In question 20, notice that h, a, and a/2 form a right triangle. From Pythagoras you know that a^2 = (1/2)^2 + h^2, so rearrange this to get 'h' by itself. What do you get? Now use that value to calculate the areas for questions 21 and 22, remembering that the formula for area of a triangle is 1/2 base times height. As for question 28, note that the line segment to the left of 'x' is part of a right triangle, and you can find its length from the dimensions given. You can also find the total length of the base of the large triangle, and 'x' is the difference between these two lengths.
Thanks for the reply! I am basically lost at squarroot(a^2-b^2) from my attempt as you can see.
I am sorry for not informing but I am looking at question 29 not 28. For question 29, I am perplexed at the same issue with qn 20 and 21, which is basically squareroot(a^2-b^2). As go on, it seems that have done something wrong somewhere. Please enlighten me.
Lastly, pardon me for my poor understanding in math. I have been trying very hard to self study a subject that I have not touched for years and poor at it.
$\displaystyle \begin{align*} a^2 - \left( \frac{a}{2} \right) ^2 &= \frac{4a^2}{4} - \frac{a^2}{4} \\ &= \frac{4a^2 - a^2}{4} \\ &= \frac{3a^2}{4} \end{align*}$
So when you take the square root:
$\displaystyle \begin{align*} \sqrt{\frac{3a^2}{4}} &= \frac{\sqrt{3}\sqrt{a^2}}{\sqrt{4}} \\ &= \frac{\sqrt{3}\,a}{2} \end{align*}$
In the large triangle, by Pythagoras we have
$\displaystyle \begin{align*} \left( \sqrt{3} \right) ^2 + 1^2 &= h^2 \\ 3 + 1 &= h^2 \\ 4 &= h^2 \\ h &= 2 \end{align*}$
So the entire length of the hypotenuse is 2 units.
Now looking at the triangle that has $\displaystyle \begin{align*} \sqrt{3} \end{align*}$ as its hypotenuse. Call the unknown length l and we have
$\displaystyle \begin{align*} x^2 + l^2 &= \left( \sqrt{3} \right) ^2 \\ x^2 + l^2 &= 3 \\ l^2 &= 3 - x^2 \\ l &= \sqrt{ 3 - x^2 } \end{align*}$
and in the triangle that has 1 as its hypotenuse, call the unknown length k and we have
$\displaystyle \begin{align*} x^2 + k^2 &= 1^2 \\ x^2 + k^2 &= 1 \\ k^2 &= 1-x^2 \\ k &= \sqrt{ 1 - x^2 } \end{align*}$
Since these two unknown lengths together make the largest hypotenuse, that gives
$\displaystyle \begin{align*} k + l &= 2 \\ \sqrt{1 - x^2} + \sqrt{ 3 - x^2 } &= 2 \end{align*}$
So now all that's left is solving this equation for x