For AD:
AD=root((c-0)^2 + (d-0)^2)=root((c)^2 + (d)^2)
For BC:
BC=root(((b+c) - b)^2+(d-0)^2)=root((c)^2+(d)^2)
For AB:
AB=root((b-0)^2 + (0-0)^2)=root((b)^2 + (0)^2)=root((b)^2)
For CD:
CD=root((c-(b+c))^2 + (d-d)^2)
CD=root((b)^2 + (0)^2)
CD=root((b)^2)
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.
Answer:
It will still be a frame
Step-by-step explanation:
Answer:

Explanation: We need to find the area of the triangle, We know that the area of any right triangle is:

Therefore in the information given we have the following:

Therefore the Area of this triangle is as follows:

Note! Units are centimeter-squared
Answer:
A) 6Gn + 12 (or 6n+12, if that’s what you meant)
B) 4p²+9p
C) 5r²+20r
Step-by-step explanation:
A. If you meant 6(n+2)...
Multiply 6 by n and also 6 by 2:
6n + 12
B. Multiply p by 4p and also p by 9
4p²+9p
C. Multiply 5r by r and also 5r by 4
5r²+20r
Hope this helped!