Question

NCERT Solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2 -Maths 9th

Answer 1

1. ABCD is a quadrilateral in which P, Q, R and S are midpoints of the sides AB, BC, CD and DA (see figure). AC is a diagonal.  Show that: (i) SR || AC and SR = (1/2) AC  (ii) PQ = SR (iii) PQRS is a parallelogram. We have P as the mid-point of AB, Q as the mid-point of BC, R as the mid-point of CD, S as the mid-point of DA, and AC as the diagonal of quadrilateral ABCD. (i) To prove that SR =(1/2) AC and SR || AC. In ΔACD, we have S as the mid-point of AD, R as the mid-point of CD. ∵ The line segment joining the mid-point of any two sides of a triangle is parallel to the third side and half of it. ∴ SR = (1/2)AC and SR || AC (ii) To prove that PQ = SR. In ΔABC, we have P is the mid-point of AB, Q is the mid-point of BC. ∴ PQ = (1/2) AC            …(1) Also, SR = (1/2) AC            [Proved]             …(2) From (1) and (2), PQ = SR (iii) To prove that PQRS is a parallelogram. In ΔABC, P and Q are the mid-points of AB and BC. ∴ PQ = (1/2)AC and PQ || AC             ...(3) In ΔACD, S and R are the mid-points of DA and CD. ∴ SR = (1/2)AC and SR || AC             …(4) From (3) and (4), we get PQ =(1/2)AC = SR and PQ || AC || SR ⇒ PQ = SR and PQ || SR i.e. One pair of opposite sides in quadrilateral PQRS is equal and parallel. ∴ PQRS is a parallelogram. 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.  Show that the quadrilateral PQRS is a rectangle.  We have P as the mid-point of AB, Q as the midpoint of BC, R as the mid-point of CD, S as the mid-point of DS. We have to prove that PQRS is a rectangle. Let us join AC. ∵ In ΔABC, P and Q are the mid-points of AB and BC. ∴ PQ =(1/2)AC and PQ || AC             …(1) Also in ΔADC, R and S are the mid-points of CD and DA. ∴ SR = (1/2)AC and SR || AC From (1) and (2), we get PQ =(1/2) AC = SR and PQ || AC || SR ⇒ PQ = SR and PQ || SR i.e. One pair of opposite sides of quadrilateral PQRS is equal and parallel. ∴ PQRS is a parallelogram. Now, in ΔERC and ΔEQC, ∠1= ∠2            [∵ The diagonal of a rhombus bisects the opposite angles] CR = CQ            [Each is equal to(1/2) of a side of rhombus]CE = CE [Common] ∴ ΔERC ≌ ΔEQC            [SAS criteria] ⇒ ∠3= ∠4[c.p.c.t.] But ∠3 + ∠4 = 180º [Linear pair] ⇒ ∠3= ∠4 = 90° But ∠5= ∠3            [Vertically opposite angles] ∴ ∠5 = 90º PQ || AC ⇒ PQ || EF ∴ PQEF is a quadrilateral having a pair of opposite sides parallel and one of the angles is 90º. ∴ PQEF is a rectangle. ⇒ ∠RQP = 90º ∴ One angle of parallelogram PQRS is 90º. Thus, PQRS is a rectangle. 3. ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.  In a rectangle ABCD, P is the mid-point of AB, Q is the midpoint of BC, R is the mid-point of CD, S is the mid-point of DA AC is the diagonal. Now, in ΔABC, PQ =(1/2)AC and PQ || AC [Mid-point theorem]             …(1) Similarly, in ΔACD, SR =(1/2)AC and SR || AC             …(2) From (1) and (2), we get PQ = SR and PQ || SR Similarly, by joining BD, we have PS = QR and PS || QR i.e. Both pairs of opposite sides of quadrilateral PQRS are equal and parallel. ∴ PQRS is a parallelogram. Now, in ΔPAS and ΔPBQ, ∠A= ∠B            [Each = 90º] AP = BP [Each = (1/2)AB] AS = BQ             [Each =(1/2) of opposite sides of a rectangle] ∴ ΔPAS ≌ ΔPBQ             [SAS criteria] ∴ Their corresponding parts are equal. ⇒ PS = PQ Also PS = QR               [Proved] and PQ = SR                        [Proved] ∴ PQ = QR = RS = SP i.e. PQRS is a parallelogram having all of its sides equal. ⇒ PQRS is a rhombus. 