Question : If the areas of two isosceles triangles with equal corresponding angles are in the ratio of $x^2:y^2$, then the ratio of their corresponding heights is:
Option 1: $x: y$
Option 2: $\sqrt{x}: \sqrt{y}$
Option 3: $x^3: y^3$
Option 4: $x^2: y^2$
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Correct Answer: $x: y$
Solution : The area of a triangle is directly proportional to the square of its height. Given, Ratio of areas = $x^2 : y^2$ $\therefore$ The ratio of their heights = $x:y$ Hence, the correct answer is $x:y$.
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Question : If the ratio of the area of two similar triangles is $\sqrt{3}:\sqrt{2}$, then what is the ratio of the corresponding sides of the two triangles?
Option 1: 9 : 4
Option 2: 3 : 2
Option 3: $\sqrt[3]{3}: \sqrt[3]{2}$
Option 4: $\sqrt[4]{3}: \sqrt[4]{2}$
Question : If the ratio of corresponding sides of two similar triangles is $\sqrt{5}: \sqrt{7},$ then what is the ratio of the area of the two triangles?
Option 1: $\sqrt[3]{5}: \sqrt{7}$
Option 2: $25: 49$
Option 3: $\sqrt{5}: \sqrt{7}$
Option 4: $5: 7$
Question : If $x=\sqrt{–\sqrt{3}+\sqrt{3+8 \sqrt{7+4 \sqrt{3}}}}$ where $x > 0$, then the value of $x$ is equal to:
Option 1: 3
Option 2: 4
Option 3: 1
Option 4: 2
Question : If $(\sqrt 3 +1)^{2}=x+\sqrt 3 y$, then the value of $(x+y)$ is:
Option 1: 2
Option 3: 6
Option 4: 8
Question : If $\cos 21^{\circ}=\frac{x}{y}$, then $(\operatorname{cosec21^{\circ}}-\cos 69^{\circ})$ is equal to:
Option 1: $\frac{x^{2}}{y\sqrt{y^{2}-x^{2}}}$
Option 2:
$\frac{y^{2}}{x\sqrt{y^{2}-x^{2}}}$
Option 3:
$\frac{y^{2}}{x\sqrt{x^{2}-y^{2}}}$
Option 4: $\frac{x^{2}}{y\sqrt{x^{2}-y^{2}}}$
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