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Selina ICSE Solutions for Class 9 Maths Chapter 14 Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]
Exercise 14(A)
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In this linear equation n and k must be integer. Therefore to satisfy this equation the minimum value of k must be 6 to get n as integer. Hence the number of sides are: 5 + 6 = 11.
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Exercise 14(B)
Solution 1: (i)True. This is true, because we know that a rectangle is a parallelogram. So, all the properties of a parallelogram are true for a rectangle. Since the diagonals of a parallelogram bisect each other, the same holds true for a rectangle. (ii)False This is not true for any random quadrilateral. Observe the quadrilateral shown below.
Clearly the diagonals of the given quadrilateral do not bisect each other. However, if the quadrilateral was a special quadrilateral like a parallelogram, this would hold true. (iii)False Consider a rectangle as shown below.
It is a parallelogram. However, the diagonals of a rectangle do not intersect at right angles, even though they bisect each other. (iv)True Since a rhombus is a parallelogram, and we know that the diagonals of a parallelogram bisect each other, hence the diagonals of a rhombus too, bisect other. (v)False This need not be true, since if the angles of the quadrilateral are not right angles, the quadrilateral would be a rhombus rather than a square. (vi)True
A parallelogram is a quadrilateral with opposite sides parallel and equal. Since opposite sides of a rhombus are parallel, and all the sides of the rhombus are equal, a rhombus is a parallelogram. (vii)False This is false, since a parallelogram in general does not have all its sides equal. Only opposite sides of a parallelogram are equal. However, a rhombus has all its sides equal. So, every parallelogram cannot be a rhombus, except those parallelograms that have all equal sides. (viii)False This is a property of a rhombus. The diagonals of a rhombus need not be equal. (ix)True A parallelogram is a quadrilateral with opposite sides parallel and equal. A rhombus is a quadrilateral with opposite sides parallel, and all sides equal. If in a parallelogram the adjacent sides are equal, it means all the sides of the parallelogram are equal, thus forming a rhombus. (x)False
Observe the above figure. The diagonals of the quadrilateral shown above bisect each other at right angles, however the quadrilateral need not be a square, since the angles of the quadrilateral are clearly not right angles.
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Solution 10: We know that AQCP is a quadrilateral. So sum of all angles must be 360. ∴ x + y + 90 + 90 = 360 x + y = 180 Given x:y = 2:1 So substitute x = 2y 3y = 180 y = 60 x = 120 We know that angle C = angle A = x = 120 Angle D = Angle B = 180 – x = 180 – 120 = 60 Hence, angles of parallelogram are 120, 60, 120 and 60.
Exercise 14(C)
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More Resources for Selina Concise Class 9 ICSE Solutions
Each member of the quadrilateral family will describe its specific properties.
Quadrilateral
I have exactly four sides.
The sum of the interior angles of all quadrilaterals is 360º.
A quadrilateral is any four sided figure. Do not assume any additional properties for a quadrilateral unless you are given additional information.
Trapezoid
I have only one set of parallel sides. [The median of a trapezoid is parallel to the bases and equal to one-half the sum of the bases.]
A trapezoid has ONLY ONE set of parallel sides. When proving a figure is a trapezoid, it is necessary to prove that two sides are parallel and two sides are not parallel.
Isosceles Trapezoid
I have: only one set of parallel sides
base angles congruent
legs congruent
diagonals congruent
opposite angles supplementary
Never assume that a trapezoid is isosceles unless you are given (or can prove) that information.
Parallelogram
I have: 2 sets of parallel sides
2 sets of congruent sides
opposite angles congruent
consecutive angles supplementary
diagonals bisect each other
diagonals form 2 congruent triangles
Notice how the properties of a parallelogram come in sets of twos: two properties about the sides; two properties about the angles; two properties about the diagonals. Use this fact to help you remember the properties.
Rectangle
I have all of the properties of the parallelogram PLUS
4 right angles
diagonals congruent
If you know the properties of a parallelogram, you only need to add 2 additional properties to describe a rectangle.
Rhombus
I have all of the properties of the parallelogram PLUS
4 congruent sides
diagonals bisect angles
diagonals perpendicular
A rhombus is a slanted square. It has all of the properties of a parallelogram plus three additional properties.
Square
I have all of the properties of the parallelogram AND the rectangle AND the rhombus.
The square is the most specific member of the quadrilateral family. It has the largest number of properties.
A quadrilateral is a figure bounded by four line segments such that no three of them are parallel.
