Answer:
Use the Pythagorean theorem for right triangles to solve this
a^2 + b^2 = c^2 where 'a' and 'b' are the legs of the triangle and 'c' is the hypotenuse
so:
8^2 + b^2 = 17^2 solve for the second leg, b
64 + b^2 = 289
b^2 = 289 - 64= 225
b = 15
- Here, both the triangles have equal sides.
- So, the two triangles are related by SSS.
- If all the sides of a triangle are equal to all the sides of another triangle, so the triangles are congruent to each other.
<u>Answer</u><u>:</u>
<u>SSS,</u><u> </u><u>congruent.</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.
Answer:
3/16
Step-by-step explanation:
Monica has a 24-inch square frame.
There are 4 equal sides.
Hence, length of each side=24/4 inch
= 6 inch
She paints 3/4 of one side of the frame blue.
i.e. she paints
inch
Total part to be painted=24 inch
Hence, fraction of the frame Monica painted blue is:

Answer:
y= x + 4 is the answer
Step-by-step explanation:
y-int= 4
slope = 1
So it the first choice
Answer:
Part a) The lateral area is 
Part b) The area of the two bases together is 
Part c) The surface area is 
Step-by-step explanation:
we know that
The surface area of a right cylinder is equal to

where
LA is the lateral area
B is the area of the base of cylinder
we have


Part a) Find the lateral area
The lateral area is equal to

substitute the values


Part b) Find the area of the two bases together
The area of the base B is equal to

so
the area of the two bases together is

Part c) Find the surface area of the cylinder

we have


substitute
