Question

can u explain this q ​

Answer 1

Answer:

(i) area of the rectangle = l*b

= 16*12

=192 sq. cm

length of diagonal of the rectangle=root(l^2+b^2)

= root(256+144)

=root(400)

=20 cm

(ii) area of the rectangle=17.28 sq.m

diagonal of the rectangle=6 m

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Answer 2

Answer:

SOLUTION :

Find the (a) area (b) length of diagonal of a rectangle whose :

(i) length = 16 cm and breadth = 12 cm

Finding the area of rectangle by substituting the values in the formula :-

${\longrightarrow{\sf{Area_{(Rectangle)} = Length \times Breadth}}}$

${\longrightarrow{\sf{Area_{(Rectangle)} = 16 \times 12}}}$

${\longrightarrow{\sf{\underline{\underline{\red{Area_{(Rectangle)} = 192 \: {cm}^{2}}}}}}}$

Hence, the area of rectangle is 192 cm².

Finding the diagonal of rectangle by substituting the values in the formula :-

${\longrightarrow{\sf{Diagonal_{(Rectangle)} = \sqrt{{(Length)}^{2} + {(Breadth)}^{2}}}}}$

${\longrightarrow{\sf{Diagonal_{(Rectangle)} = \sqrt{{(16)}^{2} + {(12)}^{2}}}}}$

${\longrightarrow{\sf{Diagonal_{(Rectangle)} = \sqrt{{(16 \times 16)} + {(12 \times 12)}}}}}$

${\longrightarrow{\sf{Diagonal_{(Rectangle)} = \sqrt{(256) + (144)}}}}$

${\longrightarrow{\sf{Diagonal_{(Rectangle)} = \sqrt{256 + 144} }}}$

${\longrightarrow{\sf{Diagonal_{(Rectangle)} = \sqrt{400 \: {cm}^{2}}}}}$

${\longrightarrow{\sf{\underline{\underline{\red{Diagonal_{(Rectangle)} =20 \: cm}}}}}}$

Hence, the diagonal of rectangle is 20 cm.

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(ii) length = 4.8 m and breadth = 3.6 m

Finding the area of rectangle by substituting the values in the formula :-

${\longrightarrow{\sf{Area_{(Rectangle)} = Length \times Breadth}}}$

${\longrightarrow{\sf{Area_{(Rectangle)} =4.8 \times 3.6}}}$

${\longrightarrow{\sf{\underline{\underline{\red{Area_{(Rectangle)} =17.28 \: {m}^{2}}}}}}}$

Hence, the area of rectangle is 17.28 m².

Finding the diagonal of rectangle by substituting the values in the formula :-

${\longrightarrow{\sf{Diagonal_{(Rectangle)} = \sqrt{{(Length)}^{2} + {(Breadth)}^{2}}}}}$

${\longrightarrow{\sf{Diagonal_{(Rectangle)} = \sqrt{{(4.8)}^{2} + {(3.6)}^{2}}}}}$

${\longrightarrow{\sf{Diagonal_{(Rectangle)} = \sqrt{{(4.8 \times 4.8)}+ {(3.6 \times 3.6)}}}}}$

${\longrightarrow{\sf{Diagonal_{(Rectangle)} = \sqrt{{(23.04)}+ {(12.96)}}}}}$

${\longrightarrow{\sf{Diagonal_{(Rectangle)} = \sqrt{23.04 + 12.96}}}}$

${\longrightarrow{\sf{Diagonal_{(Rectangle)} = \sqrt{36 \: {m}^{2} }}}}$

${\longrightarrow{\sf{\underline{\underline{\red{Diagonal_{(Rectangle)} = 6 \: m}}}}}}$

Hence, the diognal of rectangle is 6 m.

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Learn More :

$\begin{gathered}\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}\end{gathered}$

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