Two sides of quadrilateral are consecutive or adjacent sides, if they have a common point (vertex).
Two sides of a quadrilateral are opposite sides, if they have no common end-point (vertex).
The consecutive angles of a quadrilateral are two angles which include a side in their intersection.
In other words, two angles are consecutive, if they have a common arm.
Two angles of a quadrilateral are said to be opposite angles if they do not have a common arm.
The sum of the four angles of a quadrilateral is 360º.
A quadrilateral having exactly one pair of parallel sides, is called a trapezium.
A trapezium is said to be an isoscels trapezium, if its non-parallel sides are equal.
A quadrilateral is a parallelogram if its both pairs of opposite sides are parallel.
A parallelogram having all sides equal is called a rhombus.
A parallelogram whose each angle is a right angle, is called a rectangle.
A square is a rectangle with a pair of adjacent sides equal.
A quadrilateral is a kite if it has two pairs of equal adjacent sides and unequal opposite sides.
A diagonal of a parallelogram divides it into two congruent triangles.
In a parallelogram, opposite sides are equal.
The opposite angles of a parallelogram are equal.
The diagonals of a parallelogram bisect each other.
In a parallelogram, the bisectors of any two consecutive angles intersect at right angle.
If diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angle.
The angle bisectors of a parallegram form a rectangle.
A quadrilateral is a parallelogam if its opposite sides are equal.
A quadrilateral is a parallelogram if its opposite angles are equal.
If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.
A quadrilateral is a parallelogram, if its one pair of opposite sides are equal and parallel.
Each of the four angles of a rectangel is a right angle.
Each of the four sides of a rhombus is of the same length.
Each of the angles of a square is a right angle and each of the four sides is of the same length.
The diagonals of a rectangle are of equal length.
If the two diagonals of parallelogram are equal, it is a rectangle.
The diagonals of a rhombus are perpendicular to each other.
If the diagonals of a parallelogram are perpendicular, then it is a rhombus.
The diagonals of a square are equal and perpendicular to each other.
If the diagonals of a parallelogram are equal and intersect at right angles then the parallelogram is a square.
A diagonal of a parallelogram divides it into two triangles of equal area.
For each base of a parallelogram, the corresponding altitude is the line segment from a point on the base, perpendicular to the line containing the opposite side.
Parallelograms on the same base and between the same parallels are equal in area.
A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
The area of a parallelogram is the product of its base and the corresponding altitude.
Parallelograms on equal bases and between the same parallels are equal in area.
Types Of Quadrilaterals Example Problems With Solutions
Example 1: In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 2 : 4 : 5 : 7. Find the measure of each angles of the quadrilateral. Solution: We have ∠A : ∠B : ∠C : ∠D = 2 : 4 : 5 : 7. So, let ∠A = 2xº, ∠B = 4xº, ∠C = 5xº, ∠D = 7xº. ∴ ∠A + ∠B + ∠C + ∠D = 360º ⇒ x + 4x + 5x + 7x = 360º ⇒ 18x = 360º ⇒ x = 20º Thus, the angles are: ∠A = 40º, ∠B = (4 × 20)º = 80º, ∠C = (5 × 20)º = 100º and, ∠D = (7x)º = (7 × 20)º = 140º
Example 2: The sides BA and DC of a quadrilateral ABCD are produced as shown in fig. Prove that a + b = x + y. Solution: Join BD. In ∆ABD, we have
∠ABD + ∠ADB = bº ….(i) In ∆CBD, we have ∠CBD + ∠CDB = aº ….(ii) Adding (i) and (ii), we get (∠ABD + ∠CBD) + (∠ADB + ∠CDB) = aº + bº ⇒ xº + yº = aº + bº Hence, x + y = a + b
Example 3: In a quadrilateral ABCD, AO and BO are the bisectors of ∠A and ∠B respectively. Prove that ∠AOB = 1/2 (∠C + ∠D). Solution: In ∆AOB, we have
∠AOB + ∠1 + ∠2 = 180º
Example 4: In figure bisectors of ∠B and ∠D of quadrilateral ABCD meet CD and AB produced at P and Q respectively. Prove that ∠P + ∠Q = 1/2 (∠ABC + ∠ADC)
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Example 5: In a parallelogram ABCD, prove that sum of any two consecutive angles is 180º. Solution: Since ABCD is a parallelogram. Therefore, AD || BC.
