NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.4

Question 6

Step 1: Recall distance formula.

If two points are \((x_1,y_1)\) and \((x_2,y_2)\), then

\[ \,\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \]

Step 2: Calculate distance from House to Bank (H to B).

H(2,4), B(5,8)

\[ HB = \sqrt{(5-2)^2 + (8-4)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\;\text{km} \]

Step 3: Calculate distance from Bank to School (B to S).

B(5,8), S(13,14)

\[ BS = \sqrt{(13-5)^2 + (14-8)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\;\text{km} \]

Step 4: Calculate distance from School to Office (S to O).

S(13,14), O(13,26)

Here, the x-coordinates are the same, so the line is vertical.

\[ SO = |26 - 14| = 12\;\text{km} \]

Step 5: Add up the three parts.

\[ \text{Total distance via Bank and School} = HB + BS + SO = 5 + 10 + 12 = 27\;\text{km} \]

Step 6: Calculate direct distance from House to Office (H to O).

H(2,4), O(13,26)

\[ HO = \sqrt{(13-2)^2 + (26-4)^2} = \sqrt{11^2 + 22^2} = \sqrt{121 + 484} = \sqrt{605}\;\text{km} \]

Using calculator: \(\sqrt{605} \approx 24.596\;\text{km}\).

Step 7: Find the extra distance travelled.

\[ \text{Extra distance} = (27 - \sqrt{605})\;\text{km} \approx (27 - 24.596)\;\text{km} = 2.404\;\text{km} \]

Final Answer: Ayush travels about 2.404 km more when he goes through Bank and School instead of going directly from House to Office.