Question : X, Y, and Z are three equilateral triangles. The sum of the areas of X and Y is equal to the area of Z. If the side lengths of X and Y are 6 cm and 8 cm respectively, then what is the side length of Z?
Option 1: 10 cm
Option 2: 10.5 cm
Option 3: 9.5 cm
Option 4: 9 cm
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Correct Answer: 10 cm
Solution : X, Y, and Z are three equilateral triangles. The side of X = 6 cm The side of Y = 8 cm Area of a equilateral triangle = $\frac{\sqrt{3}}{4}a^2$, where $a$ = side According to the question, AreaX + AreaY = AreaZ $⇒ (\frac{\sqrt{3}}{4} \times 6^2) + (\frac{\sqrt{3}}{4} \times 8^2) = (\frac{\sqrt{3}}{4} \times x^2)$ $⇒ 36 + 64 = x^2$ $⇒ 100 = x^2$ $⇒ x = 10$ cm Hence, the correct answer is 10 cm.
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Question : In $\triangle X Y Z, P$ is a point on side YZ and XY = XZ. If $\angle X P Y=90°$ and $Y P=9\ \text{cm}$, then what is the length of $YZ$?
Option 1: 17 cm
Option 2: 18 cm
Option 3: 12 cm
Option 4: 15 cm
Question : Simplify the given expression. $\frac{x^3+y^3+z^3-3 x y z}{(x-y)^2+(y-z)^2+(z-x)^2}$
Option 1: $\frac{1}{3}(x+y+z)$
Option 2: $(x+y+z)$
Option 3: $\frac{1}{4}(x+y+z)$
Option 4: $\frac{1}{2}(x+y+z)$
Question : If $x(x+y+z)=20$, $y(x+y+z)=30$, and $z(x+y+z)=50$, then the value of $2(x+y+z)$ is:
Option 1: 20
Option 2: –10
Option 3: 15
Option 4: 18
Question : x, y, and z are 3 values, such that x + y = 12, y + z = 17 and z + x = 19. What is the average of x, y, and z?
Option 1: 10
Option 2: 8
Option 3: 6
Option 4: 4
Question : The lengths of the two sides of a triangle are 14 cm and 9 cm. Which of the options below can be the length of the third side?
Option 2: 23 cm
Option 3: 24 cm
Option 4: 5 cm
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