Now, AD || BC and transversal AB intersects them at A and B respectively. ∴ ∠A + ∠B = 180º [∵ Sum of the interior angles on the same side of the transversal is 180º] Similarly, we can prove that ∠B + ∠C = 180º, ∠C + ∠D = 180º and ∠D + ∠A = 180º.
Example 6: In a parallelogram ABCD, ∠D = 115º, determine the measure of ∠A and ∠B. Solution: Since the sum of any two consecutive angles of a parallelogram is 180º. Therefore, ∠A + ∠D = 180º and ∠A + ∠B = 180º Now, ∠A + ∠D = 180º ⇒ ∠A + 115º = 180º [∵ ∠D = 115º (given)] ⇒ ∠A = 65º and ∠A + ∠B = 180º ⇒ 65º + ∠B = 180º ⇒ ∠B = 115º Thus, ∠A = 65º and ∠B = 115º
Example 7: In figure, AB = AC, ∠EAD = ∠CAD and CD || AB. Show that ABCD is a parallelogram.
Solution: In ∆ABC, AB = AC [Given] ⇒ ∠ABC = ∠ACB ….(1) (Angles opposite the equal sides are equal) ∠EAD = ∠CAD[Given] ….(2) Now, ∠EAC = ∠ABC + ∠ACB [An exterior angle is equal to sum of two interior opposite angles of a triangles] ⇒ ∠EAD + ∠CAD = ∠ABC + ∠ACB ⇒ ∠CAD + ∠CAD = ∠ACB + ∠ACB By (1) and (2) ⇒ 2∠CAD = 2∠ACB ⇒ ∠CAD = ∠ACB ⇒ BC | | AD Also, CD | | AB [Given] Thus, we have both pairs of opposite sides of quadrilateral ABCD parallel. Therefore, ABCD is a parallelogram.
Example 8: ABCD is a parallelogram and line segments AX,CY are angle bisector of ∠A and ∠C respectively then show AX || CY.
Solution: Since opposite angles are equal in a parallelogram. Therefore, in parallelogram ABCD, we have ∠A = ∠C ⇒ 1/2 ∠A = 1/2 ∠C ⇒ ∠1 = ∠2 ….(i) [∵ AX and CY are bisectors of ∠A and ∠C respectively] Now, AB || DC and the transversal CY intersects them. ∴ ∠2 = ∠3 …(ii) [∵ Alternate interior angles are equal] From (i) and (ii), we get ∠1 = ∠3 Thus, transversal AB intersects AX and YC at A and Y such that ∠1 = ∠3 i.e. corresponding angles are equal. ∴ AX || CY
Example 9: In the adjoining figure, a point O is taken inside an equilateral quad. ABCD such that OB = OD. Show that A, O and C are in the same straight line.
Solution: Given a quad. ABCD in which AB = BC = CD = DA and O is a point within it such that OB = OD. To prove ∠AOB + ∠COB = 180º Proof In ∆OAB and OAD, we have AB = AD (given) OA = OA (common) and OB = OD (given) ∴ ∆OAB ≅ ∆OAD ∴ ∠AOB = ∠AOD ….(i) (c.p.c.t.) Similarly, ∆OBC ≅ ∆ODC ∴∠COB = ∠COD ….(ii) Now, ∠AOB + ∠COB + ∠COD + ∠AOD = 360º [∠ at a point] ⇒ 2(∠AOB + ∠COB) = 360º ⇒ ∠AOB + ∠COB = 180º
Example 10: In figure AN and CP are perpendiculars to the diagonal BD of a parallelogram ABCD. Prove that: (i) ∆ADN ≅ ∆CBP (ii) AN = CP
Solution: Since ABCD is a parallelogram. ∴ AD || BC Now, AD || BC and transversal BD intersects them at B and D. ∴ ∠1 = ∠2 [∵ Alternate interior angles are equal] Now, in ∆s ADN and CBP, we have ∠1 = ∠2 ∠AND = ∠CPD and, AD = BC [∵ Opposite sides of a ||gm are equal] So, by AAS criterion of congruence ∆ADN ≅ ∆CBP AN = CP [∵ Corresponding parts of congruent triangles are equal]
Example 11: In figure, ABCD is a trapezium such that AB || CD and AD = BC.
