Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
The 4.2 inches base length and 36 in.² area of the given triangle and the 5.6 inches base length of the similar triangle gives the area of the similar triangle as 64 square inches
Which method can be used to find the area of the similar triangle given the dimensions?
Area of a triangle = (Base length × Height)/2
Area of the given triangle = 36 in.²
Base length of the given triangle = 4.2 inches
Base length of the similar triangle = 5.6 inches
Therefore;
Area of the given triangle = (Base length × Height)/2
Which gives;
36 = (4.2 × h)/2
Where;
h = Height of the given triangle
36 × 2 = 4.2 × h
[tex]h = \mathbf{\frac{36 \times 2}{4.2}} = 17 \frac{1}{7} [/tex]
Height of the given triangle, h = 17+ 1/7
The ratio of corresponding sides of similar triangles are the same, which gives;
[tex] \frac{5.6}{4.2} = \frac{h'}{17 \frac{1}{7}} [/tex]
Where;
h' = The height of the similar triangle
Which gives;
[tex] h' = \frac{5.6}{4.2} \times 17 \frac{1}{7} = 22 \frac{6}{7} [/tex]
The area, A', of the similar triangle is therefore;
[tex] A' = \frac{1}{2} \times 5.6 \times 22 \frac{6}{7} = 64 [/tex]
- The area of the similar triangle A' = 64 in.²
The area can also be obtained using the scale factor of area as follows;
- (4.2/5.6)² = 36/A'
Which gives;
- A' = 64 square inches
Learn more about scale factors of calculations here:
https://brainly.com/question/25722260
#SPJ1
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.