Question

1. what are the perimeter and area of △xyz? show your work here. options: p = 28; a = 540; p = 42; a = 84; p = 50; a = 216; p = 54; a = 57

Explanation:

Step1: Find the lengths of XY, YZ, and XZ

From the diagram, the diagonals of the kite (which is related to the triangle) intersect at right angles? Wait, actually, in triangle XYZ, we can see that the segments from the intersection: for XY, we have a right triangle with legs 5 and 12. So by Pythagoras, \( XY = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \). Similarly, YZ: legs 9 and 12, so \( YZ = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \). And XZ: Wait, no, triangle XYZ has sides XY, YZ, and XZ? Wait, no, looking at the diagram, XZ is composed of 5 + 9 = 14? Wait, no, maybe I misread. Wait, the diagonals: one diagonal is split into 5 and 9, the other into 12 and 12 (so total 24). Wait, triangle XYZ: let's see, points X, Y, Z. The distance from X to Y: right triangle with legs 5 and 12, so 13. Y to Z: right triangle with legs 9 and 12, so 15. X to Z: 5 + 9 = 14? Wait, no, that can't be. Wait, maybe the triangle is isoceles? Wait, no, let's check the options. Wait, the perimeter options: let's recalculate.

Wait, maybe the triangle XYZ has sides XY, YZ, and XZ. Wait, XY: from X to Y, the horizontal segment is 12, vertical is 5, so \( XY = \sqrt{5^2 + 12^2} = 13 \). YZ: from Y to Z, horizontal is 12, vertical is 9, so \( YZ = \sqrt{9^2 + 12^2} = 15 \). XZ: from X to Z, vertical segment is 5 + 9 = 14? Wait, no, that's not a right triangle. Wait, maybe the diagonals are perpendicular? Wait, the diagram shows that the diagonals (the lines crossing) are perpendicular? So the triangle XYZ: the base is the length of the horizontal diagonal (12 + 12 = 24), and the height is 5 + 9 = 14? No, area of a triangle is (base * height)/2. Wait, maybe the triangle is composed of two right triangles? Wait, no, let's look at the options.

Wait, let's recalculate perimeter: if XY = 13, YZ = 15, and XZ: wait, maybe XZ is 14? No, 13 + 15 + 14 = 42, but that's one of the options, but area would be (24 14)/2 = 168, which is not 84. Wait, maybe the base is 24, and the height is 7? No. Wait, maybe the diagonals are 12 (half is 12) and 14 (5 + 9). Wait, no, the horizontal diagonal is 12 + 12 = 24, vertical is 5 + 9 = 14. Then area of triangle XYZ: (24 14)/2 = 168. Not matching. Wait, maybe the triangle is XYZ with sides XY, YZ, and XZ, where XY and YZ are equal? No, 13 and 15 are not equal. Wait, maybe I made a mistake. Let's check the options again.

Wait, another approach: the triangle XYZ. Let's see, the two right triangles: X to the intersection point (let's call it O) is 5, O to Y is 12, so XY = 13. Y to O is 12, O to Z is 9, so YZ = 15. X to O is 5, O to Z is 9, so XZ = 14? No, 5 + 9 = 14. Then perimeter is 13 + 15 + 14 = 42. Area: the triangle can be divided into two triangles: XYO and ZYO. Area of XYO: (12 5)/2 = 30. Area of ZYO: (12 9)/2 = 54. Total area: 30 + 54 = 84. Ah! That's it. So perimeter is 13 + 15 + 14? Wait, no, XZ is not 14. Wait, X to Z: is that a straight line? Wait, O is the intersection point, so XO is 5, OZ is 9, so XZ is 5 + 9 = 14. Then XY is 13, YZ is 15, XZ is 14. Perimeter: 13 + 15 + 14 = 42. Area: (12 5)/2 + (12 9)/2 = 30 + 54 = 84. Which matches the option P = 42; A = 84.

Step2: Verify the calculations

• XY: \( \sqrt{5^2 + 12^2} = 13 \)
• YZ: \( \sqrt{9^2 + 12^2} = 15 \)
• XZ: 5 + 9 = 14 (since XO = 5 and OZ = 9, and XOZ is a straight line)
• Perimeter: 13 + 15 + 14 = 42
• Area: Area of \( \triangle XYO \) + Area of \( \triangle ZYO \) = \( \frac{1}{2} \times 12 \times 5 + \frac{1}{2} \times 12 \times 9 = 30 + 54 = 84 \)

Answer:

P = 42; A = 84 (the option with P = 42; A = 84)