Question

In a quadrilateral ABCD $\angle B={90}^{\circ }andA{D}^2=A{B}^2+B{C}^2+C{D}^2.then\angle ACDis$

Answer 1

total surface area of cuboid -$2(lb+bh+bl)$

$2(7\times 4+3\times 4+21)=2(28+12+21)$

$=2(40+21)$

$=2\left(61\right)=122$

Archit = $\frac{120}{100}$ Deepak

$\frac{\text{-Deepak + Archit}}{Archit}\times 100=\frac{\text{1.2Deepak-Deepak}}{\text{1.2Deepak}}$

$=\frac{0.2}{1.2}=\times 100=16.3\%$

Refer above image

$A{C}^2=A{B}^2+B{C}^2$

$A{Q}^2=A{B}^2+B{C}^2+C{D}^2$

$A{D}^2=A{C}^2+C{D}^2$

$ACD={90}^0$