{"id":585426,"date":"2022-03-16T07:52:49","date_gmt":"2022-03-16T07:52:49","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=585426"},"modified":"2022-03-17T05:44:28","modified_gmt":"2022-03-17T05:44:28","slug":"maharashtra-board-solutions-class-8-maths-practice-set-8-2-chapter-8-quadrilateral-constructions-and-types","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-8-maths-practice-set-8-2-chapter-8-quadrilateral-constructions-and-types\/","title":{"rendered":"Maharashtra Board Solutions Class 8-Maths (Practice Set 8.2): Chapter 8- Quadrilateral: Constructions and Types"},"content":{"rendered":"\n

Class 8: Maths Chapter 8 solutions. Complete Class 8 Maths Chapter 8 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions Class 8-Maths (Practice Set 8.2): Chapter 8- Quadrilateral: Constructions and Types<\/h2>\n\n\n\n

Maharashtra Board 8th Maths Chapter 8, Class 8 Maths Chapter 8 solutions<\/p>\n\n\n\n

Question 1.
Draw a rectangle ABCD such that l(AB) = 6.0 cm and l(BC) = 4.5 cm.
Solution:<\/strong>
\"Maharashtra<\/p>\n\n\n\n

Question 2.
Draw a square WXYZ with side 5.2 cm.
Solution:<\/strong>
\"Maharashtra
\"Maharashtra<\/p>\n\n\n\n

Question 3.
Draw a rhombus KLMN such that its side is 4 cm and m\u2220K = 75\u00b0.
Solution:<\/strong>
\"Maharashtra<\/p>\n\n\n\n

Question 4.
If diagonal of a rectangle is 26 cm and one side is 24 cm, find the other side.
Solution:<\/strong>
\"Maharashtra
Let \u20b9ABCD be the rectangle.
l(BC) = 24cm, l(AC) = 26cm
In \u2206ABC,
m\u2220ABC = 90\u00b0 \u2026[Angle of a rectangle]
\u2234[l(AC)]\u00b2 = [l(AB)]2 + [l(BC)]\u00b2
\u2026[Pythagoras theorem]
\u2234 (26 )\u00b2 = [l(AB)]\u00b2 + (24)\u00b2
\u2234(26)\u00b2 \u2013 (24)\u00b2 = [l(AB)]\u00b2
\u2234(26 + 24) (26 \u2013 24) = [l(AB)]\u00b2
\u2026[\u2235 a\u00b2 \u2013 b\u00b2 = (a + b)(a \u2013 b)]
\u223450 x 2 = [l(AB)]\u00b2
\u2234100 = [l(AB)]\u00b2
i.e. [l(AB)]\u00b2 = 100
\u2234l(AB) = \u221a100
\u2026[Taking square root of both sides]
\u2234l(AB) =10 cm
\u2234The length of the other side is 10 cm.<\/p>\n\n\n\n

Question 5.
Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.
Solution:<\/strong>
In rhombus ABCD,
\"Maharashtra
l(AC) = 16 cm and l(BD) = 12 cm.
Let the diagonals of rhombus ABCD intersect at point O.
l(AO) = 12 l(AC)
\u2026[Diagonals of a rhombus bisect each other]
\u2234l(AO) = 12 \u00d7 16
\u2234l(AO) = 8 cm
Also, l(DO) = 12 l(BD)
\u2026[Diagonals of a rhombus bisect each other]
\u2234l(DO) = 12 \u00d7 12
\u2234l(DO) = 6 cm
In \u2206DOA,
m\u2220DOA = 90\u00b0
..[Diagonals of a rhombus are perpendicular to each other]
[l(AD)]\u00b2 = [l(AO)]\u00b2 + [l(DO)]\u00b2
\u2026[Pythagoras theorem]
\"Maharashtra
= (8)\u00b2 + (6)\u00b2
= 64 + 36
\u2234[l(AD)]\u00b2 = 100
\u2234l(AD) = \u221a100
\u2026 [Taking square root of both sides]
\u2234l(AD) = 10 cm
\u2234l(AB) = l(BC) = l(CD) = l(AD) = 10 cm
\u2026[Sides of a rhombus are congruent]
Perimeter of rhombus ABCD
= l(AB) + l(BC) + l(CD) + l(AD)
= 10+10+10+10
= 40 cm
\u2234The side and perimeter of the rhombus are 10 cm and 40 cm respectively.<\/p>\n\n\n\n

