https://file.aiquickdraw.com/mathbot/file/172562682260082x8h3ut.jpg

右の図において、四角形ABCDは正方形である。直線CDについて点Aと反対側に点を、直線CEの上側に2点E，Gを、四角形CEFGが正方形になるようにとる。
これについて、次の問いに答えなさい。
﻿﻿AB=4cm、点Gが辺AD上にあり、CG=5cmであるとき、辺CDを点Dの方に延長した直線上に点pを、PC=CEとなるようにとる。このとき、APDEの面積を求めなさい。 ﻿ 三角形DCE=三角形BCG=2分の1
✕4x4=8（cm） 
PC
=CE= 5cmより，CD：DP=4：（5-4）=4：1だから、三角形PDE=4分の1 
三角形DCE=2（cm）

が答えの解説だが、これになる理由

Let's break down the given problem to clarify the reasoning behind the solution.

### Key Information from the Problem:
- \(ABCD\) is a square with side \(AB = 4 \, \text{cm}\).
- Points \(E\) and \(G\) are positioned on line \(CE\), making \(CEFG\) a square.
- \(CG = 5 \, \text{cm}\), and point \(G\) lies on side \(AD\).
- Point \(P\) is chosen such that \(PC = CE\), and we are asked to find the area of quadrilateral \(APDE\).

### Clarification of Steps:

1. **Identification of Triangles:**
   - Triangle \(DCE\) is mentioned as having half the area of the square \(ABCD\).
   - Similarly, triangle \(BCG\) is also said to have an area of \(8 \, \text{cm}^2\) (half of the square).

2. **Understanding Proportions:**
   - We are told that \(PC = CE = 5 \, \text{cm}\). Since triangle \(DCE\) is isosceles and proportional to triangle \(PDE\), a key relationship between their areas is established.
   - The ratio \(CD:DP = 4:1\) comes from the given geometry, since \(P\) divides \(CD\) in this proportion. Therefore, the area of triangle \(PDE\) is \( \frac{1}{4} \) of triangle \(DCE\).

3. **Area of Triangle \(DCE\):**
   - From the geometry of the problem and knowing that triangle \(DCE\) is half of the area of square \(ABCD\), we can calculate:
     \[
     \text{Area of triangle } DCE = \frac{1}{2} \times 4 \times 4 = 8 \, \text{cm}^2.
     \]

4. **Area of Triangle \(PDE\):**
   - Since triangle \(PDE\) is \( \frac{1}{4} \) of triangle \(DCE\), its area is:
     \[
     \text{Area of triangle } PDE = \frac{1}{4} \times 8 = 2 \, \text{cm}^2.
     \]

5. **Total Area of Quadrilateral \(APDE\):**
   - To find the area of quadrilateral \(APDE\), we sum the areas of triangle \(APD\) and triangle \(PDE\

右の図において、四角形ABCDは正方形である。直線CDについて点Aと反対側に点を、直線CEの上側に2点E，Gを、四角形CEFGが正方形になるようにとる。これについて、次の問いに答えなさい。AB=4cm、点Gが辺AD上にあり、CG=5cmであるとき、辺CDを点Dの方に延長した直線上に点Pを、PC=CEとなるようにとる。このとき、APDEの面積を求めなさい。

This problem involves finding the area of quadrilateral APDE using key geometry concepts. Given a square ABCD, with AB = 4 cm, CG = 5 cm, and triangle DCE, the problem uses triangle properties and proportions to calculate areas. Suitable for students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation