In square ABCD, e is the point on AC, EF ⊥ AB, eg ⊥ ad, ab = 6, AE: EC = 2:1. Find the area of quadrilateral afeg
In square ABCD, ∵ DAB = 90 °, DAC = 45 ° and ∵ AFE = ∵ age = 90 °, the quadrangle afeg is rectangle, ∵ AEG = 90 ° - ∵ DAC = 45 °, the ∵ gae = ∵ AEG = 45 °, Ge = AG, ∵ rectangular afeg is square, ∵ square afeg ∵ square ABCD, ∵ s Square afegs square ABCD = (aeac) 2 = (23) 2 = 49, ∵ s Square afeg = 49S square AFE G=49×62=16.
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- 1. In square ABCD, e is on AC, EF ⊥ AB is on F, eg ⊥ ad is on G, ab = 8cm, AE ratio EC = 3:1, calculate the area of quadrilateral afeg Draw your own picture I can't make it easy
- 2. In square ABCD, e is a point on AC, EF is perpendicular to AB, BG is perpendicular to ad, ab = 6, AE: EC = 2:1, find the area of afeg
- 3. In the quadrilateral ABCD, point E is the midpoint of BC, point F is the midpoint of CD, and AE is perpendicular to AB, AF ⊥ CD, connecting EF to prove AB = ad When AB, CD and BC are related, the △ AEF is an equilateral triangle
- 4. In square ABCD, point F is on ad, point E is on AB, AE = EB, AF = 1 / 4AD Connecting CE, EF (Pythagorean theorem, inverse theorem)
- 5. In the square ABCD, the midpoint e of BC connects AE, and a point F on CD makes EF ⊥ AE connect AF. verify that AF = AD + FC
- 6. Given square ABCD, de ratio EC = 3:2, EF vertical AE, then FC ratio EC = -, EF ratio AE = -, de ratio AE = -
- 7. In rectangular ABCD, points E and F are on line AB and ad respectively, AE = EB = AF = 2 / 3fd = 4. Along the straight line EF, the triangle AEF is folded into triangle a'ef, Make plane a'ef perpendicular to plane bef Finding a '- fd-c cosine of dihedral angle If the points m and N are on the line segments FD and BC respectively, and the quadrilateral mncd is folded upward along the line Mn, so that C and a 'coincide, the length of the line segment FM is calculated
- 8. In the arithmetic sequence {an}, A1 = 3, the sum of the first n terms is Sn, the items of the arithmetic sequence {BN} are all positive, B1 = 1, the common ratio is Q, and B2 + S2 = 12, q = s2b2. (1) find an and BN; (2) let the sequence {CN} satisfy the first n terms and TN of CN = 1sn, {CN}, and prove that TN < 23
- 9. As shown in the figure, in quadrilateral ABCD, angle B is equal to angle D is equal to 90 degrees, CF bisects angle BCD. If AE is parallel to CF, please determine whether AE bisects angle bad
- 10. In the quadrilateral ABCD, ∠ B = ∠ d = 90 ° and AE and CF are bisectors of ∠ DAB and ∠ BCD, respectively It's a parallel CF...
- 11. In square ABCD, e is the point on AC, EF ⊥ AB, eg ⊥ ad, ab = 6, AE: EC = 2:1. Find the area of quadrilateral afeg
- 12. Given that the straight line of square ABCD passing through point C intersects the extension lines of AD and ab with points E and f respectively, and AE = 10 and AF = 15, the side length of square ABCD is calculated
- 13. Given the square ABCD, the straight line passing through C intersects the extension lines of AD and ab at points E and f respectively, and AE = 15 and AF = 10, the side length of square ABCD is calculated
- 14. Given the square ABCD, the straight line passing through C intersects the extension lines of AD and ab at points E and f respectively, and AE = 15 and AF = 10, the side length of square ABCD is calculated
- 15. Given the square ABCD, the straight line passing through C intersects the extension lines of AD and ab at points E and f respectively, and AE = 15 and AF = 10, the side length of square ABCD is calculated
- 16. Please read the following materials: problem: as shown in Figure 1, in diamond ABCD and diamond befg, points a, B and E are on the same line, P is the midpoint of line DF, connecting PG and PC. if ∠ ABC = ∠ bef = 60 °, explore the position relationship between PG and PC and the value of pgpc. XiaoCong's idea is: extend GP intersection DC at point h, construct congruent triangle, and solve the problem through reasoning. Please refer to XiaoCong Students' ideas, explore and solve the following problems: (1) write the position relationship between PG and PC and the value of pgpc in the above problem; (2) rotate the diamond BFG in Figure 1 clockwise around point B, so that the diagonal BF of diamond BFG is in the same line with the edge ab of diamond ABCD, and other conditions in the original problem remain unchanged (as shown in Figure 2). Are the two conclusions you get in (1) correct What's the change? Write down your conjecture and prove it. (3) if ∠ ABC = ∠ bef = 2 α (0 °< α< 90 °) in Figure 1, rotate the diamond BFG clockwise at any angle around point B, and other conditions in the original problem remain unchanged, please write down the value of pgpc directly (expressed by the formula containing α)
- 17. Can the four vertices of diamond ABCD be on the same circle? If they are on the same circle, what shape should it be
- 18. As shown in the figure, the area ratio of triangle ABC to triangle ade is 3:4, and the area of triangle ABF is 10 square centimeter larger than that of triangle FCE
- 19. In trapezoidal ABCD, ab ∩ CD, CE bisection ∩ BCD intersects point E, and CE ⊥ ad, de = 2ae, if s △ CDE = 1, then the area of quadrilateral abce is
- 20. In trapezoidal ABCD, ab ‖ CD, CE bisects ∠ BCD, and CE ⊥ ad is at point E, de = 2ae, the area of △ CDE is 8, then the area of trapezoidal ABCD is () A.16 B.15 C.14 D.13 The answer seems to be B, but please tell me why