Question

Find the area of quadrilateral ABCD in which right angle B=90°,AB=6cm,BC=8cm,and CD=AD=13cm

Answer 1

Answer:

I have divided the quadrilateral for the sake of simplicity of calculations.

In △ABC,

By Pythagoras’s Theorem ,

AC=10cm and area =12×8×6

⇒24cm2

In △ADC,

By Heron’s Theorem ,

S=13+13+102

⇒18cm

∴ Area =√18[18−13][18−13][18−10]

⇒ Area =60cm2

Since Area of quadrilateral ABCD= Area of triangle ABC+ Area of triangle ABC

∴Area of ABCD=60cm2+24cm2=84cm2

Answer 2

in ABD

BD2 = AB2 + AD2

=5 `2 + 6`2

= 64 + 36

= 100

BD 2 = 100/100

BD = 10 cm

in BCD

c o Is the altitude of an insceles triangle

Bo = od = 5 cm

in COB

CD 2 = ( b2+od 2

B2 = ( o2+5`2

co 2 = 169 - 25 = 144 = 122

co = 12 cm

or ( ABD ) = 1/2 AB × ABD

= 1/2 ×8×6 = 24 sq.cm

or ( ABCD = 1/2 BD × co

= 1/2 ×10×12 = 60 sq. cm

area of quadrilateral ABCD

or ( ABD ) + or ( BCD )

=24 sq. cm + 60 sq. cm

= 84 sq. cm