In the triangle ABC, AC = 13, BC = 15, height CD = 12, then its area is?

In the triangle ABC, AC = 13, BC = 15, height CD = 12, then its area is?

AD=sqrt(13^2-12^2)=5,
BD=sqrt(15^2-12^2)=9,
Area s = 1 / 2 * CD * (AD + BD) = 84

Isosceles triangle ABC, ab = AC, BC as the bottom, D as a point on the edge of AB, CD = ad = BC, find the area ratio of triangle ABC and triangle BCD

Triangle ABC is similar to triangle BCD
So: AB / CD = BC / BD
Where BC = CD, BD = ab-ad = ab-cd
So: AB / CD = CD / (ab-cd)
AB^2-AB*CD-CD^2=0
(AB/CD)^2-(AB/CD)-1=0
So: AB / CD = ((radical 5) + 1) / 5
(AB / CD) ^ 2 = (3 + (radical 5)) / 2
Because triangle ABC is similar to triangle BCD
So: the area ratio of triangle ABC to triangle BCD = (AB / CD) ^ 2 = (3 + (radical 5)) / 2

As shown in the figure, in the known isosceles triangle ABC, the bottom edge BC = 20, D is the point on AB, CD = 16, BD = 12, find the area of △ ABC

Because 20 * 20 = 16 * 16 + 12 * 12, the triangle CDB is a right triangle, so CD is perpendicular to AB, so triangle CAD is also a right triangle. Let ad = a, so AB = a + 12, AC = a + 12. According to Pythagorean theorem, (a + 12) * (a + 12) = 16 * 16 + A * a24a = 256-144 = 112a = 4.667, so the area of triangle s = (12 + 4.667

How to deduce triangle area = 1 / 2absin C = 1 / 2acsin B = 1 / 2bcsina?

When drawing, asinc is the height of B, and then use the area formula s = half bottom multiplied by height

Two identical right triangles form a rectangle with an area of 12 square decimeters
How many decimeters is the other right angle?

Length: 12 / 8 = 1.5 decimeters

The area of a right triangle is 1.68 square meters. Its two sides are 2.1 meters and 3 meters. How much is the third side?

According to the trilateral relation of right triangle, a ^ 2 = B ^ 2 + C ^ 2
The solution of C ^ 2 = a ^ 2-B ^ 2 is C = 2.14
The area formula of right triangle s = b * C / 2
c=1.6
So there is something wrong with this question
Because the right triangle right angle two words should be removed

A right triangle a (2,2) B (2,4), if the area is 6 square meters, the position of C is ()

All four points are OK
（-4,2）（8,2）（-4,4）（8,4）
If AB is 2, it can only be a right angle side, then the other right angle side is 6, and BC or AC is perpendicular to ab

The area of the largest right triangle in a circle is 40 square meters, so what is the area of the circle

The shape of the largest right triangle in a circle is isosceles right triangle, and the area of the circle is 40 * π = 3.14 * 40 = 125.6 square meters

The three vertices of an isosceles right triangle with an area of 25 square meters are on a circle, and one side passes through the center of the circle. What is the area of the circle~

It's an isosceles right triangle, for sure
25=2r×r÷2
So R & # 178; = 25
Area of circle = 3.14 × 25 = 78.5 square meters

The difference between the two right sides of a right triangle is 5 cm, and the area is 7 square meters. Find the length of the hypotenuse (accurate to 0.1 cm)

Let one side x, then the other side x + 5
x(x+5)/2=7
X = 2, x = - 7 (rounding off)
Then the two right angles are 2 and 7 respectively
Hypotenuse = under the root sign (the square of 2 and the square of 7) = 7.3