Answer:

A=48cm²

l Length

8

cm

d Diagonal

10

cm

Using the formulas

A=wl

d=w2+l2

Solving forA

A=ld2﹣l2=8·102﹣82=48cm²

Verified answer

Answer:

[tex]\boxed{\sf \: \: Area_{(Rectangle)} = 48 \: {cm}^{2} \: \: } \\ \\ [/tex]

Step-by-step explanation:

Given that, Length of a rectangle is 8 cm and one of its diagonal is 10 cm,

We know, length, breadth and diagonal of a rectangle are connected by the relationship

[tex]\sf \: {(Diagonal)}^{2} = {(Length)}^{2} + {(Breadth)}^{2} \\ \\ [/tex]

[tex]\sf \: {10}^{2} = {8}^{2} + {(Breadth)}^{2} \\ \\ [/tex]

[tex]\sf \:100 = 64 + {(Breadth)}^{2} \\ \\ [/tex]

[tex]\sf \:100 - 64 = {(Breadth)}^{2} \\ \\ [/tex]

[tex]\sf \:36 = {(Breadth)}^{2} \\ \\ [/tex]

[tex]\sf \: {6}^{2} = {(Breadth)}^{2} \\ \\ [/tex]

[tex]\implies\sf \: Breadth = 6 \: cm \\ \\ [/tex]

Now,

[tex]\sf \: Area_{(Rectangle)} = Length \times Breadth \\ \\ [/tex]

[tex]\sf \: Area_{(Rectangle)} = 8 \times 6 \\ \\ [/tex]

[tex]\implies\sf \: \sf \: Area_{(Rectangle)} = 48 \: {cm}^{2} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

[tex]\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \large\underline{\sf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}[/tex]