Answer: Option C. -> 13 cm
:
C

Given AB and CD are two chords of a circles on opposite sides of the centre.
Construction: Draw perpendiculars OE and OF onto AB and CD respectively from centre O.
AE = EB = 5cm and CF = FD = 12 cm
[ Perpendicular drawn to a chord from center bisects the chord]
Given,
Distance between two chords = 17 cm
Let distance between O and F $=x$ cm
And distance between O and E $=(17-x)cm$
In ΔOEB,
$O{B}^2=O{E}^2+E{B}^2$
[Pythagoras theorem]
$=(17-x{)}^2+5^2$---(1)
In ΔOFD,
$OD2=O{F}^2+F{D}^2$
[Pythagoras theorem]
$=(x{)}^2+{12}^2$----------→(2)
But OB = OD ( radii of the same circle).
From 1 & 2,
$(17-x{)}^2+5^2=(x{)}^2+{12}^2$
⇒ $289+x^2-34x+25=x^2+144$
⇒ $34x=170$
∴ $x=5$
Subsitute x in equation (2);
$O{D}^2=(5{)}^2+{12}^2=169$
$OD=13$
∴ Radius of the circle is 13 cm.

Answer: Option D. -> 60∘
:
D

$\angle ADB={90}^{\circ }\text{[Angle in a semi-circle]}$
$\angle ABD=\angle ACD={30}^{\circ }\text{[Angles in the same segment]}$
In $△ADB$, we have
$\angle BAD+\angle ADB+\angle ABD={180}^{\circ }\text{(Sum of anglesof a triangleis}{180}^{\circ })$
$x+{90}^{\circ }+{30}^{\circ }={180}^{\circ }$
$\Rightarrow x=({180}^{\circ }-{120}^{\circ })={60}^{\circ }$
$\text{Hence},x={60}^{\circ }$

Answer: Option D. -> 40∘
:
D
Since an angle in a semicircle is a right angle, $\angle ACB={90}^{\circ }.$
But, $\angle BCD=\angle ACD+\angle ACB={50}^{\circ }+{90}^{\circ }={140}^{\circ }.$
Consider the cyclic quadrilateral ABCD. Since opposite angles are supplementary,
$\angle BCD+\angle DAB={180}^{\circ }.$
$⟹\angle BAD={180}^{\circ }-{140}^{\circ }={40}^{\circ }$

Answer: Option A. -> True
:
A
Draw an arcof smaller radius and join its ends at its centre. Now draw an arc of bigger radius and try to join its ends at its centre, now draw a straight line and try to join its ends at its centre we can see that it will meet its centre at infinity, so from this we can conclude that a line is a circle of infinite radius.

Take a triangle and add it to another triangle it will form a quadrilateral, add another triangle to this quadrilateral it will form a pentagon, keep adding triangles to this polygon at one point you will realize that a polygon with infinite sides will be a circle, with each side being infinitesimally small.

Answer: Option A. -> 30∘
:
A
Given thatABCD is a cyclic quadilateral such that$\angle BCD={100}^{\circ }$and $\angle ABD={70}^{\circ }.$

Since opposite angles of a cyclic quadrilateral are supplementary,
$\angle DAB+\angle DCB={180}^{\circ }.$
$⟹\angle DAB={180}^{\circ }-\angle DCB$
$={180}^{\circ }-{100}^{\circ }={80}^{\circ }$
Consider $△ADB$,since sum of angles of a triangleis${180}^{\circ },$ we have
$\angle ADB+\angle DAB+\angle ABD={180}^{\circ }$
$i.e.,\angle ADB+{80}^{\circ }+{70}^{\circ }={180}^{\circ }$
$⟹\angle ADB={180}^{\circ }-{150}^{\circ }={30}^{\circ }$

Answer: Option B. -> Concentric circles
:
B
Circles that have the same center but different radii are called concentric circles.

Answer: Option B. -> Concentric circles
:
B
${180}^{\circ }$ of a circle is half of a circle (that is half of ${360}^{\circ }={180}^{\circ }$)

The above image is half of a circle. That is ${180}^{\circ }$ of a circle.

Answer: Option A. -> 9 cm
:
A
We know that the radius of the circle is half of the diameter.
Therefore, if diameter is 18, then radius is half of 18, which is 9 cm.

Answer: Option C. -> Half of Circumference
:
C
The perimeter of a circle is called the circumference.
Semi-circle is half of the circle and its circumference is half of a circle’s circumference.

Answer: Option A. -> longest chord
:
A
Diameter is the longest chord of the circle. It divides the circle into exactly two halves.