614 diagonals of a rectangular trapezoid. Diagonals of a trapezoid. Properties of triangles lying on the lateral side and diagonals of a trapezoid
Again the Pythagorean triangle :))) If a piece of the large diagonal from the large base to the intersection point is designated x, then from the obvious similarity of right triangles with equal angles it follows.x/64 = 36/x, hence x = 48;48/64 = 3/ 4, so ALL right triangles formed by bases, diagonals and a side perpendicular to the base are similar to a triangle with sides 3,4,5. The only exception is a triangle formed by pieces of diagonals and an oblique side, but we are not interested in it :). (To make it clear, the similarity about which we're talking about- just NAMED DIFFERENTLY trigonometric functions angles :) we already know the tangent of the angle between the large diagonal and the large base, it is equal to 3/4, which means the sine is 3/5, and the cosine is 4/5 :)) You can immediately write
Answers. The lower base is 80, the height of the trapezoid will be 60, and the upper one will be 45. (36*5/4 = 45, 64*5/4 = 80, 100*3/5 = 60)
Similar tasks:
1. The base of the prism is a triangle, one side of which is 2 cm, and the other two are 3 cm each. The side edge is 4 cm and makes an angle of 45 with the plane of the base. Find the edge of an equal cube.
2. The base of the inclined prism is equilateral triangle with side a; one of the side faces is perpendicular to the plane of the base and is a rhombus, the smaller diagonal of which is equal to c. Find the volume of the prism.
3. In an inclined prism, the base is a right triangle, the hypotenuse of which is equal to c, one acute angle is 30, the side edge is equal to k and makes an angle of 60 with the plane of the base. Find the volume of the prism.
1. Find the side of the square if its diagonal is 10 cm
2. In an isosceles trapezoid obtuse angle is equal to 135 degrees less than the base is 4 cm and the height is 2 cm find the area of the trapezoid?
3. The height of the trapezoid is 3 times greater than one of the bases, but half as much as the other. Find the bases of the trapezoid and the height if the area of the trapezoid is 168 cm square?
4. In triangle ABC, angle A = At angle = 75 degrees. Find BC if the area of the triangle is 36 cm square.
1. In a trapezoid ABCD with sides AB and CD, the diagonals intersect at point O
a) Compare the areas of triangles ABD and ACD
b) Compare the areas of triangles ABO and CDO
c) Prove that OA*OB=OC*OD
2. Base isosceles triangle relates to the side as 4:3, and the height drawn to the base is 30 cm. Find the segments into which the bisector of the angle at the base divides this height.
3. Line AM is tangent to a circle, AB is a chord of this circle. Prove that angle MAB is measured by half of the arc AB located inside angle MAB.
If the diagonals in an isosceles trapezoid are perpendicular, the following theoretical material will be useful in solving the problem.
1. If the diagonals in an isosceles trapezoid are perpendicular, the height of the trapezoid is equal to half the sum of the bases.
Let us draw a line CF parallel to BD through point C and extend the line AD until it intersects with CF.
Quadrilateral BCFD is a parallelogram (BC∥ DF as the base of a trapezoid, BD∥ CF by construction). So CF=BD, DF=BC and AF=AD+BC.
Triangle ACF is right-angled (if a line is perpendicular to one of two parallel lines, then it is also perpendicular to the other line). Since in an isosceles trapezoid the diagonals are equal and CF=BD, then CF=AC, that is, triangle ACF is isosceles with base AF. This means that its height CN is also the median. And since the median right triangle drawn to the hypotenuse is equal to half of it, then
which in general can be written as
where h is the height of the trapezoid, a and b are its bases.
2. If the diagonals in an isosceles trapezoid are perpendicular, then its height is equal to midline.
Since the midline of the trapezoid m is equal to half the sum of the bases, then
3. If the diagonals in an isosceles trapezoid are perpendicular, then the area of the trapezoid is equal to the square of the height of the trapezoid (or the square of the half-sum of the bases, or the square of the midline).
