Question

plzzzzzzz.............

Answer 1

given :-

angle BAD = 90°

angle BDC = 90°

AB = 6cm

AD = 8cm

BC = 28cm

by Pythagoras theorem, we get BD = 10cm (u can see in the hint mentioned there.. BD² = 100 => BD = √100 => BD = 10cm)

now in ∆BDC,

BD is the base, DC is the height and BC is the hypotenuse.

again we'll use Pythagoras theorem.

=> h² = b² + p²

=> 26² = 10² + DC²

=> 676 = 100 + DC²

=> 676 - 100 = DC²

=> 576 = DC²

=> DC = √576

=> DC = 24cm

hence, the height of ∆BDC is 24cm.

therefore area of ∆BDC = 1/2 × b × h

= 1/2 × 10 × 24

= 5 × 24

= 120cm²

and area of ∆ABD = 1/2 × b × h

= 1/2 × 6 × 8

= 3 × 8

= 24cm²

hence, total area of the quadrilateral ABCD = ar(∆BDC) + ar(∆ABD)

= 120 + 24

= 144cm²

Answer 2

Here it is..
By pythagoras the third side is 100(given)
By heron's formula..
Ar. of tri.=1/2 ·b·h
=1/2 · 6· 8=24 units.
Ar.(BCD)=BY PYTHAGORAS TH.
=_/26^2+24^2=10
=1/2·24·120=120
Total ar.=120+24=144