How can a right angled trapezoid be divided into four figures of equal size and same shape

According to the nature of the similar shape, we can divide the sloping waist into two parts, two small similar trapezoids, and then divide the remaining part into two parts

An isosceles trapezoid with a base angle of 60 degrees and an upper base equal to the waist, please divide it into four figures of equal size and shape

The painting is not standard, ha ha

In an isosceles trapezoid with a base angle of 60 degrees, it is known that the upper edge and the two sides of l have the same waist. Please divide them into four figures with the same size and shape

Take three 4 equal points on the top and connect the bottom with each other

Draw three lines and divide the isosceles trapezoid into four figures with equal area and same shape Note that there are only three lines The first problem of the fifth part of the fourth grade Wuhan mathematics summer homework

Connect the middle points of the upper and lower sides, and then take a point above the connecting line to make the two quadrilateral forms congruent

As shown in the figure, in the RT triangle ABC, the angle ACB is equal to 90 degrees, AC is equal to 10 cm, and BC is equal to 15 cm As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, AC = 10cm, BC = 15cm, point P starts from a and moves along AC to point C at a uniform speed of 1cm / S; point Q starts from C and moves along CB to point B at a speed of 2cm / s, and points P and qmin start from the starting point at the same time, and the time required to move to a certain position is T seconds (1) When t = 4, find the length of PQ; (2) When t is the value, the area of △ PQC is equal to 16cm2? (3) Point O is the midpoint of AB, connecting OC, can PQ ⊥ OC? If so, calculate t value; if not, explain the reason Mainly write the third question

(1) When t = 4,
∵ point P moves at a constant speed of 1 cm / s along AC to point C, and point Q moves at a uniform speed of 2 cm / s from C to point B along CB,
∴AP=4cm,PC=AC-AP=6cm、CQ=2×4=8cm,
ν PQ = PC square under root number + CQ square = 10cm;
（2）∵AP=t,PC=AC-AP=10-t、CQ=2t,
 s △ PQC = 1 / 2
PC×CQ=t（10-t）=16,
∴t1=2,t2=8,
When t = 8, CQ = 2T = 16 ﹥ 15,  omitted,
When t = 2, the area of △ PQC is equal to 16cm2
(3) ∵ point O is the midpoint of AB, ∵ ACB = 90 °,
/ / OA = ob = OC (center line theorem on the hypotenuse of a right triangle),
∴∠A=∠OCA,
And ∠ OCA + ∠ QPC = 90 °, a + ∠ B = 90 °,
Ψ B = ∠ QPC, and ∠ ACB = ∠ PCQ = 90 °,
∴△ABC∽△QPC,
ν CP of CB = CQ of Ca,
ν 10 out of 15 − t = 2 out of 10,
∴t=2.5s．
When t = 2.5s, PQ ⊥ OC

In the triangle a B C, how to find the circumference of the triangle ABC and the value of Tana

BC=ABsinA=4/5*15=12
According to Pythagorean theorem, AC square = AB square - BC square
Then square root to find AC = 9
Perimeter = 12 + 15 + 9 = 36
tanA=BC/AC=4/3

As shown in the figure, in the triangle ABC, ab = AC, CD ⊥ AB at point D. given AB = 10 and Tana = 3 / 4, find the length of CD and the value of cosa

∵ Tana = 3 / 4 ᙽ set CD = 3x ad = 4x
In the triangle ACD, ad 2 + CD 2 = AC 2, i.e. (3 x) 2 + 4 x) 2 = 100
X = 2  CD = 6, cosa = 4 / 5

As shown in the figure, in the triangle ABC, ab = 15, BC = 14, and the triangle area is 84, calculate the values of Tana and Tanc The picture is an ordinary acute triangle. He didn't give his feet how to remove special feet

Remember a formula for the area of a triangle: S = (1 / 2) AB * ac * sin ∠ a = (1 / 2) AB * BC * sin ∠ B = (1 / 2) ac * BC * sin ∠ C, i.e., area = (1 / 2) angle between two sides * sin ∵ AB = 15, BC = 14 ? area s = (1 / 2) AB * BC * SINB = (1 / 2) * 15 * 14 * SINB = 84, ? sin? B + cos? B = 1

In the RT triangle ABC, the angle c is equal to 90 degrees, the angle a is equal to 30 degrees, BD is the bisector of angle ABC, CD is equal to 5cm? In the triangle ABC, the angle c is equal to 90 degrees, the angle B is equal to 60 degrees, D is a point on AC, De is perpendicular to AB and E, and CD is equal to 2, De is equal to 1. What is the length of BC segment of the line?

From CD = 5, we can see that BC = 5 √ 3 BD = 10. The properties of triangles with 30 ° angle can be determined by the Pythagorean theorem AB? = AC? + BC? = 15? + (5 √ 3)? = 300 ν AB = 10 √ 3 BC = (4 √ 3) / 3 can only be marked 1

In RT △ ABC, ∠ C = 90 ° AB = 2Ac, the degree of ∠ A and ∠ B is obtained

As shown in the figure, ∵ C = 90 °, ab = 2Ac,
∴∠B=30°，
∴∠A=90°-∠B=90°-30°=60°．

How can a right angled trapezoid be divided into four figures of equal size and same shape