As shown in the figure, it is known that the quadrilateral ABCD is a square with a side length of 5cm. The area of the triangle ECF is 5 square centimeters larger than that of the triangle ADF. Find the length of the line CE
The area of ABCD is 25, so the area of triangle Abe is 30. Be = 12, so CE = 12-5 = 7
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- 1. If the areas of △ CEF, △ Abe and △ ADF are 3, 4 and 5 respectively, the area of △ AEF can be obtained
- 2. The quadrilateral ABCD is a square, and △ AEF is an equilateral triangle, where the points E and F are on BC and CD respectively. The detailed explanation of s △ CEF = s △ Abe + s △ ADF is obtained The area of s △ Abe and s △ ADF is equal to half ab. s △ CEF = half (a-b) 178; after that, how can we do it? We can't do it any more. Let's do it as soon as possible
- 3. Given that ab = 8cm, ad = 12cm, the area of triangle Abe and triangle ADF are respectively 13% of rectangle ABCD, the area of triangle AEF is calculated
- 4. Given that ab = 8cm, ad = 12cm, the area of triangle Abe and triangle ADF are respectively 13% of rectangle ABCD, the area of triangle AEF is calculated
- 5. It is known that ab = 8 cm, ad = 12 cm, the area of triangle Abe and triangle ADF are respectively One third of the rectangle ab-cd, find the area of triangle AEF
- 6. If the perimeter of parallelogram ABCD is 44cm and ab is 2cm larger than ad, then AB is equal to?
- 7. In a parallelogram ABCD with an area of 15, if AB = 5, BC = 6, then CE + CF=
- 8. In the parallelogram ABCD, the diagonal lines AC and BD intersect at point O, if AC = 8, BD = 6 What is the value range of AB?
- 9. As shown in the figure, the side length of square ABCD is 4, which is the midpoint of BC side, f is the point on DC side, and DF = 14dc, AE and BF intersect at G point. Find the area of △ ABG
- 10. As shown in the figure, ABCD is a square, point E is on BC, and DF ⊥ AE is on F. please determine a point G on AE to make △ ABG ≌ △ DAF, and explain the reason
- 11. It is known that the quadrilateral ABCD is a square with a side length of 6 cm. The area of the triangle ECF is 3 square cm larger than that of the triangle ADF
- 12. In square ABCD, e and F are on BC and CD, respectively, ∠ EAF = 45 °, try to explain s △ AEF = s △ Abe + s △ ADF
- 13. As shown in the figure, in square ABCD, e and F are on BC and CD, respectively, ∠ EAF = 45 ° to prove s △ AEF = s △ Abe + s △ ADF
- 14. As shown in the figure, in square ABCD, e and F are on BC and CD, respectively, ∠ EAF = 45 ° to prove s △ AEF = s △ Abe + s △ ADF
- 15. As shown in the figure, ABCD is a rectangle, e and F are the points on BC and CD respectively, and s △ Abe = 3, s △ CEF = 2, s △ ADF = 2, then s △ AEF = ()
- 16. Rectangle ABCD, e and F are on BC and CD respectively. The area of triangle Abe, CEF and ADF are 2, 3 and 4 respectively
- 17. Let E and f be on the edge BC and CD of rectangular ABCD respectively, and the areas of △ Abe, △ ECF and △ FDA are a, B and C respectively. Find the area s of △ AEF
- 18. As shown in the figure, in rectangular ABCD, e is the point on BC and F is the point on CD. Given s △ Abe = s △ ADF = 13sabcd, then the value of s △ AEF: s △ CEF is equal to () A. 2B. 3C. 4D. 5
- 19. In the isosceles trapezoid ABCD, ab ∥ DC, ad = BC = 5, DC = 7, ab = 13, point P starts from point a and moves to the midpoint C along ad → DC at the speed of 2 unit lengths per second At the same time, point Q starts from point B and moves along Ba to terminal a at the speed of 1 unit length per second. Let the movement time be T seconds (1) When the value of T is, the quadrilateral pqbc is a parallelogram (2) In the whole movement process, when t is the value, the quadrilateral with points B, C, P and Q as the vertex is isosceles trapezoid
- 20. Cut AE = BF = CG = DH on each side of square ABCD, connect AF, BG, CH and De, and intersect a ` B ` c ` d 'in turn. It is proved that quadrilateral a ` B ` c ` d' is a square