Question 1

Question 1If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Accepted Answer

Let ABCD be a parallelogram with equal diagonals.From figure,In ∆ ABC and ∆ DCB,⇒ AB = DC (Opposite sides of a parallelogram are equal)⇒ BC = BC (Common side)⇒ AC = DB (Diagonals of parallelogram are equal)∴ ∆ ABC ≅ ∆ DCB (By S.S.S. Congruence rule)We know that,Corresponding parts of congruent triangles are equal.⇒ ∠ABC = ∠DCB (By C.P.C.T.) ......(1)We know that,Sum of co-interior angles equal to 180°.⇒ ∠ABC + ∠DCB = 180° (AB || CD)⇒ ∠ABC + ∠ABC = 180° [From equation (1)]⇒ 2∠ABC = 180°⇒ ∠ABC = 180°2\dfrac{180°}{2}2180°​⇒ ∠ABC = ∠B = 90°⇒ ∠DCB = ∠C = 90°Since,Opposite angles of parallelogram are equal.∴ ∠D = ∠B = 90° and ∠A = ∠C = 90°.∴ ∠B = ∠D = ∠C = ∠A = 90°.Since, opposite sides of ABCD are equal in length and each interior angle equals to 90°.Hence, proved that if the diagonals of a parallelogram are equal, then it is a rectangle.

Question 2

Question 2Show that the diagonals of a square are equal and bisect each other at right angles.

Accepted Answer

Let ABCD be the square.In Δ ABC and Δ BAD,⇒ AB = AB (Common side)⇒ BC = AD (Each side of a square is equal in length)⇒ ∠ABC = ∠BAD = 90° (Each interior angle in a square is a right angle)∴ Δ ABC ≅ Δ BAD (By S.A.S. congruence rule)We know that,Corresponding parts of congruent triangles are equal.∴ AC = BD (By C.P.C.T.)In Δ OAD and Δ OCB,⇒ AD = CB (Each side of a square is equal in length)⇒ ∠OAD = ∠OCB (Alternate angles are equal)⇒ ∠ODA = ∠OBC (Alternate angles are equal)∴ Δ OAD ≅ Δ OCB (By A.S.A. congruence rule)We know that,Corresponding parts of congruent triangles are equal.⇒ OA = OC (By C.P.C.T.) ......(1)⇒ OB = OD (By C.P.C.T.) .....(2)In Δ OBA and Δ ODA,⇒ OB = OD ....[From equation (2)]⇒ OA = OA (Common side)⇒ BA = DA (Each side of a square is equal in length)∴ Δ OBA ≅ Δ ODA (By S.S.S. congruence rule)We know that,Corresponding parts of congruent triangles are equal.⇒ ∠AOB = ∠AOD = x (let) (By C.P.C.T.)⇒ ∠AOB + ∠AOD = 180° (Linear pair)⇒ x + x = 180°⇒ 2x = 180°⇒ x = 180°2\dfrac{180°}{2}2180°​ = 90°.⇒ ∠AOB = 90° and ∠AOD = 90°.Thus, AC and BD bisect each other at right angles.Hence, proved that the diagonals of a square are equal and bisects each other at right angles.

Question 3

Question 3Diagonal AC of a parallelogram ABCD bisects ∠A. Show that(i) it bisects ∠C also,(ii) ABCD is a rhombus.(Ex.8.1Q3.png)

Accepted Answer

(i) We know that,ABCD is a parallelogram.So, opposite sides are parallel and equal.⇒ ∠DAC = ∠BCA (Alternate interior angles are equal) ......(1)⇒ ∠BAC = ∠DCA (Alternate interior angles are equal) .......(2)Given,⇒ AC bisects ∠A⇒ ∠DAC = ∠BAC .......(3)From equations (1), (2), and (3), we get :⇒ ∠DCA = ∠BAC = ∠DAC = ∠BCA .........(4)∴ ∠DCA = ∠BCA∴ AC bisects ∠C.Hence, proved that AC bisects ∠C.(ii) In Δ ABC,⇒ ∠BAC = ∠BCA (Proved above)⇒ BC = AB (Side opposite to equal angles are equal) .....(5)⇒ DA = BC and AB = CD (Opposite sides of a parallelogram are equal) ....(6)From equations (5) and (6), we get :⇒ AB = BC = CD = DA.Since, all sides of quadrilateral ABCD are equal.Hence, proved that ABCD is a rhombus.

Question 4

Question 4ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that :(i) ABCD is a square(ii) diagonal BD bisects ∠B as well as ∠D.

