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geometry hon-parkes (lms) topic 7: readiness assessment (lms graded) du…

Question

geometry hon-parkes (lms) topic 7: readiness assessment (lms graded) due jan 21 - 11:59 pm triangle def has coordinates d(1, 1), e(1, 5), and f(5, 1). triangle def is the image of \\(\triangle def\\) by a dilation with center (0, 0) and a scale factor of 7. what is df?

Explanation:

Step1: Recall dilation formula

For a dilation with center \((0,0)\) and scale factor \(k\), the image of a point \((x,y)\) is \((kx,ky)\).

Step2: Find coordinates of \(D'\) and \(F'\)

Given \(D(1,1)\) and scale factor \(7\), so \(D'(7\times1,7\times1)=(7,7)\).
Given \(F(5,1)\) and scale factor \(7\), so \(F'(7\times5,7\times1)=(35,7)\).

Step3: Calculate distance \(D'F'\)

Use distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here, \(x_1 = 7,y_1 = 7,x_2 = 35,y_2 = 7\).
\[

$$\begin{align*} D'F'&=\sqrt{(35 - 7)^2+(7 - 7)^2}\\ &=\sqrt{(28)^2+0^2}\\ &=\sqrt{784}\\ & = 28 \end{align*}$$

\]

Answer:

\(28\)