Mathematics Solutions Solutions for Class 8 Math Chapter 8 Quadrilateral : Constructions And Types are provided here with simple step-by-step explanations. These solutions for Quadrilateral : Constructions And Types are extremely popular among class 8 students for Math Quadrilateral : Constructions And Types Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of class 8 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class 8 Math are prepared by experts and are 100% accurate.
Page No 43:
Question 1:
Construct the following quadrilaterals of given measures.
(1) In â MORE, l(MO) = 5.8 cm, l(OR) = 4.4 cm, m∠M = 58, m∠O = 105, m∠R = 90.
(2) Construct â DEFG such that l(DE) = 4.5 cm, l(EF) = 6.5 cm, l(DG) = 5.5 cm, l(DF) = 7.2 cm, l(EG) = 7.8 cm.
(3) In â ABCD l(AB) = 6.4 cm, l(BC) = 4.8 cm, m∠A = 70, m∠B = 50, m∠C = 140.
(4) Construct â LMNO such that l(LM) = l(LO) = 6 cm, l(ON) = l(NM) = 4.5 cm, l(OM) = 7.5 cm.
Answer:
(1)
Steps of Construction:
Step 1: Draw MO = 5.8 cm.
Step 2: Draw ∠MOX = 105.
Step 3: With O as centre and radius 4.4 cm, draw an arc cutting ray OX at R.
Step 4: Draw ∠ORY = 90.
Step 5: Draw ∠OMZ = 58 such that ray MZ and ray RY intersect each other at E.
Here, MORE is the required quadrilateral.
(2)
Steps of Construction:
Step 1: Draw DE = 4.5 cm.
Step 2: With D as centre and radius 7.2 cm, draw an arc.
Step 3: With E as centre and radius 6.5 cm, draw an arc cutting the previous arc at F.
Step 4: Join EF.
Step 5: With D as centre and radius 5.5 cm, draw an arc.
Step 6: With E as centre and radius 7.8 cm, draw an arc cutting the previous arc at G.
Step 7: Join DG and GF.
Here, DEFG is the required quadrilateral.
(3)
Steps of Construction:
Step 1: Draw AB = 6.4 cm.
Step 2: Draw ∠ABX = 50.
Step 3: With B as centre and radius 4.8 cm, draw an arc cutting ray BX at C.
Step 4: Draw ∠BCY = 140.
Step 5: Draw ∠BAZ = 70 such that ray AZ and ray CY intersect each other at D.
Here, ABCD is the required quadrilateral.
(4)
Steps of Construction:
Step 1: Draw LM = 6 cm.
Step 2: With L as centre and radius 6 cm, draw an arc.
Step 3: With M as centre and radius 7.5 cm, draw an arc cutting the previous arc at O.
Step 4: Join OL.
Step 5: With O as centre and radius 4.5 cm, draw an arc.
Step 6: With M as centre and radius 4.5 cm, draw an arc cutting the previous arc at N.
Step 7: Join ON and MN.
Here, LMNO is the required quadrilateral.
Page No 46:
Question 1:
Draw a rectangle ABCD such that l(AB) = 6.0 cm and l (BC) = 4.5 cm.
Answer:
Steps of Construction:
Step 1: Draw AB = 6 cm.
Step 2: Draw ∠ABX = 90º.
Step 3: With B as centre and radius 4.5 cm, draw an arc cutting the ray BX at C.
Step 4: With C as centre and radius 6 cm, draw an arc.
Step 5: With A as centre and radius 4.5 cm, draw an arc cutting the previous arc at D.
Step 6: Join AD and CD.
Here, ABCD is the required rectangle.
Page No 46:
Question 2:
Draw a square WXYZ with side 5.2 cm.
Answer:
Steps of Construction:
Step 1: Draw WX = 5.2 cm.
Step 2: Draw ∠WXP = 90º.
Step 3: With X as centre and radius 5.2 cm, draw an arc cutting the ray XP at Y.
Step 4: With Y as centre and radius 5.2 cm, draw an arc.
Step 5: With W as centre and radius 5.2 cm, draw an arc cutting the previous arc at Z.
Step 6: Join YZ and WZ.
âHere, WXYZ is the required square.
Page No 46:
Question 3:
Draw a rhombus KLMN such that its side is 4 cm and m∠K = 75.
Answer:
Steps of Construction:
Step 1: Draw KL = 4 cm.
Step 2: Draw ∠LKX = 75º.
Step 3: With K as centre and radius 4 cm, draw an arc cutting the ray KX at N.
