Question

13. The parallel sides of a trapezium are 20 cm
and 30 cm and its non-parallel sides are
6 cm and 8 cm. Find the area of the
trapezium​

Answer 1

Consider a trapezium ABCD with AB=20 m, BC = 6 m, CD = 30 m and DA = 8 m.

Drop a perpendicular from A on CD to meet that line at E. We consider the figure ABCD as a combination of a rectangle ABCE and a triangle AED. <B = <C = 90.

Area of ABCE = 20*6 = 120 sq m.

Area of AED can EB found by Heron’s formula where 2s =6+10+8 = 24, or s = 12

Area AED = [12(12–6)(12–10)(12–8)]^0.5

= [12*6*2*4]^0.5

= 24 sq m.

Total area of ABCE + AED = 120+24 = 144 sq m.

Answer 2

Question:-

The parallel sides of a trapezium are 20 cm and 30 cm and its non-parallel sides are 6 cm and 8 cm. Find the area of the trapezium.

$\huge{\dag\:{\underline{\boxed{\mathfrak{\purple{Answer}}}}}}$

• Draw BE ll AD and then AB ll DE
• Hence ABCD is a parallelogram.

___________________

➻ AB = DE = 20m

➻ AD = BE = 8cm

➻ EC = DC - DE = 30 - 20

➻ EC = 10m

-------------------–----------

➯Now ar∆ BCE

$➦ \sqrt{s(s - a)(s - b)(s - c)}$

$➦ \large \frac{8 + 6 + 10}{2} = 12cm$

_-------------------------------

➯ Now ar∆ BCE

$➦ \sqrt{s(s - a)(s - b)(s - c)}$

$➦ { \sqrt{12(12 - 8)(12 - 6)(12 - 10)} }$

$➦ \sqrt{12 \times 4 \times 6 \times 2} = 12 \times 2 = 24cm2$

•••••••••••••••••••••••••••••••••••

$▸▸\frac{1}{2} BC \times BF = 24m²$

$▸▸BF = \frac{24 \times 2}{10} = 4.8m$

======================

$▷ar ABCD = \frac{1}{2} (AB + CD) \times BF$

$▷ \frac{1}{2} \times (20 + 30) \times 4.8$

$▷50 \times 2.4$

________________________

• ABCD + AED = 144cm²

Area of trapezium is 144m²