Free Math Worksheets

- A median divides a triangle into two equal-area triangles.

1. Segment \(AM\) is a median, so it divides \(\triangle ABC\) into two triangles with equal areas. 2. Therefore, the total area is \(2\times18=36\,\text{cm}^2\).

Triangle \(ABC\) has area \(36\,\text{cm}^2\).

- What height do the two triangles share when their bases lie on \(AB\)? - What is the area formula for a triangle? - When the height stays the same, how does changing the base affect the area?

1. The area of triangle \(ABC\) is \(A_{ABC}=\frac{1}{2}\times12\times5=30\,\text{cm}^2\). 2. Triangles \(ABC\) and \(ADC\) have the same height to line \(AB\). Therefore, \(A_{ADC}=\frac{1}{2}\times3\times5=7.5\,\text{cm}^2\). 3. The fraction is \(\frac{7.5}{30}=\frac{1}{4}\).

\(A_{ABC}=30\,\text{cm}^2\), \(A_{ADC}=7.5\,\text{cm}^2\), and triangle \(ADC\) has \(\frac{1}{4}\) of the area of triangle \(ABC\).

- What is the area formula for a triangle? - What is true about the distance between parallel lines? - How does the distance from \(C\) to the base affect the area?

1. The area is \(A=\frac{1}{2}bh=\frac{1}{2}\times10\times4=20\,\text{cm}^2\). 2. Every point on a line parallel to \(AB\) is the same perpendicular distance from \(AB\). 3. Therefore, both the base \(AB\) and the height remain constant, so the area remains \(20\,\text{cm}^2\).

a) \(20\,\text{cm}^2\) b) The height stays constant because the moving point remains on a line parallel to the base. Since the base also stays fixed, the area does not change.

- How are the perimeter and the three side lengths related? - What is the area formula for a triangle? - Which side should be used as the base to find the requested height?

1. The third side has length \(32-10-10=12\,\text{cm}\). 2. Let \(h\) be the height to the \(12\,\text{cm}\) side. From \(48=\frac{1}{2}\times12\times h\), \(48=6h\). 3. Therefore, \(h=48\div6=8\,\text{cm}\).

The third side is \(12\,\text{cm}\) long, and its corresponding height is \(8\,\text{cm}\).

- For the rectangle, use length times width. - For the triangle, use one-half times the base times the height. - Split the step-shaped figure into smaller rectangles. - Compare the three numerical areas after you calculate them.

1. Figure A is a rectangle with width \(3\) units and height \(2\) units, so its area is \(3 \times 2 = 6\) square units. 2. Figure B is a right triangle with base \(4\) units and height \(3\) units. Its area is \(\frac{1}{2} \times 4 \times 3 = 6\) square units. 3. Figure C can be split into three vertical rectangles with areas \(3\), \(2\), and \(1\) square units. Its total area is \(3 + 2 + 1 = 6\) square units. 4. All three figures have the same area: \(6\) square units.

a) Figure A: \(6\) square units Figure B: \(6\) square units Figure C: \(6\) square units b) All three figures have the same area even though they have different shapes.

- Find the area of each figure separately. - Use one-half times the base times the height for the triangle. - Split the L-shaped figure into rectangles. - Compare the three results.

1. Figure A is a square with side length \(2\) units, so its area is \(2 \times 2 = 4\) square units. 2. Figure B is a right triangle with base \(4\) units and height \(2\) units, so its area is \(\frac{1}{2} \times 4 \times 2 = 4\) square units. 3. Figure C can be split into two nonoverlapping rectangles with areas \(2\) square units and \(2\) square units, so its total area is \(2 + 2 = 4\) square units. 4. Therefore, all three figures have the same area.

All three figures have the same area: \(4\) square units each.

- Use length times width for the rectangle. - For the triangle, imagine a rectangle with the same base and height. - What fraction of that rectangle is the triangle?

1. Figure A is a \(4 \times 2\) rectangle, so its area is \(4 \times 2 = 8\) square units. 2. Figure B is a triangle with base \(4\) units and height \(2\) units, so its area is \(\frac{1}{2} \times 4 \times 2 = 4\) square units.