Step 4: With N as centre and radius 4 cm, draw an arc.
Step 5: With L as centre and radius 4 cm, draw an arc cutting the previous arc at M.
Step 6: Join LM and NM.
âHere, KLMN is the required rhombus.
Page No 46:
Question 4:
If diagonal of a rectangle is 26 cm and one side is 24 cm, find the other side.
Answer:
Suppose ABCD is a rectangle.
Here, segment AC is a diagonal and segment AD is one side of the rectangle ABCD.
l(AC) = 26 cm and l(AD) = 24 cm.
In right âACD,
l(AC)2 = l(AD)2 + l(CD)2 (Pythagoras theorem)
⇒ l(CD)2 = l(AC)2 − l(AD)2
⇒ l(CD)2 = (26)2 − (24)2
⇒ l(CD)2 = 676 − 576 = 100
⇒ l(CD) = = 10 cm
Thus, the other side of the rectangle is 10 cm.
Page No 47:
Question 5:
Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.
Answer:
ABCD is a rhombus.
Here, segment AC and segment BD are the diagonals of the rhombus ABCD.
l(AC) = 16 cm and l(BD) = 12 cm.
Diagonals of a rhombus are perpendicular bisectors of each other.
∴ m∠AOD = 90º
Also, l(OA) = l(AC) = × 16 = 8 cm
l(OD) = l(BD) = × 12 = 6 cm
In right âAOD,
l(AD)2 = l(OA)2 + l(OD)2 (Pythagoras theorem)
⇒ l(AD)2 = (8)2 + (6)2
⇒ l(AD)2 = 64 + 36 = 100
⇒ l(AD) = = 10 cm
All sides of a rhombus are equal.
∴ Perimeter of the rhombus ABCD = 4 × Side of a rhombus = 4 × 10 = 40 cm
Thus, the side and perimeter of the rhombus are 10 cm and 40 cm, respectively.
Page No 47:
Question 6:
Find the length of diagonal of a square with side 8 cm.
Answer:
Suppose ABCD is a square of side 8 cm.
Here, segment AC is a diagonal of square ABCD.
In âABC,
l(AC)2 = l(AB)2 + l(BC)2 (Pythagoras theorem)
⇒ l(AC)2 = (8)2 + (8)2
⇒ l(AC)2 = 64 + 64 = 128
⇒ l(AC) = cm
Thus, the length of diagonal of square is cm.
Page No 47:
Question 7:
Measure of one angle of a rhombus is 50. find the measures of remaining three angles.
Answer:
Suppose ABCD is a rhombus.
Let m∠A = 50º.
Opposite angles of a rhombus are congruent.
∴ m∠C = m∠A = 50º and m∠B = m∠D
Now,
m∠A + m∠B + m∠C + m∠D = 360º
∴ 50º + m∠B + 50º + m∠D = 360º
⇒ m∠B + m∠D = 360º − 100º = 260º
⇒ 2 m∠B = 260º (m∠B = m∠D)
⇒ m∠B = = 130º
∴ m∠D = m∠B = 130º
Thus, the measures of the remaining angles of the rhombus are 130º, 50º and 130º.
Page No 49:
Question 1:
Measures of opposite angles of a parallelogram are (3x − 2) and (50 − x). Find the measure of its each angle.
Answer:
Let ABCD be the parallelogram.
Suppose ∠A = (3x − 2)º and ∠C = (50 − x)º.
We know that the opposite angles of a parallelogram are congruent.
∴ m∠A = m∠C
⇒ 3x − 2 = 50 − x
⇒ 3x + x = 50 + 2
⇒ 4x = 52
⇒ x = 13
∴ m∠A = (3x − 2)º = (3 × 13 − 2)º= (39 − 2)âº= 37º
So, m∠C = m∠A = 37º
Also, the adjacent angles of a parallelogram are supplementary.
∴ m∠A + m∠D = 180º
⇒ 37º + m∠D = 180º
⇒ m∠D = 180º − 37º = 143º
Now,
m∠B = m∠D = 143º (Opposite angles of a parallelogram are congruent)
Thus, the measure of the angles of the parallelogram are 37º, 143º, 37º and 143º.
Page No 50:
Question 2:
Referring the adjacent figure of a parallelogram, write the answer of questions given below.
(1) If l(WZ) = 4.5 cm then l(XY) = ?
(2) If l(YZ) = 8.2 cm then l(XW) = ?
(3) If l(OX) = 2.5 cm then l(OZ) = ?
