Weekly Dose 22 - Maths Olympiad Practice - Area

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