This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Haya hapa masuluhisho ya maswali uliyotoa: 13. Determine: (i) The length of XY Step 1: Identify the relevant triangle. From the previous problem, we know the pyramid has a square base PQRS of side 10 cm. Y is the apex, and X is the midpoint of RS. YS is a slant edge with length 13 cm. We need to find the length of XY, which is the slant height of the triangular face YRS. Consider the triangle YRS. Since YR = YS = 13 cm (slant edges of a right pyramid), YRS is an isosceles triangle. X is the midpoint of RS, so YX is the altitude from Y to RS. Therefore, YXS is a right-angled triangle at X. Step 2: Apply the Pythagorean theorem. In right-angled triangle YXS: YS^2 = YX^2 + XS^2 We know YS = 13 cm and XS = (1)/(2) × RS = (1)/(2) × 10 = 5 cm. 13^2 = YX^2 + 5^2 169 = YX^2 + 25 YX^2 = 169 - 25 YX^2 = 144 YX = sqrt(144) YX = 12 cm The length of XY is 12 cm. (ii) The angle between the planes RSY and QRST. Step 1: Identify the line of intersection and perpendiculars. The plane RSY is a triangular face of the pyramid. The plane QRST is the base of the pyramid. The line of intersection of these two planes is RS. To find the angle between the planes, we need to find two lines, one in each plane, that are perpendicular to the line of intersection (RS) at the same point. In plane RSY, YX is perpendicular to RS (as calculated in part (i), YX is the slant height to the base edge RS). In plane QRST (the base), OX is perpendicular to RS. (O is the center of the square base, and X is the midpoint of RS. OX is parallel to PQ and perpendicular to RS). Therefore, the angle between the planes RSY and QRST is YXO. Step 2: Determine the lengths of the sides of YXO. OY is the perpendicular height of the pyramid. From the previous problem (or by calculation): The diagonal of the base is PR = sqrt(10^2 + 10^2) = sqrt(200) = 10sqrt(2) cm. OS = (1)/(2) PR = (1)/(2) × 10sqrt(2) = 5sqrt(2) cm. In right-angled triangle OYS: OY^2 + OS^2 = YS^2 OY^2 + (5sqrt(2))^2 = 13^2 OY^2 + 50 = 169 OY^2 = 119 OY = sqrt(119) cm. OX is half the side length of the square base: OX = (1)/(2) × 10 = 5 cm. YX is the slant height calculated in part (i): YX = 12 cm. Step 3: Calculate the angle YXO. Consider the right-angled triangle YXO (right-angled at O). We can use the cosine ratio: ( YXO) = AdjacentHypotenuse = (OX)/(YX) ( YXO) = (5)/(12) YXO = ((5)/(12)) YXO ≈ 65.3756^ Rounding to one decimal place: The angle between the planes RSY and QRST is 65.4^. 14. ABCD is a square of side 5cm. A point P moves inside this square so that APB 90^. Show by construction, the locus of P. Step 1: Understand the condition. The condition APB 90^ means that point P forms an angle with points A and B that is greater than or equal to 90^. The locus of points P such that APB = 90^ is a circle with AB as its diameter. If APB > 90^, point P lies inside the circle with diameter AB. Therefore, the locus of P satisfying APB 90^ is the region on or inside the circle with diameter AB. Step 2: Combine with the square boundary. The point P must also move inside the square ABCD. So, the locus of P is the region within the square that is also on or inside the circle with diameter AB. This region is a segment of the circle. Step 3: Construction steps. 1. Draw the square ABCD with side length 5 cm. 2. Locate the midpoint of the side AB. Let's call this point M. 3. Using M as the center and MA (or MB) as the radius, draw a semicircle that passes through A and B and lies inside the square. 4. The locus of P is the region bounded by the side AB and the arc of the semicircle drawn. This region should be shaded or clearly indicated. **15. Without using tables, find the value of x in the equation. $ x^3 +