Question-219540

Question Number 219540 by BaliramKumar last updated on 28/Apr/25

Answered by som(math1967) last updated on 28/Apr/25

CD^2 +BC^2 =2×{(((25)/2))^2 +(((39)/2))^2 } CD=(√(289))=17 Ar △BCD=(√(42(42−28)(42−17)(42−39))) =14×5×3=210squ ar of ABCD=2×210=420Squ

$${CD}^{\mathrm{2}} +{BC}^{\mathrm{2}} =\mathrm{2}×\left\{\left(\frac{\mathrm{25}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{39}}{\mathrm{2}}\right)^{\mathrm{2}} \right\} \\ $$$$\:{CD}=\sqrt{\mathrm{289}}=\mathrm{17} \\ $$$${Ar}\:\bigtriangleup{BCD}=\sqrt{\mathrm{42}\left(\mathrm{42}−\mathrm{28}\right)\left(\mathrm{42}−\mathrm{17}\right)\left(\mathrm{42}−\mathrm{39}\right)} \\ $$$$=\mathrm{14}×\mathrm{5}×\mathrm{3}=\mathrm{210}{squ} \\ $$$${ar}\:{of}\:{ABCD}=\mathrm{2}×\mathrm{210}=\mathrm{420}{Squ} \\ $$

Answered by mr W last updated on 28/Apr/25

cos θ=((25^2 +39^2 −(2×28)^2 )/(2×25×39))=−((33)/(65)) sin θ=(√(1−(−((33)/(65)))^2 ))=((56)/(65)) A_(parallelogram) =(1/2)×25×39×((56)/(65))=420 ✓

$$\mathrm{cos}\:\theta=\frac{\mathrm{25}^{\mathrm{2}} +\mathrm{39}^{\mathrm{2}} −\left(\mathrm{2}×\mathrm{28}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{25}×\mathrm{39}}=−\frac{\mathrm{33}}{\mathrm{65}} \\ $$$$\mathrm{sin}\:\theta=\sqrt{\mathrm{1}−\left(−\frac{\mathrm{33}}{\mathrm{65}}\right)^{\mathrm{2}} }=\frac{\mathrm{56}}{\mathrm{65}} \\ $$$${A}_{{parallelogram}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{25}×\mathrm{39}×\frac{\mathrm{56}}{\mathrm{65}}=\mathrm{420}\:\checkmark \\ $$

Commented by Wuji last updated on 28/Apr/25

$${you}\:{make}\:{maths}\:{looks}\:{simple},{Sir}.\:{i}\:{appreciate}\:{that} \\ $$

Commented by mr W last updated on 28/Apr/25

$${thanks}! \\ $$

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