Example Question - rectangle
Here are examples of questions we've helped users solve.
Volume and Surface Area of a Box
<p>Để tính diện tích toàn phần của hình hộp chữ nhật, ta sử dụng công thức:</p> <p>S = 2(lw + lh + wh)</p> <p>Trong đó:</p> <p>l = chiều dài = 65 cm</p> <p>w = chiều rộng = 40 cm</p> <p>h = chiều cao = 55 cm</p> <p>Thay vào công thức:</p> <p>S = 2(65 \cdot 40 + 65 \cdot 55 + 40 \cdot 55)</p> <p>S = 2(2600 + 3575 + 2200)</p> <p>S = 2(8375)</p> <p>S = 16750 cm²</p> <p>Vậy diện tích toàn phần của hình hộp chữ nhật là 16750 cm².</p>
Calculating Surface Area of a Room
<p>Chiều dài phòng: 4.5 m</p> <p>Chiều rộng phòng: 3.5 m</p> <p>Chiều cao phòng: 4 m</p> <p>Diện tích tường là:</p> <p>2 \times (chiều dài + chiều rộng) \times chiều cao = 2 \times (4.5 + 3.5) \times 4 = 2 \times 8 \times 4 = 64 m^2</p> <p>Diện tích trần nhà là:</p> <p>chiều dài \times chiều rộng = 4.5 \times 3.5 = 15.75 m^2</p> <p>Tổng diện tích cần quét là:</p> <p>Diện tích tường + Diện tích trần = 64 + 15.75 = 79.75 m^2</p>
Calculate the Area of a Shaded Region in a Combined Shape
The shaded area is made up of a rectangular part and a quarter of a circular sector (since we're given a right angle and the arc of a circle with center P). The rectangle has one side that is 7 m and shares the other side with the radius of the sector, which is 7 m. The area of the rectangle \( A_{rectangle} \) is: \[ A_{rectangle} = \text{length} \times \text{width} \] \[ A_{rectangle} = 7 \times 7 \] \[ A_{rectangle} = 49 \ m^2 \] The area of the circular sector \( A_{sector} \) is a quarter of the area of a circle with radius 7 m: \[ A_{sector} = \frac{1}{4} \pi r^2 \] \[ A_{sector} = \frac{1}{4} \pi (7)^2 \] \[ A_{sector} = \frac{1}{4} \pi (49) \] \[ A_{sector} = 12.25 \pi \ m^2 \] Therefore, the total shaded area \( A \) is the sum of the rectangle area and the sector area: \[ A = A_{rectangle} + A_{sector} \] \[ A = 49 + 12.25 \pi \] \[ A \approx 49 + 38.48 \] \[ A \approx 87.48 \ m^2 \] So, the area of the shaded region is approximately 87.48 square meters.
Perimeter of a Composite Figure
<p>El perímetro \( P \) de la figura compuesta se puede calcular sumando las longitudes de todos los lados exteriores de la figura.</p> <p>La figura está compuesta por dos rectángulos que comparten un lado. Para evitar contar este lado compartido dos veces, calcularemos el perímetro como si fuera un rectángulo grande.</p> <p>El lado largo del rectángulo grande es \( x + x + 4 \).</p> <p>El lado corto del rectángulo grande (que se muestra en la imagen) es \( x \).</p> <p>El perímetro total es por lo tanto \( P = 2 \cdot (x + x + 4) + 2 \cdot x \).</p> <p>Simplificando, \( P = 2 \cdot (2x + 4) + 2x \).</p> <p>Entonces, \( P = 4x + 8 + 2x \).</p> <p>Finalmente, \( P = 6x + 8 \).</p>
Finding Algebraic Expressions for the Areas of Geometric Shapes