BE || AD and BE meets BC at E. Show that (i) ABED is a parallelogram. (ii) ∠A + ∠C = ∠B + ∠D = 180º. Solution: Here, AB || CD (Given) ⇒ AB || DE ….(1) Also, BE || AD (Given) ….(2) From (1) and (2), ABED is a parallelogram ⇒ AD = BE ….(3) Also, AD = BC (Given) ….(4) From (3) and (4), BE = BC ⇒ ∠BEC = ∠BCE ….(5) Also, ∠BAD = ∠BED (opposite angles of parallelogram ABED) i.e., ∠BED = ∠BAD ….(6) Now, ∠BED + ∠BEC = 180º (Linear pair of angles) ⇒ ∠BAD + ∠BCE = 180º By (5) and (6) ⇒ ∠A + ∠C = 180º Similarly, ∠B + ∠D = 180º
Example 12: In figure ABCD is a parallelogram and ∠DAB = 60º. If the bisectors AP and BP of angles A and B respectively, meet at P on CD, prove that P is the mid-point of CD.
Solution: We have, ∠DAB = 60º ∠A + ∠B = 180º ∴ 60º + ∠B = 180º ⇒ ∠B = 120º Now, AB || DC and transversal AP intersects them. ∴ ∠PAB = ∠APD ⇒ ∠APD = 30º [∵ ∠PAB = 30º] Thus, in ∆APD, we have ∠PAD = ∠APD [Each equal to 30º] ⇒ AD = PD …. (i) [∵ Angles opposite to equal sides are equal] Since BP is the bisector of ∠B. Therefore, ∠ABP = ∠PBC = 60º Now, AB || DC and transversal BP intersects them. ∴ ∠CPB = ∠ABP ⇒ ∠CPB = 60º [∵ ∠ABP = 60º] Thus, in ∆CBP, we have ∠CBP = ∠CPB [Each equal to 60º] ⇒ CP = BC ∵ [Sides opp, to equal angles are equal] ⇒ CP = AD …. (ii) [∵ ABCD is a parallelogram ∴ AD = BC] From (i) and (ii), we get PD = CP ⇒ P is the mid point of CD.
Example 13: Prove that the line segments joining the mid-point of the sides of a quadrilateral forms a parallelogram. Solution: Points E, F, G and H are the mid-points of the sides AB, BC, CD and DA respectively, of the quadrilateral ABCD. We have to prove that EFGH is a parallelogram.
Join the diagonal AC of the quadrilateral ABCD. Now, in ∆ABC, we have E and F mid-points of the sides BA and BC. ⇒ EF || AC and EF = 1/2 AC …. (1) Similarly, from ∆ADC, we have GH || AC and GH = 1/2 AC ….(2) Then from (1) and (2), we have EF || GH and EF = GH This proves that EFGH is a parallelogram.
Example 14: In figure ABCD is a parallelogram and X, Y are the mid-points of sides AB and DC respectively. Show that AXCY is a parallelogram. Solution: Since X and Y are the mid-points of AB and DC respectively. Therefore, AX = 1/2 AB and CY = 1/2 DC … (i) But, AB = DC [∵ ABCD is a parallelogram]
⇒ 1/2 AB = 1/2 DC ⇒ AX = CY …. (ii) Also, AB || DC ⇒ AX || YC …. (iii) Thus, in quadrilateral AXCY, we have AX || YC and AX = YC [From (ii) and (iii)] Hence, quadrilateral AXCY is a parallelogram.
Example 15: Prove that the line segments joining the mid-points of the sides of a rectangle forms a rhombus. Solution: P, Q, R and S are the mid-points of the sides AB, BC, CD and DA of the rectangle ABCD.
Example 16: In figure ABCD is a parallelogram and X and Y are points on the diagonal BD such that DX = BY. Prove that (i) AXCY is a parallelogram (ii) AX = CY, AY = CX (iii) ∆AYB ≅ ∆CXD Solution: Given : ABCD is a parallelogram. X and Y are points on the diagonal BD such that DX = BY To Prove: (i) AXCY is a parallelogram (ii) AX = CY, AY = CX (iii) ∆AYB ≅ ∆CXD C Construction : join AC to meet BD at O. Proof: (i) We know that the diagonals of a parallelogram bisect each other. Therefore, AC and BD bisect each other at O.
