# Thread: triangular blocks problem

1. ## triangular blocks problem

the lowest row has 198 triangular blocks.For each of the other rows,the number of triangular blocks are 4 more than in the above row.Joe builds a triangular blocks in this way.The number of triangular blocks in the top row is 2.

a)Calculate the number of rows in this arrangement

b)If the height and the base of each triangular block is 6 cm and 8 cm respectively,calculate the area,in cm,of the triangle.

2. Hello mastermin346
Originally Posted by mastermin346
the lowest row has 198 triangular blocks.For each of the other rows,the number of triangular blocks are 4 more than in the above row.Joe builds a triangular blocks in this way.The number of triangular blocks in the top row is 2.

a)Calculate the number of rows in this arrangement

b)If the height and the base of each triangular block is 6 cm and 8 cm respectively,calculate the area,in cm,of the triangle.
(a) Let's suppose there are $n$ rows.

In row $1$ there are $198$ blocks.

In row $2$ there are $198 - 4 \times 1$ blocks.

In row $3$ there are $198 - 4\times 2$ blocks.

...

In row $n$ there are $198 -4\times(n-1)$ blocks.

So if there are $2$ blocks in the top row:
$198-4(n-1)=2$

$\Rightarrow 198 -4n +4 =2$

$\Rightarrow 4n = 200$

$\Rightarrow n = 50$
So there are $50$ rows.

(b) The area of each block $= \tfrac12 \text{ base x height} = \tfrac12\; 6\times 8 = 24 \text{ cm}^2$

So we need to find the number of blocks altogether and multiply by $24$.

That total number, $S$, is:
$S=\underbrace{198 + 194 + 190 + ... + 6 + 2}_{50 \text{ items}}$
If we write the sum backwards:
$S=\underbrace{2+6+...+194+198}_{50 \text{ items}}$
and add the two together:
$2S=\underbrace{200 + 200 + ... + 200 + 200}_{50 \text{ items}}$
$=50\times200$

$=10\,000$
$\Rightarrow S = 5\,000$
So the area $= 24\times 5\,000 = 120\,000 \text{ cm}^2$.

Grandad