Accepted Answer

Rectangle ABCD is shown in the figure below:(i) Given :ABCD is a rectangle and AC bisects ∠A and ∠C.⇒ ∠DAC = ∠CAB ....(1)⇒ ∠DCA = ∠BCA .....(2)We know that,Opposite sides of a rectangle are parallel and equal.From figure,AD || BC and AC is transversal,⇒ ∠DAC = ∠BCA ..........(3) (Alternate interior angles are equal)From equations (1) and (3), we get :⇒ ∠CAB = ∠BCA ......(4)In △ ABC,⇒ ∠CAB = ∠BCAWe know that,Sides opposite to equal angles are equal.⇒ BC = AB .....(5)We know that,Opposite sides of a rectangle are equal.⇒ BC = AD .........(6)⇒ AB = DC .........(7)From equation (5), (6) and (7), we get :⇒ AB = BC = CD = AD.Since,ABCD is a rectangle and all the sides are equal. Hence, ABCD is a square.Hence, proved that ABCD is a square.(ii) Join BD.In Δ BCD,⇒ BC = CD (Sides of a square are equal to each other)⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal) ..... (8)⇒ ∠CDB = ∠ABD (Alternate interior angles are equal) ..... (9)From equations (8) and (9), we get :⇒ ∠CBD = ∠ABD∴ BD bisects ∠B.From figure,⇒ ∠CBD = ∠ADB (Alternate interior angles are equal) .....(10)From equations (8) and (10), we get :⇒ ∠ADB = ∠CDB∴ BD bisects ∠D.Hence, proved that diagonal BD bisects ∠B as well as ∠D.

Question 5

Question 5In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ see Fig. Show that:(i) Δ APD ≅ Δ CQB(ii) AP = CQ(iii) Δ AQB ≅ Δ CPD(iv) AQ = CP(v) APCQ is a parallelogram

Accepted Answer

Given :ABCD is a parallelogram.From figure,AD || BC and BD is transversal⇒ ∠ADP = ∠CBQ (Alternate angles are equal)(i) In Δ APD and Δ CQB,⇒ AD = CB (Opposite sides of parallelogram ABCD are equal)⇒ DP = BQ (Given)⇒ ∠ADP = ∠CBQ (Proved above)∴ Δ APD ≅ Δ CQB (By S.A.S. congruence rule)Hence, proved that Δ APD ≅ Δ CQB.(ii) In Δ APD ≅ Δ CQB,We know that,Corresponding parts of congruent triangles are equal.⇒ AP = CQ (By C.P.C.T.) ........(1)Hence, proved that AP = CQ.(iii) In Δ AQB and Δ CPD,⇒ AB = CD (Opposite sides of parallelogram ABCD are equal)⇒ ∠ABQ = ∠CDP (Alternate interior angles are equal)⇒ BQ = DP (Given)⇒ Δ AQB ≅ Δ CPD (By S.A.S. congruence rule)Hence, proved that Δ AQB ≅ Δ CPD.(iv) Since Δ AQB ≅ Δ CPD,We know that,Corresponding parts of congruent triangles are equal.⇒ AQ = CP (C.P.C.T.) ............(2)Hence, proved that AQ = CP.(v) From equation (1) and (2), we get :⇒ AQ = CP and AP = CQSince both pairs of opposite sides in APCQ are equal,Hence, proved that APCQ is a parallelogram.

Question 6

Question 6ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that(i) Δ APB ≅ Δ CQD(ii) AP = CQ

Accepted Answer

Given :ABCD is a parallelogram with AP ⊥ BD and CQ ⊥ BD.From figure,AB || DC and BD is transversal.(i) In Δ APB and Δ CQD,⇒ ∠APB = ∠CQD (Both equal to 90°)⇒ AB = CD (Opposite sides of parallelogram are equal)⇒ ∠ABP = ∠CDQ (Alternate interior angles are equal)∴ Δ APB ≅ Δ CQD (By A.A.S. congruence rule)Hence, proved that Δ APB ≅ Δ CQD.(ii) Since,Δ APB ≅ Δ CQDWe know that,Corresponding parts of congruent triangles are equal.⇒ AP = CQ (By C.P.C.T.)Hence, proved that AP = CQ.

