CBSE Class 9 Mathematics Quadrilaterals Worksheets Set 05

Questions with Answers on Quadrilaterals

Question 1. Three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.
Answer:
Sum of the angles of a quadrilateral = 360°.
Fourth angle = 360° − (110° + 50° + 40°) = 360° − 200°.
Fourth angle = 160°.
In simple words: Subtract the three given angles from 360°.

Teacher's Note:
1. The angle sum of a quadrilateral is 360°.

Question 2. In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measure of each angle of the quadrilateral.
Answer:
Let the angles be k, 2k, 4k and 5k. Then 12k = 360°, so k = 30°.
The angles are 30°, 60°, 120° and 150°.
In simple words: Divide 360° in the ratio 1 : 2 : 4 : 5.

Teacher's Note:
1. Add the ratio parts first.

Question 3. In a quadrilateral ABCD, ∠A + ∠B = 150° and ∠C = 130°. Find ∠D.
Answer:
∠D = 360° − (∠A + ∠B) − ∠C = 360° − 150° − 130°.
∠D = 80°.
In simple words: Use the angle sum 360°.

Teacher's Note:
1. Treat ∠A + ∠B as one number.

Question 4. In a quadrilateral, ∠A + ∠B = 210° and ∠C : ∠D = 2 : 3. Find ∠C and ∠D.
Answer:
∠C + ∠D = 360° − 210° = 150°.
∠C = 2/5 × 150° = 60° and ∠D = 3/5 × 150° = 90°.
∠C = 60° and ∠D = 90°.
In simple words: Share 150° in the ratio 2 : 3.

Teacher's Note:
1. First find the sum of the two unknown angles.

Question 5. What is a trapezium?
Answer:
A trapezium is a quadrilateral in which one pair of opposite sides is parallel.
In simple words: A quadrilateral with exactly one pair of parallel sides.

Teacher's Note:
1. If the non-parallel sides are equal it is an isosceles trapezium.

Question 6. State the conditions for a quadrilateral ABCD to be a parallelogram.
Answer:
A quadrilateral is a parallelogram if any one of these holds:
1. Both pairs of opposite sides are equal.
2. Both pairs of opposite angles are equal.
3. The diagonals bisect each other.
4. One pair of opposite sides is equal and parallel.
In simple words: Any of these four tests proves a parallelogram.

Teacher's Note:
1. Pick the test that matches the given data.

Question 7. State the condition for a parallelogram to be a rectangle.
Answer:
A parallelogram is a rectangle if one of its angles is a right angle (or equivalently, if its diagonals are equal).
In simple words: A parallelogram with one right angle is a rectangle.

Teacher's Note:
1. Equal diagonals also work.

Question 8. State the type of quadrilateral ABCD if AB = BC = CD = DA, ∠A = ∠C and ∠B = ∠D.
Answer:
All four sides are equal, so ABCD is a rhombus. It becomes a square only if one angle is 90°.
In simple words: Four equal sides make a rhombus.

Teacher's Note:
1. If the angles were all 90° it would be a square.

Question 9. "In a parallelogram, the diagonals are of equal length." Is the statement true or false?
Answer:
False. The diagonals of a parallelogram bisect each other but are equal only in a rectangle.
In simple words: Equal diagonals are special to rectangles.

Teacher's Note:
1. Give a rhombus as a counter-example.

Question 10. Two opposite angles of a parallelogram are (3x − 2)° and (50 − x)°. Find the measure of each angle of the parallelogram.
Answer:
Opposite angles are equal: 3x − 2 = 50 − x, so 4x = 52 and x = 13.
Each of these angles = 3(13) − 2 = 37°. The adjacent angles are 180° − 37° = 143°.
The angles are 37°, 143°, 37° and 143°.
In simple words: Equate the opposite angles first.

Teacher's Note:
1. Adjacent angles are supplementary.

Question 11. If an angle of a parallelogram is two-thirds of its adjacent angle, find the angles of the parallelogram.
Answer:
Let the adjacent angles be x and 2x/3. Then x + 2x/3 = 180°, so x = 108°.
The other is 2/3 × 108° = 72°.
The angles are 108°, 72°, 108° and 72°.
In simple words: Solve the linear equation using the supplementary angles.

Teacher's Note:
1. Opposite angles are equal.

Question 12. Find the measure of all the angles of a parallelogram, if one angle is 24° less than twice the smallest angle.
Answer:
Let the smallest angle be x. The adjacent angle is 180° − x = 2x − 24°, so x = 68°.
The angles are 68°, 112°, 68° and 112°.
In simple words: Set up 180° − x = 2x − 24°.

