Question

A tunnel is in the form of a cuboid. The dimensions at it entrance are in the ratio 4:3, the length of the tunnel is 3 m. If the area of the four walls of the tunnel is 280 cm^2, find the dimensions of the tunnel's entrance. ​

Answer 1

Answer:

$4.3 \times 3 = 129 \times 280 + 100 \times 2 = 8945$

Answer 2

The dimensions of the tunnel's entrance which is in the form of a cuboid are 12.24 cm X 9.18 cm.

Step-by-step explanation:

The length of the tunnel = 3 cm

The ratio of the dimensions of the entrance of the tunnel = 4:3

Since the tunnel is given to be in the form of a cuboid, so let the breadth be “4x” and the height be “3x”.

The area of the 4 walls of the tunnel = 280 cm²

We know the formula for the area of 4 walls of a cuboid is given as,

Area = 2* height * (length + breadth)

Now, by substituting the given values in the formula, we get

280 = 2 * 3x * [3 + 4x]

⇒ 140 =   9x + 12x²

⇒ 12x² + 9x - 140 = 0

Using the formula, x = $\frac{-b +- \sqrt{b^2 - 4ac} }{2a}$ , a = 12, b = 9 & c = -140

⇒ x = $\frac{-9 +- \sqrt{81 + 6720} }{24}$

⇒ x = $\frac{-9 + \sqrt{6801}}{24} or \frac{-9 - \sqrt{6801}}{24}$

neglecting the negative value

⇒ x = $\frac{-9 + 82.46}{24}$

⇒ x = $\frac{73.46}{24}$

⇒ x = 3.06 cm

∴ 4x = 4 × 3.06 = 12.24 cm

3x = 3 × 3.06 = 9.18 cm

Thus, the dimensions of the tunnel's entrance is 12.24 cm × 9.18 cm.

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