Question

In a right-angled trapezium $ABCD$, $\angle A = 90^\circ$, $\angle D = 90^\circ$, $AB = 7a + 1$, $CD = 3a + 1$ and $AD = 3a$. If the perimeter of the trapezium is greater than 56 but less than 92, then the range of possible values of $a$ is

• A. $4 < a < \frac{11}{2}$
• B. $\frac{7}{2} < a < 5$
• C. $3 < a < \frac{9}{2}$
• D. $3 < a < 7$
• E. $3 < a < 5$

Hint:

In trapezium problems, drawing a perpendicular to form a rectangle and a right triangle is the standard trick to find an unknown non-parallel side.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
A right-angled trapezium with $\angle A = \angle D = 90^\circ$ has $AD$ perpendicular to both $AB$ and $CD$. The perimeter is the sum of all four sides.

Step 2: Detailed Explanation:
1. Side lengths given: $AB = 7a+1$, $CD = 3a+1$, $AD = 3a$.
2. Find the fourth side $BC$:
Drop a perpendicular from $C$ to $AB$. Let this point be $E$.
$CE = AD = 3a$.
$AE = CD = 3a+1$.
$EB = AB - AE = (7a+1) - (3a+1) = 4a$.
By Pythagoras theorem in $\triangle CEB$:
\[ BC = \sqrt{CE^2 + EB^2} = \sqrt{(3a)^2 + (4a)^2} = \sqrt{9a^2 + 16a^2} = \sqrt{25a^2} = 5a \]
3. Calculate the Perimeter $P$:
\[ P = AB + BC + CD + DA \]
\[ P = (7a+1) + (5a) + (3a+1) + (3a) \]
\[ P = 18a + 2 \]
4. Set up the inequalities based on the given bounds:
\[ 56 < 18a + 2 < 92 \]
Subtract 2 from all parts:
\[ 54 < 18a < 90 \]
Divide by 18:
\[ \frac{54}{18} < a < \frac{90}{18} \]
\[ 3 < a < 5 \]

Step 3: Final Answer:
The range of values for $a$ is $3 < a < 5$.