In △ABC, DE || AC, where D and E are the points on sides AB and BC, respectively. If BD = 4 cm and AD = 2\(\sqrt{2}\) cm, then what is the ratio of the area of △BDE to the trapezium ADEC?
2 : (1 + \(\sqrt{2}\))
2 : (1 + 2\(\sqrt{2}\))
2 : (1 + 3\(\sqrt{2}\))
1 : (1 + 2\(\sqrt{2}\))
We are given a triangle \(\triangle ABC\) with a line segment \(DE\) parallel to \(AC\). This implies that \(\triangle BDE\) is similar to \(\triangle BAC\). We can use the property that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. We will find the area of \(\triangle BDE\) and the area of trapezium \(ADEC\) in terms of the area of \(\triangle BAC\) and then find their ratio.
Identify similar triangles: Given that \(DE \parallel AC\), by the Basic Proportionality Theorem (Thales's Theorem) and its converse, we can conclude that \(\triangle BDE \sim \triangle BAC\).
Find the ratio of corresponding sides: We are given \(BD = 4\text{ cm}\) and \(AD = 2\sqrt{2}\text{ cm}\). Therefore, \(AB = BD + AD = 4 + 2\sqrt{2}\text{ cm}\). The ratio of corresponding sides is:
\[ \frac{BD}{AB} = \frac{4}{4 + 2\sqrt{2}} = \frac{4}{2(2 + \sqrt{2})} = \frac{2}{2 + \sqrt{2}} \]To rationalize the denominator, multiply the numerator and denominator by the conjugate \((2 - \sqrt{2})\):
\[ \frac{2}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \frac{2(2 - \sqrt{2})}{2^2 - (\sqrt{2})^2} = \frac{2(2 - \sqrt{2})}{4 - 2} = \frac{2(2 - \sqrt{2})}{2} = 2 - \sqrt{2} \]So, the ratio of sides is \(\frac{BD}{AB} = 2 - \sqrt{2}\).
Find the ratio of areas of similar triangles: The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides:
\[ \frac{\text{Area}(\triangle BDE)}{\text{Area}(\triangle BAC)} = \left(\frac{BD}{AB}\right)^2 = (2 - \sqrt{2})^2 \] \[ (2 - \sqrt{2})^2 = 2^2 - 2(2)(\sqrt{2}) + (\sqrt{2})^2 = 4 - 4\sqrt{2} + 2 = 6 - 4\sqrt{2} \]So, \(\text{Area}(\triangle BDE) = (6 - 4\sqrt{2}) \times \text{Area}(\triangle BAC)\).
Calculate the area of trapezium ADEC: The area of trapezium \(ADEC\) is the area of \(\triangle BAC\) minus the area of \(\triangle BDE\).
\[ \text{Area}(\text{trapezium } ADEC) = \text{Area}(\triangle BAC) - \text{Area}(\triangle BDE) \] \[ \text{Area}(\text{trapezium } ADEC) = \text{Area}(\triangle BAC) - (6 - 4\sqrt{2}) \times \text{Area}(\triangle BAC) \] \[ \text{Area}(\text{trapezium } ADEC) = \text{Area}(\triangle BAC) [1 - (6 - 4\sqrt{2})] \] \[ \text{Area}(\text{trapezium } ADEC) = \text{Area}(\triangle BAC) [1 - 6 + 4\sqrt{2}] \] \[ \text{Area}(\text{trapezium } ADEC) = \text{Area}(\triangle BAC) [4\sqrt{2} - 5] \]Find the ratio of Area(\(\triangle BDE\)) to Area(trapezium ADEC):
\[ \frac{\text{Area}(\triangle BDE)}{\text{Area}(\text{trapezium } ADEC)} = \frac{(6 - 4\sqrt{2}) \times \text{Area}(\triangle BAC)}{(4\sqrt{2} - 5) \times \text{Area}(\triangle BAC)} \] \[ \frac{\text{Area}(\triangle BDE)}{\text{Area}(\text{trapezium } ADEC)} = \frac{6 - 4\sqrt{2}}{4\sqrt{2} - 5} \]Let's simplify this expression. Multiply the numerator and denominator by the conjugate of the denominator, which is \((4\sqrt{2} + 5)\):
\[ \frac{6 - 4\sqrt{2}}{4\sqrt{2} - 5} \times \frac{4\sqrt{2} + 5}{4\sqrt{2} + 5} = \frac{(6 - 4\sqrt{2})(4\sqrt{2} + 5)}{(4\sqrt{2})^2 - 5^2} \]Numerator:
\[ 6(4\sqrt{2}) + 6(5) - 4\sqrt{2}(4\sqrt{2}) - 4\sqrt{2}(5) \] \[ = 24\sqrt{2} + 30 - 16(2) - 20\sqrt{2} \] \[ = 24\sqrt{2} + 30 - 32 - 20\sqrt{2} \] \[ = (24\sqrt{2} - 20\sqrt{2}) + (30 - 32) \] \[ = 4\sqrt{2} - 2 \]Denominator:
\[ (4\sqrt{2})^2 - 5^2 = 16(2) - 25 = 32 - 25 = 7 \]So the ratio is \(\frac{4\sqrt{2} - 2}{7}\). This does not match any option directly. Let's recheck the calculation.
