Let m = 4x ^ 2-12x + 9y ^ 2 + 30y + 35, then a.m > 0, B.M ≥ 0, C.M < 0, D.M ≤ 0

Let m = 4x ^ 2-12x + 9y ^ 2 + 30y + 35, then a.m > 0, B.M ≥ 0, C.M < 0, D.M ≤ 0

M=4x^2-12x+9y^2+30y+35
=4x^2-12x+9+9y^2+30y+25+1
=(2x-3)²+(3y+5)²+1
≥1
Option a

How to calculate the area of trapezoid, square, rectangle and triangle

Trapezoid: (upper bottom + lower bottom) * height / 2
Square: side length * side length
Rectangle: length * width
Triangle: bottom * height / 2

Triangle trapezoid parallelogram rectangle square relation

1: It's all geometry
2: All are quadrilateral except triangle
3: Rectangles and squares are special parallelograms

Judge right or wrong! The circumference of rectangle, square and circle is 12 or 56 cm, and the area of circle is the smallest

With the same perimeter, the circle has the largest area

The circumference of the triangle is 46 cm. The distance between a point P and three sides is 4 cm. The area of the triangle is () square cm
I am a pupil, so please explain the process in detail

Hello!
P is the intersection point of the three bisectors. The three bisectors divide the triangle into three triangles. Each triangle can be regarded as a large triangle with a bottom edge and a height of 4 (that is, the distance between P and the bottom edge). The area of the three triangles can be expressed and added together
1 / 2 × perimeter × 4 = 1 / 2 × 46 × 4 = 92 square centimeter

The circumference of a triangle is 30 cm, and the distance between a point in the triangle and three sides is 6 cm. What is the area of the triangle in square centimeter?

The sum of the bottom edges is 30, and the height is 6. According to the area formula, s = 30 * 6 / 2 = 90 (cm ^ 2)

The perimeter of a triangle is 28 cm. The distance between a point in the triangle and three sides is 6 cm. The area of the triangle is______ Square centimeter

As shown in the figure, s △ ABC = s △ PAC + s △ PAB + s △ PBC = 12ac × pf + 12ab × PD + 12bc × PE, because pf = PD = PE = 6cm, substituting the above formula, s △ ABC = 12 × 6 × (AC + AB + BC) = 3 × 28 = 84 (square centimeter); a: the area of this triangle is 84square centimeter, so the answer is: 84

If the perimeter and area of two isosceles triangles are equal respectively, then are the two triangles congruent

Not necessarily congruent. Because it can be proved by quadratic function. The steps are more complicated
First, the perimeter is set as C, the area is set as s, the height of the bottom edge is set as h, and the waist length is set as a
C=2a+√(a^2-h^2)
S=2h√(a^2-h^2)
Obviously, when we look at these two formulas representing general isosceles triangles, it is not difficult to see,
There must be s = 2H (C-2A)
If "the perimeter and area of two isosceles triangles are respectively equal, then the two triangles must be congruent", then a and h must be fixed. But it is obviously impossible

If the sides of an isosceles triangle are 3 and 7, the perimeter of the triangle is——
If the sides of an isosceles triangle are 3 and 7, the perimeter of the triangle is——

According to the rule that the sum of two sides of triangle is greater than the third side, C = 3 + 7 + 7 = 17

Are triangles of equal perimeter and area always congruent?
Please explain why

Not necessarily
A triangle with equal perimeter and area is congruent, which is a false proposition and difficult to prove,
Examples are as follows:
Triangle 1: perimeter 18, three sides 8,5,5, area = √ [9 (9-8) (9-5) & sup2;] = 12
Triangle 2: perimeter 18. Three sides 6,6 + √ 33 / 3,6 - √ 33 / 3,
Area = √ [9 (9-6) (9-6 - √ 33 / 3) (9-6 + √ 33 / 3)] = 12
Obviously, the two triangles are not congruent