Question 6.
Find the length of diagonal of a square with side 8 cm.
Solution:<\/strong>
Let \u20b9XYWZ be the square of side 8cm.
\"Maharashtra
seg XW is a diagonal.
In \u2206 XYW,
m\u2220XYW = 90\u00b0
\u2026 [Angle of a square]
\u2234 [l(XW)]\u00b2 = [l(XY)]\u00b2 + [l(YW)]\u00b2
\u2026[Pythagoras theorem]
= (8)\u00b2 + (8)\u00b2
= 64 + 64
\u2234 [l(XW)]\u00b2 = 128
\u2234 l(XW) = \u221a128
\u2026[Taking square root of both sides]
= \u221a64 \u00d7 2
= 8 \u221a2 cm
\u2234 The length of the diagonal of the square is 8 \u221a2 cm.<\/p>\n\n\n\n

Question 7.
Measure of one angle of a rhombus is 50\u00b0, find the measures of remaining three angles.
Solution:<\/strong>
Let \u20b9ABCD be the rhombus.
\"Maharashtra
m\u2220A = 50\u00b0
m\u2220C = m\u2220A
\u2026.[Opposite angles of a rhombus are congruent]
\u2234 m\u2220C = 50\u00b0
Also, m\u2220D = m\u2220B \u2026(i)
\u2026.[Opposite angles of a rhombus are congruent]
In \u20b9ABCD,
m\u2220A + m\u2220B + m\u2220C + m\u2220D = 360\u00b0
\u2026.[Sum of the measures of the angles of a quadrilateral is 360\u00b0]
\u2234 50\u00b0 + m\u2220B + 50\u00b0 + m\u2220D = 360\u00b0
\u2234 m\u2220B + m\u2220D + 100\u00b0 = 360\u00b0
\u2234 m\u2220B + m\u2220D = 360\u00b0 \u2013 100\u00b0
\u2234 m\u2220B + m\u2220B = 260\u00b0 \u2026[From (i)]
\u2234 2m\u2220B = 260\u00b0
\u2234 m\u2220B = 2602
\u2234 m\u2220B = 130\u00b0
\u2234 m\u2220D = m\u2220B = 130\u00b0 \u2026[From (i)]
\u2234 The measures of the remaining angles of the rhombus are 130\u00b0, 50\u00b0 and 130\u00b0.<\/p>\n\n\n\n

Intext Questions and Activities<\/strong><\/p>\n\n\n\n

Question 1.
Construct a rectangle PQRS by taking two convenient adjacent sides. Name the point of intersection of diagonals as T. Using divider and ruler, measure the following lengths.<\/strong>
i. lengths of opposite sides, seg QR and seg PS.
ii. lengths of seg PQ and seg SR.
iii. lengths of diagonals PR and QS.
iv. lengths of seg PT and seg TR, which are parts of the diagonal PR.
v. lengths of seg QT and seg TS, which are parts of the diagonal QS.
Observe the measures. Discuss about the measures obtained by your classmates. (Textbook pg. no. 44)
Solution:
Draw a rectangle PQRS such that, l(PQ) = 3 cm and l(QR) = 4 cm.
\"Maharashtra
Steps of construction:
i. As shown in the rough figure, draw seg QR of length 4 cm.
ii. Placing the centre of the protractor at point Q, draw ray QW making an angle of 90\u00b0 with seg QR.
iii. By taking a distance of 3 cm on the compass and placing it at point Q, draw an arc on ray QW. Name the point as P.
iv. Draw ray PV and ray RU making an angle of 90\u00b0 with seg PQ and seg QR respectively.
v. Name the point of intersection of ray PV and ray RU as S.
\u20b9PQRS is the required rectangle.
\"Maharashtra
From the figure,
i. l(QR) = l(PS) = 4 cm
ii. l(PQ) = l(SR) = 3 cm
iii. l(PR) = l(QS) = 5 cm
iv. l(PT) = l(TR) = 2.5 cm
v. l(QT) = l(TS) = 2.5 cm<\/p>\n\n\n\n

From the above measures, we can say that for any rectangle,
i. Opposite sides are congruent.
ii. Diagonals are congruent.
iii. Diagonals bisect each other.<\/p>\n\n\n\n

