As shown in the figure, in the known trapezoidal ABCD, ad ‖ BC, ∠ B = 30 °, C = 60 °, ad = 4, ab = 33, then the length of the bottom BC is______ ．

As shown in the figure, in the known trapezoidal ABCD, ad ‖ BC, ∠ B = 30 °, C = 60 °, ad = 4, ab = 33, then the length of the bottom BC is______ ．

As shown in the figure, through a, make AE ‖ CD intersect BC at point E, ∵ ad ‖ BC, ∵ quadrilateral AECD is a parallelogram, ∵ CE = ad = 4, ∵ B = 30 °, ∵ C = 60 °, ∵ BAE = 90 °, ∵ AE = 12be (the right side of a right triangle with 30 ° angle is equal to half of the hypotenuse). In RT △ Abe, be2 = AB2 + AE2, namely be2 = (33) 2 + (12be) 2, be2 = 27 + 14be2, be2 = 36, be = 6, ∵ BC = be + EC = 6 + 4 = 10 The answer is: 10

In the trapezoidal ABCD, as shown in the figure, if ad ‖ BC, ∠ B = 90 °, C = 50 °, ad = 1, BC = root 3, then the length of AB is__
In trapezoidal ABCD, as shown in the figure, if ad ∥ BC, ∠ B = 90 °, C = 50 °, ad = 1, BC = root 3, then the length of AB is?

Extend the intersection point of Ba and CD, e ∫ B = 80 °, C = 50 ∫ e = 50 ∫ be = BC = √ 3 ∫ ad ∥ BC ∫ ead ∫ EBC ∫ AE = ad = 1 ∫ AB = eb-ea = √ 3-1
You can try. You can choose,

As shown in the figure, ladder ABCD, ad ‖ BC, ∠ B = 45 & ordm;, ∠ C = 120 & ordm;, ab = 8, find the length of CD

In trapezoidal ABCD, AD / / BC, ∠ B = 45 ° and ab = 8
The height of trapezoid = ab × sin45 ° = 8 ×√ 2 / 2 = 4 √ 2
CD=4√2÷sin（180°-120°）=4√2÷（√3/2)=8√6/3

As shown in the figure, in ladder ABCD, AB / / DC, ad = BC = 8, ab = 10, CD = 6, find the area of ladder ABCD

The area of trapezoid is (CD + AB) * H / 2
Therefore, the key is to seek high H
Make a vertical line from point d to AB, and the perpendicular foot is e. since ad = BC, AE = (ab-cd) / 2 = 2
h^2＝AD^2-AE^2=64-4=60
H = 2 times root sign 15
So the area of trapezoid is: 16 times the root 15

Pythagorean circle large square area 13 small square area 1 right triangle a is short side B long side (a + b) how many square

From the meaning of the title, we know that a & # 178; + B & # 178; = 13, (B-A) &# 178; = 1
Then (a + b) &# 178; = 2 (A & # 178; + B & # 178;) - (B-A) &# 178; = 25

As shown in the figure, a, B, C and D are the four vertices of the rectangle, ab = 16cm and ad = 6cm. The moving points P and Q start from point a and C at the same time. Point P moves to point B at the speed of 3cm / s until it reaches B, and point Q moves to D at the speed of 2 & nbsp; cm / S. (1) how many seconds does P and Q start from the start? The area of quadrilateral pbcq is 33cm2; (2) from the start of P and Q to a few seconds? The distance between point P and point q is 10 cm

(1) Let the area of the quadrilateral pbcq be 33cm2 from the starting point of P and Q to x seconds, then Pb = (16-3x) cm, QC = 2xcm. According to the trapezoidal area formula, we get 12 (16-3x + 2x) × 6 = 33, and the solution is x = 5. (2) let the distance between P and Q be 10cm from the starting point of P and Q to T seconds

As shown in the figure, it is known that a, B, C and D are the four vertices of the rectangle, ab = 16cm and ad = 6cm. The moving points P and Q start from point a and C respectively. Point P moves to point B at the speed of 3cm / s until point B, and point Q moves to point d at the speed of 2cm / s
(1) When P and Q are a few seconds from the start, the distance between P and Q is 10 cm
(2) P, Q two points from the start to a few seconds, O, Q, D triangle is isosceles triangle

1) The first distance is x seconds, PQ ^ 2 = (16-5x) ^ 2 + 6 ^ 2 = 10 ^ 2x = 1.6 seconds (the second time is 4.8 seconds) 2) point "P", Q, D triangle is isosceles triangle? 1) PD = pq3x + 3x + 2x = 16, x = 2 seconds? 2) DP = DQ6 ^ 2 + (3x) ^ 2 = (16-2x) ^ 2, x = (6 √ 59-32) / 5 ≈ 0.28 seconds? 3) DQ = PQ (16-2x) ^ 2 = (16-5x) ^ 2

In rectangular ABCD, ad = 16cm, ab = 6cm, the moving points P and Q start from a and C respectively, and point P moves to point d at the speed of 3cm / s,
Point d moves towards point B at the speed of 2cm / s. when one point reaches the end point, the other point stops
1. How long does P and Q move? The area of quadrilateral abqp is three fifths that of rectangular ABCD?
2. When P and Q exercise for a long time, PQ = six five

Let t be the time, then 3T + (16-2t) * 6 / 2 = 16 * 6 * 3 / 5. The solution t should be 3.22 ab.cd ）If we make QE perpendicular to CD and perpendicular to e, if we want to make it six times root three, we should make EP 12 (satisfying Pythagorean theorem). If we set time x, then we have 16-5x = 12 and the solution should be 4-5

As shown in the figure, the quadrilateral ABCD is rectangular, ad = 16cm, ab = 6cm. The moving points P and Q start from a and C at the same time. Point P moves to D at the speed of 3cm / s until D, and Q moves to B at the speed of 2cm / S. (1) a few seconds after P and Q start from the start, the area of quadrilateral abqp is 35 times of the rectangular area product? When is the area of quadrilateral abqp the largest and what is the largest? (2) A few seconds after P and Q start, PQ = 65cm?

(1) The area of rectangle ABCD is s = 16 × 6 = 96cm2, 35S rectangle = 35 × 96 = 57.6cm2. After x seconds, the area of quadrilateral abqp is 35 times that of rectangle, that is 12 (3x + 16-2x) × 6 = 35 × 96, and the solution is x = 3.2 seconds

As shown in the figure, in rectangular ABCD, ab = 16cm, ad = 6cm, the moving points P and Q start from a and C at the same time. Point P moves to B at the speed of 3cm per second until it reaches B, and point Q moves to D at the speed of 2cm per second. (1) how many seconds after P and Q start, the area of quadrilateral PBC q is 36cm2? (2) Is there a time to make pbcq square? If there is, find the moment; if not, explain the reason

(1) Let P and Q start at T seconds, the area of quadrilateral pbcq is 36cm2. From rectangular ABCD, ∠ B ＝ C = 90 ° ab ‖ CD is obtained, so quadrilateral pbcq is right angled trapezoid, so s trapezoid pbcq = 12 (CQ + Pb) · BC. And s trapezoid pbcq = 36, so 12 (2t + 16-3t) · 6 = 36. The solution is t = 4 (seconds