Let's recheck the ratio of sides. \(BD = 4\) and \(AD = 2\sqrt{2}\). So \(AB = 4 + 2\sqrt{2}\). Ratio of sides \[\frac{BD}{AB} = \frac{4}{4+2\sqrt{2}} = \frac{2}{2+\sqrt{2}}\]. Area ratio \[\frac{\text{Area}(\triangle BDE)}{\text{Area}(\triangle BAC)} = \left(\frac{2}{2+\sqrt{2}}\right)^2 = \frac{4}{(2+\sqrt{2})^2} = \frac{4}{4 + 4\sqrt{2} + 2} = \frac{4}{6 + 4\sqrt{2}}\]. This can be simplified by dividing numerator and denominator by 2: \(\frac{2}{3 + 2\sqrt{2}}\). Let \(A_{BDE}\) denote \(\text{Area}(\triangle BDE)\) and \(A_{BAC}\) denote \(\text{Area}(\triangle BAC)\). So, \(A_{BDE} = \frac{2}{3 + 2\sqrt{2}} A_{BAC}\). Area of trapezium \[ADEC = A_{BAC} - A_{BDE} = A_{BAC} - \frac{2}{3 + 2\sqrt{2}} A_{BAC} = A_{BAC} \left(1 - \frac{2}{3 + 2\sqrt{2}}\right)\]. \[= A_{BAC} \left(\frac{3 + 2\sqrt{2} - 2}{3 + 2\sqrt{2}}\right) = A_{BAC} \left(\frac{1 + 2\sqrt{2}}{3 + 2\sqrt{2}}\right)\]. Now, the ratio \[\frac{A_{BDE}}{\text{Area}(\text{trapezium } ADEC)}\] is:
\[ \frac{\frac{2}{3 + 2\sqrt{2}} A_{BAC}}{\frac{1 + 2\sqrt{2}}{3 + 2\sqrt{2}} A_{BAC}} = \frac{2}{1 + 2\sqrt{2}} \]The ratio is \(2 : (1 + 2\sqrt{2})\).
B) The calculated ratio is \(2 : (1 + 2\sqrt{2})\).
All 50 students of a class are standing in a straight row facing north. Rakesh is 17th from the right end, while Kamlesh is 22nd from the left end. How many students are standing between Rakesh and Kamlesh?
To find the number of students between Rakesh and Kamlesh, we first need to determine their positions from a common reference point, for example, the left end of the row. We are given Rakesh's position from the right end and Kamlesh's position from the left end. We will convert Rakesh's position to be from the left end and then calculate the number of students between them.
The formula to convert position from one end to the other in a row is:
\[ \text{Position from Left End} = \text{Total Students} - \text{Position from Right End} + 1 \]For Rakesh:
\[ \text{Rakesh's position from Left End} = 50 - 17 + 1 = 33 + 1 = 34 \]So, Rakesh is 34th from the left end.
Both positions are now from the left end:
Since Rakesh (34th) is further from the left end than Kamlesh (22nd), we can find the number of students between them by subtracting their positions and then subtracting 1 (because we are counting the students *between* them, not including them).
\[ \text{Students between them} = \text{Rakesh's position from Left} - \text{Kamlesh's position from Left} - 1 \] \[ \text{Students between them} = 34 - 22 - 1 = 12 - 1 = 11 \]A) 11 is the correct number of students standing between Rakesh and Kamlesh.
To find the number of sides of a regular polygon given its exterior angle, we use the property that the sum of the exterior angles of any convex polygon is 360 degrees. For a regular polygon, all exterior angles are equal.
The measure of each exterior angle of a regular polygon with \(n\) sides is given by the formula:
\[ \text{Exterior Angle} = \frac{360^\circ}{n} \]where \(n\) is the number of sides of the polygon.
