, you would save about 21.74 meters.

🎯 Exam Tip: Always calculate both the indirect (around the obstacle) and direct (shortest path) distances separately. The question asks for the 'meters saved', which implies finding the difference between these two calculated values.

Question 4. In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm. Calculate the length and breadth of the rectangle?
Answer:
Let the length of the rectangle be 'a' and the breadth of the rectangle be 'b'. A rectangle has two pairs of equal sides, so we can use these variables to represent the sides. The diagonals of a rectangle are equal in length, which is a key property to solve this problem.Given: \( XY + YZ = 17 \) cm Since WXYZ is a rectangle, XY is the length (a) and YZ is the breadth (b). So, \( a + b = 17 \) ......(1) Also given: \( XZ + YW = 26 \) cm In a rectangle, the diagonals are equal, so \( XZ = YW \). Therefore, \( XZ + XZ = 26 \) \( 2XZ = 26 \) \( XZ = 13 \) cm Now, consider the right-angled triangle \( \Delta WXZ \). By Pythagoras theorem: \( XZ^2 = WX^2 + WZ^2 \) Here, WX is the breadth (b) and WZ is the length (a). \( 13^2 = b^2 + a^2 \) \( a^2 + b^2 = 169 \) We know that \( (a+b)^2 = a^2 + b^2 + 2ab \). From (1), \( a+b = 17 \), so \( (a+b)^2 = 17^2 = 289 \). Substitute \( a^2 + b^2 = 169 \) into the equation: \( 289 = 169 + 2ab \) \( 2ab = 289 - 169 \) \( 2ab = 120 \) \( ab = 60 \) ......(2) Now we have two equations: 1. \( a + b = 17 \) 2. \( ab = 60 \) From (1), \( a = 17 - b \). Substitute this into (2): \( (17 - b)b = 60 \) \( 17b - b^2 = 60 \) \( b^2 - 17b + 60 = 0 \) This is a quadratic equation. We can factorize it: We need two numbers that multiply to 60 and add up to -17. These numbers are -12 and -5. \( (b - 12)(b - 5) = 0 \) So, \( b = 12 \) or \( b = 5 \) Case 1: If \( b = 12 \), then from \( a + b = 17 \), \( a = 17 - 12 = 5 \). Case 2: If \( b = 5 \), then from \( a + b = 17 \), \( a = 17 - 5 = 12 \). Since length is typically considered greater than breadth, we choose: Length \( (a) = 12 \) cm Breadth \( (b) = 5 \) cm
In simple words: We used the given information to create two equations: one for the sum of length and breadth, and one for their product. By solving these equations like a puzzle, we found that the length of the rectangle is 12 cm and the breadth is 5 cm.

🎯 Exam Tip: Remember that the diagonals of a rectangle are equal in length. This property is crucial for setting up the equations correctly when diagonal information is provided. Quadratic equations are often used to solve for dimensions.

Question 5. The hypotenuse of a right triangle is 6 m more than twice of the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.
Answer:
Let the shortest side of the right-angled triangle be \( x \) meters. We can express the other two sides in terms of \( x \) using the given relationships. Once all sides are defined with \( x \), we can use the Pythagoras theorem to form an equation and solve for \( x \). This allows us to find the actual lengths of all three sides.Let the shortest side of the right \( \Delta \) be \( x \). So, \( AB = x \) The hypotenuse is 6 m more than twice the shortest side. Hypotenuse \( AC = 2x + 6 \) The third side is 2 m less than the hypotenuse. Third side \( BC = (2x + 6) - 2 = 2x + 4 \) In the right triangle ABC, by Pythagoras theorem: \( AC^2 = AB^2 + BC^2 \) \( (2x + 6)^2 = x^2 + (2x + 4)^2 \) Expand the terms: \( (4x^2 + 24x + 36) = x^2 + (4x^2 + 16x + 16) \) Rearrange the terms to form a quadratic equation: \( 4x^2 + 24x + 36 = x^2 + 4x^2 + 16x + 16 \) \( 4x^2 + 24x + 36 = 5x^2 + 16x + 16 \) Move all terms to one side to set the equation to zero: \( 0 = 5x^2 - 4x^2 + 16x - 24x + 16 - 36 \) \( 0 = x^2 - 8x - 20 \) Factorize the quadratic equation: We need two numbers that multiply to -20 and add to -8. These are -10 and 2. \( (x - 10)(x + 2) = 0 \) This gives two possible values for \( x \): \( x - 10 = 0 \implies x = 10 \) \( x + 2 = 0 \implies x = -2 \) Since the side length cannot be negative, we omit \( x = -2 \). So, \( x = 10 \) m. Now, calculate the lengths of the sides: Shortest side \( AB = x = 10 \) m Third side \( BC = 2x + 4 = 2(10) + 4 = 20 + 4 = 24 \) m Hypotenuse \( AC = 2x + 6 = 2(10) + 6 = 20 + 6 = 26 \) m The sides of the triangle are 10 m, 24 m, and 26 m.
In simple words: We set up the problem by calling the shortest side 'x' and then used the given rules to write the other two sides as expressions of 'x'. Using the Pythagoras theorem, we solved for 'x', found it to be 10 meters, and then calculated all the side lengths: 10m, 24m, and 26m.

