The formula for area of a trapezoid is A =1/2 h(b1 + b2). Express h in terms of A, b1 and b2
1 answer:
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Step-by-step explanation:
Angles in the same segment. The angles at the circumference subtended by the same arc are equal. More simply, angles in the same segment are equal.
Picture 1 a° = a° they are the same
Picture 2 p= 52° q= 40° Angles in the same segment are equal. if they ask you to calculate you just show both angles p= 52 and q=40.
Picture 3 + 4 Let the obtuse angle MOQ = 2x.
Using the circle theorem, the angle at the centre is twice the angle at the circumference.
Therefore picture 4 tells us and proves;
Angle MNQ = x and angle MPQ = x.
Picture 5
A cyclic quadrilateral is a quadrilateral drawn inside a circle. Every corner of the quadrilateral must touch the circumference of the circle.
The second shape is not a cyclic quadrilateral. One corner does not touch the circumference.
Picture 6
The opposite angles in a cyclic quadrilateral add up to 180°.
a + c = 180°
b + d = 180°
Picture 7
Example
Calculate the angles a and b.
The opposite angles in a cyclic quadrilateral add up to 180°.
This is a picture that couldn't be added but has a quadrilateral occupying half the triangle consisting of 3 arcs the middle one was 140 and proved part of one triangle. The x (a) missing angle showed below a = 180-60 = 120 °
So hopefully this will help you remember the format here for quadrilateral shapes within a circle.
b = 180 - 140 = 40°
a = 180 - 60 = 120°
Answer:
Step-by-step explanation:
A vector perpendicular to the plane ax+by+cz+d=0 is of the form .
So, a vector perpendicular to the plane x − y + 2z = 7 is .
The parametric equations of a line through the point and parallel to the vector
are as follows:
Put and
Therefore,
xy-plane:
Put z = 0 ⇒ t = -2 ⇒x = - 1 , y = 6
So, at point (-1,6,0)
yz-plane:
Put x = 0 ⇒ t = -1 ⇒ y = 5, z =2
So, at point (0,5,2)
xz-plane:
Put y = 0 ⇒ t = 4 ⇒ x = 5, z = 12
So, at point (5,0,12)