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OO in trapezoidal ABCD, AB / / CD, diagonal AC and BD intersect at O, the area of triangle AOB is 4, the area of triangle cod is 9, find the quadrilateral of ABCD
∵ ab ∥ CD, ∥ AOB ∥ cod ∥ s ∥ AOB: s ∥ cod = OA ∥ 178;: OC ∥ 178; = 4:9 ∥ OA: OC = 2:3 because the height of OA on △ AOB side is equal to that on OC on △ BOC side ∥ s ∥ AOB: s ∥ BOC = OA: OC = 2:3 ∥ s ∥ BOC = 3 / 2S ∥ AOB = 6
In known trapezoidal ABCD, ab ‖ CD, diagonal lines AC and BD intersect at point O, the area of triangle AOB is m, and the area of triangle cod is n, so s-abcd can be obtained
Did not learn similar, is the second semester of the trapezoidal problem!
Let s ⊿ oad = s ⊿ OBC = P, then Bo / do = m / P = P / N P = √ (MN)
S (ABCD) = m + N + 2 √ (MN)
The area of isosceles trapezoid ABCD is 68, the diagonal lines AC and BD of AB and CD intersect with O, and the height is 8. The area of triangle AOB is 10, and the area of triangle cod is calculated
Let AB = A and CD = B get a + B = 17 from (a + b) * 8 / 2 = 68, because triangle AOB is similar to cod. From the ratio of corresponding sides of similar triangle equal to the ratio of corresponding height, we can know that AOB height / COD height = A / B, so AOB height = 10A. / (a + b), so AOB area = (1 / 2) * a * [10A / (a + b)] =
Known: as shown in the figure, in ladder ABCD, ad is parallel to BC, AC and BD intersect with point O, indicating that s △ AOB is equal to s △ cod
It is proved that AE ⊥ BC and DF ⊥ BC pass through point a and D respectively
∵AE⊥BC,DF⊥BC,AD∥BC
∴AE＝DF
∵S△ABC＝BC×AE/2,S△DBC＝BC×DF/2
∴S△ABC＝S△DBC
∵S△ABC＝S△AOB+ S△BOC,S△DBC＝S△COD+ S△BOC
∴S△AOB＝S△COD
As shown in the figure, in the trapezoidal ABCD, ad ‖ BC, diagonal AC and BD intersect at point O, asking whether △ AOB and △ COD are similar? There is a student's answer: ∵ ad ∥ BC, ∵ ADO = ∥ CBO, ∥ Dao = ∥ BCO. ∥ AOD ∥ BOC. ∥ Aobo = DOCO. And ∵ AOB = ∥ doc, ∥ AOB ∥ cod. Please judge whether the student's answer is correct and explain the reason
Incorrect. The reason for the error is that Aobo = DOCO is obtained from △ AOD ∽ BOC, and the positive solution is: ∵ AOD ∽ BOC, ∽ AOCO = dobo, so △ AOB ∽ cod can not be further deduced
Isosceles trapezoid ABCD, AD / / BC, ab = DC, the intersection of diagonals is O, the area of triangle AOD is 10, and the area of triangle cod is 20. It is proved that the angle ABC is 60 degrees
The history of painted pottery is isosceles trapezoid ABCD, AD / / BC, ab = DC, the intersection of diagonal is O, the area of triangle AOD is 10, and the area of triangle cod is 20, which proves that the angle ABC = 60 degrees
In the trapezoidal ABCD, AD / / BC, AC and BD intersect with o point, if s △ AOD = 1, s △ cod = 4, the area of the trapezoidal ABCD is calculated
S △ ADB = s △ DAC (same base and same height) minus s △ AOD = 1
=> S△COD=S△AOB=4
S△ COD:S △AOD=CO:OA=S△ BOC:S △AOB=4:1=S△ BOC:4 = >S△BOC=16
so S[ABCD]=1+4+4+16=25
A round flower bed with a diameter of 16m is surrounded by a 2m wide path. How many square meters is the area of the path?
The area of the path is the ring part
Big circle R: 16 divided by 2 + 2 = 10m
Small circle R: 16 divided by 2 = 8M
S ring = 3.14 * 10 * 10-3.14 * 8 * 8
=3.14*36
=113.04 square meters
The circular flower bed is surrounded by a circular path with a diameter of 10 meters and a width of 2 meters. How many square meters does this circular path cover?
3.14 × (10 ／ 2 + 2) 2-3.14 × (10 ／ 2) 2, = 3.14 × 49-3.14 × 25, = 3.14 × (49-25), = 75.36 square meters; a: this Circular Road covers an area of 75.36 square meters
A circular flowerbed is surrounded by a circular path. The diameter of the flowerbed is 10m and the width of the path is 2m. How many square meters do the flowerbed and the path cover?
S flower bed = 3.14 × (10 △ 2) &# 178; = 78.5m & nbsp; &# 178;
S Road = 3.14 × (12 △ 2) &# 178; & nbsp; - & nbsp; 68.5 = 34.54m & nbsp; &# 178;
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