Option 1-16.
In a regular quadrangular pyramid SABCD, the base ABCD is a square with side 6, and side rib is equal to 9. Point M is marked on the edge SA so that SM=6. a) Construct a section of the pyramid with a plane passing through points B, C and M. b) Find the distance from the top to the plane VSM.
Solution. The base of our pyramid is a square ABCD with side 6, vertex S is projected into the center of the square - point O, the faces are equal isosceles triangles with bases 6 and sides 9.
a) Let's construct a cross section of the HSM. If two planes have common point, then they intersect along a straight line passing through this point. The cutting plane will intersect the base of the pyramid along the straight line BC, the face SAB - along the straight line VM (see Fig. 1). The straight line of intersection of the cutting plane with the face SAD will pass through point M. How? Parallel to AD. Why? If straight line MN were not parallel to AD, then it would have to intersect AD at a point belonging to straight line BC (after all, all points of intersection of the cutting plane with the plane of the base lie on straight line BC), but this is impossible, because AD II BC. We draw MN II AD and connect point N to point C. A section of the pyramid with a plane passing through points B, C and M has been constructed and represents isosceles trapezoid BMNC with bases BC and MN.
b) Let's find the distance from the vertex S to the plane of the HSM. Let's carry out the construction: draw EF II AB (note that E is the middle of AD, F is the middle of BC). Let us connect points E and F with vertex S. The plane SEF will intersect the trapezoid BMNC along the straight line KF, which is the axis of symmetry of the trapezoid (Fig. 2). The distance from point S to the section plane will be the height of the triangle SKF drawn to side KF. The construction of this perpendicular will depend on the value of the angle SKF (we denote it by α). If angle α is acute, then the altitude of triangle SKF will lie inside the triangle. If the angle α is obtuse, then it is outside the triangle. Using the cosine theorem, we determine the cosine of the angle α in the triangle SKF.
SF – height and median of an isosceles Δ SBC with side SB=9 and base BC=6. SF 2 = SB 2 - BF 2 = 8 1- 9 = 72. The faces SAD and SBC are equal, therefore:
Let's find SK (Fig. 3). Let's determine the angle φ.
In a rectangular ΔMKS the hypotenuse is SM = 6, then MK = SM ∙ cosφ; MK = 2. SK = SM ∙ sinφ (can also be found using the Pythagorean theorem).
From ΔMAB we use the cosine theorem to find MV.
MV 2 = MA 2 + AB 2 – 2 ∙ MA ∙ AB ∙ cos∠MAB; note that cos∠MAB=φ.
Consider the trapezoid BMNC (Fig. 4). Let's do MP⟘BC. MK=2, BF=3, BP=1. From the right triangle BPM according to the Pythagorean theorem:
MP 2 = MV 2 – BP 2 = 33-1 = 32.
Consequently, angle α is obtuse, and the height ΔSKF will lie outside the triangle. Let's construct ST ⊥ KF and find the length ST - the leg of the right triangle TKS opposite the angle (π-α). ST = SK ∙ sin(π-α) = SK ∙ sinα. Knowing the cosine α, we find the sine α.
Option 1-17.
Solve the inequality: log 3 (9 x +16 x -9∙4 x +8)≥2x.
Solution. Let's represent the right side as a logarithm to base 3:
log 3 (9 x +16 x -9∙4 x +8)≥log 3 3 2x . This inequality will be true if the following conditions are met:
9 x +16 x -9∙4 x +8≥3 2 x and 9 x +16 x -9∙4 x +8>0. Since 3 2 x >0, only the first of the inequalities can be solved. Let's write it in the form: 3 2 x +4 2 x -9∙4 x +8≥3 2 x ; transform and get: 4 2 x -9∙4 x +8≥0. Let's make a replacement: 4 x =y. Let's solve the inequality: y 2 -9y+8≥0. The zeros of the trinomial y 2 -9y+8 are y 1 =1, y 2 =8. The inequality will be true for y<1 и y>8. But y=4 x, and 4 x >0 for any x. Therefore, 4 x will belong to the union of intervals (0; 1] and and, "en":["8-yYYSdzGpE"],"de":["WdBtNfapbHk","l4g4yGpfqIc"],"es":["fc4otil8UTE" ],"pt":["QruxFE8Ouno","eIt-AzICZrM","_EVKxSIiQHg"],"pl":["iplrpLjcmnw","TVgaedyu_zc"],"ro":["7COIPYhkWn8","7COIPYhkWn8"], "lt":["xDsdCWxakxk","7ieqsOukivc","xBJq0QP6o0A","SvpFoOsSxjk","551rtBJyYe0"],"el":["KhipRx4bM4Y"])