Para el rectángulo de la izquierda, el área \( A \) es el producto de la longitud y la altura: <p>\( A_{\text{rectángulo izquierdo}} = 3z \cdot z \)</p> <p>\( A_{\text{rectángulo izquierdo}} = 3z^2 \)</p> Para el triángulo en el medio, el área \( A \) es un medio del producto de la base por la altura: <p>\( A_{\text{triángulo}} = \frac{1}{2} \cdot x \cdot h \)</p> Para el rectángulo de la derecha, el área \( A \) es el producto de la longitud y la altura: <p>\( A_{\text{rectángulo derecho}} = x \cdot (2x + 1) \)</p> <p>\( A_{\text{rectángulo derecho}} = 2x^2 + x \)</p>
Finding the Area of a Quadrilateral Within a Rectangle
<p>Let the length of BE be \( x \). Hence, \( EG = \frac{5}{6} AD \) and \( BF = \frac{5}{2} x \).</p> <p>Since \( BE = \frac{3}{4} AE \) and \( EG = \frac{5}{6} AD \), let \( AE = 4x \), so \( AD = 6x \) and \( EG = 5x \).</p> <p>The area of rectangle \( ABCD \) is given as 112 \( cm^2 \), which is equal to \( AD \times AB \). Since \( AD = 6x \), we have:</p> <p>\( AB \cdot 6x = 112 \)</p> <p>\( AB = \frac{112}{6x} \)</p> <p>Now, \( AF = AE + EF = AE + BF - BE = 4x + \frac{5}{2}x - x = \frac{11}{2}x \).</p> <p>So the area of triangle \( AEF \) is:</p> <p>\( \frac{1}{2} AF \cdot BE = \frac{1}{2} \cdot \frac{11}{2}x \cdot x = \frac{11}{4}x^2 \)</p> <p>The area of triangle \( EGF \) is:</p> <p>\( \frac{1}{2} EG \cdot BF = \frac{1}{2} \cdot 5x \cdot \frac{5}{2}x = \frac{25}{4}x^2 \)</p> <p>To find the area of \( EGBF \), we sum the areas of \( AEF \) and \( EGF \):</p> <p>\( \frac{11}{4}x^2 + \frac{25}{4}x^2 = \frac{36}{4}x^2 = 9x^2 \)</p> <p>Now, the area of \( ABCD \) is the length times the width, \( AB \times AD \), which is \( \frac{112}{6x} \times 6x = 112 \).</p> <p>To find \( x \), we set up the equation:</p> <p>\( AB \cdot AD = \frac{112}{6x} \cdot 6x = 112 \)</p> <p>\( 112 = 112 \)</p> <p>This is always true, hence we can say that \( x \) can take any value. Let's find \( x \) in terms of \( AB \):</p> <p>\( AB = \frac{112}{6x} \)</p> <p>\( x = \frac{112}{6AB} \)</p> <p>Now, substitute \( x \) in terms of \( AB \) into the area of \( EGBF \):</p> <p>\( Area \;of\; EGBF = 9x^2 = 9 \left( \frac{112}{6AB} \right)^2 \)</p> <p>\( Area \;of\; EGBF = 9 \cdot \frac{112^2}{36 \cdot AB^2} \)</p> <p>\( Area \;of\; EGBF = \frac{112^2}{4 \cdot AB^2} \)</p> <p>Since \( Area \;of\; ABCD = AB \cdot AD = AB \cdot 6x = 112 \), \( x = \frac{112}{6AB} \).</p> <p>Put \( x \) into \( Area \;of\; EGBF \):</p> <p>\( Area \;of\; EGBF = \frac{112^2}{4 \cdot (\frac{112}{6x})^2} \)</p> <p>\( Area \;of\; EGBF = \frac{112^2 \cdot 36x^2}{4 \cdot 112^2} \)</p> <p>\( Area \;of\; EGBF = \frac{36x^2}{4} \)</p> <p>\( Area \;of\; EGBF = 9x^2 \)</p> <p>Since we know that \( 6x = AD \) and \( AD \cdot AB = 112 \), it follows that:</p> <p>\( x^2 = \left( \frac{AD}{6} \right)^2 \)</p> <p>\( x^2 = \left( \frac{112}{6AB} \right)^2 \)</p> <p>\( x^2 = \frac{112^2}{36AB^2} \)</p> <p>The requested area is \( EGBF = 9x^2 \), which after substitution becomes:</p> <p>\( 9 \cdot \frac{112^2}{36AB^2} \)</p> <p>Reducing the fraction gives us:</p> <p>\( \frac{112^2}{4AB^2} \)</p> <p>Since \( AB \times AD = 112 \), let \( AB = a \) and \( AD = \frac{112}{a} \).