please solve this problem please
Answers
Step-by-step explanation:
1.
since L P is 90°
we can consider PQ= 10cm as perpendicular
PR= 24cm as base.
Now according to Pythagoras theorem
QR²= PQ²+ QR²
QR²= (10cm)² +(24cm)²
QR²= 100cm²+ 576cm²
QR= ✓676cm²= 26cm
2.
original height= current height + the broken length
current height = 4m = perpendicular
distance of the top touching the ground= 3m= base
therefore, the length of the tree from where it's broken is,
H²= P²+B²
H²= (4m)²+(3m)²
H²= 16m²+9m²
H = ✓25m²= 5m
So the original height = 4m+5m = 9m.
Step-by-step explanation:
1.∆ABC is a right angle trangle,
where m∠P=90°
b=PQ=10cm
p=PR=24cm
h=QR=
We know that
h²=p²+b²
QR² =PQ²+PR³
QR=√PQ²+PR²
QR=√10²+24²
QR=√2²×5²+2²×12²
QR=√2²{5²+12²}
QR=2√25+144
QR=2√169
QR=2×13=26cm
2 .
The tree is broken at a hight of 3mfrom the ground.p= AB=4m
The distance between the top of tree touches at the ground and base base of tree is 3m. b=BC=3
The other part of tree is h= AC=
The angle between the ground and the tree broken part is 90°.m∠B=90°
We know that
h²=p²+b²
AC²=AB²+BC²
AC=√AB²+AC²
AC=√4²+3²
AC=√16+9
AC=√25
AC=5
The original hight of tree is AB+AC=4m+5m=9m