∴ OB = OD But, BY = DX ∴ OB – BY = OD – DX ⇒ OY = OX Thus, in quadrilateral AXCY diagonals AC and XY are such that OX = OY and OA = OC i.e. the diagonals AC and XY bisect each other. Hence, AXCY is a parallelogram. (ii) Since AXCY is a parallelogram ∴ AX = CY and AY = CX (iii) In triangles AYB and CXD, we have AY = CX [From (ii)] AB = CD [∵ ABCD is a parallelogram] BY = DX [Given] So, by SSS-criterion of congruence, we have ∆AYB ≅ ∆CXD
Example 17: In fig. ABC is an isosceles triangle in which AB = AC. CP || AB and AP is the bisector of exterior ∠CAD of ∆ABC. Prove that ∠PAC = ∠BCA and ABCP is a parallelogram. Solution: Given : An isosceles ∆ABC having AB = AC.AP is the bisector of ext ∠CAD and CP || AB. To Prove: ∠PAC = ∠BCA and ABCP Proof: In ∆ABC, we have AB = AC [Given] ⇒ ∠1 = ∠2 …. (i) ∵ Angles opposite to equal sides in a ∆ are equal Now, in ∆ ABC, we have ext ∠CAD = ∠1 + ∠2
⇒ ext ∠CAD = 2∠2 [∵ ∠1 = ∠2 (from (i))] ⇒ 2∠3 = 2∠2 [∵ AP is the bisector of ext.∠CAD ∴∠CAD = 2∠3] ⇒ ∠3 = ∠2 Thus, AC intersects lines AP and BC at A and C respectively such that ∠3 = ∠2 i.e., alternate interior angles are equal. Therefore, AP || BC. But, CP || AB [Gvien] Thus, ABCP is a quadrilateral such that AP || BC and CP || AB. Hence, ABCP is a parallelogram.
Example 18: In the given figure, ABCD is a square and ∠PQR = 90º. If PB = QC = DR, prove that
(i) QB = RC, (ii) PQ = QR, (iii) ∠QPR = 45º. Solution: BC = DC, CQ = DR ⇒ BC – CQ = ∆CDR ⇒ QB = RC From ∆CQR, ∠RQB = ∠QCR + ∠QRC ⇒ ∠RQP + ∠PQB = 90º + ∠QRC ⇒ 90º + ∠PQB = 90º + ∠QRC Now, ∆RCQ ≅ ∆QBP and therefore, QR = PQ PQ = QR ⇒ ∠QPR = ∠PRQ Bur, ∠QPR + ∠PRQ = 90º. So, ∠QPR = 45º
Example 19: Prove that in a parallelogram (i) opposite sides are equal (ii) opposite angles are equal (iii) each diagonal bisects the parallelogram Solution:Given: A ||gm ABCD in which AB || DC and AD || BC. To prove: (i) AB = CD and BC = AD; (ii) ∠B = ∠D and ∠A = ∠C, (iii) ∆ABC = ∆CDA and ∆ABD = ∆CDB Construction join A and C. In ∆ABC and CDA, we have,
∠1 = ∠2 [Alt. int. ∠, as AB || DC and CA cuts them] ∠3 = ∠4 [Alt. int. ∠, as BC || AD and CA cuts them] AC = CA (common) ∴ ∆ABC ≅ ∆CDA [AAS-criterial] (i) ∆ABC ≅ ∆CDA (proved) ∴ AB = CD and BC = AD (c.p.c.t.) (ii) ∆ABC ≅ ∆CDA (proved) ∴ ∠B = ∠D (c.p.c.t.) Also, ∠1 = ∠2 and ∠3 = ∠4 ∠1 + ∠4 = ∠2 + ∠3 ⇒ ∠A = ∠C Hence, ∠B = ∠D and ∠A = ∠C (iii) Since ∆ABC ≅ ∆CDA and congruent triangles are equal in area, So we have ∆ABC = ∆CDA Similarly, ∆ABD = ∆CDB
Example 20: If the diagonals of a parallelogram are perpendicular to each other, prove that it is a rhombus. Solution: Since the diagonals of a ||gm bisect each other,
we have, OA = OC and OB = OD. Now, in ∆AOD and COD, we have OA = OC, ∠AOD = ∠COD =90° and OD is common ∴ ∆AOD ≅ ∆COD ∴ AD = CD (c.p.c.t.) Now, AB = CD and AD = BC (opp. sides of a ||gm) and AD = CD (proved) ∴ AB = CD = AD = BC Hence, ABCD is a rhombus.
Example 21: PQRS is a square. Determine ∠SRP. Solution: PQRS is a square. ∴ PS = SR and ∠PSR = 90° Now, in ∆ PSR, we have PS = SR
Example 22: In the adjoining figure, ABCD is a rhombus. If ∠A = 70º, find ∠CDB Sol.
