Question

please solve this problem please​

Answer 1

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.

Answer 2

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