Question

1. make sense and persevere the base and height of a triangle are each extended 2 cm. what is the area of the shaded region? how do you know? (mp.1)

Explanation:

Step1: Assume Original Triangle Dimensions

Let the original base of the triangle be \( b \) cm and the original height be \( h \) cm. The area of a triangle is given by the formula \( A = \frac{1}{2}bh \).

Step2: Determine New Dimensions

After extending the base and height by 2 cm each, the new base is \( b + 2 \) cm and the new height is \( h + 2 \) cm. The area of the new triangle (if the shaded region is the new triangle or the difference, but since the problem is about the shaded region with extended base and height, we assume we need to find the area change or the new area. Wait, the problem statement is a bit unclear, but likely we need to find the area of the shaded region which is the difference between the new triangle (with base \( b + 2 \) and height \( h + 2 \)) and the original triangle, or maybe the shaded region is a specific shape. But since the problem is about triangle area and extension, let's assume we need to find the area of the shaded region which is the area of the new triangle minus the original. Wait, no, maybe the shaded region is a parallelogram or another shape, but since the problem is about triangle base and height extended by 2 cm, let's re - read.

Wait, the problem says "The base and height of a triangle are each extended 2 cm. What is the area of the shaded region? How do you know?"

Let's assume the original triangle has base \( b \) and height \( h \), area \( A_1=\frac{1}{2}bh \). The new triangle (after extending base and height by 2 cm) has base \( b + 2 \) and height \( h + 2 \), area \( A_2=\frac{1}{2}(b + 2)(h + 2)=\frac{1}{2}(bh+2b + 2h + 4) \). The shaded region area \( A = A_2 - A_1=\frac{1}{2}(bh + 2b+2h + 4)-\frac{1}{2}bh=\frac{1}{2}(2b + 2h + 4)=b + h+2 \). But since we don't have the original base and height, maybe the original triangle is a right triangle or has specific values? Wait, maybe the original triangle has base and height such that when extended, the shaded region is a trapezoid or a parallelogram. Wait, maybe the problem is missing some initial values of the base and height of the original triangle. Since the problem is not fully clear, but if we assume the original triangle has base \( b \) and height \( h \), and the shaded region is the area between the original triangle and the new triangle (with base \( b + 2 \) and height \( h + 2 \)), then the area of the shaded region is \( A=\frac{1}{2}(b + 2)(h + 2)-\frac{1}{2}bh=\frac{1}{2}(bh+2b + 2h + 4)-\frac{1}{2}bh=b + h + 2 \).

But if we assume the original triangle has base \( b = 0 \) (which is not possible) or maybe the original triangle is a right triangle with base \( b \) and height \( h \), and the shaded region is a rectangle or a parallelogram. Wait, maybe the problem is from a textbook where the original triangle has specific dimensions, like base \( b \) and height \( h \), and when we extend base and height by 2 cm, the shaded region is a trapezoid with bases \( b \) and \( b + 2 \) and height 2, or height \( h \) and \( h + 2 \) and base 2. No, this is getting too confusing. Wait, maybe the key is to use the formula for the area of a triangle and the change in area when base and height are increased.

Alternatively, if we assume the original triangle has base \( b \) and height \( h \), area \( A=\frac{1}{2}bh \). After extending, new base \( B=b + 2 \), new height \( H=h + 2 \), new area \( A'=\frac{1}{2}(b + 2)(h + 2)=\frac{1}{2}(bh+2b + 2h + 4) \). The shaded region area (if it's the area added) is \( A'-A=\frac{1}{2}(bh + 2b+2h + 4)-\frac{1}{2}bh=b + h + 2 \).

But since the problem is not providing the original base and height, maybe there is a diagram (which is not shown here) where the original triangle has base and height such that \( b \) and \( h \) are known, or the shaded region is a specific shape like a parallelogram with base 2 and height equal to the original height plus 2, or something else. Since the problem is about triangle area and extension, and the discipline is Mathematics (subfield Geometry, since it's about triangle area and geometric shapes), we can say that to find the area of the shaded region, we use the formula for the area of a triangle \( A=\frac{1}{2}\times base\times height \). First, find the area of the original triangle, then find the area of the triangle with extended base and height (base + 2, height + 2), and then the area of the shaded region is the difference between these two areas (if the shaded region is the area added) or the area of the new triangle (if the shaded region is the new triangle).

Since the problem is incomplete without the original base and height values or the diagram, but in a typical problem like this, if the original triangle has base \( b \) and height \( h \), the area of the shaded region (the area added) is \( \frac{1}{2}(b + 2)(h + 2)-\frac{1}{2}bh=b + h+2 \) square centimeters.

Answer:

To find the area of the shaded region, we use the formula for the area of a triangle \( A=\frac{1}{2}\times base\times height \). Let the original base be \( b \) and original height be \( h \). The area of the original triangle is \( A_1 = \frac{1}{2}bh \). The new base is \( b + 2 \) and new height is \( h + 2 \), so the area of the triangle with extended base and height is \( A_2=\frac{1}{2}(b + 2)(h + 2)=\frac{1}{2}(bh + 2b+2h + 4) \). If the shaded region is the area added (the difference between the new and original triangle areas), then the area of the shaded region \( A=A_2 - A_1=\frac{1}{2}(bh + 2b + 2h + 4)-\frac{1}{2}bh=b + h + 2 \) square centimeters. (Note: This answer assumes the shaded region is the area between the new and original triangles. If there is a diagram with specific values for \( b \) and \( h \), substitute those values into the formula \( b + h + 2 \) to get the numerical area.)