Question

can you help me for solve it before 3.00pm ​

Answer 1

$\large\underline{\sf{Solution-}}$

From the given figure,

↝ ABCD is a cyclic quadrilateral.

We know, Sum of the opposite pair of angles of a cyclic quadrilateral is supplementary.

So, using this property, we have

$\rm :\longmapsto\:\angle ABC + \angle ADC = 180 \degree$

$\rm :\longmapsto\:90\degree + \angle ADC = 180 \degree$

$\rm :\longmapsto\:\angle ADC = 180\degree - 90\degree$

$\rm \implies\:\boxed{ \tt{ \: \angle ADC = 90\degree \: }}$

Further, We know that,

Angle in semi-circle is right angle.

Now, given that,

$\rm :\longmapsto\:\angle ABC \: = \: 90\degree$

$\bf\implies \:AC \: is \: diameter \: of \: circle.$

Now, In right triangle ABC, given that

↝ AB = 6 cm

↝ BC = 8 cm

↝ By Pythagoras theorem,

$\begin{gathered}\star\;{\boxed{\sf{\pink{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}}}}\\\end{gathered}*$

So, on substituting the values, we get

$\rm :\longmapsto\: {AC}^{2} = {AB}^{2} + {BC}^{2}$

$\rm :\longmapsto\: {AC}^{2} = {6}^{2} + {8}^{2}$

$\rm :\longmapsto\: {AC}^{2} = 36 + 64$

$\rm :\longmapsto\: {AC}^{2} = 100$

$\rm :\longmapsto\: {AC}^{2} = {10}^{2}$

$\rm \implies\:\boxed{ \tt{ \: AC \: = \: 10 \: cm \: }}$

Since, AC is diameter,

$\rm :\longmapsto\:Radius = \dfrac{1}{2} \times AC$

$\rm :\longmapsto\:Radius = \dfrac{1}{2} \times 10$

$\rm \implies\:\boxed{ \tt{ \: \: Radius \: = \: 5 \: cm \: \: }}$

More to know :-

1. Angle in same segments are equal.

2. Angle subtended at the centre by an arc is double the angle subtended on the circumference of a circle by the same arc

3. Exterior angle of a cyclic quadrilateral is equals to interior opposite angle.

4. Equal chords subtends equal angles at the centre.

5. Equal chords are equidistant from the center.