In the diagram, PQRS is a circle with centre O and radius 7cm. SQ and PR intersect at K and < SKR = 90°. If the length of the arc SR is four times that of a...

Answer Details

Reading the diagram. S is at the top and Q at the bottom, with centre O lying on SQ, so SQ is a vertical chord passing through the centre. P is on the left and R on the right, with PR a horizontal chord. The two chords SQ and PR cross at the interior point K, where \(\angle SKR = 90^\circ\). The radius is \(7\text{ cm}\).

Step 1: Apply the intersecting-chords angle rule. When two chords cross inside a circle, the angle between them equals half the sum of the two arcs it intercepts. The angle \(\angle SKR\) intercepts arc \(SR\), while its vertically opposite angle intercepts arc \(PQ\). Hence

\[\angle SKR = \tfrac{1}{2}\left(\text{arc } SR + \text{arc } PQ\right).\]

Step 2: Form an equation in the arc measures. Let arc \(PQ = x^\circ\). Since arc \(SR\) is four times arc \(PQ\), arc \(SR = 4x^\circ\). With \(\angle SKR = 90^\circ\):

\[90 = \tfrac{1}{2}(4x + x) \;\Rightarrow\; 180 = 5x \;\Rightarrow\; x = 36.\]

So the arc \(SR\) subtends an angle of \(4x = 4(36) = 144^\circ\) at the centre.

Step 3: Convert the arc angle to a length. Using \(\text{arc length} = \dfrac{\theta}{360}\times 2\pi r\) with \(\theta = 144^\circ\), \(r = 7\), \(\pi = \tfrac{22}{7}\):

\[\text{arc } SR = \frac{144}{360}\times 2\times\frac{22}{7}\times 7 = \frac{144}{360}\times 44 = 0.4\times 44 = 17.6\text{ cm}.\]

Answer: The length of arc \(SR\) is \(\mathbf{17.6\text{ cm}}\).