the area of triangle pqr is 25.the median of trapezoid pqrs is 14.base rs measures 18. find length of pr.the height to base pq of triangle pqs. the height of trapezoid of pqrs. the area of trapezoid pqrs

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the area of triangle pqr is 25.the median of trapezoid pqrs is 14.base rs measures 18. find length of pr.the height to base pq of triangle pqs. the height of trapezoid of pqrs. the area of trapezoid pqrs

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|dw:1436890183946:dw|

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|dw:1436917730699:dw|
@GuidenceHelp Do you still need help with this problem?
Yes I still don't understand....
To solve this problem, you need to do one part at a time in the order they are asked. a. find the length of base PQ A trapezoid has 4 sides. Two opposite sides are parallel and are called the bases. The other two sides are not parallel and are called the legs. In a trapezoid, the median is the segment that connects the midpoints of the legs. Also, the length of the median is the average of the length of the legs.
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The figure above shows trapezoid ABCD with parallel sides AD and BC. Point M is the midpoint of leg AB, and point N is the midpoint of leg CD. Segment MN is the median of trapezoid ABCD. The length of the median, MN, is the average of the lengths of the bases, AD and BC. \(MN = \dfrac{AD + BC}{2} \)
Now we are ready to answer part a.
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We are told that the median of trapezoid PQRS is 14. Base SR is 18. We can find the length of base PQ. \(MN = \dfrac{SR + PQ}{2} \) \(14 = \dfrac{18 + PQ}{2} \) \(28 = 18 + PQ\) \(10 = PQ\) \(PQ = 10\) This is the answer to part a.
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a. find the height of triangle PQS to base PQ A height of a triangle is a segment and the length of the segment that has as endpoints a vertex of the triangle and the opposite side or the line that contains the opposite side. The height is also perpendicular to the opposite side or the line that contains it.
Every triangle has three heights. A height can be inside the triangle, outside the triangle, or a side of the triangle. Here are three triangles. Each shows one of the three heights. |dw:1437101900495:dw|
For part b, we need the height of triangle PQS that goes from vertex S to side PQ. The height is perpendicular to line PQ. |dw:1437102107225:dw|
h in the above figure shows the height of triangle PQS drawn to side PQ.
Now we need to find its length. We recall the formula for the area of a triangle: \(A= \dfrac{bh}{2} \) Notice that the problem gives the area of triangle PQS. Using the area of triangle PQS and the length of the base PQ, we can find the height of triangle PQS. \(A= \dfrac{(PQ)h}{2} \) \(25= \dfrac{10h}{2} \) \(50 = 10h\) \(5 = h\) \(h = 5\) The height of triangle PQS drawn to side PQ is 5. That answers part b.
c. the height of the trapezoid The height of a trapezoid is the segment and the length of the segment that joins both bases and is perpendicular to the bases. In this problem, it turns out that the height of triangle PQ from part b is also the height of the trapezoid. The height of the trapezoid is 5. This answers part c.
d. the area of a trapezoid is given by this formula \(\large A_{trapezoid} = \dfrac{B + b}{2} h\) where B = length of larger base b = length of shorter base h = height \(A = \dfrac{18 + 10}{2} \times 5\) \(A = 14 \times 5\) \(A = 70\)

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