Given: △UEP, m∠E = 90°

OM
⊥
EP
, OM=r, UP=c
Find: Perimeter of △EPU

Triangles and Congruency

The perimeter of ΔEPU = 2 × (c+ r )

Step-by-step explanation:

Given OM = r and UP = c  and OM⊥EP

⇒ EM = r ( As OM ⊥EP, OM = r So OM = EM)

If we draw a line joining OP , then we get two triangles ΔOMP and ΔOBC

where B is point on UP and OB is radius

Both ΔOMP and ΔOBC will be congruent, OM and OB are radius.

EP = EM  + MP

⇒ MP = EP -r

If we draw a line joining OU , then we get two triangles ΔOUB and ΔOUA

where A is point on UE and OA is radius

Both ΔOUA and ΔOUB will be congruent, OA and OB are radius.

UE = UA + AE

EA = r ( As OA ⊥ UE, OM = r So OA= EM)

UP = UB + BP

c = UB + BP

Since ΔOUA and ΔOUB are  congruent and ΔOMP and ΔOBC are congruent

⇒ UB = UA and MP = BP

⇒ c = UA + MP

so the perimeter of ΔEPU = EU + UP + EP

⇒ Perimeter = EU + c + EP

⇒ Perimeter = UA + AE + c + EM + MP

⇒ Perimeter = UA + r + c + r + MP

⇒ Perimeter = c+ r + c + r

⇒ Perimeter =  2 × (c+ r )

Hence , the perimeter of ΔEPU = 2 × (c+ r )

ddd

step-by-step explanation:

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92