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Question Number 183886 by Acem last updated on 31/Dec/22

Answered by HeferH last updated on 31/Dec/22

$$A=\left(\frac{BC+AD}{2}\right)15$$$$AE=\sqrt{{17}^2-{15}^2}=\sqrt{2\cdot 32}=8$$$$B^{\prime }D=\sqrt{{39}^2-{15}^2}=36$$$$AD=AE+ED=8+36-BC\Rightarrow$$$$A=\left(\frac{8+36}{2}\right)15=66\cdot 5=330u$$

Commented by Acem last updated on 31/Dec/22

$$VerywellSir!$$

Answered by mr W last updated on 31/Dec/22

Commented by mr W last updated on 31/Dec/22

$$\left[ABCD\right]=\left[\mathrm{\Delta }ABF\right]=\frac{AF\times BG}{2}$$$$=\frac{(\sqrt{{17}^2-{15}^2}+\sqrt{{39}^2-{15}^2})\times 15}{2}$$$$=\frac{44\times 15}{2}=330✓$$

Commented by manxsol last updated on 31/Dec/22

$$theformulawouldthenbe$$$$Areatrap(d1,d2,h)=$$$$[\sqrt{d{1}^2-h^2}+\sqrt{d{2}^2-h^2]}\times \frac{h}{2}$$

Commented by manxsol last updated on 31/Dec/22

Commented by Acem last updated on 31/Dec/22

$$VerywellSir!$$

Answered by mr W last updated on 31/Dec/22

$$alternative:$$$$\cos \alpha =\frac{15}{17}\Rightarrow \sin \alpha =\frac{\sqrt{{17}^2-{15}^2}}{17}=\frac{8}{17}$$$$\cos \beta =\frac{15}{39}\Rightarrow \sin \alpha =\frac{\sqrt{{39}^2-{15}^2}}{39}=\frac{36}{39}$$$$\left[ABCD\right]=\frac{1}{2}\times 17\times 39\times \sin (\alpha +\beta )$$$$=\frac{1}{2}\times 17\times 39\times (\frac{8}{17}\times \frac{15}{39}+\frac{15}{17}\times \frac{36}{39})$$$$=\frac{15\times 44}{2}=330✓$$

Commented by Acem last updated on 31/Dec/22

$$VerywellSir!$$$$andpleaseexplanationabout\sin (\alpha +\beta )thanks$$

Commented by mr W last updated on 31/Dec/22

Commented by mr W last updated on 31/Dec/22

$$A=\frac{1}{2}{d}_1\times d_2\times \sin \theta$$$$\theta =\alpha +\beta$$

Commented by Acem last updated on 31/Dec/22

$$Excellent!Thankyoudearfriend$$

Answered by Acem last updated on 31/Dec/22

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