We have ∠C = ∠A = 70º (opposite ∠ of a ||gm) Let ∠CDB = xº In ∆CDB, we have CD = CB ⇒∠CBD = ∠CDB = xº ∴ ∠CDB + ∠CBD + ∠DCB = 180º (angles of a triangle) ⇒ xº + xº + 70º = 180º ⇒ 2x = 110, i.e., x = 55 Hence, ∠CDB = 55º
Example 23: ABCD is a rhombus with ∠ABC = 56°. Determine ∠ACD. Solution: ABCD is a parallelogram
Example 24: Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half the difference of these sides. Solution:Given: A trapezium ABCD in which AB || DC and P and Q are the mid-points of its diagonals AC and BD respectively.
To Prove: (i) PQ || AB or DC (ii) PQ = 1/2 (AB – DC) Construction: Join DP and produce DP to meet AB in R. Proof: Since AB || DC and transversal AC cuts them at A and C respectively. ∠1 = ∠2 …. (i) [∴ Alternate angles are equal] Now, in ∆s APR and DPC, we have ∠1 = ∠2 [From (i)] AP = CP [∵ P is the mid-point of AC] and, ∠3 = ∠4 [Vertically opposite angles] So, by ASA criterion of congruence ∆APR ≅ ∆DPC ⇒ AR = DC and PR = DP ….(ii) [∵ Corresponding parts of congruent triangles are equal] In ∆DRB, P and Q are the mid-points of sides DR and DB respectively. ∴ PQ || RB ⇒ PQ || AB [∵ RB is a part of AB] ⇒ PQ || AB and DC [∵ AB || DC (Given)] This proves (i). Again, P and Q are the mid-points of sides DR and DB respectively in ∆DRB. ∴ PQ = 1/2 RB ⇒ PQ = 1/2 (AB – AR) ⇒ PQ = 1/2 (AB – DC) [From (ii), AR = DC] This proves (ii).
Example 25: In the adjoining figure, ABCD is parallelogram and X, Y are the points on diagonal BD such that DX = BY. Prove that CXAY is a parallelogram.
Solution: Join AC, meeting BD at O. Since the diagonals of a parallelogram bisect each other, we have OA = OC and OD = OB. Now, OD = OB and DX = BY ⇒ OD – DX = OB – BY ⇒ OX = OY Now, OA = OC and OX = OY ∴ CXAY is a quadrilateral whose diagonals bisect each other. ∴ CXAY is a ||gm
Example 26: Prove that the four triangles formed by joining in pairs, the mid-points of three sides of a triangle, are concurrent to each other. Solution:Given: A triangle ABC and D,E,F are the mid-points of sides BC, CA and AB respectively. To Prove: ∆ AFE ≅ ∆FBD ≅ ∆EDC ≅ ∆DEF. Proof: Since the segment joining the mid-points of the sides of a triangle is half of the third side. Therefore,
DE = 1/2 ⇒ DE = AF = BF …. (i) EF = 1/2 ⇒ EF = BD = CD …. (ii) DF = 1/2 ⇒ DF = AE = EC ….(iii) Now, in ∆s DEF and AFE, we have DE = AF [From (i)] DF = AE [From (ii)] and, EF = FE [Common] So, by SSS criterion of congruence, ∆ DEF ≅ ∆AFE Similarly, ∆DEF ≅ ∆FBD and ≅ DEF ≅ ∆EDC Hence, ∆ AFE ≅ ∆FBD ≅ ∆EDC ≅ ∆DEF.
Example 27: In fig, AD is the median and DE || AB. Prove that BE is the median. Solution: In order to prove that BE is the median, it is sufficient to show that E is the mid-point of AC. Now, AD is the median in ∆ABC ⇒ D is the mid-point of BC.
Since DE is a line drawn through the mid-point of side BC of ∆ABC and is parallel to AB (given). Therefore, E is the mid-point of AC. Hence, BE is the median of ∆ABC.
Example 28: Let ABC be an isosceles triangle with AB = AC and let D,E,F be the mid-points of BC, CA and AB respectively. Show that AD ⊥ FE and AD is bisected by FE. Solution: Given: An isosceles triangle ABC with D, E and F as the mid-points of sides BC, CA and AB respectively such that AB = AC. AD intersects FE at O. To Prove: AD ⊥ FE and AD is bisected by FE. Constructon: Join DE and DF. Proof: Since the segment joining the mid-points of two sides of a triangle is parallel to third side and is half of it. Therefore, DE || AB and DE = 1/2 AB Also, DF || AC and DF = 1/2 AC
But, 1/2 AB = 1/2 AC [Given] ⇒ AB = AC ⇒ DE = DF …. (i) Now, DE = 1/2 AB ⇒ DE = AF …. (ii) and, DF = 1/2 AC ⇒ DF = AE …(iii) From (i), (ii) and (iii) we have DE = AE = AF = DF ⇒ DEAF is a rhombus. ⇒ Diagonals AD and FE bisect each other at right angle. AD ⊥ FE and AD is bisected by FE.