Figure A has an area of \(8\) square units, and Figure B has an area of \(4\) square units.

- Use length times width for the rectangle. - Imagine completing the triangle to make a rectangle. - What fraction of that rectangle is the triangle?

1. Rectangle a) has side lengths \(4\) units and \(1\) unit, so its area is \(4 \times 1 = 4\) square units. 2. Triangle b) has base \(2\) units and height \(4\) units, so its area is \(\frac{1}{2} \times 2 \times 4 = 4\) square units. 3. The two figures have equal areas.

The two figures have the same area: \(4\) square units each.

- Find the area of each figure separately. - For each triangle, use one-half times the base times the height. - Compare all four results.

1. Figure a) is a \(2 \times 2\) square, so its area is \(4\) square units. 2. Figure b) is a \(4 \times 1\) rectangle, so its area is \(4\) square units. 3. Figure c) is a triangle with base \(3\) units and height \(2\) units, so its area is \(\frac{1}{2} \times 3 \times 2 = 3\) square units. 4. Figure d) is a triangle with base \(4\) units and height \(2\) units, so its area is \(\frac{1}{2} \times 4 \times 2 = 4\) square units. 5. Figures a), b), and d) have the same area.

Figures a), b), and d) have the same area: \(4\) square units each.

- Use the triangle area formula with a base and its corresponding height. - Count the spaces between pegs to determine lengths. - For triangle a), the two sides that form the right angle can be used as the base and height.

1. Triangle a) is a right triangle with perpendicular side lengths \(4\) units and \(5\) units. Its area is \(A_a=\frac{1}{2}\times4\times5=10\) square units. 2. Triangle b) has a base of \(5\) units and a height of \(4\) units. Its area is \(A_b=\frac{1}{2}\times5\times4=10\) square units. 3. The areas are equal, so neither triangle covers more area.

Both triangles have an area of \(10\) square units, so neither is larger.

- Use the grid to find the base and corresponding height. - Half of one grid square has an area of \(0.5\,\text{cm}^2\).

1. The horizontal base is \(5\,\text{cm}\) long. 2. The corresponding height is \(3\,\text{cm}\). 3. The area is \(A=\frac{1}{2}\times5\times3=7.5\,\text{cm}^2\).

The area is \(7.5\,\text{cm}^2\).

- Complete the triangle mentally to form a rectangle. - What fraction of the rectangle is the triangle?

1. The triangle is half of a \(7\,\text{cm} \times 3\,\text{cm}\) rectangle. 2. The rectangle area is \(7 \times 3=21\,\text{cm}^2\). 3. The triangle area is \(21 \div 2=10.5\,\text{cm}^2\).

\(10.5\,\text{cm}^2\)

- How many congruent right triangles make the completed rectangle? - Find the rectangle's area first. - Divide that area by the number of triangles.

1. The completed rectangle has area \(8\times 5=40\,\text{cm}^2\). 2. The right triangle is half of that rectangle, so its area is \(40\div 2=20\,\text{cm}^2\).

\(20\,\text{cm}^2\)

- Which two measurements are needed to find a triangle's area? - Recall the triangle area formula. - A triangle with a given base and height has half the area of the corresponding rectangle.

1. Use the triangle area formula: \(A=\frac{1}{2}bh\). 2. Substitute the measurements: \(A=\frac{1}{2}\times15\times10=75\,\text{cm}^2\).

The area is \(75\,\text{cm}^2\).

- Substitute the known values into the triangle area formula. - Rearrange the equation so the unknown height is alone.

1. Use \(A=\frac{1}{2}bh\): \(30=\frac{1}{2}\times12\times h=6h\). 2. Divide by \(6\): \(h=30\div6=5\,\text{cm}\).

The height is \(5\,\text{cm}\).

- Which side is shown as the base? - Recall the triangle area formula. - Use the height that is perpendicular to the chosen base.

1. The base is \(8\,\text{cm}\), and its corresponding height is \(4\,\text{cm}\). 2. The area is \(A=\frac{1}{2}\times8\times4=16\,\text{cm}^2\).

