Question

Can I Plss get some help on number 15 &16 plsss

Answer 1

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.

`\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}`

From (1)

`\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}`

Length cannot be negative.

Therefore,

`x=5`

Therefore, DC = 5

From (2)

`\begin{gathered} x^2+h^2=15^2 \\ \\ h^2=15^2-5^2 \\ \\ h^2=(20)(10) \\ \\ h=10√(2) \end{gathered}`

The Area of the Triangle is

`\begin{gathered} Area=(1)/(2)* base* height \\ \\ Area=(1)/(2)*(40+5)*10√(2) \\ \\ Area=45*5√(2) \\ \\ Area=225√(2)\text{ square units} \end{gathered}`

Question 16

`112^2+15^2=113^2`

The third side is 112