Example 29: ABCD is a parallelogram. P is a point on AD such that AP = 1/3 AD and Q is a point on BC such that CQ = 1/3 BP. Prove that AQCP is a parallelogram.
Solution: ABCD is a parallelogram. ⇒ AD = BC and AD || BC ⇒ 1/3 AD = 1/3 BC and AD || BC ⇒ AP = CQ and AP || CQ Thus, APCQ is a quadrilateral such that one pair of opposite side AP and CQ are parallel and equal. Hence, APCQ is a parallelogram.
Example 30: In fig. D,E and F are, respectively the mid- points of sides BC, CA and AB of an equilateral triangle ABC. Prove that DEF is also an equilateral triangle. Solution: Since the segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively.
Example 31: P,Q and R are, respectively, the mid-points of sides BC, CA and AB of a triangle ABC. PR and BQ meet at X. CR and PQ meet at Y. Prove that XY = 1/4 BC Solution:Given: A ∆ABC with P,Q and R as the mid-points of BC, CA and AB respectively. PR and BQ meet at X and CR and PQ meet at Y. Construction: Join “X and Y.
Proof: Since the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it. Therefore, Q and R are mid-points of AC and AB respectively.
Similarly, Y is the mid-point of PQ. Now, consider ∆PQR. XY is the line segment joining the mid-points of sides PR and PQ. ∴ XY = 1/2 RQ …. (i) But RQ = 1/2 BC [From (i)] Hence, XY = 1/4 BC
Example 32: Show that the quadrilateral, formed by joining the mid-points of the sides of a square, is also a square. Solution:Given: A square ABCD in which P, Q, R, S are the mid-points of sides AB, BC, CD, DA respectively. PQ, QR, RS and SP are joined. To Prove: PQRS is a square. Construction: Join AC and BD.
Proof : In ∆ABC, P and Q are the mid-points of sides AB and BC respectively. ∴ PQ || AC and PQ = 1/2 AC …. (i) In ∆ADC, R and S are the mid-points of CD and AD respectively. ∴ RS || AC and RS = 1/2 AC ….(ii) From (i) and (ii), we have PQ || RS and PQ = RS ….(iii) Thus, in quadrilateral PQRS one pair of opposite sides are equal and parallel. Hence, PQRS is a parallelogram. Now, in ∆s PBQ and RCQ, we have PB = RC
Now, PQ || AC [From (i)] ⇒ PM || NO ….(vi) Since P and S are the mid-points of AB and AD respectively. PS || BD ⇒ PM || MO ….(vii) Thus, in quadrilateral PMON, we have PM || NO [From (vi)] PN || MO [From (vii)] So, PMON is a parallelogram. ⇒ ∠MPN = ∠MON ⇒ ∠MPN = ∠BOA [∵ ∠MON = ∠BOA] ⇒ ∠MPN = 90° ⇒ ∠QPS = 90° Thus, PQRS is a quadrilateral such that PQ = QR = RS = SP and ∠QPS = 90°. Hence, PQRS is a square.
Example 33: ∆ABC is a triangle right angled at B ; and P is the mid-point of AC. Prove that PB = PA = 1/2 AC. Solution: Given : ∆ABC right angled at B, P is the mid-point of AC. To Prove: PB = PA = 1/2 AC. Construction: Through P draw PQ || BC meeting AB at Q.
Proof: Since PQ || BC. Therefore, ∠AQP = ∠ABC [Corresponding angles] ⇒ ∠AQP = 90° [∵ ∠ABC = 90°] But, ∠AQP + ∠BQP = 180° [∵ ∠AQP & ∠BQP are angles of a linear pair] ∴ 90° + ∠BQP = 180° ⇒ ∠BQP = 90° Thus, ∠AQP = ∠BQP = 90° Now, in ∆ABC, P is the mid-point of AC and PQ || BC. Therefore, Q is the mid-point of AB i.e, AQ = BQ. Consider now ∆s APQ and BPQ. we have, AQ = BC [Proved above] ∠AQP = ∠BQP [From (i)] and, PQ = PQ So, by SAS cirterion of congruence ∆APQ ≅ ∠BPQ ⇒ PA = PB Also, PS = 1/2 AC, since P is the mid-point of AC Hence, PA = PB = 1/2 AC.