We are given that the exterior angle is \(40^\circ\). So, we have:
\[ 40^\circ = \frac{360^\circ}{n} \]To find the number of sides \(n\), we can rearrange the equation:
\[ n = \frac{360^\circ}{40^\circ} \] \[ n = 9 \]The number of sides in the regular polygon is 9.
Correct Option: C) 9
This question tests knowledge about the types of epithelial tissues lining different organs in the human body. Specifically, it asks to identify the organ that is NOT lined by squamous epithelium. We need to recall the primary function and location of squamous epithelium and compare it with the lining of each given option.
Correct Option: D) Kidney tubules
The problem involves decoding a word based on a given coding pattern. We need to identify the rule used to transform the letters of the original words into their coded forms. Once the rule is found, we apply it to the target word to find its code.
Correct Option: D) VMGIB
The question asks to identify which of the given mountain ranges is NOT a part of the Himalayan mountains. This requires knowledge of the major mountain ranges in and around the Indian subcontinent and their classification.
Correct Option: C) Mahadeo Range — The Mahadeo Range is part of the Satpura Range in central India, which is geographically and geologically distinct from the Himalayan mountain system.
The problem presents a proportion, which is an equation stating that two ratios are equal. We are given three terms of the proportion and need to find the fourth term, denoted by 'x'. The fundamental property of proportions states that the product of the means is equal to the product of the extremes.
Understand the proportion notation: The notation \(a : b :: c : d\) means that the ratio \(a/b\) is equal to the ratio \(c/d\). This can be written as:
\[ \frac{a}{b} = \frac{c}{d} \]Apply the given values to the proportion: We are given \(9.6 : 16.8 :: 18 : x\). So, we can write this as:
\[ \frac{9.6}{16.8} = \frac{18}{x} \]Cross-multiply to solve for x: To solve for \(x\), we cross-multiply the terms:
\[ 9.6 \times x = 16.8 \times 18 \]Calculate the product on the right side:
\[ 16.8 \times 18 = 302.4 \]Set up the equation with the calculated product:
\[ 9.6x = 302.4 \]Isolate x by dividing both sides by 9.6:
\[ x = \frac{302.4}{9.6} \]Perform the division: To simplify the division, we can multiply both the numerator and the denominator by 10 to remove the decimal points:
\[ x = \frac{3024}{96} \]Divide 3024 by 96:
\[ 3024 \div 96 = 31.5 \]Therefore, the value of x is 31.5.
D) 31.5 is the correct value for x, as derived from the proportion calculation.
This problem can be solved by setting up a system of linear equations based on the given information. We will assign variables to the cost of one table and one chair, form two equations, and then solve them simultaneously to find the cost of one chair. Finally, we will calculate the cost of 5 chairs.
Let \(T\) be the cost of one table and \(C\) be the cost of one chair.
From the first statement, "The cost of 2 tables and 3 chairs is Rs 540", we can form the equation:
\[ 2T + 3C = 540 \quad \text{(Equation 1)} \]From the second statement, "that of 2 tables and 1 chair is ₹470", we can form the equation:
\[ 2T + 1C = 470 \quad \text{(Equation 2)} \]To find the cost of one chair (\(C\)), we can subtract Equation 2 from Equation 1:
\[\begin{array}{rcl} (2T + 3C) - (2T + 1C) &=& 540 - 470 \ 2T + 3C - 2T - C &=& 70 \ 2C &=& 70 \ C &=& \frac{70}{2} \ C &=& 35 \end{array}\]So, the cost of one chair is Rs 35.
The question asks for the cost of 5 chairs. We multiply the cost of one chair by 5:
\[ \text{Cost of 5 chairs} = 5 \times C = 5 \times 35 = 175 \]Correct Option: A) 175
The problem requires us to find the value of 'a' in the given trigonometric identity. We will expand the left-hand side of the equation using algebraic identities and fundamental trigonometric identities to simplify it and match it with the right-hand side.
