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If the waist length of an isosceles triangle is 10 and its area is 30, then the tangent of the base angle of the isosceles triangle is________
Let the base of an isosceles triangle be a and the height above the base is h. then ah = 60, (1) a ^ 2 + 4H ^ 2 = 400. (2) then (a + 2H) ^ 2 = a ^ 2 + 4H ^ 2 + 4ah = 640, so a + 2H = 8 √ 10. (3) the following equation can be understood by Wei Dading: x ^ 2-8 √ 10 * x + 120 = 0x = 6 √ 10 or x = 2 √ 10
The waist of an isosceles triangle is known to be 2 6 cm, bottom edge 4 2cm, find the area of this isosceles triangle
∵ the base of the isosceles triangle is 4
2cm，
The bottom half is 2
2cm，
The height on the bottom edge=
(2
6)2−(2
2)2=4cm，
So, the area of this isosceles triangle is 1
2×4
2×4=8
2cm2．
Let the base length of the isosceles triangle be 6cm, and the midlines of the two waists are perpendicular to each other, so as to find the area of the quadrilateral aegd emergency
That's it,
There are midlines perpendicular to each other,
Then the triangle BGC is an isosceles right triangle
So BG = GC = radical 3
S triangle BGC = 3 / 2
Because be, GC is the midline,
So s triangle ABG = s triangle AGC = s triangle BGC = 3 / 2
And s triangle ADG = s triangle age = s triangle ABG / 2
So the area of the quadrilateral aegd is 3 / 2
I wish you progress in your study!
If the waist length of an isosceles triangle is root 24 and its base is 2 times root 6, what is its area? Why is Gao root 18
Height = √ {(√ 24) 2 - [(2 √ 6) / 2]} = √ (24-6) = √ 18 = 3 √ 2 (unit)
Area = 1 / 2 × (3 √ 2) × (2 √ 6) = 3 √ 12 = 6 √ 3 (square unit)
Given that the base edge of an isosceles triangle is 2 and the area is the root 2, find the waist length? Process and result
1 / 2 (bottom * height) = area: height = root 2
Square of waist length = square of height + square of (1 / 2 bottom) (Pythagorean theorem)
The square of waist length = 3, waist length = root 3
An isosceles triangle has an area of 40, a side length of 10, and a circumference of 10_____ I know that there are three kinds of 2 root numbers 89 + 10 20 + 4 root numbers 3
(1) Bottom edge = 10
Bottom height = 40 * 2 / 10 = 8
Waist = root (8 ^ 2 + 5 ^ 2) = root 89
Circumference = 10 + 2 * root number 89 = 10 + 2 root number 89
(2) Waist = 10
Waist height = 40 * 2 / 10 = 8
The height above the waist divides the waist into two parts, and the distance from the apex = root (10 ^ 2-8 ^ 2) = 6
Then the distance to the bottom edge = 10-6 = 4
Base = root (8 ^ 2 + 4 ^ 2) = 4 Radix 5
Perimeter = 10 * 2 + 4 root mark 5 = 20 + 4 root number 5
The area of a'b'c'd 'diagram is as follows: CD = 1, ad = CB = root 2, bottom AB = 3, parallel to the top and bottom, and taking the x-axis? The answer is the root of two. How do you get it?
When viewed directly, the length on the y-axis is regarded as the height, and the length becomes 1 / 2 of the original length. There is also a 45 angle. Therefore, the height is the root of 4 / 2
So it turns out to be the root of two
ACD = 1, ACD = 1= The area of a ′ B ′ C ′ D ′ of the intuitionistic graph a ′ B ′ C ′ D ′ is______ .
ACD = 1, ACD = 1=
2, bottom AB = 3, so the trapezoid height is: 1,
If the X axis is taken parallel to the upper and lower bottom edges, the height of a ′ B ′ C ′ D ′ is 1
2sin45°＝
Two
Four
So the area of the visual graph is: 1
2×(1+3)×
Two
4=
Two
Two
So the answer is:
Two
Two
As shown in the figure, in the isosceles trapezoid ABCD, the upper bottom ad = 3 root number 2cm, the bottom bottom 8 root number 2, and the height AE = 4 root number 2. Find the circumference l of trapezoidal ABCD and the area s of ABCD
Isosceles trapezoid, be = (8 √ 2-3 √ 2) / 2 = 5 / 2 √ 2
AB=√(AE^2+BE^2)=√(25/2+32)=1/2√178
CD=AB
Perimeter: 3 √ 2 + 8 √ 2 + 2 * 1 / 2 √ 178 = 11 √ 2 + √ 178
Area: (3 √ 2 + 8 √ 2) * 4 √ 2 * 1 / 2 = 44
In isosceles trapezoid ABCD, ad is parallel to BC and AC is perpendicular to BD. if AD + BC = four and root two cm, calculate the length of diagonal AC and the area of trapezoid ABCD
If a parallelogram of AC is made through D, and the extension line of BC intersects at point E, then the quadrilateral aced is a parallelogram,
∴AD=CE,DE=AC,
The diagonals of isosceles trapezoid are equal, that is, BD = AC = De,
∵AC⊥BD
The △ BDE is an isosceles right triangle,
∵BE=BC+CE=BC+AD=4√2
The hypotenuse of an isosceles right triangle is √ 2 times that of the right angle,
∴BD=DE=4
The height of the trapezoid is the height on the hypotenuse of the isosceles right angle △ BDE, i.e. H = (1 / 2) be = 2 √ 2
The area of trapezoid is s = (AD + BC) H / 2 = 8
RELATED INFORMATIONS
- 1. The cosine value of an isosceles triangle is root 2 / 2, waist length is 3, root 2. Find the area of this isosceles triangle
- 2. Isosceles trapezoid's upper, lower bottom and waist length are 4cm, 10cm, 6cm respectively. How many upper and lower corners should be reminded
- 3. The area of an isosceles trapezoid is known to be 6.18.10 for the upper, lower, and waist
- 4. The circumference of the right angle trapezoid is 48 cm. The ratio of the sum of the two bottoms to the sum of the two waist is 2:1, and the ratio of one waist to the other is 3:5
- 5. It is known that the length of the upper and lower bottom of isosceles trapezoid is root 3 and root 12 respectively, and the height is root 6
- 6. The length of the trapezoid is 5, the length of the bottom is 10, and the length of the waist is 3 and 4 respectively A. 18 B. 22.5 C. 26.25 D. 30
- 7. The circumference of an isosceles trapezoid is 80 cm. The upper and lower heights are 20 cm and 30 cm respectively. How many centimeters is the waist length of this isosceles trapezoid?
- 8. A piece of wire is surrounded by a square with an edge length of 8 cm. If it is surrounded by a rectangle with a length of 12 cm, what is the width of the rectangle?
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- 13. As shown in the figure, the four points a, B, C, D are on the circle O. if an external angle DCE of the quadrilateral ABCD is 70 °, calculate the size of the angle bad and the angle BOD
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- 17. As shown in the figure, three straight lines AB, CD and EF intersect at the same point O, if ∠ AOE = 2 ∠ AOC, ∠ COF = 3 2 ∠ AOE, then ∠ doe=______ .
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