Question

The sum of the length, breadth and depth of a cuboid is 19 cm and its diagonal is $5\sqrt{5$ cm. Its surface area is :}

• A. $361 \text{ cm}^2$
• B. $125 \text{ cm}^2$
• C. $236 \text{ cm}^2$
• D. $486 \text{ cm}^2$

Hint:

This problem cleverly uses the algebraic identity $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$ to connect the sum of dimensions, the diagonal, and the surface area of a cuboid. Recognizing this identity is key to a straightforward solution.

Answer 1

The Correct Option is C

Step 1: Define variables and write down the given information.
Let the length, breadth, and depth (height) of the cuboid be $l$, $b$, and $h$ respectively.
Given:
1. Sum of length, breadth, and depth: $l + b + h = 19$ cm
2. Length of the diagonal: $D = 5\sqrt{5}$ cm Step 2: Recall the formulas for the diagonal and surface area of a cuboid.
The formula for the diagonal of a cuboid is $D = \sqrt{l^2 + b^2 + h^2}$.
The formula for the total surface area of a cuboid is $SA = 2(lb + bh + hl)$. Step 3: Use the diagonal information.
We are given $D = 5\sqrt{5}$ cm.
So, $\sqrt{l^2 + b^2 + h^2} = 5\sqrt{5}$.
Squaring both sides:
$l^2 + b^2 + h^2 = (5\sqrt{5})^2$
$l^2 + b^2 + h^2 = 25 \times 5$
$l^2 + b^2 + h^2 = 125$
Step 4: Use the sum of sides information to find the surface area.
We know the algebraic identity:
$(l + b + h)^2 = l^2 + b^2 + h^2 + 2(lb + bh + hl)$ We have:
$l + b + h = 19$
$l^2 + b^2 + h^2 = 125$ Substitute these values into the identity:
$(19)^2 = 125 + 2(lb + bh + hl)$
$361 = 125 + 2(lb + bh + hl)$
We recognize $2(lb + bh + hl)$ as the surface area ($SA$) of the cuboid. $361 = 125 + SA$ Step 5: Solve for the surface area.
$SA = 361 - 125$
$SA = 236 \text{ cm}^2$ Step 6: Compare with the given options.
The calculated surface area is $236 \text{ cm}^2$, which matches option (3). (3) $236 \, \text{cm^{2}$}