Question

calculate the area of a quadrilateral abcd in the given figure​

Answer 1

Answer:

For Quadrilateral ABCD:

In triangle DBC,

{bc}^{2} + {bd}^{2} = {dc}^{2}

{bd}^{2} = {17}^{2} - {8}^{2} \\ bd = 15cm

In triangle ABD,

{ad}^{2} + {ab}^{2} = {bd}^{2}

{ab}^{2} = {15}^{2} - {9}^{2} \\ ab = 12cm

Now just apply 1/2base×height

Area of triangle ABD,

1/2 ×9×12=54cm^{2}

Area of second triangle,

1/2×15×8=60cm^{2}

Add both areas,

Area of the Quad.=114cm^2

Now for trapezium,

Apply pythagoras theorem in Triangle RTQ to find RT=15cm

Then Apply area of trapezium=1/2(sum of parallel sides)×height