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Isosceles trapezoid ABCD, AD / / BC, ab = DC, P is any point of BC, PM / / CD, intersects BD with m, PN / / AB intersects AC with N, what is the relationship between PM + PN and ab?
∵PM//CD
∴BP/BC=PM/CD
PM= BP/BC×CD
∵AB=CD
∴PM= BP/BC×AB
∵PN//BA
∴CP/BC=PN/AB
PN= CP/BC×AB
∴PM+PN= BP/BC×AB+ CP/BC×AB
PM+PN=AB
The image shape of quadratic function is the same as y = x & # 178; and the symmetry axis is x = - 1 / 2, which intersects with y axis at the point (0, - 1)
The image shape of quadratic function is the same as y = x & # 178; and the symmetry axis is x = - 1 / 2
Let the analytic expression be y = (x + 1 / 2) &# 178; + K
X = 0, y = - 1
-1=1/4+k
k=-5/4
The analytic formula is y = (x + 1 / 2) &# 178; - 5 / 4
As shown in Figure 4, in isosceles trapezoid ABCD, ad ∥ BC, ab = DC, P is any point on BC, PM ∥ CD intersects BD with m, PN ∥ AB intersects AC with N, let's ask PM + PN and ab
What is the relationship between PM + PN and ab
Because PM ∥ CD PN ∥ ab
So pN / AB = PC / BC = (bc-bp) / BC 1
PM/DC=BP/BC 2
And because AB = CD
So 1 + 2 is (PN + PM) / AB = (bc-bp + BC) / BC = 1
So PN + PM = ab
From the same shape of parabola, can we get the same a or the same absolute value of a
The problem description should be detailed
The absolute value of a should be the same, because a determines the opening direction of the parabola. At this time, because its shape is already the same, whether upward or downward, it meets the problem
As shown in the figure, in ladder ABCD, ad ∥ BC, ab = CD, P is the point on BC, PM ⊥ AB in M, PN ⊥ CD in N, be ⊥ CD in E. verification: PM + PN = be
∵ ad ∥ BC, ab = CD ∥ quadrilateral ABCD is isosceles trapezoid ∥ ABC = ∥ DCB ∥ PM ⊥ AB, PN ⊥ CD, be ⊥ CD ∥ RT Δ BMP ∥ RT Δ CEB ∥ RT Δ CNP ∥ BP / BC = PM / be, CP / BC = pN / be, then (BP + CP) / BC = (PM + PN) / be → (PM + PN) / be = 1 (because BP + CP = BC) ∥ PM + PN = be
The same shape of two parabolas means that a is the same, or the absolute value of a is the same
Absolute value of a
As shown in the figure, in the isosceles triangle ABC, point P is a moving point on the bottom edge AC, and m and N are the midpoint of AB and BC respectively. If the minimum value of PM + PN is 2, then the perimeter of △ ABC is___ ．
Make the symmetry point m 'of m point about AC, connect M'n, and the intersection point with AC is the position of P point, ∵ m, n are the midpoint of AB, BC respectively, ∵ Mn is the median line of △ ABC, ∵ Mn ∥ AC, Mn = 12ac, ∵ PM ′ PN = km ′ km = 1, ∵ PM ′ PN, ∵ MP = PN, ∵ in △ MBP and △ NBP, BN = bmbp = BPPN = PM
When the image of quadratic function is similar in shape but different in size, what is the characteristic of the value of a?
By the way, does it have anything to do with the values of B and C?
The value of a is the opening direction and size of the absolute parabola
(when A0, the opening is upward)
(the larger the absolute value of a, the smaller the opening)
(the smaller the absolute value of a, the larger the opening)
2. The values of a and B determine together whether the symmetry axis of the parabola is on the left or right side of the y-axis
(when a and B are of the same sign, the axis of symmetry is on the left side of the y-axis)
(when a and B are different, the axis of symmetry is on the right side of the y-axis)
(when B = 0, the axis of symmetry is Y-axis)
[this rule can be called "left same, right different", which is very important for judging the sign of parabola a, B, C in the future]
3. The value of C determines the position of the intersection of the parabola and the y-axis
(when C0, the Y-axis of the parabola intersects the positive half axis)
Hope to help the landlord!
In an isosceles right triangle, ∠ ABC = 90 degrees, point P is a moving point on the bottom edge AC, and Mn is the midpoint of AB and BC respectively, if the minimum value of PM + PN is 2,
Find the circumference of triangle ABC
Because the triangle ABC is an isosceles right triangle, and m and N are the middle points of two right angles, when p is the middle point of AC, PM + PN gets the minimum value. So PM = PN = 1, ab = BC = 2, AC = 2, root 2
So we can get the perimeter of triangle ABC = 4 + 2, root 2
When p is the midpoint of AC, the minimum value of PM + PN is obtained
(1) Let n be a symmetric point n 'with respect to AC, connected with Mn', and intersected with AC at P, then PM + PN is the smallest;
(2) Prove Mn '/ / BC
In the isosceles triangle ABC, the angle ABC is 120 degrees, the point P is the moving point on the base BC, and m.n is the moving point AB.BC The minimum value of PM + PN is 3, and the perimeter of triangle is 0
There are no points,
The minimum value of PM + PN is 3 = = > BP = PC = = > AB = AC = 3
In isosceles △ ABC, ABC = 120 ° = = > BC = √ 3aC = 3 √ 3
Then perimeter = 3 √ 3 + 6
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