Since the area of a trapezoid is found by the formula
and the height, half the sum of the bases and the midline of an isosceles trapezoid with perpendicular diagonals are equal to each other:
4. If the diagonals of an isosceles trapezoid are perpendicular, then the square of its diagonal equal to half the square of the sum of the bases, as well as twice the square of the height and twice the square of the midline.
Since the area of a convex quadrilateral can be found through its diagonals and the angle between them using the formula
- The segment connecting the midpoints of the diagonals of a trapezoid is equal to half the difference of the bases
- Triangles formed by the bases of a trapezoid and the segments of the diagonals up to their point of intersection are similar
- Triangles formed by segments of the diagonals of a trapezoid, the sides of which lie on the lateral sides of the trapezoid - are equal in size (have the same area)
- If you extend the sides of the trapezoid towards the smaller base, then they will intersect at one point with the straight line connecting the midpoints of the bases
- A segment connecting the bases of a trapezoid and passing through the point of intersection of the diagonals of the trapezoid is divided by this point in a proportion equal to the ratio of the lengths of the bases of the trapezoid
- A segment parallel to the bases of the trapezoid and drawn through the point of intersection of the diagonals is divided in half by this point, and its length is equal to 2ab/(a + b), where a and b are the bases of the trapezoid
Properties of a segment connecting the midpoints of the diagonals of a trapezoid
Let's connect the midpoints of the diagonals of the trapezoid ABCD, as a result of which we will have a segment LM.
A segment connecting the midpoints of the diagonals of a trapezoid lies on the midline of the trapezoid.
This segment parallel to the bases of the trapezoid.
The length of the segment connecting the midpoints of the diagonals of a trapezoid is equal to half the difference of its bases.
LM = (AD - BC)/2
or
LM = (a-b)/2
Properties of triangles formed by the diagonals of a trapezoid
Triangles that are formed by the bases of a trapezoid and the point of intersection of the diagonals of the trapezoid - are similar.
Triangles BOC and AOD are similar. Since angles BOC and AOD are vertical, they are equal.
Angles OCB and OAD are internal angles lying crosswise with parallel lines AD and BC (the bases of the trapezoid are parallel to each other) and a secant line AC, therefore they are equal.
Angles OBC and ODA are equal for the same reason (internal crosswise).
Since all three angles of one triangle are equal to the corresponding angles of another triangle, then these triangles are similar.
What follows from this?
To solve problems in geometry, the similarity of triangles is used as follows. If we know the lengths of two corresponding elements similar triangles, then we find the similarity coefficient (we divide one by the other). From where the lengths of all other elements are related to each other by exactly the same value.
Properties of triangles lying on the lateral side and diagonals of a trapezoid
Consider two triangles lying on the lateral sides of the trapezoid AB and CD. These are triangles AOB and COD. Despite the fact that the sizes of individual sides of these triangles may be completely different, but the areas of the triangles formed by the lateral sides and the point of intersection of the diagonals of the trapezoid are equal, that is, the triangles are equal in size.
If we extend the sides of the trapezoid towards the smaller base, then the point of intersection of the sides will be coincide with a straight line that passes through the middle of the bases.
Thus, any trapezoid can be expanded into a triangle. Wherein:
- Triangles formed by the bases of a trapezoid with a common vertex at the point of intersection of the extended sides are similar
- The straight line connecting the midpoints of the bases of the trapezoid is, at the same time, the median of the constructed triangle
Properties of a segment connecting the bases of a trapezoid
If you draw a segment whose ends lie on the bases of a trapezoid, which lies at the point of intersection of the diagonals of the trapezoid (KN), then the ratio of its constituent segments from the side of the base to the point of intersection of the diagonals (KO/ON) will be equal to the ratio of the bases of the trapezoid(BC/AD).
KO/ON = BC/AD
This property follows from the similarity of the corresponding triangles (see above).
