Math Problem Statement

Solution

The problem shows a right triangle with sides labeled as 48, 20, and the hypotenuse (unlabeled). The perimeter of the triangle is given as P=112P = 112. You are asked to show that the calculation (1122)×(112248)×10\left(\frac{112}{2}\right) \times \left(\frac{112}{2} - 48\right) \times 10 gives the correct area for this triangle.

Step-by-Step Solution:

  1. Find the hypotenuse using the Pythagorean theorem: c=482+202=2304+400=2704=52c = \sqrt{48^2 + 20^2} = \sqrt{2304 + 400} = \sqrt{2704} = 52 So, the hypotenuse c=52c = 52.

  2. Check the perimeter: The perimeter PP is the sum of all sides: P=48+20+52=120P = 48 + 20 + 52 = 120 However, the problem states P=112P = 112. Therefore, there might be a mistake in the problem statement, but we'll proceed with the correct perimeter as P=120P = 120.

  3. Halve the perimeter: P2=1202=60\frac{P}{2} = \frac{120}{2} = 60

  4. Find the area: The calculation provided is: (1202)×(120248)×10\left(\frac{120}{2}\right) \times \left(\frac{120}{2} - 48\right) \times 10 Substituting the values: 60×(6048)×10=60×12×10=720060 \times (60 - 48) \times 10 = 60 \times 12 \times 10 = 7200 But this expression doesn't seem to represent the area directly.

  5. Recalculate using Heron's formula: Using Heron's formula, the area AA is: A=s(sa)(sb)(sc)A = \sqrt{s(s-a)(s-b)(s-c)} where s=P2s = \frac{P}{2} is the semi-perimeter. For the correct perimeter P=120P = 120: s=60s = 60 The area can be calculated as: A=60(6048)(6020)(6052)=60×12×40×8=230400=480A = \sqrt{60(60-48)(60-20)(60-52)} = \sqrt{60 \times 12 \times 40 \times 8} = \sqrt{230400} = 480

The original calculation doesn’t match Heron’s formula directly but seems to involve a simplification or alternate approach which isn’t standard for calculating the area. However, the method given seems to be incorrect or incomplete for directly finding the area based on the given data.

Would you like to explore this further, or do you have any questions?


5 Questions for Further Exploration:

  1. What is the significance of Heron's formula in triangle area calculations?
  2. How does the Pythagorean theorem help in verifying the sides of a right triangle?
  3. Can the formula (P2)×(sa)×(sb)×(sc)\left(\frac{P}{2}\right) \times (s - a) \times (s - b) \times (s - c) be simplified in some specific cases?
  4. How might errors in a problem statement, like the incorrect perimeter here, affect the solution process?
  5. What other methods exist to find the area of a triangle if you only have side lengths?

Tip:

Always double-check given values (like the perimeter in this case) against your calculations, as even small mistakes can lead to incorrect assumptions.

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Math Problem Analysis

Mathematical Concepts

Geometry
Pythagorean Theorem
Heron's Formula

Formulas

Pythagorean theorem: a^2 + b^2 = c^2
Heron's formula: A = sqrt(s(s-a)(s-b)(s-c))

Theorems

Pythagorean theorem
Heron's formula

Suitable Grade Level

Grades 9-12