2020/007) We know for a right angled triangle $a^2+b^2=c^2$ where a,b are shorter sides and c is hypotenuse. What is $\frac{1}{a^2}+ \frac{1}{b^2}$ for a right angled triangle

We have $a^2+b^2=c^2$

Hence deviding by $a^2b^2$ we get

$\frac{1}{a^2}+ \frac{1}{b^2} = \frac{c^2}{a^2b^2}\cdots(1)$

Now area of the triangle $A = \frac{1}{2}ab\cdots(2)$

If h is the altitude drawn from the right angle to the hypotenuse then area of
the triangle $A = \frac{1}{2}ch\cdots(3)$

So from (2) and (3) ab = ch and putting in (1)

$\frac{1}{a^2}+ \frac{1}{b^2} = \frac{c^2}{a^2b^2} = \frac{c^2}{c^2h^2}= \frac{1}{h^2}$

Or  $\frac{1}{a^2}+ \frac{1}{b^2} = \frac{1}{h^2}$

Where h is the altitude drawn from the right angle to the hypotenuse

2020/006) How many squares and rectangles are there on a standard chess board

First let us calculate the number of squares


For the square of side n we can choose in rows in 9-n ways(side 1 8 ways, side 2
7 ways son on). In the column in 9-n ways. so the number of ways the
square of side n can be chosen in $(9-n)^2$ ways and as we have number of sides from 1 to 8
so number of ways = $$\sum_{k=1}^{8}(9-k)^2= \sum_{n=1}^{8}(n)^2 = 204$$

Now for calculation of number of rectangles.


For a rectangle we need to choose 2 lines in rows $9 \choose 2$ ways and in
columns in $9 \choose 2$ ways so total number of ways ${9 \choose 2  }^2$ or
$(\frac{9 * 8}{2})^2$ ways that is 1296 ways. As there are 204 squares
so number of rectangles = 1296-204 = 1092