Test site: Pythagorean theorem; The judgement and properties of rectangle.

Analysis: If AE⊥BC, DE⊥BC and AG⊥DF are used, the quadrilateral AEFG is a rectangle and AE = fg. EF = AG, because △ADG is a right triangle, so AD= AG square +DG square under the radical sign, and AE, BE, CF and CDF can be obtained according to the right angle △AEB and right angle △CDF.

Solution: solution: AE⊥BC, DF⊥BC, AG⊥DF,

The four internal angle of that quadrilateral AEFG are right angles,

∴ quadrilateral AEFG is a rectangle, AE = fg. ef = ag。

∠ABE= 180 - 135 =45，∠DCF= 180 - 120 =60，

∴ae=eb= 6×2/2 = 3 under radical number, CF= 1/2×CD=3 under radical number, FD = 3 = 3 under radical number.

∴AG=EF=8, DG=DF-AE=2, under the root number 3,

∴AD= AG square +DG square = 19 under root number 2,

So the answer under two numbers is 19.

Comments: This topic examines the judgment of rectangle and the equality of opposite sides of rectangle, and the application of Pythagorean theorem in right triangle. The key to solve this problem is to construct rectangular AEFG.

28、

Test site: the nature of isosceles trapezoid; Determination of equilateral triangle; Triangle midline theorem.

Analysis: (1) Because the trapezoid ABCD is an isosceles trapezoid ∠ ACD = 60, we can know that △OCD and △OAB are equilateral triangles. Connecting CS and BP, we can know that △BCS and △BPC are right triangles according to the properties of equilateral triangles, and then we can know that △PQS=BP= 1/2 BC by using the properties of right triangles.

(2) According to the properties of isosceles trapezoid and ∠ ACD = 60, find out the side length of equilateral triangle and get the answer.

Answer:

(1) Proof: Lian CS, BP.

∵ABCD is an isosceles trapezoid, AC and BD intersect at O,

∴AO=BO,CO=DO.

∫∠ACD = 60, ∴△OCD and △OAB are equilateral triangles.

∫s is the midpoint of OD, ∴ cs ⊥ do.

In Rt△BSC, q is the midpoint of BC and SQ is the midline of oblique BC.

∴SQ= 65438 BC.

Similarly, BP Branch.

In Rt△BPC, PQ = 1/2bc.

SP is the center line of δ△OAD,

∴SP= 1/2AD= 1/2 BC。

∴SP=PQ=SQ.

So △SPQ is an equilateral triangle.

(2) The problem (2) is relatively simple, because it has been found in the problem (1) that △SPQ is an equilateral triangle.

So you only need one side to get the answer. . .

Comments: The knowledge of isosceles trapezoid and equilateral triangle is difficult to examine. Pay attention to mastering some basic knowledge, which is the key to solving comprehensive problems.

29、

Test center: the application of a quadratic equation.

Analysis: (1) Let the area of quadrilateral PBCQ be 32cm2 (square) from p and q to a few seconds, then PB=( 16-3x)cm and QC=2xcm. According to the trapezoid area formula, the equation can be listed as: 1/2 (18).

(2) Let QE⊥PB pass Q as E, because BC=AD=6cm, and when PE=8cm, PQ = 10 cm. Because PE=PB-QC, it is the same as (1) countable equation (16-3x).

Solution: Solution: (1) Let the area of quadrilateral PBCQ at p and q be 32cm2 from the beginning to x seconds.

PB=( 16-3x)cm，QC=2xcm，

According to the area formula of trapezoid,1/2 (16-3x+2x) × 6 = 32,

X= 16/3，

(2) Let QE⊥PB be in E after passing Q, because BC=AD=6cm. According to Pythagorean theorem, when PE=8cm, PQ = 10cm.

Suppose that the distance from point P to point Q is 10 cm from departure to x seconds, because PE=PB-QC,

Get ( 16-3x)-2x=8，

X= 1.6，

Answer: (1) The area of quadrilateral PBCQ is 32cm2 from the starting point of P and Q to 16/3 seconds;

(2) The distance from point P to point Q is 10cm. From departure to 1.6 seconds.

Remarks: (1) The trapezoidal area formula is mainly used: S= 1/2 (upper bottom+lower bottom) × height;

(2) Making auxiliary lines is the key. After forming a right triangle, use Pythagorean theorem.