Question

find the area of trapezium in which parallel sides are 25 cm and 10 non parallel sides are 14 cm and 30 cm

Answer 1

${\huge{\red{\mathfrak{\underline{Question:}}}}}$
​


Find the area of trapezium in which parallel sides are 25cm and 10cm and non-parallel sides are 14cm and 30cm.

${\huge{\red{\mathfrak{\underline{Answer:}}}}}$
​

Given : AB \\ DC, AB = 10cm,
DC = 25cm, AD = 14cm, BC=30cm.


To find : Area of trapezium ABCD.


Construction : Draw a line AE \\ BC.


Solution :


Area of trapezium = ar(ABCE) + ar(ADC).

We construct AE \\ BC and it is given AB \\ DC. Thus, quadrilateral ABCE is a parallelogram. As opposite sides are parallel.



Area of parallelogram = base × height

=> 10× 30 = 300cm²



Now, for finding the area of Δ ADC, finding height of triangle that is equal to height of parallelogram.


Join E to B.


Area of Δ BEC = half of area of parallelogram AECB. As diagonal of parallelogram divides it into two triangles of equal area.


ar(ΔBEC) =

$\frac{1}{2} \times 300 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - (1)$


and


ar(ΔBEC) =

$\frac{1}{2} \times base \times height$

$\frac{1}{2} \times 10 \times height \:\:\:\:\:\:\:\:\:\:-(2)$


by putting (1) and (2) equal we get height = 10 cm


Now area of Δ ADE =

$\frac{1}{2} \times height \times DE$

$\frac{1}{2} \times 30 \times 15$

$= 125 {cm}^{2}$



Area of trapezium = (300 + 125)cm² = 425cm²