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The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs. 58 per hectare is Rs. 783, Find its base and height.

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To solve the problem step by step, we will follow these instructions: ### Step 1: Define Variables Let the altitude of the triangular field be \( x \) meters. According to the problem, the base of the triangular field is three times the altitude. Therefore, the base \( b \) can be expressed as: \[ b = 3x \] ### Step 2: Calculate the Area of the Triangle The area \( A \) of a triangle is given by the formula: \[ A = \frac{1}{2} \times \text{base} \times \text{height} \] Substituting the values we have: \[ A = \frac{1}{2} \times (3x) \times x = \frac{3x^2}{2} \] ### Step 3: Determine the Cost of Sowing The total cost of sowing the field is given as Rs. 783, and the cost per hectare is Rs. 58. To find the area in hectares, we can use the formula: \[ \text{Area in hectares} = \frac{\text{Total cost}}{\text{Cost per hectare}} \] Substituting the values: \[ \text{Area in hectares} = \frac{783}{58} \] Calculating this gives: \[ \text{Area in hectares} = 13.5 \text{ hectares} \] ### Step 4: Convert Area to Square Meters Since 1 hectare = \( 10^4 \) square meters, we convert the area: \[ \text{Area in square meters} = 13.5 \times 10^4 = 135000 \text{ m}^2 \] ### Step 5: Set Up the Equation Now, we set the area calculated using the altitude and base equal to the area calculated from the cost: \[ \frac{3x^2}{2} = 135000 \] ### Step 6: Solve for \( x \) To solve for \( x \), we first multiply both sides by 2: \[ 3x^2 = 270000 \] Now, divide by 3: \[ x^2 = 90000 \] Taking the square root of both sides gives: \[ x = 300 \text{ meters} \] ### Step 7: Find the Base Now that we have the altitude, we can find the base: \[ b = 3x = 3 \times 300 = 900 \text{ meters} \] ### Final Answer Thus, the altitude of the triangular field is \( 300 \) meters and the base is \( 900 \) meters. ---

To solve the problem step by step, we will follow these instructions: ### Step 1: Define Variables Let the altitude of the triangular field be \( x \) meters. According to the problem, the base of the triangular field is three times the altitude. Therefore, the base \( b \) can be expressed as: \[ b = 3x \] ...
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RS AGGARWAL-AREAS OF TRIANGLES AND QUADRILATERALS-Exercise 14
  1. Find the area of the triangle whose base measures 24 cm and the corres...

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  2. The base of a triangular field is three times its altitude. If the cos...

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  3. Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm i...

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  4. Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30...

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  5. Find the area of a triangular field whose sides are 91 m, 98 m and 105...

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  6. The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter...

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  7. The perimeter of a triangular field is 540 m and its sides are in the ...

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  8. Two sides of a triangular field are 85 m and 154 m in length and its p...

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  9. Find the area of an isosceles triangle each of whose equal sides measu...

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  10. The base of an isosceles triangle measures 80 cm and its area is 360 c...

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  11. The perimeter of an isosceles triangle is 32 cm. The ratio of the equa...

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  12. The perimeter of a triangle is 50 cm. One side of the triangle is 4 cm...

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  13. The triangular side walls of a flyover have been used for advertisemen...

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  14. The perimeter of an isosceles triangle is 42 cm and its base is 1(1)/(...

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  15. The area of an equilateral triangle is 36 sqrt(3) cm^(2). Its perimete...

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  16. If the area of an equilateral triangle is 81 sqrt(3) cm^(2), find its ...

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  17. Each side of an equilateral triangle measures 8 cm. Find (i) the area ...

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  18. The height of an equilateral triangle measures 9 cm. Find its area, co...

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  19. The base of a right-angled triangle measures 48 cm and its hypotenuse ...

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  20. The sides of a quadrilateral ABCD taken in order are 6 cm, 8 cm, 12 cm...

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