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC. In trapezium ABCD, AB || DC. E is the mid-point of AD. EF is drawn parallel to AB. We have to prove that F is the mid-point of BC. Join BD. In DDAB, ∵ E is the mid-point of AD            [Given] and EG || AB             [∵ EF || AB] ∴ Using the converse of mid-point theorem, we get that G is the mid-point BD. Again in DBDC, ∵ G is the mid-point of BD             [Proved] GF || DC            [∵ AB || DC and EF || AB and GF is a part of EF] ∴ Using the converse of the mid-point theorem, we get that F is the mid-point of BC. 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD.  We have ABCD is a parallelogram such that: E is the mid-point of AB and F is the mid-point of CD. Let us join the opposite vertices B and D. Since, the opposite sides of a parallelogram are parallel and equal. ∴ AB || DC ⇒ AE || FC            …(1) Also AB = DC or (1/2) AB =(1/2)DC ⇒ AE = FC            …(2) From (1) and (2), we can say that AECF is quadrilateral having a pair of the opposite sides as parallel and equal. ∴ AEFC is a parallelogram. ⇒ AE || CF Now, in DDBC, F is the mid-point of DC             [Given] and FP || CQ             [∵ AF || CE] ⇒ P is the mid-point of DQ             [Converse of mid-point theorem] ⇒ DP = PQ             …(3) Similarly, in DBAP, BQ = PQ             …(4) ∴ From (3) and (4), we have DP = PQ = BQ ⇒ The line segments AF and EC trisect the diagonal BD.  6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.  A quadrilateral ABCD such that the mid-points of AB, BC, CD and DA are P, Q, R and S respectively, we have to prove that diagonals of PQRS are bisected at O. Join PQ, QR, RS and SP. Let us also join PR and SQ. Now, in ΔABC, we have P and Q as the mid-points of its sides AB and BC respectively. ∴ PQ || AC and PQ =(1/2)AC Similarly, RS || AC and RS =(1/2)AC ⇒ PQRS is a quadrilateral having a pair of opposite sides (PQ and RS) as equal and parallel. ∴ PQRS is a parallelogram. But the diagonals of a parallelogram bisect each other. i.e. PR and SQ bisect each other. Thus, the line segments joining the mid-points of opposite sides of a quadrilateral ABCD bisect each other.  7. ABC is a triangle, right angled at C. A line through the midpoint M of 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 =(1/2) AB  We have a triangle ABC, such that ∠C = 90º M is the mid-point of AB and MD || BC (i) To prove that D is the mid-point of AC. In ΔACB, we have M as the mid-point of AB.             [Given] MD || BC            [Given] ∴ Using the converse of mid-point theorem, D is the mid-point of AC. (ii) To prove that MD ⊥ AC. Since, MD || BC             [Given] and AC is a transversal. ∴ ∠ MDA = ∠BCA            [Corresponding angles] But ∠BCA = 90º            [Given] ∴ ∠MDA = 90º ⇒ MD ⊥ AC. (iii) To prove that CM = MA =(1/2) AB In ΔADM and ΔCDM, we have ∠ADM = ∠CDM            [Each = 90º] MD = MD            [Common] AD = CD            [∵ M is the mid-point of AC (Proved)] ∴ ΔADM ≌ ΔCOM            [SAS criteria] ⇒ MA = MC [c.p.c.t.]             …(1) ∵ M is the mid-point AB.            [Given] ∴ MA =(1/2)AB              …(2) From (1) and (2), we have CM = MA = (1/2) AB