</p> <p>Substitute \( AB = a \) into the expression for area:</p> <p>\( \frac{112^2}{4a^2} \)</p> <p>Now, express \( AB^2 \) as \( a^2 \), knowing \( a \cdot \frac{112}{a} = 112 \)</p> <p>\( \frac{112^2}{4 \cdot \left( \frac{112}{AD} \right)^2} \)</p> <p>Simplifying, we find:</p> <p>\( \frac{112^2 \cdot AD^2}{4 \cdot 112^2} \)</p> <p>\( \frac{AD^2}{4} \)</p> <p>Since \( AD = 6x \), substitute back to find \( x \):</p> <p>\( \frac{(6x)^2}{4} = \frac{36x^2}{4} = 9x^2 \)</p> <p>Since the area values must match, solving the equation \( 9x^2 = 112 \) to find \( x \):</p> <p>\( x^2 = \frac{112}{9} \)</p> <p>\( x = \frac{\sqrt{112}}{3} \)</p> <p>Finally, calculate the area of \( EGBF \):</p> <p>\( Area \;of\; EGBF = 9 \left( \frac{\sqrt{112}}{3} \right)^2 \)</p> <p>\( Area \;of\; EGBF = 9 \cdot \frac{112}{9} \)</p> <p>\( Area \;of\; EGBF = 112 \; cm^2 \)</p> <p>Thus, the area of quadrilateral \( EGBF \) is 112 \( cm^2 \).</p>
Finding the Area of a Shaded Region in a Geometric Diagram
The problem is to find the area of region EGFB within rectangle ABCD. We're given that the area of rectangle ABCD is 112 cm². First, let's express the lengths BE, EG, and BF in terms of a single variable. Considering that BE is one-third of EG, let's say that EG = 3x, so BE = x. Furthermore, given that BF is half of 5FC and FC is equal to EG (that is, 3x), we have BF = \(\frac{5}{2}\)x. Next, we can find the side lengths of ABCD by exploiting the fact that the area of the rectangle is the product of its length and width: - The width of the rectangle is BE + EG, which is x + 3x = 4x. - The length of the rectangle is BF, which is \(\frac{5}{2}\)x. Area of ABCD is then: \[ 4x \times \frac{5}{2}x = 112 \] \[ 10x^2 = 112 \] \[ x^2 = \frac{112}{10} \] \[ x^2 = 11.2 \] \[ x = \sqrt{11.2} \] \[ x \approx 3.35 \] Now that we have x, we can calculate the length of EG (which is 3x): \[ EG = 3 \times 3.35 \approx 10.05 \] We also calculate BF: \[ BF = \frac{5}{2} \times 3.35 \approx 8.38 \] The area of triangle BEG is half of the rectangle ABCD's area because BEG cuts the rectangle diagonally in half: \[ \text{Area of BEG} = \frac{\text{Area of ABCD}}{2} \] \[ \text{Area of BEG} = \frac{112}{2} \] \[ \text{Area of BEG} = 56 \text{ cm}^2 \] Finally, we find the area of triangle BFG using its base BF and height x (remember that height is BE): \[ \text{Area of BFG} = \frac{1}{2} \times BF \times BE \] \[ \text{Area of BFG} = \frac{1}{2} \times 8.38 \times 3.35 \] \[ \text{Area of BFG} \approx \frac{1}{2} \times 28.07 \] \[ \text{Area of BFG} \approx 14.04 \text{ cm}^2 \] To find the area of the shaded region EGFB, we subtract the area of triangle BFG from the area of triangle BEG: \[ \text{Area of EGFB} = \text{Area of BEG} - \text{Area of BFG} \] \[ \text{Area of EGFB} = 56 - 14.04 \] \[ \text{Area of EGFB} \approx 41.96 \text{ cm}^2 \] So, the area of EGFB is approximately 41.96 cm².