Teacher's Note:
1. Adjacent angles are supplementary.

Question 13. The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm, what is the measure of the shorter side?
Answer:
Perimeter = 2(a + b) = 22, so a + b = 11 cm.
Shorter side = 11 − 6.5.
The shorter side is 4.5 cm.
In simple words: Half the perimeter is the sum of two adjacent sides.

Teacher's Note:
1. Opposite sides are equal.

Question 14. In a parallelogram ABCD, ∠D = 135°. Determine the measures of ∠A and ∠B.
Answer:
∠A and ∠D are adjacent, so ∠A = 180° − 135° = 45°.
∠B is opposite to ∠D, so ∠B = 135°.
∠A = 45° and ∠B = 135°.
In simple words: Adjacent angles add to 180° and opposite angles are equal.

Teacher's Note:
1. Draw a rough parallelogram to see which angles are adjacent.

Question 15. ABCD is a parallelogram in which ∠A = 70°. Compute ∠B, ∠C and ∠D.
Answer:
∠B = 180° − 70° = 110°.
∠C = ∠A = 70° and ∠D = ∠B = 110°.
∠B = 110°, ∠C = 70°, ∠D = 110°.
In simple words: Use the supplementary and opposite angle properties.

Teacher's Note:
1. Opposite angles are equal.

Question 16. ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute ∠CDB and ∠ADB.
Answer:
AD ∥ BC, so ∠ADB = ∠DBC = 60° (alternate angles).
In ΔABD, ∠ABD = 180° − 75° − 60° = 45°.
AB ∥ DC, so ∠CDB = ∠ABD = 45°.
∠CDB = 45° and ∠ADB = 60°.
In simple words: Two pairs of alternate angles give the answers.

Teacher's Note:
1. Alternate angles between parallel lines are equal.

Question 17. ABCD is a parallelogram and E is the mid-point of side BC. DE and AB, when produced, meet at F. Prove that AF = 2AB.
Answer:
In ΔDEC and ΔFEB: CE = BE (E is the mid-point), ∠DEC = ∠FEB (vertically opposite) and ∠DCE = ∠FBE (alternate angles, DC ∥ AF).
So ΔDEC ≅ ΔFEB (ASA), hence DC = BF.
DC = AB (opposite sides of a parallelogram), so BF = AB.
AF = AB + BF = 2AB.
In simple words: Show that triangles DEC and FEB are congruent.

Teacher's Note:
1. ASA congruence with vertically opposite angles.

Question 18. In a parallelogram ABCD, determine the sum of the angles ∠C and ∠D.
Answer:
∠C and ∠D are adjacent angles (consecutive angles) of the parallelogram.
∠C + ∠D = 180°.
In simple words: Adjacent angles of a parallelogram are supplementary.

Teacher's Note:
1. Co-interior angles between parallel lines.

Question 19. In a parallelogram ABCD, if ∠B = 135°, determine the measures of its other angles.
Answer:
∠D = ∠B = 135° (opposite angles).
∠A = ∠C = 180° − 135° = 45°.
∠A = ∠C = 45° and ∠B = ∠D = 135°.
In simple words: Opposite angles are equal and adjacent angles add to 180°.

Teacher's Note:
1. Check that the four angles add to 360°.

Question 20. ΔABC is a triangle right-angled at B and P is the mid-point of AC. Q is the point where the line through P parallel to BC meets AB. Prove that (i) Q is the mid-point of AB, (ii) PQ ⊥ AB, (iii) PB = PA = ½ AC.
Answer:
(i) In ΔABC, P is the mid-point of AC and PQ ∥ BC, so by the converse of the mid-point theorem Q is the mid-point of AB.
(ii) PQ ∥ BC and BC ⊥ AB, so PQ ⊥ AB.
(iii) In ΔPQA and ΔPQB: AQ = BQ, ∠PQA = ∠PQB = 90° and PQ is common, so the triangles are congruent (SAS) and PA = PB. Since P is the mid-point of AC, PA = ½ AC.
Hence PB = PA = ½ AC.
In simple words: The mid-point of the hypotenuse of a right triangle is equidistant from the three vertices.

Teacher's Note:
1. Use the converse of the mid-point theorem, then congruence.

Question 21. In ΔABC, AD and BE are two medians (D on BC, E on AC) and BE ∥ DF, where F lies on AC. Prove that CF = ¼ AC.
Answer:
In ΔBCE, D is the mid-point of BC and DF ∥ BE, so F is the mid-point of EC. Hence CF = ½ CE.
E is the mid-point of AC, so CE = ½ AC.
CF = ½ × ½ AC = ¼ AC.
In simple words: Use the mid-point theorem twice.

