SOLUTION: Could you please help with question I cannot copy it because of a picture you will have to see here is the website http://www.analyzemath.com/middle_school_math/grade_8/problems.ht
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Question 558597: Could you please help with question I cannot copy it because of a picture you will have to see here is the website http://www.analyzemath.com/middle_school_math/grade_8/problems.html
Look at question number 11 and 13 or you could just to either.
thank you Answer by Edwin McCravy(20086) (Show Source):
The size of the perimeter of the square ABCD is equal to 100 cm. The length of
the segment MN is equal to 5 cm and the triangle MNC is isosceles. Find the
area of the pentagon ABNMC.
(The picture on that site isn't to scale. This is.)
The square's perimeter is 100 cm, so each side of the square is 25 cm.
The area of the square is (25 cm)² or 625 cm²
We need to subtract the area of the triangle. It is a right triangle, and
it isosceles, so MC = CN. We use the Pythagorean theorem to find MC and CN.
MC² + CN² = MN²
MC² + MC² = 5²
2MC² = 25
MC² = 12.5
MC = = CN
Area of the triangle = BASE×HEIGHT = MC×CN = × = = (12.5) = 6.25 cm²
Area of pentagon = Area of square - Area of triangle
= 625 cm² - 6.25 cm² = 618.75 cm²
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I won't draw the other one.
Initially the rectangular prism on the left was full of water.
It measures 2cm×4cm×10cm
Volume = length × width × height = 2cm × 4cm × 10cm = 80cm³
so the volume of water in the original container is 80cm³
The first container into which part of it was poured to a height of h
is a rectangular prism like the original except that it is only
filled to a height of h cm.
Volume of water = length × width × height = 2cm × 4cm × h cm = 8h cm³
The second container into which the other part of it was poured to a height
of h cm is a cylinder with radius 1 cm.
Volume of water = p × (radius)² × (height) = p(1)²h = ph = (3.14)h
so the volume of water in the cylinder is 3.14h cm³.
water in original = water in 1st + water in 2nd
80 = 8h + 3.14h
80 = 11.14h
= h
7.2 = h
So the height is 7.2 cm to the nearest tenth of a centimeter.
Edwin