Question

In quadrilateral ABCD, AB = 10 cm, BC = 6 cm,CD = 2.8 cm, DA = 6 cm and its diagonals are 8 cm each.
(i) Are angle ADB and ACB right angles?
(ii) Is ABCD a trapezium? If yes, find its height.
(iii) Find the area of the quadrilateral.

with an attachment.!
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Answer 1

Step-by-step explanation:

1) yes it's right angle triangle

by Pythagoras theorem

if D is the right angle then AB is hypotenuse

so AB^2 = AD^2+DB^2

10^2= 6^2+8^2

100= 36+64

100=100

hence D is the right angle

similarly

ACB as well

2)

if you look at the triangles ADC and BDC their areas are equal and their having same base

so they should be lie inbetween same parallel lines

so ABCD is a trapezium with AB||CD

3)so height of the trapezium = 6^2-3.6^2 = 4.8

so area = (10+2.8)/2 *4.8

= 30.72