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

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.

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