<h3><u>Given </u><u>:</u><u>-</u><u> </u></h3>
- <u>Rectangle </u><u>1</u><u> </u><u> </u><u>has </u><u>length </u><u>x</u><u> </u><u>and </u><u>width </u><u>y</u>
- <u>Rectangle</u><u> </u><u>2</u><u> </u><u>is </u><u>made </u><u>by </u><u>multiplying </u><u>each </u><u>dimensions </u><u>of </u><u>rectangle </u><u>1</u><u> </u><u>by </u><u>a </u><u>factor </u><u>of </u><u>k </u>
- <u>Where</u><u>, </u><u>k </u><u>></u><u> </u><u>0</u><u> </u><u> </u>
<h3><u>Answer </u><u>1</u><u> </u><u>:</u><u>-</u></h3>
Yes, The rectangle 1 and rectangle 2 are similar .
<h3><u>According </u><u>to </u><u>the </u><u>similarity </u><u>theorem </u><u>:</u><u>-</u></h3>
- If the ratio of length and breath of both the triangles are same then the given triangles are similar.
<u>Let's </u><u>Understand </u><u>the </u><u>above </u><u>theorem </u><u>:</u><u>-</u>
The dimensions of rectangle 1 are x and y
<u>Now</u><u>, </u>
- Rectangle 2 is made by multiplying each dimensions of rectangle 1 by a factor of k .
Let assume the value of K be 5
<u>Therefore</u><u>, </u>
The dimensions of rectangle 2 are
<u>Now</u><u>, </u><u> </u><u>The </u><u>ratios </u><u>of </u><u>dimensions </u><u>of </u><u>both </u><u>the </u><u>rectangle </u><u>:</u><u>-</u>
<u>From </u><u>above</u><u>, </u>
We can conclude that the ratios of both the rectangles are same
Hence , Both the rectangles are similar
<h3>
<u>Answer </u><u>2</u><u> </u><u>:</u><u>-</u><u> </u></h3>
<u>Here</u><u>, </u>
We have to proof that, the
- Perimeter of rectangle 2 = k(perimeter of rectangle 1 )
In the previous questions, we have assume the value of k = 5
<h3><u>Therefore</u><u>, </u></h3>
<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>
Perimeter of rectangle 1
Thus, The perimeter of rectangle 1
Perimeter of rectangle 2
Thus, The perimeter of rectangle 2
<u>According </u><u>to </u><u>the </u><u>given </u><u>condition </u><u>:</u><u>-</u>
- Perimeter of rectangle 2 = k( perimeter of rectangle 1 )
<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>
<u>From </u><u>Above</u><u>, </u>
We can conclude that the, Perimeter of rectangle 2 is 5 times of perimeter of rectangle 1 and we assume the value of k = 5.
Hence, The perimeter of rectangle 2 is k times of rectangle 1
<h3>
<u>Answer 3 :</u></h3>
<u>Here</u><u>, </u>
We have to proof that ,
- <u>The </u><u>area </u><u>of </u><u>rectangle </u><u>2</u><u> </u><u>is </u><u>k²</u><u> </u><u>times </u><u>of </u><u>the </u><u>area </u><u>of </u><u>rectangle </u><u>1</u><u>.</u>
<u>That </u><u>is</u><u>, </u>
- Area of rectangle 1 = k²( Area of rectangle)
<h3><u>Therefore</u><u>, </u></h3>
<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>
<u>Area </u><u>of </u><u>rectangle </u><u>1</u>
<u>Area </u><u>of </u><u>rectangle </u><u>2</u>
<u>According </u><u>to </u><u>the </u><u>given </u><u>condition </u><u>:</u><u>-</u>
- Area of rectangle 1 = k²( Area of rectangle)
<u>From </u><u>Above</u><u>, </u>
We can conclude that the, Area of rectangle 2 is (5)² times of area of rectangle 1 and we have assumed the value of k = 5
Hence, The Area of rectangle 2 is k times of rectangle 1 .