Maths Class 8 RS Aggarwal Solutions

RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18A
• RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18B
• RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18C

Tick the correct answer in each of the following :

Question 1.
Solution:
Parallel sides 14 cm and 18 cm
Distance between parallel sides (h) = 9cm

Question 2.
Solution:
Length of parallel sides are 19 cm and 13 cm
Area of trapezium = 128 cm²
Distance between then

Question 3.
Solution:
Ratio in parallel sides = 3:4
Perpendicular distance (h) = 12 cm
Area of trapezium = 630 cm²

Question 4.
Solution:
Area of trapezium = 180 cm²
and height (h) = 9 cm

Question 5.
Solution:
In the figure, AB || DC, DA ⊥ AB
DC = 7 cm, BC = 10 cm, AB = 13 cm
CL ⊥ AB
AD = DC = 7 cm
and LB – 13 – 7 = 6 cm

Hope given RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18C are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18A
• RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18B
• RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18C

Question 1.
Solution:
In quad. ABCD
AC = 24 cm, BL ⊥ AC and DM ⊥ AC
BL = 8 cm and DM = 7 cm

Question 2.
Solution:
In quad. ABCD, diagonal BD = 36 m
AL ⊥ BD and CM ⊥ BD
AL = 19 m and CM = 11 m

Question 3.
Solution:
In the given pentagon ABCDE,
BL ⊥ AC, DM ⊥ AC, EN ⊥ AC
AC = 18 cm, AM = 14 cm, AN = 6 cm,
BL = 4 cm, DM = 12 cm and EN = 9 cm

Question 4.
Solution:
In hexagon ABCDEF, there are triangles and trapeziums
AP = 6 cm, PL = 2 cm, LN = 8 cm,
NM = 2 cm, MD = 3 cm, FP = 8 cm,
EN = 12 cm, BL = 8 cm and CM = 6 cm

Question 5.
Solution:
In the given pentagon ABCDE,
BL ⊥ AC, CM ⊥ AD, EN ⊥ AD
AC = 10 cm, D = 12 cm, BL = 3 cm,
CM = 7 cm and EN = 5 cm

Question 6.
Solution:
In the figure, ABCF is 0 square and CDEF is a trapezium
Now area of sq. ABCF
= (side)² = (20)² = 400 cm²
area of trap. CDEF
= \(\\ \frac { 1 }{ 2 } \) (ED + FC ) x height

Question 7.
Solution:
In the right ∆ABC
AB² = BC² + AC²
=> (5)² = (4)² + AC²
25 = 16 + AC²
AC² = 25 – 16 = 9 = (3)²
AC = 3 cm

= 32 + 36
= 68 cm²

Question 8.
Solution:
AD = 23 cm, LM = 13 cm
AL = MD = \(\\ \frac { 23-13 }{ 2 } \) = \(\\ \frac { 10 }{ 2 } \) = 5 cm

Hope given RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18A
• RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18B
• RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18C

Question 1.
Solution:
In trapezium ABCD,
Length of parallel sides
AB = 24 cm, DC = 20 cm
and distance between them = 15 cm

Question 2.
Solution:
Parallel sides of a trapezium ABCD are
l1 = 38.7 cm. and l2 = 22.3 cm

Question 3.
Solution:
Parallel sides of the trapezium = 1 m, 1.4 m

Question 4.
Solution:
Area of trapezium = 1080 cm²
Lengths of parallel sides are
l1 = 55 cm and l2 = 35 cm
Let h be the distance between them

Question 5.
Solution:
Area of trapezium shaped field = 1586m²
Distance between parallel sides = 26 m
Sum of the parallel sides = \(\frac { Area\times 2 }{ Altitude }\)
= \(\frac { 1586\times 2 }{ 26 } \) = 122 m
One side = 84 m
Second side = 122 – 84
= 38 m

Question 6.
Solution:
Area of trapezium = 405 cm²
Ratio in parallel sides = 4:5
and distance between them = 18 cm

Question 7.
Solution:
Area of trapezium = 180 cm²
Height (h) = 9 cm.
Let l1 and l2 be the parallel sides,

