Question

find the length of a diagonal of a rectangle whose length and breadth are 16 cm and 12 CM respectively. Explain.​

Answer 1

$\large \sf \purple{hi \: \: friend : )}$

$\large \sf \star \underline{ \: \red{ solution : }}$

Given :

• Length = 16cm
• Breadth = 12cm

To Find :

• Length of diagonal

Construction :

• Rectangle ABCD
• Refer to attached picture

$\large \star \sf \: in \: \triangle \: bcd :$

We know,

• angles of rectangle = 90°

Therefore, Triangle BCD is a

• Right angled triangle

$\large \star \sf \: \underline{ pythagores \: \: theorem : }$

$\large \looparrowright \pink{ \sf \: {h}^{2} = {b}^{2} + {p}^{2} }$

• h = Hypotenuse (diagonal)
• b = base
• p = perpendicular

$\\ \large \implies \tt \: {h}^{2} = {16}^{2} + {12}^{2} \\ \\ \large \implies \tt \: {h}^{2} = 256 + 144 \\ \\ \large \implies \tt \: {h}^{2} = 400\\ \\ \large \implies \tt \: {h}^{} = \sqrt{ 400} \\ \\ \large \therefore \green{ \sf \: h = 20}$

Diagonal = 20cm

Answer 2

Answer:

Given :

• ➝ Lenght of rectangle = 16 cm
• ➝ Breadth of rectangle = 12 cm

$\begin{gathered}\end{gathered}$

To Find :

• ➝ Lenght of diagonal

$\begin{gathered}\end{gathered}$

Using Formula :

$\longrightarrow{\underline{\boxed{\sf{d = \sqrt{{(l)}^{2} + {(w)}^{2}}}}}}$

• ➝ d = diognal
• ➝ l = length
• ➝ w = width

$\begin{gathered}\end{gathered}$

Solution :

Finding the diognal of rectangle by substituting the values in the formula :

$\begin{gathered} \qquad\implies{\sf{d = \sqrt{{(l)}^{2} + {(w)}^{2}}}} \\ \\ \qquad\implies{\sf{d = \sqrt{{(16)}^{2} + {(12)}^{2}}}} \\ \\ \qquad\implies{\sf{d = \sqrt{{(16 \times 16)} + {(12 \times 12)}}}} \\ \\ \quad\implies{\sf{d = \sqrt{{(256)} + {(144)}}}} \\ \\ \quad\implies{\sf{d = \sqrt{{256} + {144}}}} \\ \\ \quad\implies{\sf{d = \sqrt{{256} + {144}}}} \\ \\ \quad\implies{\sf{d = \sqrt{400}}} \\ \\ \quad\implies{\sf{d = 20 \: cm}} \\ \\ \quad\bigstar \: {\underline{\boxed{\sf{\pink{Diagonal = 20 \: cm}}}}} \end{gathered}$

• Hence, the diagonal of rectangle is 20 cm.

$\begin{gathered}\end{gathered}$

Learn More :

$\begin{gathered}\begin{gathered} \boxed{\begin{array}{l}\\ \large\dag\quad\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star \: \: \sf Circle = \pi r^2 \\ \\ \star \: \; \sf Square=(side)^2\\ \\ \star\; \; \sf Rectangle=Length\times Breadth \\\\ \star \: \: \sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \: \: \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \: \: \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star \: \: \sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star \: \: \sf Parallelogram =Breadth\times Height\\\\ \star \: \: \sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star \: \: \sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

$\rule{220pt}{3.5pt}$