Answer:
r = 12
Step-by-step explanation:
From the figure attached,
QP is a tangent to the circle O at the point P.
Therefore, by the property of tangency,
OP ⊥ QP
By applying Pythagoras theorem In right triangle QPO,
(Hypotenuse)² = (Leg 1)² + (Leg 2)²
(OQ)² = (OP)² + (PQ)²
(25 + r)² = (35)² + r²
625 + r² + 50r = 1225 + r²
50r = 1225 - 625
50r = 600
r = 12
Therefore, r = 12 units is the answer.
The greatest number of the sets you can make are 32 sets
Answer:
All 3 are right triangles.
Step-by-step explanation:
In a right triangle c^2 = a^2 + b^2 where c is the longest side. ( by Pythagoras).
a) 3^2 = 9, 6^2 = 36 and √27 ^ 2 = 27
36 = 9 + 27
So this is a right triangle.
b)
17^2 = 289, 15^2 = 225 and 8^2 = 64
289 = 225 + 64
Also right triangle.
c) √50 ^ 2 = 50, 5^2 = 25 and 5^2 = 25
So also right triangle.
Answer: A 24 cm piece of string is cut into two pieces , one piece is used to form a circle and the other piece is used to form a square.
How should this string be cut so that the sum of the areas is a minimum .
:
Let x = the circumference of the circle
then
(24-x) = the perimeter of the square
:
Find the area of the circle
find r
2*pi*r = x
r =
Find the area of the circle
A =
A =
A = sq/cm, the area of the circle
:
Find the area of the square
A = sq/cm the area of the square
The total area
At = +
Graph this equation, find the min
Min occurs when x=10.6 cm
cut string 10.6 cm from one end
Step-by-step explanation: Hope I help out alot (-: :-)