Question 2.
Draw a square by taking convenient length of side. Name the point of intersection of its diagonals as E. Using the apparatus in a compass box, measure the following lengths.<\/strong>
i. lengths of diagonal AC and diagonal BD.
ii. lengths of two parts of each diagonal made by point E.
iii. all the angles made at the point E.
iv. parts of each angle of the square made by each diagonal, (e.g. \u2220ADB and \u2220CDB).
Observe the measures. Also observe the measures obtained by your classmates and discuss about them. (Textbook pg. no. 44)
Solution:
Draw a square ABCD such that its side is 5cm
\"Maharashtra
Steps of construction:
i. As shown in the rough figure, draw seg BC of length 5 cm.
ii. Placing the centre of the protractor at point B, draw ray BP making an angle of 90\u00b0 with seg BC.
iii. By taking a distance of 5 cm on the compass and placing it at point B, draw an arc on ray BP. Name the point as A.
iv. Placing the centre of the protractor at point C, draw ray CQ making an angle of 90\u00b0 with seg BC.
v. By taking a distance of 5 cm on the compass and placing it at point C, draw an arc on ray CQ. Name the point as D.
vi. Draw seg AD.
\u20b9ABCD is the required square.
\"Maharashtra
From the figure,
i. l(AC) = l(BD) \u2245 7cm
ii. l(AE) = l(EC) \u2245 3.5cm,
l(BE) = l(ED) \u2245 3.5cm
iii. m\u2220AED = m\u2220BEC = m\u2220CED = m\u2220BEA = 90\u00b0
iv. Angles made by diagonal AC:
m\u2220BAC = m\u2220DAC = 45\u00b0
m\u2220BCA = m\u2220DCA = 45\u00b0
Angles made by diagonal BD:
m\u2220ABD = m\u2220CBD = 45\u00b0
m\u2220ADB = m\u2220CDB = 45\u00b0<\/p>\n\n\n\n

From the above measures, we can say that for any square,
i. Diagonals are congruent.
ii. Diagonals bisect each other.
iii. Diagonals are perpendicular to each other.
iv. Diagonals bisect the opposite angles.<\/p>\n\n\n\n

Question 3.
Draw a rhombus EFGH by taking convenient length of side and convenient measure of an angle.
Draw its diagonals and name their point of Intersection as M.<\/strong>
i. Measure the opposite angles of the quadrilateral and angles at the point M.
ii. Measure the two parts of every angle made by the diagonal.
iii. Measure the lengths of both diagonals. Measure the two parts of diagonals made by point M.
Observe the measures. Also observe the measures obtained by your classmates and discuss about them. (Textbook pg. no. 45)
Solution:
<\/strong>Draw a rhombus EFGH such that its side is 5 cm and m\u2220F = 60\u00b0.
\"Maharashtra
Steps of construction:
i. As shown in the rough figure, draw seg FG of length 5 cm.
ii. Placing the centre of the protractor at point F, draw ray FX making an angle 60\u00b0 with seg FG.
iii. By taking a distance of 5 cm on the compass and placing it at point F, draw an arc on ray FX. Name the point as E.
iv. By taking a distance of 5 cm on the compass and placing it at point E and point G, draw arcs. Name the point of intersection of arcs as H. \u20b9EFGH is the required rhombus.
\"Maharashtra
From the figure,
i. Opposite angles:
m\u2220EFG = m\u2220GHE = 60\u00b0,
m\u2220FEH = m\u2220HGF = 120\u00b0
Angles at the point M:
m\u2220EMF = m\u2220FMG = m\u2220GMH = m\u2220HME = 90\u00b0<\/p>\n\n\n\n

ii. Angles made by diagonal FH:
m\u2220EFH = m\u2220GFH = 30\u00b0 m\u2220EHF = m\u2220GHF = 30\u00b0
Angles made by diagonal EG:
m\u2220FEG = m\u2220HEG = 60\u00b0 m\u2220FGE = m\u2220HGE = 60\u00b0<\/p>\n\n\n\n

iii. l(FH) \u2248 8.6 cm
l(EG) = 5 cm
l(FM) = l(HM) \u2248 4.3 cm
l(EM) = l(GM) \u2248 2.5 cm<\/p>\n\n\n\n

From the above measures, we can say that for any rhombus,
i. Opposite angles are congruent.
ii. Diagonals bisect the opposite angles.
iii. Diagonals bisect each other and they are perpendicular to each other.<\/p>\n\n\n\n

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Maharashtra Board Solutions Class 8-Maths (Practice Set 8.2): Chapter 8- Quadrilateral: Constructions and Types<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions Class 8-Maths (Practice Set 8.2): Chapter 8- Quadrilateral: Constructions and Types PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 8 Maths :<\/strong><\/h2>\n\n\n\n