(4) If l(WO) = 3.3 cm then l(WY) = ?
(5) If m∠WZY = 120 then m∠WXY = ? and m∠XWZ = ?
Answer:
WXYZ is a parallelogram.
(1)
l(XY) = l(WZ) = 4.5 cm (Opposite sides of a parallelogram are congruent)
(2)
l(XW) = l(YZ) = 8.2 cm (Opposite sides of a parallelogram are congruent)
(3)
l(OZ) = l(OX) = 2.5 cm (Diagonals of parallelogram bisect each other)
(4)
l(WY) = 2 × l(WO) = 2 × 3.3 = 6.6 cm (Diagonals of parallelogram bisect each other)
(5)
m∠WXY = m∠WZY = 120º (Opposite angles of a parallelogram are congruent)
Now,
m∠WZY + m∠XWZ = 180º (Adjacent angles of a parallelogram are supplementary)
⇒ 120º + m∠XWZ = 180º
⇒ m∠XWZ = 180º − 120º = 60º
Page No 50:
Question 3:
Construct a parallelogram ABCD such that l(BC) = 7 cm, m∠ABC = 40, l(AB) = 3 cm.
Answer:
Steps of Construction:
Step 1: Draw AB = 3 cm.
Step 2: Draw ∠ABX = 40.
Step 3: With B as centre and radius 7 cm, draw an arc cutting the ray BX at C.
Step 4: With C as centre and radius 3 cm, draw an arc.
Step 5: With A as centre and radius 7 cm, draw an arc cutting the previous arc at D.
Step 6: Join AD and CD.
Here, ABCD is the required parallelogram.
Page No 50:
Question 4:
Ratio of consecutive angles of a quadrilateral is 1:2:3:4. Find the measure of its each angle. Write, with reason, what type of a quadrilateral it is.
Answer:
Suppose PQRS is a quadrilateral.
Let m∠P : m∠Q : m∠R : m∠S = 1 : 2 : 3 : 4
So, m∠P = k, m∠Q = 2k, m∠R = 3k and m∠S = 4k, where k is some constant
Now,
m∠P + m∠Q + m∠R + m∠S = 360º
∴ k + 2k + 3k + 4k = 360º
⇒ 10k = 360º
⇒ k = 36º
∴ m∠P = 36º
m∠Q = 2k = 2 × 36º = 72º
m∠R = 3k = 3 × 36º = 108º
m∠S = 4k = 4 × 36º = 144º
Now, m∠P + m∠S = 36º + 144º = 180º
We know if two lines are intersected by a transversal such that the sum of interior angles on the same of the transversal are supplementary, then the two lines are parallel.
∴ Side PQ || Side SR
Also, m∠P + m∠Q = 36º + 72º = 108º ≠ 180º
So, side PS is not parallel to side QR.
In quadrilateral PQRS, only one pair of opposite sides is parallel. Therefore, quadrilateral PQRS is a trapezium.
Page No 50:
Question 5:
Construct â BARC such that l(BA) = l(BC) = 4.2 cm, l(AC) = 6.0 cm, l(AR) = l(CR) = 5.6 cm
Answer:
Steps of Construction:
Step 1: Draw BA = 4.2 cm.
Step 2: With B as centre and radius 4.2 cm, draw an arc.
Step 3: With A as centre and radius 6 cm, draw an arc cutting the previous arc at C.
Step 4: Join BC.
Step 5: With A as centre and radius 5.6 cm, draw an arc.
Step 6: With C as centre and radius 5.6 cm, draw an arc cutting the previous arc at R.
Step 7: Join AR and CR.
Here, BARC is the required quadrilateral.
Page No 50:
Question 6:
Construct â PQRS, such that l(PQ) = 3.5 cm, l(QR) = 5.6 cm, l(RS) = 3.5 cm, m∠Q = 110, m∠R = 70.
If it is given that â PQRS is a parallelogram, which of the given information is unnecessary?
Answer:
Steps of Construction:
Step 1: Draw PQ = 3.5 cm.
Step 2: Draw ∠PQX = 110.
Step 3: With Q as centre and radius 5.6 cm, draw an arc cutting ray QX at R.
Step 4: Draw ∠QRY = 70.
Step 5: With R as centre and radius 3.5 cm, draw an arc cutting ray RY at S.
Step 6: Join PS.
Here, PQRS is the required quadrilateral.
If it is given that quadrilateral PQRS is a parallelogram, then the information l(RS) = 3.5 cm and m∠R = 70 is unnecessary.
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