Question

A trapezium has vertices marked as P, Q, R, and S (in that order anticlockwise). The side PQ is parallel to side SR. Further, it is given that, PQ = 11 cm, QR = 4 cm, RS = 6 cm, and SP = 3 cm. What is the shortest distance between PQ and SR (in cm)?

• A. 1.80
• B. 2.40
• C. 4.20
• D. 5.76

Hint:

The shortest distance between parallel sides in a trapezium is the perpendicular distance between them, which can be derived from the geometry of the figure.

Answer 1

The Correct Option is B

The shortest distance between two parallel sides in a trapezium is the perpendicular distance between them. To find this, we can use the formula for the area of the trapezium and equate it to the sum of the areas of two triangles and a rectangle formed by the given dimensions.
First, calculate the area of the trapezium using the formula:
\[ A = \frac{1}{2} \times (b_1 + b_2) \times h \] where \( b_1 \) and \( b_2 \) are the lengths of the parallel sides and \( h \) is the perpendicular height (the shortest distance). We are given:
- \( b_1 = PQ = 11 \, \text{cm} \)
- \( b_2 = SR = 6 \, \text{cm} \)
- The total length of the non-parallel sides \( QR + SP = 4 + 3 = 7 \, \text{cm} \)
Next, use the fact that the area of the trapezium can also be expressed as the area of the rectangle plus the two triangular areas formed by the slant sides. After solving the geometry and using the trapezium area formula, the shortest distance (height) is found to be:
\[ h = 2.40 \, \text{cm} \]