Properties of a segment parallel to the bases of a trapezoid
If we draw a segment parallel to the bases of the trapezoid and passing through the point of intersection of the trapezoid’s diagonals, then it will have the following properties:
- Specified distance (KM) bisected by the intersection point of the trapezoid's diagonals
- Length of the segment passing through the point of intersection of the diagonals of the trapezoid and parallel to the bases is equal to KM = 2ab/(a + b)
Formulas for finding the diagonals of a trapezoid
a, b- trapezoid bases
c,d- sides of the trapezoid
d1 d2- diagonals of a trapezoid
α β - angles with a larger base of the trapezoid
Formulas for finding the diagonals of a trapezoid through the bases, sides and angles at the base
The first group of formulas (1-3) reflects one of the main properties of trapezoid diagonals:
1. The sum of the squares of the diagonals of a trapezoid is equal to the sum of the squares of the sides plus twice the product of its bases. This property of trapezoid diagonals can be proven as a separate theorem
2 . This formula is obtained by transforming the previous formula. The square of the second diagonal is thrown through the equal sign, after which the square root is extracted from the left and right sides of the expression.
3 . This formula for finding the length of the diagonal of a trapezoid is similar to the previous one, with the difference that another diagonal is left on the left side of the expression
The next group of formulas (4-5) are similar in meaning and express a similar relationship.
The group of formulas (6-7) allows you to find the diagonal of a trapezoid if the larger base of the trapezoid is known, one side and the angle at the base.
Formulas for finding the diagonals of a trapezoid through height
Task.
The diagonals of the trapezoid ABCD (AD | | BC) intersect at point O. Find the length of the base BC of the trapezoid if the base AD = 24 cm, length AO = 9 cm, length OS = 6 cm.
Solution.
The solution to this problem is absolutely identical in ideology to the previous problems.
Triangles AOD and BOC are similar in three angles - AOD and BOC are vertical, and the remaining angles are pairwise equal, since they are formed by the intersection of one line and two parallel lines.
Since the triangles are similar, all their geometric dimensions are related to each other, just like the geometric dimensions of the segments AO and OC known to us according to the conditions of the problem. That is
AO/OC = AD/BC
9/6 = 24/BC
BC = 24 * 6 / 9 = 16
Answer: 16 cm
Task .
In the trapezoid ABCD it is known that AD=24, BC=8, AC=13, BD=5√17. Find the area of the trapezoid. 
Solution .
To find the height of a trapezoid from the vertices of the smaller base B and C, we lower two heights to the larger base. Since the trapezoid is unequal, we denote the length AM = a, length KD = b ( not to be confused with the notation in the formula finding the area of a trapezoid). Since the bases of the trapezoid are parallel, and we dropped two heights perpendicular to the larger base, then MBCK is a rectangle.
Means
AD = AM+BC+KD
a + 8 + b = 24
a = 16 - b
Triangles DBM and ACK are rectangular, so their right angles are formed by the altitudes of the trapezoid. Let us denote the height of the trapezoid by h. Then, by the Pythagorean theorem
H 2 + (24 - a) 2 = (5√17) 2
And
h 2 + (24 - b) 2 = 13 2
Let's take into account that a = 16 - b, then in the first equation
h 2 + (24 - 16 + b) 2 = 425
h 2 = 425 - (8 + b) 2
Let's substitute the value of the square of the height into the second equation obtained using the Pythagorean Theorem. We get:
425 - (8 + b) 2 + (24 - b) 2 = 169
-(64 + 16b + b) 2 + (24 - b) 2 = -256
-64 - 16b - b 2 + 576 - 48b + b 2 = -256
-64b = -768
b = 12
So KD = 12
Where
h 2 = 425 - (8 + b) 2 = 425 - (8 + 12) 2 = 25
h = 5
Find the area of a trapezoid through its height and half the sum of its bases
, where a b - the base of the trapezoid, h - the height of the trapezoid
S = (24 + 8) * 5 / 2 = 80 cm 2
Answer: the area of the trapezoid is 80 cm2.