🎯 Exam Tip: When solving for lengths, always discard negative solutions for 'x' as physical distances cannot be negative. Double-check your algebraic expansion and factorization to avoid errors in the quadratic equation.

Question 6. 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall?
Answer:
This problem involves two scenarios of a ladder leaning against a wall, both forming right-angled triangles. The ladder's length remains constant. First, we find the initial distance of the ladder's foot from the wall. Then, we adjust this distance as the foot is moved, and use Pythagoras theorem again to find the new height the ladder reaches. The difference in heights will tell us how much the top slid upwards.**Initial position of the ladder (right \( \Delta ABC \)):** Let 'A' be the initial position of the top of the ladder on the wall. Let 'C' be the initial position of the foot of the ladder on the ground. Let 'B' be the base of the wall. The ladder length \( AC = 5 \) m. The height it reaches on the wall \( AB = 4 \) m. By Pythagoras theorem in \( \Delta ABC \): \( BC^2 = AC^2 - AB^2 \) \( BC^2 = 5^2 - 4^2 \) \( BC^2 = 25 - 16 \) \( BC^2 = 9 \) \( BC = \sqrt{9} = 3 \) m. So, the initial distance of the foot of the ladder from the wall is 3 m. **When the foot of the ladder is moved 1.6 m towards the wall:** The new position of the foot of the ladder is E. The new distance of the foot from the wall \( BE = BC - 1.6 \) m. \( BE = 3 - 1.6 = 1.4 \) m. Let 'D' be the new position of the top of the ladder on the wall. Let 'h' be the distance the top of the ladder slid upward from A. So, the new height it reaches on the wall \( BD = AB + h = 4 + h \) m. The ladder length \( DE \) remains 5 m. Now, consider the new right-angled triangle \( \Delta BED \). By Pythagoras theorem: \( DE^2 = BD^2 + BE^2 \) \( 5^2 = (4 + h)^2 + (1.4)^2 \) \( 25 = (4 + h)^2 + 1.96 \) Subtract 1.96 from both sides: \( 25 - 1.96 = (4 + h)^2 \) \( 23.04 = (4 + h)^2 \) Take the square root of both sides: \( 4 + h = \sqrt{23.04} \) \( 4 + h = 4.8 \) Solve for h: \( h = 4.8 - 4 \) \( h = 0.8 \) m. The distance moved upward on the wall is 0.8 m.
In simple words: First, we used Pythagoras to find how far the ladder's foot was initially from the wall (3m). Then, we moved the foot closer by 1.6m, so it was 1.4m away. Using Pythagoras again with the new distance and the ladder's length, we found the new height reached. The difference in heights showed the ladder slid up by 0.8 meters.

🎯 Exam Tip: Always break down ladder problems into two separate right-angled triangles, one for the initial position and one for the final position. The key is that the ladder's length (hypotenuse) remains constant in both scenarios.

Question 7. The perpendicular PS on the base QR of a \( \Delta PQR \) intersects QR at S, such that QS = 3 SR. Prove that \( 2PQ^2 = 2PR^2 + QR^2 \).
Answer:
This problem requires us to use the Pythagoras theorem in the right-angled triangles formed within \( \Delta PQR \), specifically \( \Delta PQS \) and \( \Delta PSR \), because PS is perpendicular to QR. We need to express QS and SR in terms of QR using the given relationship, then substitute these into the Pythagoras equations to manipulate them algebraically until we reach the desired proof.Given: PS is perpendicular to QR, so \( \angle PSQ = \angle PSR = 90^\circ \). Also, \( QS = 3 SR \). We know that \( QR = QS + SR \). Substitute \( QS = 3 SR \) into this equation: \( QR = 3SR + SR \) \( QR = 4SR \)
\( \implies SR = \frac{1}{4} QR \) ......(1) From \( QS = 3SR \), we can also write \( SR = \frac{QS}{3} \) ......(2) From (1) and (2), we can express QS in terms of QR: \( \frac{QS}{3} = \frac{1}{4} QR \)
\( \implies QS = \frac{3}{4} QR \) ......(3) **In the right-angled triangle \( \Delta PQS \):** By Pythagoras theorem: \( PQ^2 = PS^2 + QS^2 \) ......(4) **In the right-angled triangle \( \Delta PSR \):** By Pythagoras theorem: \( PR^2 = PS^2 + SR^2 \) ......(5) Subtract equation (5) from equation (4): \( PQ^2 - PR^2 = (PS^2 + QS^2) - (PS^2 + SR^2) \) \( PQ^2 - PR^2 = PS^2 + QS^2 - PS^2 - SR^2 \) \( PQ^2 - PR^2 = QS^2 - SR^2 \) Now, substitute the values of QS and SR from (3) and (1) into this equation: \( PQ^2 - PR^2 = \left(\frac{3}{4} QR\right)^2 - \left(\frac{1}{4} QR\right)^2 \) \( PQ^2 - PR^2 = \frac{9QR^2}{16} - \frac{QR^2}{16} \) \( PQ^2 - PR^2 = \frac{9QR^2 - QR^2}{16} \) \( PQ^2 - PR^2 = \frac{8QR^2}{16} \) \( PQ^2 - PR^2 = \frac{1}{2} QR^2 \) Multiply the entire equation by 2: \( 2(PQ^2 - PR^2) = 2 \left(\frac{1}{2} QR^2\right) \) \( 2PQ^2 - 2PR^2 = QR^2 \) Rearrange the terms to match the required proof: \( 2PQ^2 = 2PR^2 + QR^2 \) Hence, the statement is proved.
In simple words: We used the Pythagoras theorem in the two small right triangles formed by the perpendicular line. Then, we used the given relationship between QS and SR to write them in terms of QR. By subtracting the Pythagoras equations and substituting these new expressions, we simplified the equation step by step until we arrived at the proof required.

