Question

find the length of a diagonal of rectangle of length 9cm and width 4cm​

Answer 1

Answer:

root 97 / 9.848

Step-by-step explanation:

Length=9cm

Width=4cm

To find a diagonal of rectangle we use pythagorous property.

Altitude^2+Base^2=Hypoteneuse^2

Base=4cm

Altitude=9cm

Hypoteneuse=x

9^2+4^2=x^2

81+16=x^2

97=x^2

=root 97

Answer 2

GivEn:

• Length of Rectangle = 9cm

• Breadth of Rectangle = 4cm

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To find:

• Diagonal of Rectangle.

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SoluTion:

$\bf Here = \begin{cases} & \text{Length (l) = \bf{9 cm} } \\ & \text{Breadth (b) = \bf{4 cm} } \end{cases}$

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As we know that,

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★ Formula to find Diagonal (d) of Rectangle is,

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$\star\;{\boxed{\sf{\purple{ d = \sqrt{b^2 + l^2}}}}}$

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$\;\;\;\small\sf \underline{Now,\;put\;the\;given\;values\;:}$

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$:\implies\sf d = \sqrt{(4)^2 + (9)^2}$

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$:\implies\sf d = \sqrt{16 + 81}$

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$:\implies\sf d = \sqrt{97}$

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$:\implies{\underline{\boxed{\sf{\pink{d = 9.84\;cm\;(aprrox.)}}}}}\;\bigstar$

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⋆ DIAGRAM:

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$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\tt\large{A}}\put(7.7,1){ \tt\large{B}}\put(9.5,0.7){\sf{\large{9 cm}}}\put(11.5,1){ \tt\large{C}}\put(8,1){\line(1,0){3.5}}\put(8,1){\line(0,2){2}}\put(11.5,1){\line(0,3){2}}\put(8,3){\line(3,0){3.5}}\put(11.6,2){\sf{\large{4 cm}}}\qbezier(8,1)(8,1)(11.5,3)\put(11.5,3){ \tt\large{D}}\put(11.3,1){\line(0,2){0.2}}\put(11.3,1.2){\line(2,0){0.2}}\end{picture}$

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$\therefore$ Here, BD is the diagonal of Rectangle which is 9.84 cm.

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★ Additional Information:

• Area of Rectangle = l × b

• Perimeter of Rectangle = 2(l + b)