If the diagonals in an isosceles trapezoid are perpendicular, the following theoretical material will be useful in solving the problem.
1. If the diagonals are perpendicular in an isosceles trapezoid, the height of the trapezoid is half the sum of the bases.
Let us draw line CF through point C parallel to BD and extend line AD until it intersects 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 the diagonals in an isosceles trapezoid are equal, and CF=BD, then CF=AC, that is, triangle ACF is isosceles with base AF. Hence, its height CN is also the median. And since the median of a right triangle drawn to the hypotenuse is equal to half of it, then
what in general view can be written as
where h is the height of the trapezoid, a and b are its bases.
2. If in an isosceles trapezoid the diagonals are perpendicular, then its height is equal to middle line.
Since the midline of the trapezoid m is equal to half the sum of the bases, then
3. If the diagonals are perpendicular in an isosceles trapezoid, 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 in an isosceles trapezoid are perpendicular, then the square of its diagonal is 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
- The triangles formed by the bases of the trapezoid and the segments of the diagonals up to the point of their intersection are similar
- Triangles formed by segments of the diagonals of a trapezoid, the sides of which lie on the sides of the trapezoid - equal area (have the same area)
- If we 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
- The segment connecting the bases of the 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 intersection point of the diagonals is bisected 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
Connect the midpoints of the diagonals of the trapezoid ABCD, as a result of which we will have a segment LM.
A line segment that joins the midpoints of the diagonals of a trapezoid lies on the midline of the trapezium.
This segment parallel to the bases of the trapezium.
The length of the segment connecting the midpoints of the diagonals of a trapezoid is equal to the half-difference of its bases.
LM = (AD - BC)/2
or
LM = (a-b)/2
Properties of triangles formed by the diagonals of a trapezoid
The triangles that are formed by the bases of the trapezoid and the point of intersection of the diagonals of the trapezoid - are similar.
Triangles BOC and AOD are similar. Because the angles BOC and AOD are vertical, they are equal.
Angles OCB and OAD are internal crosswise lying at parallel lines AD and BC (the bases of the trapezium are parallel to each other) and the secant line AC, therefore, they are equal.
Angles OBC and ODA are equal for the same reason (internal cross-lying).
Since all three angles of one triangle are equal to the corresponding angles of another triangle, 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 the two corresponding elements of 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 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 can be completely different, but the areas of the triangles formed by the sides and the point of intersection of the diagonals of the trapezoid are, that is, the triangles are equal.
If the sides of the trapezoid are extended towards the smaller base, then the point of intersection of the sides will be coincide with a straight line that passes through the midpoints of the bases.
Thus, any trapezoid can be extended to a triangle. Wherein:
- The 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 the trapezoid, which lies at the intersection point of the diagonals of the trapezoid (KN), then the ratio of its constituent segments from the side of the base to the intersection point 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 you draw a segment parallel to the bases of the trapezoid and passing through the intersection point of the diagonals of the trapezoid, then it will have the following properties:
- Preset distance (KM) bisects the point of intersection of the diagonals of the trapezoid
- Cut length, 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- bases of a trapezoid
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 the 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 the diagonals of a trapezoid can be proved as a separate theorem
2 . This formula is obtained by transforming the previous formula. The square of the second diagonal is thrown over 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) is similar in meaning and expresses a similar relationship.
The group of formulas (6-7) allows you to find the diagonal of a trapezoid if you know the larger base of the trapezoid, one side and the angle at the base.
Formulas for finding the diagonals of a trapezoid in terms of height
A 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 of this task is absolutely identical to the previous tasks in terms of ideology.
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, then all their geometric dimensions are related to each other, as the geometric dimensions of the segments AO and OC known to us by the condition of the problem. That is
AO/OC=AD/BC
9 / 6 = 24 / B.C.
BC = 24 * 6 / 9 = 16
Answer: 16 cm
A 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 onto the larger base. Since the trapezoid is unequal, we denote the length AM = a, the length KD = b ( not to be confused with the symbols in the formula finding the area of a trapezoid). Since the bases of the trapezoid are parallel and we have omitted 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 right-angled, so their right angles are formed by the heights of the trapezoid. Let's denote the height of the trapezoid as h. Then by the Pythagorean theorem
H 2 + (24 - a) 2 \u003d (5√17) 2
and
h 2 + (24 - b) 2 \u003d 13 2
Consider that a \u003d 16 - b, then in the first equation
h 2 + (24 - 16 + b) 2 \u003d 425
h 2 \u003d 425 - (8 + b) 2
Substitute the value of the square of the height into the second equation, obtained by 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
Thus, KD = 12
Where
h 2 \u003d 425 - (8 + b) 2 \u003d 425 - (8 + 12) 2 \u003d 25
h = 5
Find the area of a trapezoid using its height and half the sum of the bases
, where a b - the bases of the trapezoid, h - the height of the trapezoid
S \u003d (24 + 8) * 5 / 2 \u003d 80 cm 2
Answer: the area of a trapezoid is 80 cm2.
Again, the Pythagorean triangle :))) If a piece of the large diagonal from the large base to the intersection point is denoted by x, then from the obvious similarity of right-angled triangles with the same 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 be clear, the likeness in question is just NAMED BY ANOTHER trigonometric functions angles :) we already know the tangent of the angle between the large diagonal and the large base, it is 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)
Related tasks:
1. The base of the prism is a triangle, in which one side is 2 cm, and the other two are 3 cm each. The lateral edge is 4 cm and makes an angle of 45 with the base plane. Find the edge of an equal cube.
2. The base of the inclined prism is an equilateral triangle with side a; one of the side faces is perpendicular to the plane of the base and is a rhombus whose smaller diagonal is c. Find the volume of the prism.
3. In an inclined prism, the base is right triangle, whose hypotenuse is c, one sharp corner 30, the lateral edge is equal to and makes an angle of 60 with the base plane. Find the volume of the prism.
1. Find the side of a square if its diagonal is 10 cm
2. In an isosceles trapezoid, the obtuse angle is 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 more than one of the bases, but half the other. Find the bases of the trapezoid and the height if the area of the trapezoid is 168 cm squared?
4. In triangle ABC, angle A = In angle = 75 degrees. Find BC if the area of a triangle is 36 cm squared.
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. The base of an isosceles triangle is related to the side as 4:3, and the height drawn to the base is 30 cm. Find the segments into which this height is divided by the bisector of the angle at the base.
3. Line AM -tangent to the circle, AB-chord of this circle. Prove that angle MAB is measured by half of arc AB located inside angle MAB.