Teacher's Note:
1. A line through a mid-point parallel to another side bisects the third side.

Question 22. In ΔABC, AD is the median through A and E is the mid-point of AD. BE produced meets AC at F. Prove that AF = ⅓ AC.
Answer:
Draw DK ∥ BF meeting AC at K.
In ΔADK, E is the mid-point of AD and EF ∥ DK, so F is the mid-point of AK: AF = FK.
In ΔBCF, D is the mid-point of BC and DK ∥ BF, so K is the mid-point of CF: FK = KC.
Hence AF = FK = KC, so AF = ⅓ AC.
In simple words: Draw a parallel line through D to use the mid-point theorem twice.

Teacher's Note:
1. The construction DK ∥ BF is the key step.

Question 23. P is the mid-point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R. Prove that (i) AR = 2BC, (ii) BR = 2BQ.
Answer:
(i) In ΔABR, P is the mid-point of AB and PD ∥ BR, so D is the mid-point of AR. Hence AR = 2AD = 2BC.
(ii) D is the mid-point of AR and DQ ∥ AB (DC ∥ AB), so in ΔRAB the point Q is the mid-point of RB.
Hence BR = 2BQ.
In simple words: Use the converse of the mid-point theorem twice.

Teacher's Note:
1. Opposite sides of a parallelogram are equal and parallel.

Question 24. ABCD is a trapezium in which AB ∥ DC, E is the mid-point of AD and F is a point on BC such that EF ∥ DC. Prove that F is the mid-point of BC and EF = ½ (AB + DC).
Answer:
Let AC meet EF at G.
In ΔADC, E is the mid-point of AD and EG ∥ DC, so G is the mid-point of AC and EG = ½ DC.
In ΔABC, G is the mid-point of AC and GF ∥ AB, so F is the mid-point of BC and GF = ½ AB.
EF = EG + GF = ½ (AB + DC).
In simple words: Split EF at G and apply the mid-point theorem in two triangles.

Teacher's Note:
1. The diagonal AC is the helpful construction.

Question 25. ABCD is a square and PAB is a triangle (P on the opposite side of DC from A and B) such that PA meets DC at Q and PB meets DC at R with AQ = BR. Prove that PQR is an isosceles triangle.
Answer:
In right triangles ADQ and BCR: AD = BC, AQ = BR and ∠D = ∠C = 90°, so ΔADQ ≅ ΔBCR (RHS).
Hence ∠AQD = ∠BRC.
∠PQR = ∠AQD and ∠PRQ = ∠BRC (vertically opposite angles), so ∠PQR = ∠PRQ.
Hence PQ = PR and PQR is isosceles.
In simple words: Congruent right triangles give equal angles at Q and R.

Teacher's Note:
1. Vertically opposite angles are equal.

Question 26. ABCD is a parallelogram and diagonal AC bisects ∠A. Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus.
Answer:
(i) ∠DAC = ∠ACB and ∠CAB = ∠ACD (alternate angles). Since ∠DAC = ∠CAB, we get ∠ACB = ∠ACD, so AC bisects ∠C.
(ii) ∠DAC = ∠DCA, so in ΔACD, DA = DC.
A parallelogram with two adjacent sides equal is a rhombus. Hence ABCD is a rhombus.
In simple words: Alternate angles show the bisector property and equal sides give the rhombus.

Teacher's Note:
1. Equal adjacent sides in a parallelogram make a rhombus.

Question 27. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that (i) ΔAPD ≅ ΔCQB, (ii) AP = CQ, (iii) ΔAQB ≅ ΔCPD, (iv) AQ = CP, (v) APCQ is a parallelogram.
Answer:
(i) AD = CB, DP = BQ and ∠ADP = ∠CBQ (alternate angles, AD ∥ BC), so ΔAPD ≅ ΔCQB (SAS).
(ii) AP = CQ by CPCT.
(iii) AB = CD, BQ = DP and ∠ABQ = ∠CDP (alternate angles, AB ∥ CD), so ΔAQB ≅ ΔCPD (SAS).
(iv) AQ = CP by CPCT.
(v) In APCQ both pairs of opposite sides are equal (AP = CQ and AQ = CP), so APCQ is a parallelogram.
In simple words: Two pairs of congruent triangles give two pairs of equal opposite sides.

Teacher's Note:
1. SAS with alternate angles.

