Diagonals of an isosceles trapezoid are perpendicular to each other. the length of its leg is 26 cm. an altitude from the vertex of an obtuse angle to the base divides the longer base into two parts, such that the shorter part is 10 cm long. what is the area of this trapezoid?

square units.

Step-by-step explanation:

We draw  a ABCD isosceles trapezoid (see in the figure ), in which the length of leg is 26 cm (side AD and side BC), diagonals perpendicular to each other and an altitude (line segment DE or CF ) from the vertex of an obtuse angle to the base divides the longer base into two parts, such that the shorter part is 10 cm long.

Now in triangle ADE (see in figure ), AD =26 cm, AE =10 cm and ∠DEA =90 degree.We apply the Pythagoras theorem in triangle ADE ,then we write units

units

units

units

units.

If we see in triangle DOC the side DO and Side CO are same and angle DOC =90 degree, because the given trapezium is an isosceles trapezium and diagonals of trapezium are perpendicular to each other,then triangle DOC is an isosceles right triangle.

Again if we see in triangle DEB angle DEB =90 degree, ∠EDB and ∠EBD are same because ∠DEB =90 degree and ∠ABD =∠CDB =45 degree (see in figure), so triangle DEB is isosceles right triangle  and we say that line segment DE =line segment EB. But the line segment DE =24 units ( from above calculation ), so line segment BE = 24 units.

We write from figure,

units

units

units and

units

units.

Hence area of isosceles trapezoid = square units

or  area of isosceles trapezoid = square units

or  area of isosceles trapezoid = square units

or  area of isosceles trapezoid = units

or  area of isosceles trapezoid = square units.

when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range:

x coordinate can be written by using the formula, x= x1 + 2/5(x2 - x1)

step-by-step explanation:

formula

x = x1 + p/w(x2 - x1)

let the line segment ab with a(x1,y1) and b(x2,y2)

a point(x,y) divide ab in the ratio, 2: 3 the x coordinate can be written as

x = x1 + p/w(x2 - x1)

here p=2 and w = 2+3

therefore,

x coordinate can be written by using the formula, x= x1 + 2/5(x2 - x1)

similarly y coordinate y = y1 + 2/5(y2 - y1)