In the figure, a circle with centre 𝑂 is inscribed in a quadrilateral 𝐴𝐵𝐶𝐷 such that it touches the side 𝐵𝐶, 𝐴𝐵, 𝐴𝐷 and 𝐶𝐷 at points 𝑃, 𝑄, 𝑅 and 𝑆 respectively. If 𝐴𝐵 = 29 𝑐𝑚,
𝐴𝐷 = 23 𝑐𝑚, ∠𝐵 = 90° and 𝐷𝑆 = 5 𝑐𝑚, the radius of the circle (in cm) is:

i. 11

ii. 18

iii. 6

iv. 15

                                             (OR)

In triangle 𝐴𝐵𝐶 and 𝐷𝐸𝐹, ∠𝐵 = ∠𝐸, ∠𝐹 = ∠𝐶 and 𝐴𝐵 = 3𝐷𝐸. Then, the two
triangles are:

i. Congruent but not similar

ii. Similar but not congruent

iii. Neither congruent nor similar

iv. congruent as well as similar

It is given that 𝐴𝐵 = 29 𝑐𝑚, 𝐴𝐷 = 23 𝑐𝑚, ∠𝐵 = 90° and 𝐷𝑆 = 5 𝑐𝑚
Tangents to a circle from an external point are equal in length.

$\begin{array}{r}AQ=AR\\ DS=DR\\ CP=CS\\ PB=BQ\\ \text{ Also, }AB=AQ+BQ\end{array}$

Since DS = 5 cm, then DR = 5 cm

$\begin{array}{r}So, AR=18\mathrm{cm}\\ AQ=18\mathrm{cm}\\ BQ=29-18\\ BQ=11\mathrm{cm}\end{array}$

Now, in quadrilateral 𝑂𝑃𝐵𝑄,  ∠𝐵 = 90°
Also, 𝑂𝑃𝐵 = 𝑂𝑄𝐵 = 90°(tangent is perpendicular to radius at point of contact)
So, ∠𝑃𝑂𝑄 = 90°, that is 𝑂𝑃𝑄𝑅 is a rectangle.
Since 𝑃𝐴 = 𝑃𝐵, ,𝑂𝑃𝑄𝑅 is a square, the radius of 𝑂𝑃 = 𝐵𝑄 = 11𝑐𝑚.
Hence, option (a)11 is correct

                                           (OR)

In △𝐴𝐵𝐶 and △𝐷𝐸F,

∠𝐵 = ∠𝐸 (𝑔𝑖𝑣𝑒𝑛)
∠𝐹 = ∠𝐶(given)
By 𝐴𝐴 test of similarity △𝐴𝐵𝐶 ~ △𝐷𝐸𝐹
Since there are no congruence criteria like 𝐴𝐴.
Therefore △𝐴𝐵𝐶 and △𝐷𝐸𝐹 are similar but not congruent.

Hence, option (b) similar but not congruent is correct.