Question 8.
Solution:
Let one of parallel sides = x
Then second sides = 2x
Area = 9450 m²
Distance between them = 84 m

Question 9.
Solution:
Perimeter of trapezium ABCD = 130 m
AB ⊥ AD and BC
BC = 54 m, CD = 19 m, AD = 42 m

Question 10.
Solution:
In the given trapezium ABCD, AC is

Question 11.
Solution:
In trapezium ABCD,

Question 12.
Solution:
In trapezium ABCD,

AB || DC
AB = 25 cm DC = 1 cm
AD = 13 cm and BC = 15 cm

 

Hope given RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17A
• RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17B

Question 1.
Solution:
Steps of Construction :

(i) Draw a line segment AB = 5.2 cm.
(ii) With centre A and radius 7.6 cm. and with centre B and radius 4.7 cm. draw arcs which intersect each other at C.
(iii) Join AC and BC.
(iv) Again with centre A and radius 4.7 cm and with centre C and radius 5.2 cm draw arcs which intersect each other at D.
(v) Join AD and CD.
ABCD is the required parallelogram.

Question 2.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 4.3 cm.

(ii) With centre A and radius 4 cm. and with centre B and radius 6.8 cm., draw arcs which intersect each other at D.
(iii) Join AD and BD.
(iv) Again with centre B and radius 4 cm. and with centre D and radius 4.3 cm., draw arcs intersecting each other at C.
(v) Join DC and BC. ABCD is the required parallelogram.

Question 3.
Solution:
Steps of Construction :
(i) Draw a line segment PQ = 4 cm.
(ii) At Q, draw a ray making an angle of 60° and cut off QR = 6 cm.

(iii) With centre P and radius 6 cm. and with centre R and radius 4 cm draw arcs intersecting each other at S.
(iv) Join RS and PS.
PQRS is the required parallelogram. Q.

Question 4.
Solution:
Steps of Construction :
(i) Draw a line segment BC = 5 cm.

(ii) At C, draw a ray making an angle of 120° and cut off CD = 4.8 cm.
(iii) With centre B and radius 4.8 cm. with centre D and radius 5 cm, draw arcs intersecting each other at A.
(iv) Join AB and AD
ABCD is the required parallelogram.

Question 5.
Solution:
Steps of Construction :
We know that diagonals of a parallelogram bisect each other.
(i) Draw a line segment AB = 4.4 cm.
(ii) With centre A and radius \(\\ \frac { 5.6 }{ 2 } \) cm and with centre B and radius \(\\ \frac { 7 }{ 2 } \) = 3.5 cm. draw arcs
intersecting each other at O.
(iii) Join AO and BO and produce them to C and D respectively such that OC = 2.8 cm and OD = 3.5 cm.

(iv) Join AD, CD and BC
ABCD is the required parallelogram

Question 6.
Solution:
Steps of Construction :

(i) Draw a line segment AB = 6.5 cm.
(ii) At A, draw a perpendicular AX and cut off AL = 2.5 cm.
(iii) Through L, draw a line PQ parallel to AB.
(iv) From A, draw an arc of radius 3 -4 cm which intersects the line PQ at C.
(v) Join AC. BC
(vi) From PQ, cut off CD = AB.
(vii) Join AD
(viii) From C, draw a perpendicular CM to AB.
ABCD is the required parallelogram.

Question 7.
Solution:
Steps of Construction :

(i) Draw a line segment AC = 3.8 cm.
(ii) Bisect it at O.
(iii) At O, draw a ray making an angle of 60° and produce it both sides.
(iv) From O cut off OB = OD = 2.3 cm.
(v) Join AB, BC, CD and AD.
ABCD is the required parallelogram.

Question 8.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 11 cm.

(ii) At B, draw a perpendicular and cut off BC = 8.5 cm.
(iii) With centre A and radius 8.5 cm and with centre C and radius 11 cm, draw arcs intersecting each other at D.
(iv) Join AD and CD.
ABCD is the required rectangle.

Question 9.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 6.4 cm.

(ii) At A and B draw perpendiculars and
cut off AD = BC = AB = 6.4 Cm.
(iii) Join CD.
ABCD is the required square.