Example 34: Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus. Solution:Given: A rectangle ABCD in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined. To Prove: PQRS is rhombus. Construction: Join AC. Proof: In ∆ABC, P and Q are the mid-points of sides AB and BC respectively. ∴ PQ || AC and PQ = AC …. (i) In ∆ ADC, R and S are the mid-points of CD and AD respectively.
∴ SR || AC and SR = 1/2 AC …. (ii) From (i) and (ii), we get PQ || SR and PQ = SR ….(iii) ⇒ PQRS is a parallelogram. Now, ABCD is a rectangle. ⇒ AD = BC ⇒ 1/2 AD = 1/2 BC ⇒ AS = BQ ….(iv) In ∆s APS and BPQ , we have AP = BP [∴P is the mid-point of AB] ∠PAS = ∠PBQ [Each equal to 90°] and, AS = BQ [From (iv)] So, by SAS criterion of congruence ∆APS ≅ ∆BPQ PS = PQ ….(v) [∵ Corresponding parts of congruent triangles are equal] From (iii) and (v), we obtain that PQRS is a parallelogram such that PS = PQ i.e., two adjacent sides are equal. Hence, PQRS is a rhombus.
Parallelogram: A quadrilateral in which the pairs of opposite sides are parallel, is called a parallelogram. In a parallelogram,
Opposite sides are equal and parallel, i.e., AB = DC, BC = AD, AB || DC, and BC || AD.
Opposite angles are equal, i.e., ∠A = ∠C and ∠B = ∠D.
Diagonals bisect each other, i.e., AO = OC and BO = OD.
Diagonal divides the parallelogram into two congruent triangles, i.e., ΔADC and ΔABC are congruent and ΔADB and ΔCBD are congruent.
Rectangle: A parallelogram in which each angle is a right angle, is called a rectangle. In a rectangle,
Opposite sides are equal and parallel, i. e., AB = CD, AB || CD, BC = DA, and BC || DA.
All angles are right angles, i.e., ∠A = ∠B = ∠C = ∠D = 90°.
Both the diagonals are equal, i.e., AC = BD.
Diagonals bisect each other, i.e., AR = RC and BR = RD.
Rhombus: A parallelogram in which all four sides are equal, is called a rhombus. In a rhombus,
All sides are equal, i.e., PQ = QR = RS = SR
Opposite sides are parallel, i.e., PQ || SR and SP || RQ.
Opposite angles are equal, i.e., ∠P = ∠R and ∠Q = ∠S
Diagonals bisect each other at right angle, i.e., PO = OR, SO = OQ and ∠SOR = ∠SOP = ∠ROQ = ∠QOP = 90°.
Square: A rhombus with all its angles equal to right angle, is called a square. In other words, a square is a rectangle in which all the four sides are equal. In a square,
Opposite sides are parallel, i.e., AB || DC and AD || BC.
All the sides are equal, i.e., AB = BC = CD = DA.
All the angles are equal to 90°, i.e., ∠A = ∠B = ∠C = ∠D = 90°
The diagonals bisect each other at right angle, i. e., AO = OC and BO = OD and ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
Both the diagonals are equal in length, i.e., AC = BD.
Trapezium: A quadrilateral in which one pair of opposite sides are parallel, is called a trapezium. In a trapezium,
A pair of opposite sides is parallel.
Both the diagonals AC and BD are of different lengths.
Diagonals do not bisect each other at right angle.
Opposite angles are not equal.
Isosceles trapezium: A quadrilateral in which a pair of opposite sides is parallel and the other two sides are equal, is called an isosceles trapezium. In an isosceles trapezium,
A pair of opposite sides is parallel, i.e., AB || CD.
Non-parallel sides are equal, i.e., AD = BC.
Both the diagonals are equal, i.e., AC = BD.
∠A = ∠B and ∠D = ∠C.
Kite: A quadrilateral in which two pairs of adjacent sides are equal is called a kite. In a kite,
Two pairs of adjacent sides are equal, i.e., AB = AD and BC = CD.
Diagonals intersect each other at right angle, i.e., ∠AEB = ∠AED = ∠BEC = ∠DEC = 90°.