If the product of the slopes of two straight lines is - 1, they must be perpendicular to each other

If the product of the slopes of two straight lines is - 1, they must be perpendicular to each other

That's right
But the opposite is not necessarily true. For example, a line parallel to the x-axis is perpendicular to the line parallel to the y-axis, but a slope is 0, a slope does not exist, and the product is meaningless

If we know △ def ≌ △ ABC, ab = AC, and the perimeter of △ ABC is 23cm, BC = 4 & nbsp; cm, then one of the edges of △ def must be equal to______ ．

If the circumference of △ ABC is 23cm, BC = 4 & nbsp; cm, ab = AC, then AB = AC = 23 − 42 = 9.5cm. Because the corresponding sides of congruent triangles are equal, one of the sides of △ def must be equal to 4cm or 9.5cm

If △ ABC ≌ Δ def, ab = 2, AC = 4 and the circumference of △ DEF is even, then the length of EF is______ ．

4-2 ＜ BC ＜ 4 + 22 ＜ BC ＜ 6. If the perimeter is even, BC should also be even, so it is 4. So the length of EF is also 4. So the answer is: 4

If AB = 5, EF = 6, then AC=

7

If the perimeter of triangle DEF is 12cm, EF = 4, DF is 2cm longer than De, DB = 1, the direction of translation is indicated
Distance of translation
leave

According to the known DB = 1, the direction of triangle def translation is left down translation. Let the side length de = x, DF = x + 2, the perimeter of triangle def: x + 4 + (x + 2) = 12, and the solution is x = 3. The distance of triangle translation is 3 + 1 = 4cm

In known triangle ABC, a (1.3) B (2.5) C (3.2) how to find triangle ABC area

This is a right angle isosceles triangle, ab = radical 5, AC = radical 5, BC = radical 10, so AB2 + ac2 = BC2, so it is a right angle isosceles triangle, so area = AB * AC / 2 = 2.5

Let the vertex of ABC be a (3,3,2) B (5,3,1) C (0, - 1,3) to find the area of triangle

Try it! As shown in the picture:

In △ ABC, if the three internal angles ∠ a, ∠ B and ∠ C satisfy sin ^ 2A + sin ^ 2c-sina * sinc = sin ^ 2B, then the angle B=

From the sine theorem, the original formula can be changed into
a^2+c^2-ac=b^2
That is [(a ^ 2 + C ^ 2-B ^ 2) / 2Ac] = 0.5
That is, CoSb = 0.5
∴B=π/3

In △ ABC, sin ^ 2A = sin ^ 2B + SINB * sinc + sin ^ 2C, then a = ()
In △ ABC, sin ^ 2A = sin ^ 2B + SINB &; sinc + sin ^ 2C, then a = ()

The original formula is equivalent to (A / 2R) ^ 2 = (B / 2R) ^ 2 + (B / 2R) * (C / 2R) + (C / 2R) ^ 2, that is, a ^ 2 = B ^ 2 + b * C + C ^ 2, that is, the cosine of 2BC multiplied by a is equal to - BC, that is, the cosine of a is equal to - 1 / 2, that is, because it is in a triangle, the range of a is 0 to 180 degrees, then a is 120 degrees

In the triangle ABC, sin ^ 2B + sin ^ 2c-sin ^ 2A = sibb * sinc, if C = 3, B = 4, calculate the area

sin^2B+sin^2C-sin^2A=sibB*sinC
So we use the sine theorem to get the following results
b^2+c^2-a^2=bc
Then we use cosine theorem: A ^ 2 = B ^ 2 + C ^ 2-2bccosa to get
cosA=1/2
So a = 60 degrees
So area = (bcsina) / 2 = 12 * ((radical 3) / 2) / 2 = 3 radical 3