🎯 Exam Tip: For geometric proofs involving squares of sides, always consider applying the Pythagoras theorem to any right-angled triangles in the figure. Expressing segments in terms of a common base (like QR here) is often a crucial step in algebraic manipulation.

Question 8. In the adjacent figure, ABC is a right angled triangle with right angle at B and points D, E trisect BC. Prove that \( 8AE^2 = 3AC^2 + 5AD^2 \).
Answer:
In a right-angled triangle, when the base is trisected, we can use the Pythagoras theorem multiple times for different right-angled sub-triangles. The key is to express all segment lengths (BD, BE, BC) in terms of a common variable related to the trisection. Then, we apply Pythagoras theorem to \( \Delta ABD \), \( \Delta ABE \), and \( \Delta ABC \) and combine these equations to prove the given relationship.Given that points D and E trisect BC, it means they divide BC into three equal parts. Let \( BD = DE = EC = x \). Then, \( BE = BD + DE = x + x = 2x \). And, \( BC = BD + DE + EC = x + x + x = 3x \). In the right-angled triangle \( \Delta ABC \) (right-angled at B): By Pythagoras theorem: \( AC^2 = AB^2 + BC^2 \) \( AC^2 = AB^2 + (3x)^2 \) \( AC^2 = AB^2 + 9x^2 \) ......(1) In the right-angled triangle \( \Delta ABD \) (right-angled at B): By Pythagoras theorem: \( AD^2 = AB^2 + BD^2 \) \( AD^2 = AB^2 + x^2 \) ......(2) In the right-angled triangle \( \Delta ABE \) (right-angled at B): By Pythagoras theorem: \( AE^2 = AB^2 + BE^2 \) \( AE^2 = AB^2 + (2x)^2 \) \( AE^2 = AB^2 + 4x^2 \) ......(3) We need to prove \( 8AE^2 = 3AC^2 + 5AD^2 \). Let's work with the Right Hand Side (R.H.S.) of the equation and substitute the expressions from (1), (2), and (3): R.H.S. \( = 3AC^2 + 5AD^2 \) Substitute \( AC^2 \) from (1) and \( AD^2 \) from (2): R.H.S. \( = 3(AB^2 + 9x^2) + 5(AB^2 + x^2) \) R.H.S. \( = 3AB^2 + 27x^2 + 5AB^2 + 5x^2 \) Combine like terms: R.H.S. \( = (3AB^2 + 5AB^2) + (27x^2 + 5x^2) \) R.H.S. \( = 8AB^2 + 32x^2 \) Factor out 8: R.H.S. \( = 8(AB^2 + 4x^2) \) Now, look at equation (3): \( AE^2 = AB^2 + 4x^2 \). Substitute \( AE^2 \) into the R.H.S.: R.H.S. \( = 8AE^2 \) This is equal to the Left Hand Side (L.H.S.). Therefore, \( 8AE^2 = 3AC^2 + 5AD^2 \). Hence, the statement is proved.
In simple words: We used the Pythagoras theorem for three different right triangles within the main triangle (ABC, ABD, ABE) to write expressions for \( AC^2 \), \( AD^2 \), and \( AE^2 \). By substituting these into the right side of the equation we needed to prove and simplifying, we found that it matched the left side, thus proving the statement.

🎯 Exam Tip: When points trisect a segment, clearly define a variable (like 'x') for each equal part to simplify calculations. Systematically apply Pythagoras theorem to each relevant right-angled triangle and substitute carefully to derive the required proof.