## 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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Deva dharshan says

Very usefull for doubt clarification