The area of the triangle is \(16\,\text{cm}^2\).

- Compare the triangle with the entire rectangle. - Use the area formulas for a triangle and a rectangle with the same base and height. - Imagine making a congruent copy of the triangle. - Decide how many triangle areas make the rectangle area.

1. The triangle and rectangle have the same base and height. 2. A triangle with the same base and height as a rectangle has half the rectangle area. 3. Therefore, the rectangle area is \(2 \times 14=28\,\text{cm}^2\).

\(28\,\text{cm}^2\)

- First find the area shared by both triangles. - How long is the second triangle's base? - Use the triangle area formula to find the missing height. - Compare the two base-height pairs.

1. The first triangle has area \(A=\frac{1}{2}\times10\times6=30\,\text{cm}^2\). 2. The second triangle's base is \(10\div2=5\,\text{cm}\). 3. Set its area equal to \(30\): \(30=\frac{1}{2}\times5\times h_2=2.5h_2\). Thus, \(h_2=30\div2.5=12\,\text{cm}\). 4. When the base is halved, the height must double to keep the product of base and height constant.

a) \(30\,\text{cm}^2\) b) \(12\,\text{cm}\) c) The height must double.

- First express the measurements in compatible units. - Equal areas can be represented by an equation. - Find the first sail's area before solving for the missing height. - Convert the final height to centimeters.

1. Convert the first sail's height to meters: \(80\,\text{cm}=0.8\,\text{m}\). 2. The first sail has area \(A=\frac{1}{2}\times1.2\times0.8=0.48\,\text{m}^2\). 3. For the second sail, \(0.48=\frac{1}{2}\times1.5\times h=0.75h\). 4. Therefore, \(h=0.48\div0.75=0.64\,\text{m}=64\,\text{cm}\).

The second sail's height is \(64\,\text{cm}\).

- A triangle's area can be found using any side and its corresponding height. - The area is the same no matter which base-height pair you use. - Once two side lengths are known, use the perimeter to find the third.

1. Use the \(10\,\text{cm}\) side and its corresponding height to find the area: \(A=\frac{1}{2}\times10\times4.8=24\,\text{cm}^2\). 2. Let \(x\) be the side whose corresponding height is \(8\,\text{cm}\). Then \(24=\frac{1}{2}\times x\times8=4x\), so \(x=6\,\text{cm}\). 3. Let \(y\) be the remaining side. Use the perimeter: \(y=24-6-10=8\,\text{cm}\).

The second side is \(6\,\text{cm}\), and the remaining side is \(8\,\text{cm}\).

- Find the area of the square first. - Equal-area figures have the same numerical area. - Substitute the known base and area into the triangle area formula. - Solve the resulting equation for the height.

1. The square's area is \(12\times12=144\,\text{m}^2\). 2. The triangle must also have area \(144\,\text{m}^2\), so \(144=\frac{1}{2}\times16\times h\). 3. Simplify to \(144=8h\), then divide: \(h=144\div8=18\,\text{m}\).

The triangle must have a height of \(18\,\text{m}\).

- Use the triangle area formula. - Find Leo's base and height before calculating his area. - What happens when one factor doubles and the other is halved? - In part b), substitute the target area and solve for the height.

1. Mia's triangle has area \(A_M=\frac{1}{2}\times10\times6=30\,\text{cm}^2\). 2. Leo's second triangle has base \(20\,\text{cm}\) and height \(3\,\text{cm}\), so \(A_L=\frac{1}{2}\times20\times3=30\,\text{cm}^2\). The areas are equal. 3. The third triangle needs area \(4\times30=120\,\text{cm}^2\). 4. Solve \(120=\frac{1}{2}\times20\times h=10h\). Thus, \(h=12\,\text{cm}\).

a) Both areas are \(30\,\text{cm}^2\). b) The height must be \(12\,\text{cm}\).

- Find the area of the triangular sail first. - The replacement sail must have the same area. - Divide the rectangle's area by its known length.

1. The triangular sail has area \(\frac{1}{2}\times5.40\times3.50=9.45\,\text{m}^2\). 2. The rectangle must have the same area, so its width is \(9.45\div4.20=2.25\,\text{m}\).

