Question

in a tree the top of broken part is 24 m distant from the trees base,while the unbroken part is 7m tall
a)what is the length of the trees broken section
b)what was the length of the whole tree?
c) what is the perimeter of the triangle formed by the broken tree?

Answer 1

For diagram see the attached photo

_______________________

$\large \sf (a) \: \: What \: \: is \: \: the \: \: length \: \: of \: \: the \: \: tree's \: \: broken \: \: section \: \: ?$

Answer :-

here AD = AC = broken section

We know Pythagoras theorem,

AC² = AB² + BC²

where AC = d

AB = 7m

BC = 24 m

Now we can put these values

AC² = AB² + BC²

d² = (7)² + (24)²

d² = 49 + 576

d² = 625

d = √625

d = 25 m

Hence, Length of the tree's broken section = d = AC = 25 m = AD

_______________________

$\large \sf (b) \: \: what \: \: was \: \: the \: \: length \: \: of \: \: the \: \: whole \: \: tree \: \:?$

Answer :

whole Tree

BD = DA + AB

BD = 25 m + 7 m

BD = 32 m

_______________________

$\large \sf (c) \: \: what \: \: is \: \: the \: \: perimeter \: \: of \: \: the \: \: triangle \: \: formed \\ \\ \large \sf by \: \: the \: \: broken \: \: tree? \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Answer :

Perimeter of traingle

∆ ABC = AB + BC + AC

= ( 7 + 24 + 25 ) m

= 56 m

I hope it helps you ❤️✔️