Quadrilaterals – MBA CET questions with solutions

Approach: Use (l + b)² = l² + b² + 2lb.

Solution:
Perimeter 46 → l + b = 23.
Diagonal 17 → l² + b² = 289.
(l + b)² = 529 = 289 + 2lb → 2lb = 240 → lb = 120 cm².
(Sides 15 and 8: 15 + 8 = 23, 15² + 8² = 289 ✓.)

Why the other options are wrong:

• A, B, C, E: do not satisfy 2lb = 529 − 289.

Cetking shortcut: Spot the Pythagorean triple 8-15-17 with 8 + 15 = 23 → area 120.

Answer: D

Approach: Area of trapezium = ½ (sum of parallel sides) × height.

Solution:
½ (a + b) × 19 = 475 → a + b = 50.
a − b = 4.
a = 27 cm, b = 23 cm.

Why the other options are wrong:

• A (54), B (52), D (48), E (46): sums are not 50.

Cetking shortcut: Only the option whose two sides add to 50 works — check sums directly.

Answer: C

Approach: Diagonals of a rhombus bisect each other at right angles.

Solution:
Half-diagonals = 12 cm and 5 cm.
Side = √(12² + 5²) = √169 = 13 cm.
Perimeter = 4 × 13 = 52 cm.

Why the other options are wrong:
B, C, D, E – use full diagonals or add them instead of using Pythagoras.

Cetking shortcut: 5–12–13 triplet: half-diagonals 5 and 12 ⇒ side 13.

Answer: A

Approach: Saving = (a + b) − diagonal = b/2, where b is the longer side.

Solution:
a + b − √(a² + b²) = b/2 → √(a² + b²) = a + b/2
Square: a² + b² = a² + ab + b²/4
3b²/4 = ab → a/b = 3/4

Why the other options are wrong:
A (2/3): 2 + 3 − √13 ≈ 1.39 ≠ 1.5.
C/D/E: savings don't equal half the longer side.

Cetking shortcut: Try 3-4-5: walking 3 + 4 = 7, diagonal 5, saved 2 = half of 4 ✓.

Answer: B

Approach: Parallel sides make △EDC ∼ △EAB; the ratio of the parallel sides gives the ratio of the slant sides.

Solution:
DC ∥ AB, so △EDC ∼ △EAB with ratio DC : AB = 1 : 2.
ED : EA = 1 : 2 → D is the midpoint of EA → EA = 2 × AD.
Similarly EB = 2 × BC.
Perimeter of △AEB = EA + EB + AB = 2(AD + BC) + AB = 2 × 3 + 2 = 8 cm.

Cetking shortcut: When the smaller parallel side is half the larger, each slant side of the trapezium is exactly half of the triangle's side.

Answer: B

Approach: Turn the condition into an equation for the diagonal and apply Pythagoras.

Solution:
Let longer side = l, shorter = b, diagonal = d.
(l + b) − d = l/2 → d = l/2 + b.
Pythagoras: l² + b² = (l/2 + b)² = l²/4 + lb + b² → 3l²/4 = lb → b = 3l/4.
b : l = 3 : 4.

Cetking shortcut: Test 3-4-5: (3 + 4) − 5 = 2 = 4/2 ✓.

Answer: D