Question 7

Question 7ABCD is a trapezium in which AB || CD and AD = BC. Show that(i) ∠A = ∠B(ii) ∠C = ∠D(iii) Δ ABC ≅ Δ BAD(iv) diagonal AC = diagonal BD

Accepted Answer

Steps of construction :Extend AB.Draw a line through C parallel to DA intersecting AB produced at E.Given :ABCD is a trapezium, AB || CD and AD = BCBy construction,AD || CEandAE || DC.In AECD, both pair of opposite sides are parallel, hence, AECD is a parallelogram(i) From figure,AD = CE (Opposite sides of parallelogram are equal)As,AD = BC∴ BC = CEWe know that,Angles opposite to equal sides in a triangle are also equal.⇒ ∠CEB = ∠CBE .........(1)Consider, parallel lines AD and CE where AE is the transversal.We know that,Sum of co-interior angles = 180°⇒ ∠BAD + ∠CEB = 180°⇒ ∠BAD + ∠CBE = 180° [From equation (1)] ..... (2)From figure,⇒ ∠ABC + ∠CBE = 180° (Linear pairs) ....... (3)From equations (2) and (3), we get :⇒ ∠BAD + ∠CBE = ∠ABC + ∠CBE⇒ ∠BAD = ∠ABC.∴ ∠A = ∠BHence, proved that ∠A = ∠B.(ii) From figure,AB || CDWe know that,Sum of co-interior angles equal to 180°.⇒ ∠A + ∠D = 180° .........(4)⇒ ∠C + ∠B = 180° .........(5)From equation (4) and (5), we get :⇒ ∠A + ∠D = ∠C + ∠BSince, ∠A = ∠BWe can say that,⇒ ∠D = ∠CHence, proved that ∠C = ∠D.(iii) Join AC and BD.In ∆ ABC and ∆ BAD,⇒ AB = BA (Common side)⇒ BC = AD (Given)⇒ ∠B = ∠A (above)∴ ∆ ABC ≅ ∆ BAD (By S.A.S. congruence rule)Hence, proved that Δ ABC ≅ Δ BAD.(iv) Since ∆ ABC ≅ ∆ BAD,We know that,Corresponding parts of congruent triangles are equal.⇒ AC = BD (By C.P.C.T.)Hence, proved that diagonal AC = diagonal BD.

Question 8

Question 1ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that :(i) SR || AC and SR = 12AC\dfrac{1}{2}AC21​AC(ii) PQ = SR(iii) PQRS is a parallelogram

Accepted Answer

Given :ABCD is a quadrilateral, where P, Q, R and S are the mid points of the sides AB, BC, CD and DA.Mid-point theorem : The line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it.(i) In △ ADC,S and R are the mid-points of side AD and CD respectively.By mid-point theorem,⇒ SR || AC .......(1)⇒ SR = 12AC\dfrac{1}{2}AC21​AC ......(2)Hence, proved that SR || AC and SR = 12AC\dfrac{1}{2}AC21​AC.(ii) In Δ ABC, P and Q are mid-points of sides AB and BC.By using the mid-point theorem,⇒ PQ || AC .........(3)⇒ PQ = 12AC\dfrac{1}{2}AC21​AC ......(4)From equations (3) and (4), we get :⇒ PQ = SRHence, proved that PQ = SR.(iii) From equation (1) and (3), we get :⇒ PQ || AC || SR⇒ PQ || SRAlso,⇒ PQ = SR (Proved above)We know that,If one pair of opposite sides are equal and parallel, then the figure is parallelogram.Hence, proved that PQRS is a parallelogram.

Question 9

Question 2ABCD 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.

Accepted Answer

In Δ ABC,P and Q are mid points of AB and BCBy mid-point theorem,PQ || AC and PQ = 12AC\dfrac{1}{2}AC21​AC .....(1)In Δ ADC,S and R are mid points of AD and DC.By mid-point theorem,⇒ SR || AC and SR = 12AC\dfrac{1}{2}AC21​AC .....(2)From equations (1) and (2), we get :⇒ SR = PQ and SR || PQ.Since, one of the opposite pairs of PQRS are parallel and equal.∴ PQRS is a parallelogram.Let the diagonals AC and BD intersect at O.In Δ BAC,P and Q are mid points of AB and BC.By mid-point theorem,PQ || AC and PQ = 12AC\dfrac{1}{2}AC21​AC∴ MQ || ONIn Δ BCD,Q and R are mid points of BC and CD.By mid-point theorem,QR || BD and QR = 12BD\dfrac{1}{2}BD21​BD∴ QN || OMSince, opposite sides of quadrilateral OMQN are parallel.∴ OMQN is a parallelogram⇒ ∠MQN = ∠NOM (Opposite angles of parallelogram are equal)We know that,Diagonals of rhombus intersect at 90°∴ ∠NOM = 90°⇒ ∠PQR = ∠NOM = 90°So, ∠PQR = 90°In PQRS,One pair of opposite side is parallel and equal and one of its interior angle is 90°.Hence, proved that PQRS is a rectangle.

