The area of the trapezoidal ABCD is 80 square centimeters, BO 3DO, OC 3AO find the area of the trian

The area of the triangle AOD is 5cm. Let aod=x because ad bc

So s adb=s adc, so s aob=s doc because bo=3do, so s aob=s doc=3s aod and because ad bc, aod boc, so s boc s aod=(ob od).

Because ob=3od, so s boc=9s aod, so (1+2 3+9)x=80, x=5, that is, the area of the triangle aod is 5cm.

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The area of the trapezoidal ABCD is 32 square meters, and the OA is equal to 1 4AC, so how many square centimeters is the area of the triangular BOC.

Hello, according to the question you described, the answer is as follows: Answer: The solver contains:

In trapezoidal ABCD, the area of AD BC, AOD COB, AOD is 4, and the area of BOC is 9, OD:ob=2:3, S AOD:

s abo=2:3, s abo=6 Therefore, choose c Comment Xinhui: This question examines the determination and properties of similar triangles and the properties of trapezoids This question is not very difficult, pay attention to the application of the idea of combining numbers and shapes.

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The area relationship of the parts within the trapezoid.

Let the area of ABO be S.

The sides of aod and abo are collinear, the vertex a is common, the height on ab is the same, the triangle of the same height, the area ratio = the ratio of the base, s aod:s abo=bo:do=2:5, s aod=5s 2;

Similarly, s boc = 5s 2;

ABO is similar to CDO, similar triangle area ratio = square of similar ratio, so s CDO=25S 4

Sum of all triangle areas = trapezoidal area:

s+5s/2×2+25s/4=98

6s+25s/4=98

49s/4=98

s=98/49×4=8

s△boc=5s/2=20

The answer to this question is 50 square centimeters. The area of the triangular COD is ten square centimeters, the area of the triangular BOC is 20 square centimeters, and the area of the triangular AOD and the triangular AOB are both ten square centimeters.

Taken together, the total area is 50 square centimeters.

3. Because the bases of the triangle adb aod are equal, the area ratio is equal to the ratio of the high, and the comparable aod and boc in the problem are exactly the same because they are exactly the same. So the ratio of the two is = 2:3, that is, the total height is 5, so aod:

AOB = 2:3, AOD is 4, and the area ratio of AOD and BOC = square of the side ratio = high square ratio = 4 9, so BOC = 9

In the third step, because the height of the triangle AOB AOD (from a to the height) is equal, the area ratio is equal to the base ratio.

The fourth step is to go high from b.

AOD and COD are equal in height, so the area ratio is equal to the base ratio. AOD is similar to BOC, and the area of BOC is found.

Let aod=x, because ad bc, so s adb=s adc, (triangles of equal height are equal in area), so s aob=s doc, because bo=3do, so s aob=s doc=3s aod. (The ratio of the area of the contour triangle is equal to the ratio of the base). And because ad bc, aod boc, so s boc s aod=(ob od), because ob=3od, so s boc=9s aod, (the area ratio of similar triangles is equal to the square of the similar ratio), so (1+2 3+9)x=80, x=5, that is, the area of the triangle aod is 5cm.

s△aod=5

Let : s aod=k

Since AOD is the same height as AOB, the ratio of the area is equal to BOod=3, i.e. S AOB=3K

In the same way, AOD and COD are equal to the same height, so the ratio of the area is equal to CO OA=3

i.e. s cod=3k

AOD is similar to BOC because of ab cd, and the similarity ratio is 1:3, so the ratio of area is 1:9

i.e. s boc = 9k

The area of trapezoidal ABCD is 80 square centimeters.

So: s aod+s aob+s cod+s boc=80, i.e. k+3k+3k+9k=80

16k=80

k=5s△aod=k=5

Solution: Because ao:oc=1:

2. In the triangle AOD and DOC, the height from D to AO and OC is equal, so the area of the triangle AOD and DOC is 1:2, that is: the area of the triangle DOC = 2 square centimeters.

In the same way, the area of the triangle AOB is equal to 2 square centimeters. Since the quadrilateral ABCD is trapezoidal, the triangular AOD is similar to the triangular COB. So the area of the triangular cob:

The area of the triangle AOD = (2:1) 2=4, so the area of the triangle COB is 4 square centimeters. The sum of several gives the trapezoidal ABCD with an area of 9 square centimeters.