In this article, we will explore the solutions to exercise 46, 47, and 48 from page 84 of the mathematics textbook for grade 8. These exercises cover the topic of similar right triangles, which is part of the chapter on similar triangles. The purpose of this article is to provide a comprehensive guide with formulas, explanations, and step-by-step solutions to help students excel in mathematics.
Understanding the Concept
To solve these exercises, it is essential to have a solid understanding of the concept of similar triangles. Two right triangles are considered similar if they satisfy the following conditions:
- One acute angle of the first triangle is equal to the corresponding acute angle of the second triangle.
- The ratio of the lengths of the two legs in one triangle is equal to the ratio of the lengths of the corresponding legs in the other triangle.
Identifying Special Cases of Similar Right Triangles
Exercise 46 asks us to identify pairs of similar triangles in a given figure. By applying the concepts mentioned above, we can determine that the right triangles (ΔDEF) and (ΔD’E’F’) are similar since:
- The ratio of their legs is 1/2.
- The triangles share a common acute angle.
To further illustrate this, refer to the image below:
Understanding the Ratios of Altitudes and Areas in Similar Triangles
Exercise 48 deals with the ratios of altitudes and areas in similar triangles. According to the following principles:
- The ratio of two corresponding altitudes in similar triangles is equal to the ratio of their corresponding sides.
- The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
With these principles in mind, let’s tackle exercise 48 together.
Solving Exercise 47
Exercise 47 requires us to calculate the lengths of the sides of triangle (A’B’C’) when triangle ABC has side lengths of 3cm, 4cm, and 5cm, and the area of triangle (A’B’C’) is 54cm².
To solve this, we utilize the following formula:
From this equation, we can deduce that the length of each side of triangle (A’B’C’) is three times the length of the corresponding side of triangle ABC. Therefore, the sides of triangle (A’B’C’) are:
- A’B’ = 9cm
- A’C’ = 12cm
- B’C’ = 15cm
Solving Exercise 48
In exercise 48, we are given the height of a column’s shadow on the ground (4.5m) and the length of a vertical rod’s shadow on the ground (0.6m) when both objects are illuminated by sunlight at the same time. We are asked to find the height of the column.
To solve this exercise, we can use the concept of similar triangles. Suppose the column is represented by line segment (AB) with shadow (AC), and the rod is represented by line segment (A’B’) with shadow (A’C’).
Since both the column and the rod are perpendicular to the ground, triangles (ABC) and (A’B’C’) are both right triangles. Furthermore, the angle of elevation from the ground for both shadows is the same. Therefore, we can conclude that triangles (ABC) and (A’B’C’) are similar.
By utilizing the ratio between the sides of similar triangles, we can determine that the height of the column (AB) is equal to the product of the height of the rod (A’B’) and the length of the column’s shadow divided by the length of the rod’s shadow.
Using the given measurements, we find that the height of the column is 15.75m.
Wrapping it Up
We hope this guide has provided you with a clear understanding of how to solve exercises 46, 47, and 48 from page 84 of the mathematics textbook for grade 8. Remember to carefully read each question and apply the relevant formulas and concepts to arrive at the correct solutions.
For additional practice, you can explore the exercises on our website Kienthucykhoa.com, which offers comprehensive solutions and step-by-step explanations for a wide range of mathematical problems. Good luck with your studies, and remember that practice makes perfect!