The rectangular sail must be \(2.25\,\text{m}\) wide.

- Compare the area formulas for a triangle and a rectangle. - One side length is the same in both figures. - Picture two congruent copies of the triangle forming a rectangle.

1. A triangle has half the area of a rectangle with the same base and height. 2. Since both figures have a \(12\,\text{m}\) side, the rectangle's other side must be half of \(18\,\text{m}\) to have the same area. 3. Therefore, the missing side length is \(18\div2=9\,\text{m}\).

The rectangle's other side is \(9\,\text{m}\). The factor \(\frac{1}{2}\) in the triangle area formula means the corresponding rectangle side must be half as long when the shared side is unchanged.

- Find the area of the original triangular stage. - Apply the required change to the area before working with the rectangle. - Divide the new area by the rectangle's known length.

1. The original stage has area \(\frac{1}{2}\times15\times6.40=48\,\text{m}^2\). 2. Doubling the area gives \(2\times48=96\,\text{m}^2\). 3. The rectangle's width is \(96\div12=8\,\text{m}\).

The new rectangular stage must be \(8\,\text{m}\) wide.

- Which measurements appear in the standard triangle area formula? - Imagine changing the other two side lengths while keeping the base and perpendicular height fixed. What happens to the area?

1. Use the \(4\,\text{m}\) side as the base: \(A=\frac{1}{2}\times4\times2.2=4.4\,\text{m}^2\). 2. A triangle's area is determined by a chosen base and the perpendicular height to that base. The other two sides affect the triangle's shape but do not appear in this area calculation.

1. \(4.4\,\text{m}^2\) 2. The area formula uses the base and its perpendicular height, so the other two side lengths are not needed.

- Consider acute, right, and obtuse triangles separately. - Sketch the three altitudes for each type of triangle. - What happens when one angle is exactly \(90^\circ\)? - An altitude is perpendicular to a side or to the line containing that side.

1. True. In an acute triangle, all three altitudes lie inside. An acute triangle can have an area of \(18\,\text{cm}^2\). 2. False. An acute or right triangle has no altitudes outside, while an obtuse triangle has two altitudes outside. 3. True. A triangle with a \(110^\circ\) angle is obtuse, so the two altitudes drawn to the sides that form the obtuse angle meet extensions of those sides outside the triangle.

1. True 2. False 3. True

- Which measurements determine a triangle's area? - Does the area formula include the horizontal position of the opposite vertex or the angle type? - Recall where the altitudes lie in an obtuse triangle.

1. The area is \(A=\frac{1}{2}\times10\times4=20\,\text{cm}^2\). 2. Moving the vertex sideways does not change the base or its corresponding height, so the area remains \(20\,\text{cm}^2\). 3. An obtuse triangle has exactly two altitudes outside the triangle.

a) \(20\,\text{cm}^2\) b) No. The base and corresponding height are unchanged, so the area is unchanged. c) Two altitudes lie outside the triangle.

- What does the right angle tell you about the two legs and their corresponding heights? - Use the area you found to write a second area equation with the hypotenuse as the base. - Rearrange the triangle area formula to solve for a height.

1. The legs are perpendicular. Therefore, the height corresponding to the \(5\,\text{cm}\) leg is \(12\,\text{cm}\), and the height corresponding to the \(12\,\text{cm}\) leg is \(5\,\text{cm}\). 2. The area is \(A=\frac{1}{2}\times5\times12=30\,\text{cm}^2\). 3. Let \(h\) be the height drawn to the hypotenuse. Using the hypotenuse as the base, \(30=\frac{1}{2}\times13\times h\). Thus, \(h=\frac{60}{13}\,\text{cm}\approx4.62\,\text{cm}\).

a) The two heights are \(12\,\text{cm}\) and \(5\,\text{cm}\), respectively. b) \(30\,\text{cm}^2\) c) \(\approx4.62\,\text{cm}\)

- How are a triangle's base, height, and area related? - Rearrange the formula to find the missing measurement. - Test the claim step by step using the values from part a).

