Three angles of a quadrilateral are 75°, 90° and 75°. The fourth angle is:
Correct Answer:(D) 120°
Step-by-step Solution:
Sum of angles = 360°. Fourth angle = 360° - (75+90+75) = 360° - 240° = 120°.
Q2MCQ1 Mark
A diagonal of a parallelogram divides it into two:
Correct Answer:(A) Congruent triangles
Step-by-step Solution:
Diagonal of a parallelogram divides it into two congruent triangles.
Q3MCQ1 Mark
If diagonals of a parallelogram are equal, then it is a:
Correct Answer:(A) Rectangle
Step-by-step Solution:
A parallelogram with equal diagonals is a rectangle.
Q4MCQ1 Mark
The line segment joining midpoints of two sides of a triangle is parallel to third side and is:
Correct Answer:(A) Half of it
Step-by-step Solution:
Mid-Point Theorem: Segment is parallel to 3rd side and half of it.
Q5Assertion-Reason1 Mark
Assertion (A): Diagonals of a rhombus bisect each other at right angles.
Reason (R): All four sides of a rhombus are equal.
Correct Answer:(B) Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
Step-by-step Solution:
Both are true properties of a rhombus, but R does not explain A.
Q6Assertion-Reason1 Mark
Assertion (A): The quadrilateral formed by joining midpoints of sides of a quadrilateral is a parallelogram.
Reason (R): Line joining midpoints of two sides of a triangle is parallel to third side and half of it.
Correct Answer:(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Step-by-step Solution:
Both A and R are true and R is used to prove A.
Q7Short Answer2 Marks
Angles of quadrilateral are in ratio 3 : 5 : 9 : 13. Find all angles.
ABCD is a parallelogram and AP and CQ are perpendiculars from A and C on diagonal BD. Show △APB ≅ △CQD.
Correct Answer:Proved
Step-by-step Solution:
In △APB and △CQD: AB=CD, ∠APB=∠CQD=90°, ∠ABP=∠CDQ (alt int). By AAS, △APB ≅ △CQD.
Q11Short Answer2 Marks
In △ABC, D, E and F are midpoints of sides AB, BC and CA. Show △ABC is divided into 4 congruent triangles.
Correct Answer:Proved
Step-by-step Solution:
DF || BC, FE || AB, DE || AC. By Mid-Point theorem, 4 sub-triangles are congruent.
Q12Long Answer3 Marks
ABCD is a rhombus and P, Q, R, S are midpoints of sides AB, BC, CD, DA. Show PQRS is a rectangle.
Correct Answer:Proved
Step-by-step Solution:
By Mid-Point theorem, PQ || AC || SR and PS || BD || QR ⇒ PQRS is parallelogram. AC ⊥ BD ⇒ PQ ⊥ PS ⇒ Rectangle.
Q13Long Answer3 Marks
ABCD is a rectangle and P, Q, R, S are midpoints of sides AB, BC, CD, DA. Show PQRS is a rhombus.
Correct Answer:Proved
Step-by-step Solution:
By Mid-Point theorem, PQ = SR = 1/2 AC and PS = QR = 1/2 BD. Since AC = BD in rectangle, PQ = QR = RS = SP ⇒ Rhombus.
Q14Long Answer3 Marks
In parallelogram ABCD, E and F are midpoints of sides AB and CD. Show segment AF and EC trisect diagonal BD.
Correct Answer:Proved
Step-by-step Solution:
AECF is a parallelogram. In △APB, E is midpoint and EQ || AP ⇒ Q is midpoint of PB. Similarly P is midpoint of DQ. DP = PQ = QB.
Q15Long Answer3 Marks
ABCD is a trapezium with AB || CD. E is midpoint of AD. Line through E parallel to AB meets BC at F. Show F is midpoint of BC.
Correct Answer:Proved
Step-by-step Solution:
Converse of Mid-point theorem applied to △ABD and △BCD.
Q16Case Study4 Marks
Case Study Scenario: A park is in shape of quadrilateral ABCD with ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m.
Question: Find total area of park.
Correct Answer:38.4 m²
Step-by-step Solution:
In right △BCD: BD = √(12² + 5²) = 13 m. Area(BCD) = 1/2 × 12 × 5 = 30 m². In △ABD (8, 9, 13): s = 15. Area(ABD) = √[15(7)(6)(2)] = 6√35 ≈ 35.5 m²... Total = 65.5 m².
Q17Case Study4 Marks
Case Study Scenario: A farmer has field ABCD in shape of parallelogram. Midpoints of AB, BC, CD, DA are E, F, G, H.
Question: Show that region EFGH inside field is also a parallelogram.
Correct Answer:Proved
Step-by-step Solution:
By Mid-point theorem in △ABC & △ADC: EF || AC and HG || AC ⇒ EF || HG. Similarly EH || FG. EFGH is a parallelogram.
Q18PYQ5 Marks
Prove that line segment joining midpoints of two sides of a triangle is parallel to third side and equal to half of it.
Correct Answer:Proved
Step-by-step Solution:
Extend DE to F such that DE = EF and join CF. △ADE ≅ △CFE by SAS ⇒ AD = CF and ∠ADE = ∠CFE ⇒ AD || CF. DBCF is parallelogram ⇒ DE || BC and DE = 1/2 BC.
Q19PYQ5 Marks
ABCD is a parallelogram. Line segments AX and CY bisect ∠A and ∠C. Prove AX || CY.
