Determine: (i) The length of XY (ii) The angle between the planes RSY and QRST.

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), $\triangle YRS$ is an isosceles triangle. X is the midpoint of RS, so YX is the altitude from Y to RS. Therefore, $\triangle YXS$ 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 $XS = \frac{1}{2} \times RS = \frac{1}{2} \times 10 = 5$ cm.

  $$13^2 = YX^2 + 5^2$$ $$169 = YX^2 + 25$$ $$YX^2 = 169 - 25$$ $$YX^2 = 144$$ $$YX = \sqrt{144}$$ $$YX = 12 \text{ cm}$$

  The length of XY is $\boxed{12 \text{ cm}}$.

(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 $\angle YXO$.

• Step 2: Determine the lengths of the sides of $\triangle YXO$. OY is the perpendicular height of the pyramid. From the previous problem (or by calculation): The diagonal of the base is $PR = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}$ cm. $OS = \frac{1}{2} PR = \frac{1}{2} \times 10\sqrt{2} = 5\sqrt{2}$ cm. In right-angled triangle OYS: $OY^2 + OS^2 = YS^2$ $OY^2 + (5\sqrt{2})^2 = 13^2$ $OY^2 + 50 = 169$ $OY^2 = 119 \implies OY = \sqrt{119}$ cm. OX is half the side length of the square base: $OX = \frac{1}{2} \times 10 = 5$ cm. YX is the slant height calculated in part (i): $YX = 12$ cm.

• Step 3: Calculate the angle $\angle YXO$. Consider the right-angled triangle YXO (right-angled at O). We can use the cosine ratio:

  $$\cos(\angle YXO) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{OX}{YX}$$ $$\cos(\angle YXO) = \frac{5}{12}$$ $$\angle YXO = \arccos\left(\frac{5}{12}\right)$$ $$\angle YXO \approx 65.3756^\circ$$

  Rounding to one decimal place: The angle between the planes RSY and QRST is $\boxed{65.4^\circ}$.

14. ABCD is a square of side 5cm. A point P moves inside this square so that $\angle APB \ge 90^\circ$. Show by construction, the locus of P.

• Step 1: Understand the condition. The condition $\angle APB \ge 90^\circ$ means that point P forms an angle with points A and B that is greater than or equal to $90^\circ$. The locus of points P such that $\angle APB = 90^\circ$ is a circle with AB as its diameter. If $\angle APB > 90^\circ$, point P lies inside the circle with diameter AB. Therefore, the locus of P satisfying $\angle APB \ge 90^\circ$ 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.
$\log x^3 +