# Math Help - Triangle and Trapezium

1. ## Triangle and Trapezium

I need help in the below question.
The figure below, not drawn to scale, is made up of a right-angle triangle ABC, and a trapezium BCDE. Find the difference between the sum of the areas of the unshaded parts in the figure and the area of the shaded trapezium.

2. ## Re: Triangle and Trapezium

Originally Posted by kingman
I need help in the below question.
The figure below, not drawn to scale, is made up of a right-angle triangle ABC, and a trapezium BCDE. Find the difference between the sum of the areas of the unshaded parts in the figure and the area of the shaded trapezium.
1. Since $|AB| = |BC|$ the length of CD is $|CD| = 4$

F is the midpoint of AB, G is the midpoint of AC.

The area of $\Delta ABC$ has the same value as the area of the parallelogram CBEG.

2. Therefore the 2 red triangles have the same area.

The shaded region, the trapezium BFGC, has an area of $\frac34$ of the triangle ABC. And that is the half of a square.

3. ## Re: Triangle and Trapezium

Originally Posted by kingman
I need help in the below question.
The figure below, not drawn to scale, is made up of a right-angle triangle ABC, and a trapezium BCDE. Find the difference between the sum of the areas of the unshaded parts in the figure and the area of the shaded trapezium.
Please explain why "Since |AB| = |BC| the length of CD is |CD| = 4

F is the midpoint of AB, G is the midpoint of AC." when it is not given in the question.
I tried using the idea but could not get the difference 8 cm square which is the answer
thanks

4. ## Re: Triangle and Trapezium

Originally Posted by kingman
Please explain why "Since |AB| = |BC| the length of CD is |CD| = 4

F is the midpoint of AB, G is the midpoint of AC." when it is not given in the question.
I tried using the idea but could not get the difference 8 cm square which is the answer
thanks
Touché!

I've modified your sketch a little bit. (see attachment)

1. Since |AB| = |BC| the length of CD is |CD| = |DG| = k

The area of the shaded region (a trapezium) is

$a_{shaded\ region}=\frac{8+(8-k)}2 \cdot k = 8k-\frac12 k^2$

2. The area of the trapezium CBED is

$a_{trapezium\ CBED} = \frac{8+(8+k)}2 \cdot k = 8k+\frac12 k^2$

3. The area of the region with green border is:

$a_{trapezium\ CBED} + a_{\Delta GFA} = 8k+\frac12 k^2 + \underbrace{\frac12 (8-k)^2}_{a_{\Delta GFA}} = 32 +k^2$

4. The area of the triangle DCA is (only in case you need it to determine the difference of sums)

$a_{\Delta DCA} = \frac12 \cdot k \cdot 8 = 4k$

5. ## Re: Triangle and Trapezium

Originally Posted by earboth
Touché!

I've modified your sketch a little bit. (see attachment)

1. Since |AB| = |BC| the length of CD is |CD| = |DG| = k

The area of the shaded region (a trapezium) is

$a_{shaded\ region}=\frac{8+(8-k)}2 \cdot k = 8k-\frac12 k^2$

2. The area of the trapezium CBED is

$a_{trapezium\ CBED} = \frac{8+(8+k)}2 \cdot k = 8k+\frac12 k^2$

3. The area of the region with green border is:

$a_{trapezium\ CBED} + a_{\Delta GFA} = 8k+\frac12 k^2 + \underbrace{\frac12 (8-k)^2}_{a_{\Delta GFA}} = 32 +k^2$

4. The area of the triangle DCA is (only in case you need it to determine the difference of sums)

$a_{\Delta DCA} = \frac12 \cdot k \cdot 8 = 4k$

Everthing is blurred on the solution page ; so cannot read what is written on the page.
Thanks