
The ratio of corresponding sides for similar triangles is constant. So, we have the equation:
\[ \frac{4 \text{ in.}}{7 \text{ in.}} = \frac{x}{28 \text{ in.}} \]
Convert \( 21 \frac{3}{4} \) ft. to inches:
\( 21 \frac{3}{4} \) ft. = \( 21 \frac{3}{4} \times 12 = 28 \) in.
Now, solve the proportion:
\[ \frac{4}{7} = \frac{x}{28} \]
Cross multiply:
\[ 4 \times 28 = 7 \times x \implies 112 = 7x \implies x = \frac{112}{7} = 16 \text{ in.} \]
Thus, one possible value for \( x \) is 16 in. Therefore, the final answer is:
16 in.

In \(\triangle ABC\), \(DE \parallel BC\). If \(AE = (2x+1)\) cm, \(EC = 4\) cm, \(AD = (x+1)\) cm and \(DB = 3\) cm, then the value of \(x\) is

In the adjoining figure, PA and PB are tangents to a circle with centre O such that $\angle P = 90^\circ$. If $AB = 3\sqrt{2}$ cm, then the diameter of the circle is
In the adjoining figure, TS is a tangent to a circle with centre O. The value of $2x^\circ$ is