1. For triangle A, \(24=\frac{1}{2}\times8\times h=4h\), so \(h=6\,\text{cm}\). 2. For triangle B, let \(b\) be the base. Then \(24=\frac{1}{2}\times b\times4=2b\), so \(b=12\,\text{cm}\). 3. Doubling triangle A's base gives \(16\,\text{cm}\), and halving its height gives \(3\,\text{cm}\). 4. The new area is \(\frac{1}{2}\times16\times3=24\,\text{cm}^2\), so the claim is correct.

a) \(6\,\text{cm}\) b) \(12\,\text{cm}\) c) The claim is correct; the new area is still \(24\,\text{cm}^2\).

- Rearrange the area formula to find a missing height. - Find two positive numbers whose product is twice the area. - Imagine moving the third vertex left or right without changing its distance from the base.

1. Use \(A=\frac{1}{2}bh\): \(15=\frac{1}{2}\times6\times h=3h\), so \(h=5\,\text{cm}\). 2. Any positive base-height pair with product \(30\,\text{cm}^2\) works. For example, \(b=10\,\text{cm}\) and \(h=3\,\text{cm}\). 3. The triangles do not have to have the same shape. The third vertex can move along a line parallel to the base, changing side lengths and angles while keeping the height and area fixed.

a) \(5\,\text{cm}\) b) One possible pair is \(10\,\text{cm}\) and \(3\,\text{cm}\). c) No. The third vertex can move parallel to the base without changing the height or area.

- How can the perpendicular legs be used as a base and height? - Any side can be a base if you use its corresponding height. - What happens to a product when one factor doubles?

1. The legs are perpendicular, so \(A=\frac{1}{2}\times6\times8=24\,\text{cm}^2\). 2. Let \(h\) be the height corresponding to the hypotenuse. Then \(24=\frac{1}{2}\times10\times h=5h\), so \(h=4.8\,\text{cm}\). 3. If the \(6\,\text{cm}\) leg doubles while the other leg stays fixed, one factor in \(\frac{1}{2}bh\) doubles. Therefore, the area doubles to \(48\,\text{cm}^2\).

a) \(24\,\text{cm}^2\) b) \(4.8\,\text{cm}\) c) The area doubles to \(48\,\text{cm}^2\).

- Find each triangle's area separately. - Identify which rectangle side is halved in each case. - Compare the two base-height products.

1. In Rectangle A, the triangle has base \(5\,\text{cm}\) and height \(6\,\text{cm}\). Its area is \(\frac{1}{2}\times5\times6=15\,\text{cm}^2\). 2. In Rectangle B, the triangle has base \(3\,\text{cm}\) and height \(10\,\text{cm}\). Its area is \(\frac{1}{2}\times3\times10=15\,\text{cm}^2\). 3. The triangles have equal areas because the base-height products are equal: \(5\times6=3\times10=30\).

Both triangles have area \(15\,\text{cm}^2\). Their shapes differ, but their base-height products are equal.

- First use the triangle area formula to find the missing height. - How does halving or tripling one factor affect a product? - You may calculate directly or use proportional reasoning.

1. Solve \(24=\frac{1}{2}\times8\times h=4h\), giving \(h=6\,\text{cm}\). 2. With base \(4\,\text{cm}\), the area is \(\frac{1}{2}\times4\times6=12\,\text{cm}^2\). 3. With height \(18\,\text{cm}\), the area is \(\frac{1}{2}\times8\times18=72\,\text{cm}^2\).

a) \(6\,\text{cm}\) b) \(12\,\text{cm}^2\) c) \(72\,\text{cm}^2\)

- Which base already has a known corresponding height? - How are base, height, and area related? - Can one fixed triangle have two different areas?

1. The area is \(A=\frac{1}{2}\times14\times5=35\,\text{cm}^2\). 2. Let \(h\) be the height corresponding to the \(10\,\text{cm}\) side. Then \(35=\frac{1}{2}\times10\times h=5h\). Thus, \(h=7\,\text{cm}\). 3. The triangle itself does not change when a different side is chosen as the base. Each base must be paired with its own perpendicular height, and every valid pair gives the same enclosed area.

a) \(35\,\text{cm}^2\) b) \(7\,\text{cm}\) c) Choosing a different base does not change the triangle; the corresponding height adjusts so the area remains the same.