Question 28. In ΔABC and ΔDEF, AB = DE, AB ∥ DE, BC = EF and BC ∥ EF. Vertices A, B, C are joined to vertices D, E, F respectively. Show that (i) ABED is a parallelogram, (ii) BEFC is a parallelogram, (iii) AD ∥ CF and AD = CF, (iv) ACFD is a parallelogram, (v) AC = DF, (vi) ΔABC ≅ ΔDEF.
Answer:
(i) AB = DE and AB ∥ DE, so ABED is a parallelogram.
(ii) BC = EF and BC ∥ EF, so BEFC is a parallelogram.
(iii) AD = BE and AD ∥ BE (from (i)); BE = CF and BE ∥ CF (from (ii)). Hence AD = CF and AD ∥ CF.
(iv) One pair of opposite sides AD and CF is equal and parallel, so ACFD is a parallelogram.
(v) Opposite sides of ACFD are equal, so AC = DF.
(vi) AB = DE, BC = EF and AC = DF, so ΔABC ≅ ΔDEF (SSS).
In simple words: Show one pair of equal parallel sides at each step.

Teacher's Note:
1. One pair of equal and parallel opposite sides is enough for a parallelogram.

Question 29. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. AF and EC meet the diagonal BD at P and Q. Show that the line segments AF and EC trisect the diagonal BD.
Answer:
AE ∥ FC and AE = ½ AB = ½ CD = CF, so AECF is a parallelogram and AF ∥ EC.
In ΔDQC, F is the mid-point of DC and FP ∥ CQ, so P is the mid-point of DQ: DP = PQ.
In ΔBPA, E is the mid-point of AB and EQ ∥ AP, so Q is the mid-point of BP: BQ = PQ.
Hence DP = PQ = QB, so BD is trisected.
In simple words: Prove AECF is a parallelogram first, then use the mid-point theorem twice.

Teacher's Note:
1. Equal and parallel opposite sides give a parallelogram.

Question 30. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Answer:
Let P, Q, R, S be the mid-points of AB, BC, CD and DA of quadrilateral ABCD. Join PQ, QR, RS, SP and AC.
In ΔABC, PQ ∥ AC and PQ = ½ AC. In ΔADC, SR ∥ AC and SR = ½ AC.
So PQ ∥ SR and PQ = SR, hence PQRS is a parallelogram.
The segments PR and QS joining the mid-points of opposite sides are its diagonals, and the diagonals of a parallelogram bisect each other.
In simple words: The mid-point figure is a parallelogram, and its diagonals bisect each other.

Teacher's Note:
1. Join a diagonal and use the mid-point theorem.

Question 31. ABC is a triangle right-angled at C. A line through the mid-point M of the hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC, (ii) MD ⊥ AC, (iii) CM = MA = ½ AB.
Answer:
(i) M is the mid-point of AB and MD ∥ BC, so D is the mid-point of AC (converse of the mid-point theorem).
(ii) MD ∥ BC and BC ⊥ AC, so MD ⊥ AC.
(iii) In ΔADM and ΔCDM: AD = CD, ∠ADM = ∠CDM = 90° and DM is common, so the triangles are congruent (SAS) and MA = MC. Since M is the mid-point of AB, MA = ½ AB.
Hence CM = MA = ½ AB.
In simple words: The mid-point of the hypotenuse is at the same distance from all three vertices.

Teacher's Note:
1. Congruent triangles on the two sides of DM.

Question 32. Show that the bisectors of the angles of a parallelogram form a rectangle.
Answer:
Let ABCD be a parallelogram. The bisectors of ∠A and ∠B meet at P. ∠A + ∠B = 180°, so ∠PAB + ∠PBA = 90° and ∠APB = 90°.
Similarly the bisectors of the other adjacent pairs meet at right angles, so the quadrilateral formed by the four bisectors has all four angles equal to 90°.
Hence it is a rectangle.
In simple words: Half of two supplementary angles adds to 90°.

Teacher's Note:
1. The four corners of the enclosed figure are right angles.

Question 33. ABC is an isosceles triangle in which AB = AC. AD bisects the exterior angle PAC (P on BA produced) and CD ∥ AB. Show that (i) ∠DAC = ∠BCA and (ii) ABCD is a parallelogram.
Answer:
(i) The exterior angle ∠PAC = ∠ABC + ∠ACB = 2∠ACB, because AB = AC gives ∠ABC = ∠ACB. AD bisects it, so ∠DAC = ½ ∠PAC = ∠ACB.
(ii) ∠DAC = ∠BCA are alternate angles for AD and BC with transversal AC, so AD ∥ BC.
CD ∥ AB is given, so both pairs of opposite sides are parallel and ABCD is a parallelogram.
In simple words: The exterior angle of a triangle equals the sum of the two interior opposite angles.

Teacher's Note:
1. Alternate angles prove AD ∥ BC.