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Splitting up the area below the cut
Splitting the area below the cut into a triangle and a rectangle gives the diagram below.
The red area is twice the blue area, because the areas are in the ratio 1 : 2.
The area of the red triangle is equal to the area of the blue triangle, so the area of the red rectangle must also be equal to the area of the blue triangle.
That means that in the diagram below, the area of the red rectangle is equal to the area of one of the blue triangles, so the area of the blue rectangle below must be twice the area of the red rectangle. So $a$ must be twice $b$, so the ratio $a$ : $b$ is 2 : 1.
Showing the areas as rectangles
The two diagrams below show the rectangle with a dotted line drawn horizontally at a distance $\frac{1}{2}a$ from the top.
In the diagram on the left, the triangles between the cut and the dotted lines have equal areas. This means that the areas of the blue shapes in the two diagrams are equal, and the areas of the red shapes in the two diagrams are equal.
The blue triangle occupies $\frac{1}{3}$ of the whole rectangle, because the areas are in the ratio 1 : 2.
So the height of the blue rectangle is $\frac{1}{3}$ of the height of the whole rectangle. So $\frac{1}{2}a$ is $\frac{1}{3}$ of the height of the whole rectangle, so $a$ is $\frac{2}{3}$ of the height of the whole rectangle.
That means that $b$ is $\frac{1}{3}$ of the height of the whole rectangle, so the ratio $a$ : $b$ is 2 : 1.
Comparing the ratio of the areas
Let the length of the rectangle be $l$. Then the area of the triangle is $\frac{al}{2}$, and the area of the trapezium is $\frac{(a+b) + b}{2} \times l$, so the ratio of the area of the triangle to the area of the trapezium is $$\frac{al}{2}:\frac{(a+2b)l}{2}$$ which is equivalent to $$a:a+2b$$
Since the ratio of the two areas is given in the question as $1:2$, we must have that $2b=a$ so $a:b$ is $2:1$.
Ratio Cut
Age 14 to 16
ShortChallenge Level 
Splitting up the area below the cut
Splitting the area below the cut into a triangle and a rectangle gives the diagram below.
The red area is twice the blue area, because the areas are in the ratio 1 : 2.
The area of the red triangle is equal to the area of the blue triangle, so the area of the red rectangle must also be equal to the area of the blue triangle.
That means that in the diagram below, the area of the red rectangle is equal to the area of one of the blue triangles, so the area of the blue rectangle below must be twice the area of the red rectangle. So $a$ must be twice $b$, so the ratio $a$ : $b$ is 2 : 1.
Showing the areas as rectangles
The two diagrams below show the rectangle with a dotted line drawn horizontally at a distance $\frac{1}{2}a$ from the top.
In the diagram on the left, the triangles between the cut and the dotted lines have equal areas. This means that the areas of the blue shapes in the two diagrams are equal, and the areas of the red shapes in the two diagrams are equal.
The blue triangle occupies $\frac{1}{3}$ of the whole rectangle, because the areas are in the ratio 1 : 2.
So the height of the blue rectangle is $\frac{1}{3}$ of the height of the whole rectangle. So $\frac{1}{2}a$ is $\frac{1}{3}$ of the height of the whole rectangle, so $a$ is $\frac{2}{3}$ of the height of the whole rectangle.
That means that $b$ is $\frac{1}{3}$ of the height of the whole rectangle, so the ratio $a$ : $b$ is 2 : 1.
Comparing the ratio of the areas
Let the length of the rectangle be $l$. Then the area of the triangle is $\frac{al}{2}$, and the area of the trapezium is $\frac{(a+b) + b}{2} \times l$, so the ratio of the area of the triangle to the area of the trapezium is $$\frac{al}{2}:\frac{(a+2b)l}{2}$$ which is equivalent to $$a:a+2b$$
Since the ratio of the two areas is given in the question as $1:2$, we must have that $2b=a$ so $a:b$ is $2:1$.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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