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 $\displaystyle n$ rows.

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

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

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

...

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

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

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

$\displaystyle \Rightarrow 4n = 200$

$\displaystyle \Rightarrow n = 50$

So there are $\displaystyle 50$ rows.

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

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

That total number, $\displaystyle S$, is: $\displaystyle S=\underbrace{198 + 194 + 190 + ... + 6 + 2}_{50 \text{ items}}$

If we write the sum backwards:$\displaystyle S=\underbrace{2+6+...+194+198}_{50 \text{ items}}$

and add the two together:$\displaystyle 2S=\underbrace{200 + 200 + ... + 200 + 200}_{50 \text{ items}}$$\displaystyle =50\times200$

$\displaystyle =10\,000$

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

Grandad