Question 10.
Solution:
Steps of Construction :
(i) Draw a line segment AC = 5.8 cm.

(ii) Draw its perpendicular bisector intersecting AC at O.
(iii) From O, cut off OD = OB = 2.9 cm.
\(\qquad =\left( \frac { 1 }{ 2 } BD \right) \)
(iv) Join AB, BC, CD and DA. ABCD is the required square.

Question 11.
Solution:
Steps of Construction :

(i) Draw a line segment QR = 3.6 cm.
(ii) At Q, draw a ray QX making an angle of 90°.
(iii) With centre R and radius 6 cm. draw an arc which intersects QX at P.
(iv) Join PR.
(v) With centre P and radius equal to QR and with centre R and radius equal to QP, draw arcs intersecting each other at S.
(vi) Join PS and RS.
PQRS is the required rectangle.
The length of other side PQ = 4.8 cm.

Question 12.
Solution:
Steps of Construction :

(i)Draw a line segment AC = 8 cm.
(ii)Draw its perpendicular bisector intersecting it at O.
(iii)From O, cut off OB = OD = 3 cm.
(iv)Join AB, BC, CD and DA.
ABCD is the required rhombus.

Question 13.
Solution:
Steps of Construction :
(i)Draw a line segment AC = 6.5 cm.

(ii) With centres A and C and radius equal to 4 cm., draw arcs which intersect each other on both sides of line segment AC at B and D respectively.
(iii) Join AB, BC, CD and DA.
ABCD is the required rhombus.

Question 14.
Solution:
Steps of Construction :

(i)Draw a line segmentAB = 7.2 cm.
(ii)At A draw a ray AX making an angle of 60° and cut off AD = 7.2 cm.
(iii)With centres D and B, and radius 7.2 cm., draw arcs intersecting each other at C.
(iv)Join CD and CB.
ABCD is the required rhombus.

Question 15.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 6 cm

(ii)At B, draw a ray BX making an angle of 75° and cut off BC = 4 cm.
(iii) At C, draw a ray CY making an angle of 180° – 75° = 105°
So that CY may be parallel to AB.
(iv) From CY, Cut off CD = 3.2 cm.
(v) Join DA.
ABCD is the required trapezium.

Question 16.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 7 cm.

(ii) At B, draw a ray BX making an angle of 60° and cut off BC = 5 cm.
(iii) At C, draw a ray CY making an angle of (180° – 60°) = 120° so that CY || AB.
(iv) With centre A and radius 6.5 cm. draw an arc intersecting CY at D.
(v) Join AD,
ABCD is the required trapezium.

Hope given RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17A
• RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17B

Question 1.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 4.2 cm.

(ii) With centre A and radius 8 cm and with centre B and radius 6 cm., draw arcs intersecting each other at C.
(iii) Join AC and BC.
(iv) Again with centre A and radius 5 cm. and with centre C, radius 5 2 cm. draw arcs intersecting each other at D.
(v) Join AD and CD. ABCD is the required quadrilateral.

Question 2.
Solution:
Steps of Construction :
(i) Draw a line segment PQ = 5.4 cm.

(ii) With Centre P and radius 4 cm. and with centre Q and radius 4.6 cm., draw arcs intersecting each other at R.
(iii) Join PR and QR.
(iv) Again with centre P and radius 3.5 cm. and with centre R and radius 4.3 cm. draw arcs intersecting each other at S.
(v) Join PS and RS. PQRS is the required quadrilateral.

Question 3.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 3.5 cm.

(ii) With centre A and radius 4.5 cm. and with centre B and radius 5.6 cm. draw arcs intersecting each other at D.
(iii) Join AD and BD.
(iv) With centre B and radius 3.8 cm. and with centre D and radius 4.5 cm., draw arcs intersecting each other at C.
(v) Join BC and DC. ABCD is the required quadrilateral.

Question 4.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 3.6 cm.
(ii) With centre A and radius 4.6 cm. and with centre B and radius 3.3 cm. draw arcs intersecting each other at C.

(iii) Join AC and BC.
(iv) Again with centre A and radius 2.7 cm. and centre B and radius 4 cm., draw arcs intersecting each other at D.
(v) Join BD, AD and CD. ABCD is the required quadrilateral.