Find the Area of a Triangular Section in a Rectangle
<p>Let \( AF = x \), \( FC = 5x \), \( BF = y \), and \( BG = 3y \).</p> <p>Since \( \triangle ABF \) is similar to \( \triangle EBF \) and \( \triangle BGC \), their sides are proportional.</p> <p>The area \( A \) of \( \triangle ABF \) can be expressed with the base \( AF \) and height \( BF \):</p> <p>\[ A_{\triangle ABF} = \frac{1}{2} AF \cdot BF = \frac{1}{2} x \cdot y \]</p> <p>The area \( A \) of \( \triangle EBF \) is \( \frac{2}{5} \) of the area of \( \triangle ABF \):</p> <p>\[ A_{\triangle EBF} = \frac{2}{5} A_{\triangle ABF} = \frac{2}{5} \cdot \frac{1}{2} x \cdot y = \frac{1}{5} xy \]</p> <p>The area \( A \) of \( \triangle BGC \) is 3 times the area of \( \triangle ABF \):</p> <p>\[ A_{\triangle BGC} = 3 A_{\triangle ABF} = 3 \cdot \frac{1}{2} x \cdot y = \frac{3}{2} xy \]</p> <p>Since the rectangle \( ABCD \) has area 112 cm²:</p> <p>\[ A_{\text{rectangle}} = AF \cdot BF + BF \cdot FC = x \cdot y + 5x \cdot y = 112 \]</p> <p>Solving for \( y \), we get:</p> <p>\[ 6xy = 112 \]</p> <p>\[ y = \frac{112}{6x} \]</p> <p>Substitute \( y \) back to find \( x \):</p> <p>\[ A_{\text{rectangle}} = x \cdot \frac{112}{6x} + 5x \cdot \frac{112}{6x} \]</p> <p>\[ 112 = \frac{112}{6} + \frac{560}{6} \]</p> <p>\[ 112 = \frac{672}{6} \]</p> <p>\[ 6 \cdot 112 = 672 \]</p> <p>This is not true for any positive \( x \), hence there must be a mistake in the initial setup. Based on the image provided, the written setup does not lead to a feasible solution. Please review the problem statement or constraints and verify the setup before attempting to solve.</p>
Finding the Area of a Quadrilateral Within a Rectangle
Let's denote the length of BE as \( x \), so AE is \( 3x \) because BE is one-third of AE. Given that the area of the rectangle ABCD is \( 112 \, \text{cm}^2 \) and that AE is three times BE, the width of the rectangle is \( 4x \). Assume the height of the rectangle (BF) is \( h \). From \( 2\text{BF} = 5\text{FC} \), we have \( 2h = \frac{5}{7}(4x) \) because FC is the remainder of the length of the rectangle after removing BF, which is \( \frac{7}{2}h \). Solving for \( h \), we get \( h = \frac{5}{7} \cdot \frac{4}{2}x = \frac{10}{7}x \). Now we can find \( x \) from the area of the rectangle: \[ \text{Area} = 4x \cdot h = 112 \] \[ 4x \cdot \frac{10}{7}x = 112 \] \[ \frac{40}{7}x^2 = 112 \] \[ x^2 = \frac{112 \cdot 7}{40} \] \[ x^2 = 19.6 \] \[ x = \sqrt{19.6} \] \[ x = 4.43 \] (approx) Now we calculate \( h \): \[ h = \frac{10}{7} \cdot 4.43 = 6.33 \] (approx) To find the area of EFGF, we need to find the areas of triangles BEG and CFF and subtract them from the area of the rectangle. The triangle BEG has a base of x and height h, and the triangle CFF has a base of 3x and height \( \frac{2}{7}h \). Area of BEG: \[ \text{Area}_{\text{BEG}} = \frac{1}{2} \cdot x \cdot h = \frac{1}{2} \cdot 4.43 \cdot 6.33 \] \[ \text{Area}_{\text{BEG}} = 14.01 \] (approx) Area of CFF: \[ \text{Area}_{\text{CFF}} = \frac{1}{2} \cdot 3x \cdot \frac{2}{7}h = \frac{1}{2} \cdot 3 \cdot 4.43 \cdot \frac{2}{7} \cdot 6.33 \] \[ \text{Area}_{\text{CFF}} = 12.15 \] (approx) Total area of the two triangles: \[ \text{Area}_{\text{TotalTriangles}} = \text{Area}_{\text{BEG}} + \text{Area}_{\text{CFF}} \] \[ \text{Area}_{\text{TotalTriangles}} = 14.01 + 12.15 \] \[ \text{Area}_{\text{TotalTriangles}} = 26.16 \] (approx) Finally, the area of quadrilateral EFGF is: \[ \text{Area}_{\text{EFGF}} = 112 - \text{Area}_{\text{TotalTriangles}} \] \[ \text{Area}_{\text{EFGF}} = 112 - 26.16 \] \[ \text{Area}_{\text{EFGF}} = 85.84 \, \text{cm}^2 \] (approx) Please note that the solution is an approximation due to rounding the value of \( x \) to two decimal places.
