Math Problem Statement

A square of side length 3 meters is modified by removing a triangular region. Given that the base of the triangle is 1 meter and its hypotenuse is 5 meters, what is the remaining area of the modified square?

Solution

Let me extract the mathematical content from the image and provide an analysis. One moment, please.The question in the image is related to geometry and involves two shapes, a square and a triangle within the square. Here's a breakdown:

Extracted Question:

  • Shape 1: A square with a given side length.

  • Shape 2: A square modified by removing a triangular region, as seen in the second diagram.

  • Given:

    • AD=3m,DC=1m,AK=5m|AD| = 3 \, \text{m}, \, |DC| = 1 \, \text{m}, \, |AK| = 5 \, \text{m}
    • KCL\triangle \text{KCL} is removed, and we are to calculate the remaining area of the modified square (Shape 2).
  • The options for the remaining area are: A)52,B)1,C)12,D)112A) \frac{5}{2}, \, B) 1, \, C) \frac{1}{2}, \, D) \frac{1}{12}.

Let me solve this step-by-step.### Solution:

  1. Interpret the given dimensions:

    • AD=3m|AD| = 3 \, \text{m} indicates the side length of the large square.
    • DC=1m|DC| = 1 \, \text{m} represents the base of the triangle KCL\triangle KCL.
    • AK=5m|AK| = 5 \, \text{m} is the hypotenuse of KCL\triangle KCL.
  2. Calculate the area of the square S1S_1: The area of the square S1S_1 is: Area of square=side2=32=9m2.\text{Area of square} = \text{side}^2 = 3^2 = 9 \, \text{m}^2.

  3. Determine the dimensions of KCL\triangle KCL:

    • The triangle is right-angled (as seen from the diagram).
    • Base DC=1m|DC| = 1 \, \text{m}.
    • Hypotenuse AK=5m|AK| = 5 \, \text{m}.
    • Use the Pythagorean theorem to find the height KC|KC|: AK2=DC2+KC2    52=12+KC2.|AK|^2 = |DC|^2 + |KC|^2 \implies 5^2 = 1^2 + |KC|^2. 25=1+KC2    KC2=24    KC=24=26m.25 = 1 + |KC|^2 \implies |KC|^2 = 24 \implies |KC| = \sqrt{24} = 2\sqrt{6} \, \text{m}.
  4. Calculate the area of KCL\triangle KCL: The area of KCL\triangle KCL is: Area of KCL=12baseheight=12126=6m2.\text{Area of } \triangle KCL = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 1 \cdot 2\sqrt{6} = \sqrt{6} \, \text{m}^2.

  5. Calculate the remaining area in S2S_2: Subtract the area of KCL\triangle KCL from the area of the square: Remaining area=Area of squareArea of triangle.\text{Remaining area} = \text{Area of square} - \text{Area of triangle}. Remaining area=96.\text{Remaining area} = 9 - \sqrt{6}.

    However, none of the options directly matches this exact result. Let me review and confirm.


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Math Problem Analysis

Mathematical Concepts

Geometry
Area calculation
Right triangle properties

Formulas

Area of a square = side^2
Area of a triangle = (1/2) * base * height

Theorems

Pythagorean theorem

Suitable Grade Level

Grades 8-10