- Find a rectangle that exactly encloses the triangle. - Calculate the areas outside the triangle but inside the rectangle. - Subtract those areas from the rectangle area.

1. Enclose the triangle in a \(6 \times 4\) rectangle with area \(24\) square units. 2. The left outside triangle has area \(\frac{1}{2} \times 2 \times 4=4\) square units. The right outside triangle has area \(\frac{1}{2} \times 4 \times 4=8\) square units. 3. The triangle area is \(24-4-8=12\) square units.

\(12\) square units

- Subtract the shorter parallel side from the longer one. - The triangle and trapezoid have the same height. - Use the triangle area formula.

1. The triangle's base is \(100-70=30\,\text{cm}\). 2. Its height is the trapezoid's height, \(40\,\text{cm}\). 3. The area is \(\frac{1}{2}\times30\times40=600\,\text{cm}^2\).

The triangle has area \(600\,\text{cm}^2\).

- The standard triangle area formula also applies to obtuse triangles. - Notice where the perpendicular height meets the line containing the base. - Which two measurements determine the area?

1. The height lies outside the triangle because it is drawn to the extension of the base. 2. The area formula still applies: \(A=\frac{1}{2}\times5\times6=15\,\text{cm}^2\).

The area of the triangle is \(15\,\text{cm}^2\).

- Count the grid cells along the horizontal side. - Count the perpendicular distance from the opposite vertex to the base line. - Use the triangle area formula.

1. The horizontal base is \(3\,\text{cm}\) long. 2. The perpendicular height is \(3\,\text{cm}\). 3. The area is \(A=\frac{1}{2}\times3\times3=4.5\,\text{cm}^2\).

The area of the triangle is \(4.5\,\text{cm}^2\).

- Find \(DC\) from the total length \(BC\). - Do the two triangles have the same height to line \(BC\)?

1. Find the remaining segment: \(DC=10-3=7\,\text{cm}\). 2. The two triangles have the same perpendicular height to line \(BC\), so their areas are proportional to their bases. 3. Therefore, the ratio is \(\frac{BD}{DC}=\frac{3}{7}\).

The ratio of the area of \(\triangle ABD\) to the area of \(\triangle ADC\) is \(\frac{3}{7}\).

- First calculate the height needed to produce the required area. - Compare a triangle's height with the sides that connect the opposite vertex to the base. - Can an adjacent side be shorter than the perpendicular distance to the base line?

1. Let \(h\) be the height to the \(8\,\text{cm}\) base. From \(A=\frac{1}{2}bh\), \(20=\frac{1}{2}\times8\times h\). 2. This gives \(20=4h\), so \(h=5\,\text{cm}\). 3. The perpendicular distance from the opposite vertex to the base line cannot be longer than a segment from that vertex to an endpoint of the base. 4. Because the required height is \(5\,\text{cm}\) but the specified adjacent side is only \(4\,\text{cm}\), the triangle is impossible.

The required height is \(5\,\text{cm}\). Since an adjacent side is only \(4\,\text{cm}\), it cannot reach a point that is \(5\,\text{cm}\) from the base line. Therefore, the triangle cannot be constructed.

- What happens to a product when one factor is divided by \(2\) and the other is multiplied by \(2\)? - Find the new target area and new base separately. - Write an equation for the unknown height.

1. The original area is \(A=\frac{1}{2}\times12\times4=24\,\text{cm}^2\). 2. The changed measurements are a \(6\,\text{cm}\) base and an \(8\,\text{cm}\) height. The area is \(\frac{1}{2}\times6\times8=24\,\text{cm}^2\), so the statement is true. 3. The new base in part c) is \(12\div3=4\,\text{cm}\), and the target area is \(2\times24=48\,\text{cm}^2\). 4. Solve \(48=\frac{1}{2}\times4\times h=2h\). Thus, \(h=24\,\text{cm}\).

a) \(24\,\text{cm}^2\) b) The statement is true. c) \(24\,\text{cm}\)