Question 10

Question 3ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Accepted Answer

We know that,The diagonals of a rectangle are equal.⇒ BD = AC = x (let)In △ ABC,P and Q are the mid-points of AB and BC respectively.By mid-point theorem,⇒ PQ || AC and PQ = 12AC=x2\dfrac{1}{2}AC = \dfrac{x}{2}21​AC=2x​ ....(1)In △ ADC,S and R are the mid-points of AD and CD respectively.SR || AC and SR = 12AC=x2\dfrac{1}{2}AC = \dfrac{x}{2}21​AC=2x​ .....(2)From equation (1) and (2), we get :PQ || SR and PQ = SRIn quadrilateral PQRS, one pair of opposite sides are equal and parallel to each other.∴ PQRS is a parallelogram.In △ BCD, Q and R are the mid-points of side BC and CD respectively.By mid-point theorem,⇒ QR || BD and QR = 12BD=x2\dfrac{1}{2}BD = \dfrac{x}{2}21​BD=2x​ ....(3)In △ BAD, P and S are the mid-points of side AB and AD respectively.By mid-point theorem,PS || BD and PS = 12BD=x2\dfrac{1}{2}BD = \dfrac{x}{2}21​BD=2x​ .......(4)From equations (1), (2), (3), (4), we get :PQ = QR = SR = PSHence, proved that the quadrilateral PQRS is a rhombus.

Question 11

Question 4ABCD 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 Fig. Show that F is the mid-point of BC.

Accepted Answer

Given :ABCD is a trapezium, where AB || DC, E is the mid-point of AD and EF || AB.By converse of mid-point theorem,A line drawn through the mid-point of any side of a triangle and parallel to another side bisects the third side.Let EF intersect BD at point G.In trapezium ABCD,EF || AB and E is the mid-point of AD.By converse of mid-point theorem,G is the mid-point of DBAs, EF || AB and AB || CD,EF || CD (Two lines parallel to the same line are parallel to each other)In Δ BCD,GF || CD and G is the mid-point of line BD.By using the converse of mid-point theorem, F is the mid-point of BC.Hence, proved that F is the mid-point of BC.

Question 12

Question 5In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively see Fig. Show that the line segments AF and EC trisect the diagonal BD.

Accepted Answer

Given :ABCD is a parallelogram, where E and F are the mid-points of sides AB and CD.We know that,Opposite sides of a parallelogram are equal and parallel.∴ AB || CD and AB = CD∴ AE || FCSince, AB = CDDividing both sides of equation by 2, we get :⇒ 12AB\dfrac{1}{2}AB21​AB = 12CD\dfrac{1}{2}CD21​CD⇒ AE = FC∴ AEFC is a parallelogram.∴ AF || CEFrom figure,⇒ PF || QC⇒ EQ || APIn △ DQC,F is the mid-point of DC and FP || CQ.By converse of mid-point theorem, we get :P is the mid-point of DQ.⇒ DP = PQ .....(1)In △ APB,E is the mid-point of AB and EQ || AP.By converse of mid-point theorem, we get :Q is the mid-point of BP.⇒ PQ = QB ......(2)From equations (1) and (2) we get :⇒ PQ = QB = DP.∴ AF and EC trisect BD.Hence, proved that AF and EC trisect the diagonal BD.

Question 13

Question 6ABC is a triangle right angled at C. A line through the mid-point 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 = 12AB\dfrac{1}{2}AB21​AB

Accepted Answer

(i) In Δ ABC,It is given that M is the mid-point of AB and MD || BC.By converse of mid-point theorem, we get :Thus, D is the mid-point of AC.Hence, proved that D is the mid-point of AC.(ii) As DM || CB and AC is a transversal,We know that,Sum of co-interior angles = 180°.⇒ ∠MDC + ∠DCB = 180°⇒ ∠MDC + 90° = 180°⇒ ∠MDC = 180° - 90°⇒ ∠MDC = 90°Hence, proved that MD ⊥ AC.(iii) Join MC,In Δ AMD and Δ CMD,⇒ AD = CD (D is the mid-point of side AC)⇒ ∠ADM = ∠CDM (Since, MD ⊥ AC)⇒ DM = DM (Common side)∴ Δ AMD ≅ Δ CMD (By S.A.S. congruence rule)We know that,Corresponding parts of congruent triangles are equal.⇒ AM = CM (By C.P.C.T.)We know that,M is the mid-point of AB.∴ AM = 12AB\dfrac{1}{2}AB21​AB⇒ CM = AM = 12AB\dfrac{1}{2}AB21​AB.Hence, proved that CM = MA = 12AB\dfrac{1}{2}AB21​AB.

Quadrilaterals

Exercise 81

Exercise 82