The question presented in the image states:

Q16: A rectangle whose length is twice its breadth has an area of $128 \, \text{cm}^2$. What is the length of the rectangle?

Let's solve this step-by-step.

Step 1: Set up variables

Let:

• $l$ be the length of the rectangle
• $b$ be the breadth of the rectangle

Given that:

• $l = 2b$ (the length is twice the breadth)
• The area of the rectangle is given by $\text{Area} = l \times b = 128 \, \text{cm}^2$

Step 2: Substitute the expression for $l$

Since $l = 2b$, substitute this into the area formula: $\text{Area} = l \times b = 2b \times b = 2b^2$ $2b^2 = 128$

Step 3: Solve for $b$

Divide both sides by 2: $b^2 = \frac{128}{2} = 64$

Now, take the square root of both sides: $b = \sqrt{64} = 8 \, \text{cm}$

Step 4: Find the length $l$

Since $l = 2b$, the length is: $l = 2 \times 8 = 16 \, \text{cm}$

Answer:

The length of the rectangle is 16 cm, which corresponds to option (D).

Would you like more details or have any further questions?

Here are 5 questions for expanding on this:

1. What would be the breadth if the length was thrice the breadth for the same area?
2. How do you find the area of a rectangle if only the diagonal is given?
3. What is the perimeter of this rectangle based on the given dimensions?
4. How would the dimensions change if the area were doubled?
5. What are the units of area, and how do they relate to the units of length?

Tip: When solving geometry problems involving shapes, always write down the relationships between dimensions, like length and breadth, before diving into calculations.