Question

can someone please help​

Answer 1

Step-by-step explanation:

$bd = \sqrt{ {17}^{2} - {8}^{2} } = \sqrt{289 - 64} \\ \\ = \sqrt{225} \\ = 15 \: cm \\ ab = \sqrt{225 - 81} = \sqrt{144} = 12 \: cm \\ \\ area \: of \: bdc = \frac{1}{2} \times 8 \times 15 \\ \\ = 60 \: {cm}^{2} . \\ \\ area \: of \: abd = \frac{1}{2} \times 9 \times 12 \\ \\ = 54 \: {cm}^{2} . \\ \\ area \: of \: quad = 60 + 54 = 114 \: {cm}^{2} .$

Answer 2

\begin{gathered}bd = \sqrt{ {17}^{2} - {8}^{2} } = \sqrt{289 - 64} \\ \\ = \sqrt{225} \\ = 15 \: cm \\ ab = \sqrt{225 - 81} = \sqrt{144} = 12 \: cm \\ \\ area \: of \: bdc = \frac{1}{2} \times 8 \times 15 \\ \\ = 60 \: {cm}^{2} . \\ \\ area \: of \: abd = \frac{1}{2} \times 9 \times 12 \\ \\ = 54 \: {cm}^{2} . \\ \\ area \: of \: quad = 60 + 54 = 114 \: {cm}^{2} .\end{gathered}

bd=

17

2

−8

2

=

289−64

=

225

=15cm

ab=

225−81

=

144

=12cm

areaofbdc=

2

1

×8×15

=60cm

2

.

areaofabd=

2

1

×9×12

=54cm

2

.

areaofquad=60+54=114cm

2

.