answer:1. **Identify the Given and Required Information:** - We have a square (ABCD) with side length 12 cm. - A point (P) is chosen such that (triangle PAB) is isosceles with (overline{PA} = overline{PB} = 10) cm. - We need to find the area of (triangle PAC). 2. **Draw Perpendicular From Point (P) to Side (overline{AB}):** - Let (E) be the foot of the perpendicular from (P) to (overline{AB}). Therefore, (PE perp AB). 3. **Determine the Lengths of Segments (overline{AE}) and (overline{EB}):** - Since (triangle PAB) is isosceles and (PE perp AB), (E) is the midpoint of (overline{AB}). - Consequently, (overline{AE} = overline{EB} = frac{AB}{2} = frac{12}{2} = 6 text{ cm}). 4. **Use the Pythagorean Theorem to Find (PE):** - In the right triangle ( triangle PEA ): [ PE^2 = PA^2 - AE^2 = 10^2 - 6^2 = 100 - 36 = 64 ] - Thus, ( PE = sqrt{64} = 8 text{ cm}). 5. **Calculate the Area of (triangle PAB):** - The area of an isosceles triangle can be found using the formula for a right triangle: [ S_{triangle PAB} = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 12 times 8 = 48 text{ cm}^2 ] 6. **Calculate the Area of (triangle PBC):** - Since (overline{BC} = 12) cm and (P) is directly above the midpoint of (AB), (P) lies on the perpendicular bisector of (overline{BC}) in the middle of square: - So, the area is: [ S_{triangle PCB} = frac{1}{2} times 12 times 6 = 36 text{ cm}^2 ] 7. **Combine Areas to Find (S_{triangle PAC}):** - Applying the area subtraction and addition: [ S_{triangle PAC} = S_{triangle PAB} + S_{triangle PCB} - S_{triangle ABC} ] - Substituting the known areas: [ S_{triangle PAC} = 48 + 36 - frac{1}{2} times 12 times 12 ] - Calculate the area of the full square (S_{triangle ABC}): [ S_{triangle ABC} = frac{1}{2} times 12 times 12 = 72 text{ cm}^2 ] - Final calculation: [ S_{triangle PAC} = 48 + 36 - 72 = 12 text{ cm}^2 ] # Conclusion: The area of triangle ( triangle PAC ) is ( boxed{12 text{ cm}^2} ).

answer:Let's calculate the amount Nelly paid based on Joe's bid first. Thrice Joe's bid would be: 3 * 160,000 = 480,000 Nelly paid 2000 more than thrice Joe's bid, so: 480,000 + 2,000 = 482,000 Now let's calculate the amount Nelly paid based on Sarah's bid. Four times Sarah's bid would be: 4 * 50,000 = 200,000 Nelly paid 1500 more than four times Sarah's bid, so: 200,000 + 1,500 = 201,500 Since Nelly outbid both Joe and Sarah, her bid must be higher than both amounts calculated from Joe's and Sarah's bids. Therefore, the amount Nelly paid for the painting is the higher of the two calculated amounts, which is boxed{482,000} .

answer:1. Let's first consider the circle that intersects the sides (AB), (BC), (CD), and (DA) of the rectangle (ABCD) at the points (K_1, K_2, L_1, L_2, M_1, M_2, N_1,) and (N_2) respectively, creating four right-angled triangles. The middle points of the hypotenuses of these triangles are (A_0, B_0, C_0,) and (D_0) respectively. 2. The points (K_1) and (K_2) lie on (AB) and (BC) respectively, (L_1) and (L_2) lie on (BC) and (CD) respectively, (M_1) and (M_2) lie on (CD) and (DA) respectively, and (N_1) and (N_2) lie on (DA) and (AB) respectively. This structure implies that the segments (A_0C_0) and (B_0D_0) are formed by the diagonal intersections of these triangles. 3. Notice that (K_1K_2M_2M_1) forms an isosceles trapezoid where (AK_1 = DM_1) and (BK_2 = CM_2). Since (AK_1 + CM_2 = BK_2 + DM_1), projecting these segments onto the side (AB), we have: [ AB - frac{1}{2}(AK_1 + CM_2) ] And projecting onto side (BC), [ AB - frac{1}{2}(BK_2 + DM_1) ] These projections are equal because: [ AB - frac{1}{2}(AK_1 + CM_2) = AB - frac{1}{2}(BK_2 + DM_1) ] 4. Similarly, the projections of these segments onto (BC) are also equal. From the equality of projections, we deduce that the segments (A_0C_0) and (B_0D_0) themselves must also be equal, forming isotropic lines in the rectangular setup. Conclusion: [ boxed{A_0 C_0 = B_0 D_0} ]

answer:To solve this problem, follow the below steps: 1. **Define variables based on the problem statement**: Let (B) represent the number of black pieces and (W) represent the number of white pieces initially in the box. 2. **Analyze the first condition**: After taking out one black piece, the remaining black pieces are (B - 1), and the number of white pieces remains (W). According to the problem, the ratio of black pieces to white pieces becomes (9:7). Therefore, [ frac{B-1}{W} = frac{9}{7} ] Solving for (B), we get: begin{align*} 7(B - 1) &= 9W 7B - 7 &= 9W 7B - 9W &= 7 quad text{(1)} end{align*} 3. **Analyze the second condition**: When the black piece is returned, the number of black pieces is back to (B). Then, one white piece is taken out, so the remaining number of white pieces is (W - 1). According to the problem, the ratio of black pieces to white pieces now becomes (7:5). Therefore, [ frac{B}{W-1} = frac{7}{5} ] Solving for (B), we get: begin{align*} 5B &= 7(W - 1) 5B &= 7W - 7 5B - 7W &= -7 quad text{(2)} end{align*} 4. **Solve the system of equations**: We have obtained two linear equations: [ 7B - 9W = 7 quad text{(1)} ] [ 5B - 7W = -7 quad text{(2)} ] To solve this system, multiply equation (2) by 7 and equation (1) by 5: begin{align*} 7(5B - 7W) &= 7(-7) quad Rightarrow quad 35B - 49W = -49 quad text{(3)} 5(7B - 9W) &= 5(7) quad Rightarrow quad 35B - 45W = 35 quad text{(4)} end{align*} Subtract equation (4) from equation (3): begin{align*} (35B - 49W) - (35B - 45W) &= -49 - 35 -49W + 45W &= -49 - 35 -4W &= -84 W &= 21 end{align*} Substitute (W = 21) back into equation (1): begin{align*} 7B - 9(21) &= 7 7B - 189 &= 7 7B &= 196 B &= 28 end{align*} 5. **Determine the difference**: Calculate the initial difference in the number of black and white pieces: [ B - W = 28 - 21 = 7 ] Conclusion. [ boxed{C} ]