(a) If a hexagon PQRSTU circumscribes a circle, prove that, PQ + RS + TU = QR + ST + UP. OR (b) In the given figure, two concentric circles have radii 3 cm and 5 cm. Two tangents TR and TP are drawn to the circles from an external point T such that TR touches the inner circle at R and TP touches the outer circle at P. If TR = $4\sqrt{10}$ cm, then find the length of TP.

Step 1: Identify given information and properties of tangents

We are given the radii of the inner and outer concentric circles as $OR = 3\text{ cm}$ and $OP = 5\text{ cm}$ respectively. We are also given the length of the tangent $TR = 4\sqrt{10}\text{ cm}$. A key property of tangents is that the radius drawn to the point of tangency is perpendicular to the tangent. Therefore, $\angle ORT = 90^\circ$ and $\angle OPT = 90^\circ$.

Step 2: Calculate the length of OT using the Pythagorean theorem in $\triangle$ ORT

In the right-angled triangle $\triangle ORT$, we can use the Pythagorean theorem to find the length of $OT$. Substituting the given values, we have $OT^2 = (3)^2 + (4\sqrt{10})^2$.

Step 3: Solve for OT

Calculating the squares, $3^2 = 9$ and $(4\sqrt{10})^2 = 16 \times 10 = 160$. Adding these gives $OT^2 = 9 + 160 = 169$. Taking the square root, we find $OT = 13\text{ cm}$.

Step 4: Calculate the length of TP using the Pythagorean theorem in $\triangle$ OPT

Now consider the right-angled triangle $\triangle OPT$. We know $OT = 13\text{ cm}$ and $OP = 5\text{ cm}$. Using the Pythagorean theorem, we can find the length of the tangent $TP$.

Step 5: Solve for TP

Substituting the values, $TP^2 = 13^2 - 5^2 = 169 - 25 = 144$. Taking the square root, we get $TP = 12\text{ cm}$.

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