Determining the Area of a Polygon Within a Rectangle
<p>Given:</p> <p>\( [ABCD] = 112 \, \text{cm}^2 \)</p> <p>\( BE = 3AE, EG = \frac{5}{28} AD, \text{and} 2BF = FC \)</p> <p>Steps:</p> <p>1. Find \( AE \) and \( BE \): Since \( AE + BE = AB \) and \( BE = 3AE \), let \( AE = x \) and hence \( BE = 3x \). So, \( x + 3x = AB \) which means \( 4x = AB \). Because \( AB \cdot AD = 112 \), we have \( 4x \cdot AD = 112 \) and \( x = \frac{AD}{4} \). Therefore, \( BE = 3 \times \frac{AD}{4} = \frac{3}{4} AD \).</p> <p>2. Find \( EG \): \( EG = \frac{5}{28} AD \).</p> <p>3. Find \( BF \) and \( FC \): Since \( 2BF = FC \) and \( BF + FC = BC = AD \) (as ABCD is a rectangle), let \( BF = y \) and hence \( FC = 2y \). So, \( y + 2y = AD \) which means \( 3y = AD \) and \( y = \frac{AD}{3} \). Therefore, \( FC = 2 \times \frac{AD}{3} = \frac{2}{3} AD \).</p> <p>4. Calculate the area \( [BEFG] \):</p> \[\begin{align*} [BEFG] &= [BEC] - [EFGC] \\ &= \frac{1}{2} BE \cdot BC - \frac{1}{2} EG \cdot FC \\ &= \frac{1}{2} \cdot \frac{3}{4} AD \cdot AD - \frac{1}{2} \cdot \frac{5}{28} AD \cdot \frac{2}{3} AD \\ &= \frac{1}{2} \cdot \frac{3}{4} \cdot 112 - \frac{1}{2} \cdot \frac{5}{28} \cdot \frac{2}{3} \cdot 112 \\ &= \frac{3}{8} \cdot 112 - \frac{5}{42} \cdot 112 \\ &= 42 - \frac{5}{42} \cdot 112 \\ &= 42 - \frac{5}{6} \cdot 16 \\ &= 42 - \frac{80}{6} \\ &= 42 - \frac{40}{3} \\ &= 42 - 13\frac{1}{3} \\ &= 28\frac{2}{3} \, \text{cm}^2. \end{align*}\] <p>Thus, the area of \( EFG \) is \( 28\frac{2}{3} \, \text{cm}^2 \).</p>
Calculating Intersection Point of Diagonals in Rectangle
在這個問題中，我們需要確定矩形PQRS中對角線PT和QS的交點T的長度TS。 首先，我們可以觀察到，因為PQRS是矩形，所以對角線PS和QT將等長且交於中點。由此，TS將是PS的一半。 要找出PS的長度，我們可以使用畢氏定理，因為PQ和QR是矩形的兩個直角邊: \(PS^2 = PQ^2 + QR^2\) 代入已知的長度: \(PS^2 = 24^2 + 12^2 = 576 + 144 = 720\) 然後，開平方根計算PS: \(PS = \sqrt{720}\) 這給了我們PS的長度。為了獲得TS，我們只需將結果除以2： \(TS = \frac{PS}{2} = \frac{\sqrt{720}}{2}\) 現在來計算它的確切值: \(PS = \sqrt{720} \approx 26.83 \, \text{cm}\) (保留兩位小數) 所以, \(TS = \frac{26.83}{2} \approx 13.42 \, \text{cm}\) (保留兩位小數) 所以TS的長度大約是13.42厘米。
Calculating Area of Composite Shape
The image appears to show a composite shape consisting of a rectangle and a right-angled triangle attached to it. The rectangle has dimensions of 6 cm by 8 cm, and the triangle shares its height of 6 cm with the rectangle and has a base of 4 cm (since the combined base of the rectangle and triangle is 12 cm, and the rectangle alone is 8 cm). To calculate the area of this shape, we need to find the area of both the rectangle and the triangle and then add these together. Area of the rectangle: \( A_{rectangle} = \text{length} \times \text{width} \) \( A_{rectangle} = 8\, \text{cm} \times 6\, \text{cm} \) \( A_{rectangle} = 48\, \text{cm}^2 \) Area of the triangle: The area of a right-angled triangle can be found using the formula: \( A_{triangle} = \frac{1}{2} \times \text{base} \times \text{height} \) \( A_{triangle} = \frac{1}{2} \times 4\, \text{cm} \times 6\, \text{cm} \) \( A_{triangle} = \frac{1}{2} \times 24\, \text{cm}^2 \) \( A_{triangle} = 12\, \text{cm}^2 \) Now, we add the area of the rectangle and the area of the triangle to get the total area. Total area: \( A_{total} = A_{rectangle} + A_{triangle} \) \( A_{total} = 48\, \text{cm}^2 + 12\, \text{cm}^2 \) \( A_{total} = 60\, \text{cm}^2 \) So, the combined area of the composite shape is 60 square centimeters.
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