Math Problem Statement

41. The perimeter of a square is 70 cm. Its area (cm²) is: A. 654.75, B. 306.25, C. 4900, D. 1225 42. What is the apothem of a regular polygon having an area of 225 sq.cm. and a perimeter of 60 cm? A. 4.5, B. 5.5, C. 6.5, D. 7.5 43. Find the area of the inscribed circle of a regular hexagon whose side has length 8. A. 50π, B. 48π, C. 60π, D. 80π 44. The median of a triangle is a line drawn from a vertex: A. none of these, B. dividing the triangle into two equal areas, C. bisecting the angle to that vertex, D. to the midpoint of opposite side 45. In any triangle, a circle can be drawn tangent to the three sides. A. false, B. it depends, C. true, D. none of these 46. In any triangle, a circle can be drawn that will pass through the three vertices. A. it depends, B. true, C. it depends, D. false 47. The centroid of a triangle always passes through its: A. median, B. altitude, C. angle bisector, D. side

Solution

Let's go through the problems one by one:


41. The perimeter of a square is 70 cm. Its area (cm²) is:

  • The perimeter PP of a square is given by P=4sP = 4s, where ss is the side length. s=704=17.5cms = \frac{70}{4} = 17.5 \, \text{cm}
  • The area AA of a square is A=s2A = s^2: A=(17.5)2=306.25cm2A = (17.5)^2 = 306.25 \, \text{cm}^2
  • Answer: B. 306.25

42. What is the apothem of a regular polygon having an area of 225 sq.cm. and a perimeter of 60 cm?

  • The area AA of a regular polygon is given by A=12PaA = \frac{1}{2} P a, where PP is the perimeter and aa is the apothem. 225=12×60×a225 = \frac{1}{2} \times 60 \times a Solving for aa: 225=30aa=22530=7.5cm225 = 30a \quad \Rightarrow \quad a = \frac{225}{30} = 7.5 \, \text{cm}
  • Answer: D. 7.5

43. Find the area of the inscribed circle of a regular hexagon whose side has length 8.

  • The radius of the inscribed circle is the same as the apothem of the regular hexagon, which is given by r=32sr = \frac{\sqrt{3}}{2} s, where s=8s = 8: r=32×8=43r = \frac{\sqrt{3}}{2} \times 8 = 4\sqrt{3}
  • The area of the circle is A=πr2A = \pi r^2: A=π×(43)2=π×48=48πA = \pi \times (4\sqrt{3})^2 = \pi \times 48 = 48\pi
  • Answer: B. 48π

44. The median of a triangle is a line drawn from a vertex:

  • A median connects a vertex of the triangle to the midpoint of the opposite side.
  • Answer: D. To the midpoint of opposite side

45. In any triangle, a circle can be drawn tangent to the three sides.

  • This is true for the incircle of a triangle, which is tangent to all three sides.
  • Answer: C. True

46. In any triangle, a circle can be drawn that will pass through the three vertices.

  • This is true for the circumcircle of a triangle, where the circle passes through all three vertices.
  • Answer: B. True

47. The centroid of a triangle always passes through its:

  • The centroid is the point of intersection of the medians of a triangle.
  • Answer: A. Median

Would you like more detailed explanations on any of these questions?

Related Questions:

  1. How do you derive the formula for the area of a square?
  2. What are the properties of a regular polygon with respect to the apothem?
  3. How do you calculate the apothem for other polygons (besides a hexagon)?
  4. What is the difference between the incircle and circumcircle in a triangle?
  5. Can the centroid of a triangle be outside of the triangle?

Tip:

In any regular polygon, the apothem can be found using both the perimeter and area. It is essential to use this when calculating areas more efficiently.

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Math Problem Analysis

Mathematical Concepts

Geometry
Properties of polygons
Circles inscribed in polygons
Properties of triangles

Formulas

Perimeter of a square: P = 4s
Area of a square: A = s²
Area of a regular polygon: A = (1/2) * P * a
Radius of inscribed circle in a regular hexagon: r = (sqrt(3)/2) * s
Area of a circle: A = πr²

Theorems

Median theorem of a triangle
Incircle and circumcircle properties

Suitable Grade Level

Grades 9-12