Solution
Question 15 Solution
The image below will be of help
From triangle ABC, BCD and ABD, we respectively have the following equation respectively using pythagoras theorem.
[tex]\begin{gathered} (40+x)^2=15^2+y^2...................(1) \\ \\ 15^2=x^2+h^2...............................(2) \\ \\ y^2=40^2+h^2..............................(3) \end{gathered}[/tex]
From (1)
[tex]\begin{gathered} 40^2+80x+x^2=15^2+y^2 \\ \\ From\text{ }(3),\text{ substitute the value of }y^2=40^2+h^2\text{ into the above equation} \\ \\ 40^2+80x+x^2=15^2+40^2+h^2 \\ \\ 80x+x^2=15^2+h^2 \\ \\ Add\text{ }x^2\text{ to both sides} \\ \\ 80x+2x^2=15^2+h^2+x^2 \\ \\ From\text{ }(2),\text{ substitute the value of }x^2+h^2=15^2\text{ in the equation above} \\ \\ 80x+2x^2=15^2+15^2 \\ \\ divide\text{ through by 2} \\ \\ x^2+40x-15^2=0 \\ \\ (x+45)(x-5)=0 \\ \\ x=-45\text{ and }x=5 \end{gathered}[/tex]
Length cannot be negative.
Therefore,
[tex]x=5[/tex]
Therefore, DC = 5
From (2)
[tex]\begin{gathered} x^2+h^2=15^2 \\ \\ h^2=15^2-5^2 \\ \\ h^2=(20)(10) \\ \\ h=10\sqrt{2} \end{gathered}[/tex]
The Area of the Triangle is
[tex]\begin{gathered} Area=\frac{1}{2}\times base\times height \\ \\ Area=\frac{1}{2}\times(40+5)\times10\sqrt{2} \\ \\ Area=45\times5\sqrt{2} \\ \\ Area=225\sqrt{2}\text{ square units} \end{gathered}[/tex]
Question 16
[tex]112^2+15^2=113^2[/tex]
The third side is 112
