Example Question - triangle area
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Relationship Between Triangle Areas in a Quadrilateral
<p>Let the area of triangle \( ABD \) be \( A_{ABD} \).</p> <p>According to the problem, the area of quadrilateral \( ABCD \) is \( 6 \times A_{ABD} \)</p> <p>The area of triangle \( ABC \) can be expressed as:</p> <p> \( A_{ABC} = A_{ABD} + A_{BCD} \)</p> <p>Since \( ABCD \) is a quadrilateral with right angles, triangle \( BCD \) is congruent to triangle \( ABD \). Thus, we can say:</p> <p> \( A_{BCD} = A_{ABD} \)</p> <p>Therefore, \( A_{ABC} = A_{ABD} + A_{ABD} = 2A_{ABD} \)</p> <p>The ratio of the area of triangle \( ABC \) to the area of triangle \( ABD \) is:</p> <p> \( \frac{A_{ABC}}{A_{ABD}} = \frac{2A_{ABD}}{A_{ABD}} = 2 \)</p> <p>Hence, the answer is 2.</p>
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>
Calculating Perimeter and Area of a Triangle
Para resolver este problema, necesitamos hallar dos cosas: el perímetro y el área del triángulo. **Perímetro**: El perímetro de una figura es la suma de las longitudes de todos sus lados. En este caso, el triángulo tiene lados de 3/4 pulgadas, 1 pulgada y \(1\frac{1}{4}\) pulgadas (que es \(1 + \frac{1}{4} = \frac{5}{4}\) pulgadas). Para sumar estas fracciones, deben tener el mismo denominador. El 1 podemos escribirlo como \(4/4\), entonces el perímetro sería: \[ \frac{3}{4} + \frac{4}{4} + \frac{5}{4} = \frac{3 + 4 + 5}{4} = \frac{12}{4} = 3\text{ pulgadas} \] Así que el perímetro del triángulo es 3 pulgadas. **Área**: El área de un triángulo es \(\frac{1}{2} \times \text{base} \times \text{altura}\). En este caso, puedes tomar AD como la base, que es \(1 \frac{3}{4}\) pulgadas (que es \(1 + \frac{3}{4} = \frac{7}{4}\) pulgadas), y BC como la altura, que es \(1 \frac{1}{4}\) pulgadas (o \(\frac{5}{4}\) pulgadas). \[ \text{Área} = \frac{1}{2} \times \frac{7}{4} \times \frac{5}{4} = \frac{1}{2} \times \frac{35}{16} \] Multiplicamos los numeradores y los denominadores por separado: \[ \frac{1}{2} \times \frac{35}{16} = \frac{35}{32} = 1\frac{3}{32}\text{ pulgadas cuadradas} \] Entonces, el área del triángulo es \(1\frac{3}{32}\) pulgadas cuadradas. En resumen, el perímetro del triángulo es 3 pulgadas y el área es \(1\frac{3}{32}\) pulgadas cuadradas.
Rate of Change of Triangle Area with Changing Sides and Angle
To solve the given problem, we must determine the rate of change of the area of a triangle with sides A and B and angle B between them, given how A and B are changing with time. We are given: - The area of a triangle is given by the formula \( A = \frac{1}{2} ab \sin{\theta} \), where \( a \) and \( b \) are the sides of the triangle and \( \theta \) is the angle between them. For our problem, we can rewrite the given area formula as \( Area = \frac{1}{2} AB \sin{B} \). - Side \( A \) is increasing at 0.1 units/s. - Side \( B \) is decreasing at 0.1 units/s. - Angle \( B \) is increasing at 0.2 radians/s. We want to find the rate of change of the triangle's area, which can be found by differentiating the area formula with respect to time \( t \). This is an application of the chain rule in calculus. So we differentiate \( Area = \frac{1}{2} AB \sin{B} \) with respect to \( t \): \( \frac{d(Area)}{dt} = \frac{1}{2} \left[ B \sin{B} \frac{dA}{dt} + A \sin{B} \frac{dB}{dt} + AB \cos{B} \frac{dB}{dt} \right] \) Plugging in the rates given: \( \frac{dA}{dt} = 0.1 \, \text{units/s} \) \( \frac{dB}{dt} = -0.1 \, \text{units/s} \) \( \frac{dB}{dt} = 0.2 \, \text{radians/s} \) Substitute these into the derivative: \( \frac{d(Area)}{dt} = \frac{1}{2} \left[ B \cdot \sin{B} \cdot 0.1 + A \cdot \sin{B} \cdot (-0.1) + AB \cdot \cos{B} \cdot 0.2 \right] \) Unfortunately, the image doesn't provide the exact values of \( A \), \( B \), and angle \( B \) at the moment of calculation, so we cannot give a numerical answer. You would need to substitute the current values of \( A \), \( B \), and angle \( B \) into the derivative to find the rate of change at that particular moment.
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