Question : What is the radius of the circle that circumscribes the triangle ABC whose sides are 16, 30, and 34 units respectively?
Option 1: 16 Units
Option 2: 17 Units
Option 3: 28 Units
Option 4: 34 Units
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Correct Answer: 17 Units
Solution : Dimension of triangle = 16, 30, 34 units In triangle ABC, $\because$ 34 2 = 16 2 + 30 2 This means it is a Pythagorean triplet and the triangle ABC is a right-angle triangle. $\therefore$ Radius of circle which circumscribes the triangle ABC= $\frac{\text{Hypotenuse}}{2}$ = $\frac{34}{2}$ = 17 units. Hence, the correct answer is 17 Units.
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Question : The midpoints of sides AB and AC of the triangle ABC are, respectively, X and Y. If (BC + XY) = 12 units, then the value of (BC – XY) is:
Question : D, E, and F are the midpoints of the sides BC, CA, and AB, respectively of a $\triangle ABC$. Then the ratio of the areas of $\triangle DEF$ and $\triangle ABC$ is:
Question : $D$ and $E$ are points on the sides $AB$ and $AC$ respectively of $\triangle ABC$ such that $DE$ is parallel to $BC$ and $AD: DB = 4:5$, $CD$ and $BE$ intersect each other at $F$. Find the ratio of the areas of $\triangle DEF$ and $\triangle CBF$.
Question : ABC is an equilateral triangle. If the area of the triangle is $36 \sqrt{3}$, then what is the radius of the circle circumscribing the $\triangle ABC$?
Question : ABC is an isosceles right-angle triangle. $\angle ABC = 90 ^{\circ}$ and AB = 12 cm. What is the ratio of the radius of the circle inscribed in it to the radius of the circle circumscribing $\triangle ABC$?
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