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总结帝笔记—初三寒假班第三讲

Problem 51. Diagram showing trapezoid ABCD where AD is parallel to BC. Sides AB and DC are extended to meet at M, forming triangle MBC. Given lengths: AD=8, AB=4, DC=5, BC=10. Dashed lines MA and MD are shown. A construction is shown inside the trapezoid: DP is perpendicular to BC (P on BC). A red dashed line DQ is shown (Q on BC) labeled 4. Points Q and P divide BC such that QP=2 and PC=2.

看到“平行”,再看到周长

相似比=周长比

\therefore AD \parallel BC \therefore \triangle MAD \sim \triangle MBC

\therefore \text{相似比为} \frac{AD}{BC} = \frac{8}{10} = \frac{4}{5} = \frac{MA}{MB} = \frac{MD}{MC}

\therefore \frac{\triangle MAD \text{周长}}{\triangle MBC \text{周长}} = \frac{4}{5} \quad \text{①}

\therefore \frac{MA}{MB} = \frac{4}{5} \therefore \frac{MA}{MA+4} = \frac{4}{5} \therefore MA = 16

\frac{MD}{MC} = \frac{4}{5} \therefore \frac{MD}{MD+5} = \frac{4}{5} \therefore MD = 20

\therefore \triangle MAD \text{周长} = MA + AD + MD = 16 + 8 + 20 = 44 \quad \text{②}

\therefore \text{由①②可知} \triangle MBC \text{周长} = 55

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