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As shown in the figure, P is a point on ▱ ABCD. Given s △ ABP = 3 and s △ PCD = 1, the area of parallelogram ABCD is () A. 6B. 8C. 10d. Not sure
∵ s △ BPC = height on 12bc × BC side, area of parallelogram ABCD = height on BC × BC side, ∵ s △ BPC = 12s parallelogram ABCD, s △ BPC = area of s shadow part, so area of parallelogram ABCD = 2S, area of shadow part = 8
Known: as shown in the figure, point P is a point in ▱ ABCD, the areas of △ PAB and △ PCD are recorded as S1 and S2 respectively, and the area of ▱ ABCD is recorded as S. try to explore the relationship between S1 + S2 and s
A: S1 + S2 = 12s. It is proved that if we make ef ‖ AB, ∫ ab ‖ CD, ∫ EF ‖ CD through P point, then S1 = 12S ▱ abef, S2 = 12S ▱ EFDC, ∫ s ▱ abef + s ▱ EFDC = s, ∫ S1 + S2 = 12s
As shown in the figure, in the pyramid s-abcd whose bottom is a diamond, ABC = 60 °, SA = AB = a, sb = SD = 2Sa, point P is on SD, and SD = 3pd. (1) prove that SA ⊥ plane ABCD; (2) let e be the midpoint of SC, prove that be ∥ plane APC
Proof: (1) proof: because the bottom surface ABCD is rhombic, ∠ ABC = 60 °, so AB = AC = ad = a in △ SAB, from SA2 + AB2 = 2A2 = SB2, we know SA ⊥ AB, the same as SA ⊥ ad. so sa ⊥ plane ABCD (6 points) (2) connect BD, let BD and AC intersect o, connect OP, O obviously divide BD equally, take the midpoint m of SP
In the parallelogram ABCD, m and N are the midpoint of DC and BC respectively. The known vector am = vector C and vector an = vector D. try to use vector C and vector D to represent vector M
In the parallelogram ABCD, m and N are the midpoint of DC and BC respectively. The known vector am = vector C and vector an = vector D. try to use vector C and vector D to represent vector Mn, vector AB and vector ad
From the meaning of the title
Vector AD + vector AC = 2C
Vector AB + vector AC = 2D
And vector AB + vector ad = vector AC
Simultaneous equations
Vector ad = 2 / 3D + 4 / 3C
Vector AB = 8 / 3d-2 / 3C
RELATED INFORMATIONS
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