Question 5.
Solution:
Steps of Construction :
(i) Draw a line segment RS = 5 cm.
(ii) With centre R and S, radius 6 cm. each, draw arcs intersecting each other at R
(iii) Join PR and PS.

(iv) With centre R and radius 7.5 cm. and with centre S and radius 10 cm, draw arcs intersecting each other at Q.
(v) Join RQ, SQ and PQ. PQRS is the required quadrilateral. Measuring the fourth sides PQ, it is 4.7 cm. (approx.)

Question 6.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 3.4 cm.

(ii) With centre A and radius 5.7 cm. and with centre B and radius 4 cm., draw arcs intersecting each other at D.
(iii) Join BD and AD.
(iv) Again with centre A and radius 8 cm and with centre D and radius 3 cm., draw arcs intersecting each other at C.
(v) Join AC, BC and DC. ABCD is the required quadrilateral

Question 7.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 3.5 cm.

(ii) At B, draw a ray BX making an angle of 120° using protractor and cut off BC = 3.5 cm
(iii) With centres A and C and radius 5.2 cm, draw arcs intersecting each other at D.
(iv) Join CD and AD. ABCD is the required quadrilateral.

Question 8.
Solution:
Steps of Construction :
(i) Draw a line AB = 2.9 cm.
(ii) At A, draw a ray AX making an angle of 70° with AB. Using protractor and cut off AD = 3.4 cm.

(iii) With centre B and radius 3.2 cm and with centre D and radius 2.7 cm., draw arcs intersecting each other at C.
(iv) Join BC and DC. ABCD is the required quadrilateral.

Question 9.
Solution:
Steps of Construction

(i) Draw a line segment BC = 5 cm.
(ii) At B, draw a ray BX making an angle of 125° and cut off BA = 3.5 cm.
(iii) At C, draw a ray CY making an angle of 60° and cut off CD = 4.6 cm
(iv) Join AD. ABCD is the required quadrilateral.

Question 10.
Solution:
Steps of Construction :
(i) Draw a line segment QR = 5.6 cm.

(ii) At Q, draw a ray QX making an angle of 45° and cut off QP = 6 cm.
(iii) At R, draw a ray RY making an angle of 90° and cut off RS = 2.7 cm.
(iv) Join SP PQRS is the required quadrilateral.

Question 11.
Solution:
Steps of Construction :
∠A = 50°, ∠B = 105° and ∠D = 80°
and ∠A + ∠B + ∠C + ∠D = 360°
=> 50° + 105° + ∠C + 80° = 360°
=> ∠C + 235° = 360°
=> ∠C = 360° – 235°
=> ∠C = 125°

(i) Draw a line segment AB = 5.6 cm.
(ii) At B, draw a ray BY making an angle of 105° and cut off BC = 4 cm.
(iii) At C, draw a ray CZ making an. angle of 125° and at A, a ray AX making an angle of 50° intersecting each other at D.
then ∠D = 80°
ABCD is the required quadrilateral.

Question 12.
Solution:
∠P + ∠Q + ∠R + ∠S = 360°
100° + ∠Q + 100° + 75° = 360°
=> ∠Q + 275° = 360°
=> ∠Q = 360° – 275°
∠Q = 85°
Steps of Construction :

(i) Draw a line segment PQ = 5 cm.
(it) At Q, draw a ray QX making an angle of 85° and cut off QR = 6.5 cm.
(iii) At R, draw a ray making an angle of 100° and at P, another ray making an angle of 100° which intersect each other at S. then ∠S = 75°
PQRS is the required quadrilateral.

Question 13.
Solution:
Steps of Construction :

(i) Draw a line segment AB = 4 cm.
(ii) At B, draw a ray BX making an angle of 90°.
(iii) From A, draw an arc of 5 cm. radius intersecting BX at C.
(iv) Join AC.
(v) At C, draw a ray CY making an angle of 90°.
(vi) From A, draw an arc of radius 5.5 cm. which intersects CY at D.
(vii) Join AD.
ABCD is the required quadrilateral.

 

Hope given RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.