NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 6: Triangles - Exercise 6.4

Question 10

Step 1: Recall the property of a right triangle with altitude drawn to the hypotenuse.

• Here, \(QS\) is perpendicular to hypotenuse \(PR\).
• The property says:
  \(PQ^2 = PS \times PR\),
  \(QR^2 = RS \times PR\),
  \(QS^2 = PS \times RS\).

Step 2: Use \(PQ^2 = PS \times PR\) to find \(PR\).

\(PQ = 6\,\text{cm},\; PS = 4\,\text{cm}\).

\(PQ^2 = PS \times PR\)

\(6^2 = 4 \times PR\)

\(36 = 4PR\)

\(PR = 36/4 = 9\,\text{cm}\).

Step 3: Find \(RS\).

Total hypotenuse \(PR = 9\,\text{cm}\).

\(PS = 4\,\text{cm}\).

So, \(RS = PR - PS = 9 - 4 = 5\,\text{cm}\).

Step 4: Use \(QS^2 = PS \times RS\) to find \(QS\).

\(QS^2 = 4 \times 5 = 20\).

\(QS = \sqrt{20} = 2\sqrt{5}\,\text{cm}\).

Step 5: Use \(QR^2 = RS \times PR\) to find \(QR\).

\(QR^2 = 5 \times 9 = 45\).

\(QR = \sqrt{45} = 3\sqrt{5}\,\text{cm}\).

Final Answer:

\(QS = 2\sqrt{5}\,\text{cm},\; RS = 5\,\text{cm},\; QR = 3\sqrt{5}\,\text{cm}.\)