Math, asked by alevens9590, 4 months ago

Look at the picture of a scaffold used to support construction workers. The height of the scaffold can be changed by adjusting two slanting rods, one of which, labeled PR, is shown:(screenshot)
Part A: What is the approximate length of rod PR? Round your answer to the nearest hundredth. Explain how you found your answer, stating the theorem you used. Show all your work. (5 points)

Part B: The length of rod PR is adjusted to 16 feet. If width PQ remains the same, what is the approximate new height QR of the scaffold? Round your answer to the nearest hundredth. Show all your work.

Attachments:

Answers

Answered by kulkarninishant346
0

Answer:

Let A,B and C be vertices of a triangle.

Let D,E and F be the midpoints of the sides BC,AC and AB respectively. Let

OA

=

a

,

OB

=

b

,

OC

=

c

,

OD

=

d

=

OE

=

e

and

OF

=

f

be position vectors of points A,B,C,D,E and F respectively.

Therefore, by Midpoint formula,

d

=

2

b

+

c

,

e

=

2

a

+

c

and

f

=

2

a

+

b

∴2

d

=

b

+

c

,2

e

=

a

+

c

and 2

f

=

a

+

b

∴2

d

+

a

=

a

+

b

+

c

,

2

e

+

b

=

a

+

b

+

c

,

2

e

+

b

=

a

+

b

+

c

2

f

+

c

=

a

+

b

+

c

Now,

3

2

d

+

a

=

3

2

e

+

b

=

3

2

f

+

c

=

3

a

+

b

+

c

Let

g

=

3

a

+

b

+

c

. Then, we have

g

=

3

a

+

b

+

c

=

2+1

(2)

d

+(1)

a

=

2+1

(2)

e

+(1)

b

=

2+1

(2)

f

+(1)

c

If G is the point whose position vector is

g

, then from the above equation it is clear that the point G lies on the medians AB,BE,CF and it divides them internally in the ratio 2:1.

Hence, the medians of a triangle are concurrent.

Answered by ljparrone
0

Answer:

deciphered version of first answer, probably won't even be helpful but oh well

Step-by-step explanation:

Let A,B and C be vertices of a triangle.

Let D,E and F be the midpoints of the sides BC,AC and AB respectively. Let OA = a , OB = b , OC=  c ,OD  = d = OE = e and OF = f be position vectors of points A,B,C,D,E and F respectively. Therefore, by Midpoint formula, d = 2 b + c , e =2  a + c and f = 2 a + b ∴2 d = b + c,2 e = a + c and 2 f = a + b ∴2 d +a = a + b + c , 2 e + b = a + b + c , 2 e + b = a + b + c 2 f + c = a + b +c  Now, 3 2d + a=3 2 e+  b = 32 f +c  = 3 a+ b  c Let g = 3 a + b + c . Then, we have g =  a + b + c = 2+1 (2) d +(1) a = 2+1 (2) E +(1) B = 2+1 (2) F +(1)  C If G is the point whose position vector is g , then from the above equation it is clear that the point G lies on the medians AB,BE,CF and it divides them internally in the ratio 2:1. Hence, the medians of a triangle are concurrent.

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