1. Quadrilateral ABCD is inscribed in circle O
A quadrilateral is a four sided figure, in this case ABCD is a cyclic quadrilateral such that all its vertices touches the circumference of the circle.
A cyclic quadrilateral is a four sided figure with all its vertices touching the circumference of a circle.
2. mBCD = 2 (m∠A) = Inscribed Angle Theorem
An inscribed angle is an angle with its vertex on the circle, formed by two intersecting chords.
Such that Inscribed angle = 1/2 Intercepted Arc
In this case the inscribed angle is m∠A and the intercepted arc is MBCD
Therefore; m∠A = 1/2 mBCD
4. The sum of arcs that make up a circle is 360
Therefore; mBCD + mDAB = 360°
The circles is made up of arc BCD and arc DAB, therefore the sum angle of the arcs is equivalent to 360°
5. 2(m∠A + 2(m∠C) = 360; this is substitution property
From step 4 we stated that mBCD +mDAB = 360
but from the inscribed angle theorem;
mBCD= 2 (m∠A) and mDAB = 2(m∠C)
Therefore; substituting in the equation in step 4 we get;
2(m∠A) + 2(m∠C) = 360
Try this solution:
1. the standart form of the circle is: (x-x₀)²+(y-y₀)²=r², where point (x₀;y₀) - the cetre of the given circle, r - radius of the given circle.
2. Using r=7 and coordinates of the centre (0;0) it is possible to make up the required equation:
(x-0)²+(y-0)²=7²;
x²+y²=49.
I think 1/2 of all the flowers are yellow tulips. Hope this helps =)
