Question Video: Finding the Area of an Object, Then Finding the Area of a Fraction of That Object | Nagwa Question Video: Finding the Area of an Object, Then Finding the Area of a Fraction of That Object | Nagwa

Question Video: Finding the Area of an Object, Then Finding the Area of a Fraction of That Object Physics

A rectangular wall, shown in the diagram, has a length 𝑙 = 18 m and a height ℎ = 11 m. The part of the wall shown in orange is in the shade, and the part of the wall shown in white is in direct sunlight. What is the area of the wall? Answer to the nearest square metre. What is the area of the part of the wall in direct sunlight? Answer to the nearest square metre.

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Video Transcript

A rectangular wall shown in the diagram has a length 𝑙 equals 18 metres and a height ℎ equals 11 metres. The part of the wall shown in orange is in the shade, and the part of the wall shown in white is in direct sunlight. What is the area of the wall? Answer to the nearest square metre. What is the area of the part of the wall in direct sunlight? Answer to the nearest square metre.

All right, so we’re looking at these two separate questions. What is the area of the wall and what is the area of the part of the wall in direct sunlight? Looking for a moment at our diagram, we see the sketch of the wall. And we’re told that the orange part, this wedge right there, is the part of the wall that’s in the shade, whereas the white part, the part of the wall here shown in these dashed lines, is in direct sunlight. Along with this, we’re told what the values of the length and the height of the wall are. Based on all this, first off, we want to solve for the area of this wall.

As we work to figure out what that area is, a helpful clue is that we’re told the wall is rectangular and that it has the length and height marked out on our sketch. This means when we calculate the overall area of the wall, we’ll include the orange section, and we’ll also include the white section, the sections and shade and in sunlight. It’s this total combined area that makes up the area of the wall. Because our walls are rectangle, we can recall the mathematical equation for the area of a rectangle. We’ll call it 𝐴 sub 𝑟. That’s often written as the base of the rectangle 𝑏 multiplied by its height ℎ. Now in our case, the base 𝑏 is represented by this variable 𝑙 which stands for length.

Because it’s just a different name for the same thing, we could write that into our equation. The area of a rectangle is equal to its length times its height. And let’s give this area a name. Let’s call it capital 𝐴. Based on our equation, this area is equal to the length of a rectangle multiplied by its height. And both those values are given to us. The length is 18 metres and the height is 11 metres. With those values substituted in, our next step is to combine them through multiplication. Before we do that though, notice that each one of these values has both a number as well as a unit attached to that number. When we multiply the values together, we can treat these two things separately. First, we can combine the numbers, 18 times 11, and then the units, metres times metres.

When we multiply 18 times 11, that’s equal to 198. And then a metre multiplied by a metre is a metre squared. And this then is the total area of our rectangular wall. To the nearest square metre, it’s 198 square metres. Now, let’s move on to the second part of our question, which asks about the area of the part of the wall that’s in direct sunlight. Going back to our problem statement, we’re told that the area that’s shown in white is the area in direct sunlight. So that’s this area here on our sketch. Now, because our wall is a rectangle, that means that this angle right here is a 90-degree angle. All the interior angles of a rectangle are. So that means that this area and white, which we can see is triangular, is a right triangle. And not only that, but the length of this triangle is equal to 𝑙. And its height is equal to ℎ.

At this point, we’ll want to recall the mathematical equation for the area of a triangle. If we call that area 𝐴 sub 𝑡, it’s equal to one-half the length of the triangle 𝑙 multiplied by its height ℎ. And notice that this equation for the area of the triangle is the same as the equation for the area of the rectangle, except for now multiplying by one-half. This means that there are two ways we could go about answering this second question, about solving for the area of the part of the wall in direct sunlight. One way would be to take the answer we calculated earlier, 198 metres squared, and divide it by two.

Based on our equations for the area of a triangle and the area of a rectangle, when they share a common exterior, as our rectangle and triangle do in this case, then the triangle’s area as we saw is half the rectangles area. And a second way to solve for the triangle’s area is simply to use this equation here. If we go about answering the question the second way, then we’ll be multiplying one-half by the length of our triangle, 18 metres, and by its height, 11 metres. And if we separate out our numbers and our units, then we have one-half times 18 times 11 all multiplied by a metre times a metre. The number of part of it, one-half times 18 times 11, is equal to 99. And then like we saw earlier, a metre times a metre is a metre squared. And that is our answer. The area of the wall in direct sunlight is 99 metres squared.

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