Tiêu đề MSTE 10 Solutions.pdf ✅

[SOLUTION] From the figure, s = 32.2 → s 2 = 16.1 θ = 1 2 (180° − 360° 5 ) = 54° tan 54° = 16.1 a → a = 16.1 tan 54° sin 54° = 16.1 b → b = 16.1 sin 54° A = π(b 2 − a 2 ) A = π [( 16.1 sin 54°) 2 − ( 16.1 tan 54°) 2 ] A = π(16.1) 2 (csc2 54° − cot2 54°) A = π(16.1) 2 A = 814.332 cm2 ▣ 3. The areas of two similar polygons are 80 and 5. If a side of the smaller polygon has a length of 2, find the length of the corresponding side of a larger polygon. [SOLUTION] ( s1 s2 ) 2 = A1 A2 ( s 2 ) 2 = 80 5 s = 8 ▣ 4. A regular octagon is to be cut out from a square section having a side of 16 cm. Determine the side of the octagon.
[SOLUTION] From the figure, x √2 + x + x √2 = 16 x = 16(√2 − 1) cm or 6.6274 cm ▣ 5. A lot has a frontage of 120 m long along a road. The other sides which are both perpendicular to the road are 90 m and 60 m, respectively. It is desired to subdivide the lot into two parts by another perpendicular line to the road such that the area of the lot that adjoins the 90 m side is equal to 1/3 of the whole area. Determine the length of the dividing line. [SOLUTION]