Given the equation:
\[ (\sin\beta + \operatorname{cosec}\beta)^2 + (\sec\beta - \cos\beta)^2 = a + \tan^2\beta + \cot^2\beta \]Expand the first term \((\sin\beta + \operatorname{cosec}\beta)^2\) using the identity \((x+y)^2 = x^2 + y^2 + 2xy\):
\[ (\sin\beta + \operatorname{cosec}\beta)^2 = \sin^2\beta + \operatorname{cosec}^2\beta + 2\sin\beta\operatorname{cosec}\beta \]Since \(\operatorname{cosec}\beta = \frac{1}{\sin\beta}\), we have \[\sin\beta\operatorname{cosec}\beta = \sin\beta \cdot \frac{1}{\sin\beta} = 1\]. So the first term becomes:
\[ \sin^2\beta + \operatorname{cosec}^2\beta + 2(1) = \sin^2\beta + \operatorname{cosec}^2\beta + 2 \]Expand the second term \((\sec\beta - \cos\beta)^2\) using the identity \((x-y)^2 = x^2 + y^2 - 2xy\):
\[ (\sec\beta - \cos\beta)^2 = \sec^2\beta + \cos^2\beta - 2\sec\beta\cos\beta \]Since \(\sec\beta = \frac{1}{\cos\beta}\), we have \[\sec\beta\cos\beta = \frac{1}{\cos\beta} \cdot \cos\beta = 1\]. So the second term becomes:
\[ \sec^2\beta + \cos^2\beta - 2(1) = \sec^2\beta + \cos^2\beta - 2 \]Now, substitute these expanded forms back into the original equation:
\[ (\sin^2\beta + \operatorname{cosec}^2\beta + 2) + (\sec^2\beta + \cos^2\beta - 2) = a + \tan^2\beta + \cot^2\beta \]Combine like terms on the left-hand side:
\[ \sin^2\beta + \cos^2\beta + \operatorname{cosec}^2\beta + \sec^2\beta + 2 - 2 = a + \tan^2\beta + \cot^2\beta \]We know the fundamental trigonometric identity \(\sin^2\beta + \cos^2\beta = 1\). Also, \(2 - 2 = 0\). So the equation simplifies to:
\[ 1 + \operatorname{cosec}^2\beta + \sec^2\beta = a + \tan^2\beta + \cot^2\beta \]Use the identities \(\operatorname{cosec}^2\beta = 1 + \cot^2\beta\) and \(\sec^2\beta = 1 + \tan^2\beta\):
\[ 1 + (1 + \cot^2\beta) + (1 + \tan^2\beta) = a + \tan^2\beta + \cot^2\beta \]Simplify the left-hand side:
\[ 1 + 1 + \cot^2\beta + 1 + \tan^2\beta = a + \tan^2\beta + \cot^2\beta \]\[ 3 + \tan^2\beta + \cot^2\beta = a + \tan^2\beta + \cot^2\beta \]By comparing both sides of the equation, we can see that:
\[ a = 3 \]A) 3 — The step-by-step simplification of the given trigonometric identity leads to \(3 + \tan^2\beta + \cot^2\beta = a + \tan^2\beta + \cot^2\beta\), from which we deduce that \(a=3\).
Which of the following interchanges of numbers would make the given equation correct?
5 × 15 ÷ 5 + 22 − 18 = 25
The problem requires us to find which interchange of two numbers in the given equation \(5 \times 15 \div 5 + 22 - 18 = 25\) will make the equation correct. We will test each option by swapping the specified numbers and then evaluating the left-hand side (LHS) of the equation using the order of operations (BODMAS/PEMDAS) to see if it equals the right-hand side (RHS), which is 25.
The given equation is:
\[ 5 \times 15 \div 5 + 22 - 18 = 25 \]Interchange 5 and 15:
\[ 15 \times 5 \div 15 + 22 - 18 \]Since \(9 \neq 25\), Option A is incorrect.
Interchange 22 and 25. The equation becomes:
\[ 5 \times 15 \div 5 + 25 - 18 \]The original RHS was 25. After interchanging 22 and 25, the new RHS should be 22. Since \(22 = 22\), Option B makes the equation correct.
Interchange 22 and 18:
\[ 5 \times 15 \div 5 + 18 - 22 \]Since \(11 \neq 25\), Option C is incorrect.
Interchange 5 and 25. The equation becomes:
\[ 25 \times 15 \div 25 + 22 - 18 \]Since \(19 \neq 25\), Option D is incorrect.
Correct Option: B) Interchanging 22 and 25 makes the equation correct. The original equation is \(5 \times 15 \div 5 + 22 - 18 = 25\). If we swap 22 and 25, the equation becomes \(5 \times 15 \div 5 + 25 - 18 = 22\). Evaluating the LHS: \(75 \div 5 + 25 - 18 = 15 + 25 - 18 = 40 - 18 = 22\). Since LHS = RHS (\(22 = 22\)), this interchange is correct.