Question

Q11. Find the area of a quadrilateral ABCD whose sides are
9
cm,28cm and 15cm
and the angle between the first two sides is 90 degree

Answer 1

Answer:

AB= 9

BC= 40

CD= 28

DA= 15

join A and C to form a diagonal AC in quadrilateral ABCD

using pythagorean theorem,

AB ^(2) + BC ^(2) = AC ^(2)

9*9 + 40*40 = AC ^(2)

81 + 1600 = AC ^(2)

1681 = AC^(2)

\sqrt{1681}

1681

= AC

41 = AC

now, using herons formula, u can find the area of triangle ABC an d triangle ADC.

in triangle ABC

\sqrt{s(s-a)(s-b)(s-c)}

s(s−a)(s−b)(s−c)

= \sqrt{45(45-9)(45-40)(45-41)}

45(45−9)(45−40)(45−41)

= \sqrt{45(36) (5)(4)}

45(36)(5)(4)

= \sqrt{5*3*3*2*3*2*3*5*2*2}

5∗3∗3∗2∗3∗2∗3∗5∗2∗2

= \sqrt{5*5*3*3*2*2*3*3*2*2}

5∗5∗3∗3∗2∗2∗3∗3∗2∗2

= 5*3*2*3*2

= 5*36

= 180

in triangle ADC

\sqrt{s(s-a)(s-b)(s-c)}

s(s−a)(s−b)(s−c)

= \sqrt{42(42-28)(42-15)(42-41)}

42(42−28)(42−15)(42−41)

= \sqrt{42*14*27*1}

42∗14∗27∗1

= \sqrt{2*3*7*2*7*3*3*3}

2∗3∗7∗2∗7∗3∗3∗3

= \sqrt{2*2*7*7*3*3*3*3}

2∗2∗7∗7∗3∗3∗3∗3

=2*7*3*3*

= 14*9

= 126

area of ABC + area of ADC = area of ABCD

180 + 126 = 306