In a circle with a radius of 12 decimeters, draw a square, draw a small circle in the square, and calculate the area of the small circle

In a circle with a radius of 12 decimeters, draw a square, draw a small circle in the square, and calculate the area of the small circle

The side length of the square in the circle is 12 * 2 / √ 2 = 12 √ 2,
The diameter of the small circle in the square is 12 √ 2 of the side length of the square, so the radius is 6 √ 2,
So the area of the small circle is π R ^ 2 = π * (6 √ 2) ^ 2 = 72 π square decimeter

Draw the largest circle in a square with an area of 900 square centimeters. What is the radius of the circle?
Please answer the steps together,

If the area is 900cm, the side length is 30cm, and if the circle area is the largest, the diameter of the circle is not more than the side length of the square, that is, the diameter is 30cm, so the radius is 15cm

Draw the largest circle in a square with a side length of 5DM. The circle has a circumference of () DM and an area of () square decimeters

Perimeter: 5 × 3.14 = 15.7 decimeters
Area: 2.5 × 2.5 × 3.14 = 19.625 square decimeters

As shown in the figure, the points m and N are on the sides BC and CD of the square ABCD respectively. It is known that the perimeter of △ MCN is equal to half of the perimeter of the square ABCD, then ∠ man=______ ．

When △ adn is rotated 90 ° clockwise around point a, we can get △ Abe,  AE = an, be = DN,  Abe =  d = 90 degree,  NAE = 90 degree and  ABC = 90 degree,  points m, B and E are collinear,  me = be + BM = DN + BM,  MCN's perimeter is equal to half of square ABCD's perimeter,  Mn + NC + MC = DC + BC = DN + NC + MC + BM,  Mn = DN + BM,  Mn = me. In △ man and △ Mae, an = aemn = meam = am,  Mn = me Therefore, the answer is 45 degrees

Square ABCD, point n is on edge BC, point m is on CD, the perimeter of triangle MNC is half of the perimeter of square, find the angle man

Suppose that the angle ban is ∠ 1, the angle dam is ∠ 2, and the side length of the square is X
According to the meaning of the title:
CN=X*(1-tg1)
CM=X*(1-tg2)
MN=x*√(1-tg1)^2+(1-tg2)^2
Then CN + cm + Mn = 2x
X*(1-tg1)+X*(1-tg2)+x*√(1-tg1)^2+(1-tg2)^2=2X
(TG1 + TG2) / (1-tg1 * TG2) = 1
That is, TG (1 + 2) = (TG1 + TG2) / (1-tg1 * TG2) = 1
Because according to the meaning of the title, it must be an acute angle
Therefore, 1 + 2 = 45 degrees
So the angle man = 90 - (∠ 1 + ∠ 2) = 45 degrees

In diamond ABCD, m and N are on CD and BC respectively. Angle B = angle man = 60 degrees. Is triangle man an equilateral triangle?
In diamond ABCD, N and m are on CD and BC respectively, angle B = angle man = 60 degrees, is triangle man an equilateral triangle,

Connecting AC, proving that triangle and and AMC are congruent (AAS)

In the diamond ABCD, ∠ bad = 120 degrees, m and N are the points on BC and DC respectively. If in the triangle amn, ∠ man = 60 degrees, try to judge whether the triangle amn is equilateral

It's an equilateral triangle
Proof: connect AC,
ABCD is a diamond, so ∠ CAD = ∠ bad / 2 = 60, ∠ ACB = ∠ BCD / 2 = 60
∠D=180-∠BAD=60.
Because the adjacent sides of the diamond are equal, ad = CD, ∠ d = 60. So the triangle ACD is an equilateral triangle, AC = ad
∠MAN=∠MAC+∠CAN=60
∠CAD=∠CAN+∠NAD=60
Therefore, MAC = can
∠ACM=∠D
AC=AD
△ACM≌△ ADN.AM=AN
Because ∠ man = 60, △ amn is an equilateral triangle (an isosceles triangle with a 60 degree angle is an equilateral triangle)

In diamond ABCD, the angle bad = 120 degrees, M is the point on BC, if the angle man = 60 degrees. Proof: Triangle amn is equilateral triangle

It is proved that the connection AC ∵ bad = 120 & # 186; ∵ B = ∵ d = 60 & # 186;, and the four sides of the diamond are equal, that is ab = BC = CD = ad ∵ ABC and ⊙ ACD are equilateral triangles ∵ AB = AC, ∵ DAC = ∵ ACD = 60 & # 186; ∵ man = 60 & # 186; ∵ BAM + ∵ Dan = ∵ bad - ∵ man = 60 & # 186; ∵ can + ∵ Dan = 60 & # 186

In diamond ABCD, ∠ bad = 120 °, M is the point on BC, n is the point on CD, if there is an angle of 60 ° in △ amn, try to judge that △ amn is an equilateral triangle
thank you

Because the quadrilateral ABCD is a diamond, ad = AB = BC = DC ad parallel BC AB parallel DC angle B = D so angle ban = DNA angle dam = AMB so angle BAM = nad so triangle ABM is equal to triangle and so am = an because triangle ABM has 60 ° angle, so △ amn is an equilateral triangle

As shown in the figure, diamond ABCD, ∠ bad = 120 °, point m is the point on BC, and point n is the point on CD. If ∠ amn = 60 °, try to judge the shape of △ amn and explain the reason (please use the method of congruent triangle)

A: △ amn is an equilateral triangle. It is proved that connecting AC with Mn at point F, crossing point m as me ‖ AC with ab at point E, in ∵ rhombic ABCD, ∵ bad = 120 °, ∵ ABC and △ ACD are equilateral triangles, ∵ BCD = 120 °, ∵ AB = BC, ∵ B = 60 °, ∵ BME are equilateral triangles, ∵ EM = BM = be, ∵ BEM = 60 °, ∵ AEM = 120 °, ∵ AEM = ∵ BCD, ∵ ab-be = bc-bm, i.e., AE = MC, ? AMC are equilateral triangles In △ AEM and △ MCN, 1 ＝ 2ae = MC ∠ AEM = MCN, ｜ △ AEM ≌ △ MCN (ASA), ｜ am = Mn, ≌ amn = 60 ° and ｜ amn = 60 ° are equilateral triangles