The document contains solutions to several geometry problems involving areas of triangles, circles, rectangles, and composite shapes. The problems utilize properties of similar triangles, partitioning of areas, and relationships between parts and wholes. Diagrams are provided and calculations are shown step-by-step to arrive at the requested values.

1. Given the area for triangle ABC is 1, BE = 2AB, BC = CD, find the area for triangle BDE.
Solution:
Draw a line connecting CE, making AB and BE
as the base for △ABC and △BEC
S△BEC = 2 S△ABC = 2
And △BEC and △CED are of same base and same height
S△CED = S△BEC = 2
S△BDE = S△BEC + S△CED = 4

2. Find the area of the shaded region (diameter for big circle is 2cm).
Solution:
Flip the bottom half of the diagram to look like this:
The area of the shaded region = area of big
circle – area of small circle
Radius of big circle = 2 ÷ 2 = 1
Radius for small circle = 1 ÷ 2 = 0.5
Area of shaded region = 3.14 × (12 − 0.52) = 2.355 cm2

3. Given the area for triangle ABE, ADF and rectangle AECF are the same, find the
area for triangle AEF.
Solution:
S □ ABCD = 9× 6 = 54
Since S □ AECF = S△ABE = S△AFD,
∴ S □ AECF + S△ABE + S△AFD = area for rectangle ABCD
S □ AECF = S△ABE = S△AFD = 54 ×
1
3
= 18
S△ABE =
1
2
× 6 × (9 – EC) = 18
EC = 9 – 18 × 2 ÷ 6 = 3
S△AFD =
1
2
× 9 × (6 – FC) = 18
FC = 6 – 18 × 2 ÷ 9 = 2
S△ECF =
1
2
× 3 × 2 = 3
S△AEF = S □ AECF – S△ECF
=18 – 3 = 15

4. The diagram shows an isosceles right triangle ABC with side 10cm, The area for the
shaded region 甲and 乙 are the same. Find the area of circle which AEF is part of.
Solution:
AC = BC, ∠A = ∠B = 45º
S△ABC =
1
2
× 10 × 10 = 50
Since S甲 = S乙 ，area for AEF = S△ABC
Area of AEF
Area of the circle which it is part of
=
45º
360º
Area of the circle =
Area of AEF
45
× 360
=
S△ABC
45 × 360
= S△ABC × 8
= 400 cm2

5. Cut a rectangle with the width of ½ meter out of a square board. The area of the
remaining board is 65/18 m2. Find the area of the rectangle.
Solution:
Redraw the diagram as below (we called “弦图”in Chinese):
As you can see,
the width of the big square is x + y,
the width of the small square is x – y =
1
2
---- ①
Also, the area of the big square = (x + y)2 =
65
18
× 4 +
1
2
×
1
2
=
529
36
x + y =
23
6
---- ②
From ① + ② , x = (
23
6
+
1
2
) ÷ 2 =
13
6
The area of the remaining board =
13
6
×
1
2
= 1
1
12

6. A rectangle board, with the length cut 4m, and width cut 1m, it becomes a square, as
shown in diagram below. The new area now is 49m2 less than original area. Find the
original area of the rectangle.
Solution:
Let the width of the square as x.
(x + 4) × 1 + 4x = 49
x + 4 + 4x = 49
x = 9
The area of the rectangle = 9 × 9 + 49 = 130 m2

7. Given any rectangle ABCD, where AE =
2
3
AB, BF =
2
3
BC, CG =
2
3
CD, DH =
2
3
DA, joining
E, F, C, H. Find the ratio of area for EFGH to area of ABCD.
Solution:
Connect ED and BD.
We know S△AEH =
1
3
S△AED , S△AED =
2
3
S△ADB
∴ S△AEH=
1
3
×
2
3
S△ADB =
2
9
S△ADB
Similarly, S△CGF =
2
9
S△BCD
S△AEH + S△CGF =
2
9
S△ADB +
2
9
S△BCD =
2
9
(S△ADB + S△BCD) =
2
9
S □ ABCD
Similarly, S△BFE + S△DHG =
2
9
S □ ABCD
S△AEH + S△CGF + S△BFE + S△DHG =
4
9
S □ ABCD
Therefore, S □ EFGH = (1 -
4
9
) S □ ABCD =
5
9
S □ ABCD
S □ EFGH : S □ ABCD = 5 : 9

8. As shown in diagram, the three heights of the triangle ABC meet at point P, prove
that
𝑃𝐷
𝐴𝐷
+
𝑃𝐸
𝐴𝐸
+
𝑃𝐹
𝐴𝐹
= 1
Solution:
Since △PBC and △ABC share the same base, , the ratio of their
area is the same as the ratio of their heights
△PBC
△ABC
=
𝑃𝐷
𝐴𝐷
Similarly,
△PCA
△ABC
=
𝑃𝐸
𝐵𝐸
,
△PAB
△ABC
=
𝑃𝐹
𝐶𝐹
△PBC
△ABC
+
△PCA
△ABC
+
△PAB
△ABC
=
𝑃𝐷
𝐴𝐷
+
𝑃𝐸
𝐵𝐸
+
𝑃𝐹
𝐶𝐹
△PBC + △PCA + △PAB
△ABC
=
△ABC
△ABC
= 1
𝑃𝐷
𝐴𝐷
+
𝑃𝐸
𝐵𝐸
+
𝑃𝐹
𝐶𝐹
= 1