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Start your free trialElijah Njeri
3,175 PointsCan't seem to find an answer to this simple quiz..
give intro a unitless line-height that's 1.6 times larger than the font-size value.
The font-size is 1.25em. shouldn't the line-height be 2, as derived by multiplying 1.25 by 1.6?
<!DOCTYPE html>
<html>
<head>
<title>Lake Tahoe</title>
<link rel="stylesheet" href="page.css">
<link rel="stylesheet" href="style.css">
</head>
<body>
<header id="top" class="main-header">
<span class="title">Journey Through the Sierra Nevada Mountains</span>
<h1>Lake Tahoe, California</h1>
</header>
<div class="primary-content t-border">
<p class="intro">
Lake Tahoe is one of the most <span>breathtaking attractions</span> located in California. It's home to a number of ski resorts, summer outdoor recreation, and tourist attractions. Snow and skiing are a significant part of the area's reputation.
</p>
<a href="#more">Find out more</a>
</div>
<footer class="main-footer">
<p>All rights reserved to the state of <a href="#">California</a>.</p>
<a href="#top">Back to top »</a>
</footer>
</body>
</html>
/* Complete the challenge by writing CSS below */
.intro {
font-size: 1.25em;
line-height: 2;
}
2 Answers
Jamie Reardon
Treehouse Project ReviewerHi, the teacher has asked you to give a line-height property value that is 1.6 times larger than the font size of the element. Since the line-height property multiplies the font size value, the answer here would be 1.6 like it says in the question.
.intro {
font-size: 1.25em;
line-height: 1.6;
}
Here, the line-height property is already multiplying the font size by the given unitless value of 1.6.
Jamie Reardon
Treehouse Project ReviewerHi Elijah, the line-height property multiplies whatever value the font size of the element is. If you put 2 for the value of the line height, it is going to therefore multiply the value 1.25em of the font size by 2.
The value 1.6 is multiplying 1.25em by 1.6.
Elijah Njeri
3,175 PointsElijah Njeri
3,175 PointsHi, I haven't understood the explanation you've given, but 1.6 is the answer...at least I can now proceed. Thanks