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10. A triangle has sides

    \[\mathbf{5}\text{ }\mathbf{cm},\text{ }\mathbf{12}\text{ }\mathbf{cm}\text{ }\mathbf{and}\text{ }\mathbf{13}\text{ }\mathbf{cm}.\]

Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is

    \[\mathbf{13}\text{ }\mathbf{cm}.\]

CBSE | Class 10 | Maths | RD Sharma | Triangles | Triangles
10. A triangle has sides

    \[\mathbf{5}\text{ }\mathbf{cm},\text{ }\mathbf{12}\text{ }\mathbf{cm}\text{ }\mathbf{and}\text{ }\mathbf{13}\text{ }\mathbf{cm}.\]

Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is

    \[\mathbf{13}\text{ }\mathbf{cm}.\]

Solution:

From the fig.

    \[AB\text{ }=\text{ }5cm,\text{ }BC\text{ }=\text{ }12\text{ }cm\text{ }and\text{ }AC\text{ }=\text{ }13\text{ }cm.\]

Then,

    \[A{{C}^{2}}~=\text{ }A{{B}^{2}}~+\text{ }B{{C}^{2}}.\]

    \[\Rightarrow {{\left( 13 \right)}^{2}}~=\text{ }{{\left( 5 \right)}^{2}}~+\text{ }{{\left( 12 \right)}^{2}}~=\text{ }25\text{ }+\text{ }144\text{ }=\text{ }169\text{ }=\text{ }{{13}^{2}}\]

                                                             

    \[\]

This proves that ∆ABC is a right triangle, right angled at B.

Let BD be the length of perpendicular from B on AC.

So, area of ∆ABC = (BC x BA)/

    \[~2\]

[Taking BC as the altitude]

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> =\text{ }\left( 12\text{ }x\text{ }5 \right)/\text{ }2 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> =\text{ }30\text{ }c{{m}^{2}} \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

                                                       

    \[\]

Also, area of ∆ABC = (AC x BD)/

    \[~2\]

  [Taking BD as the altitude]

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> =\text{ }\left( 13\text{ }x\text{ }BD \right)/\text{ }2 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \Rightarrow \left( 13\text{ }x\text{ }BD \right)/\text{ }2\text{ }=\text{ }30 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

                                                       

    \[\]

BD

    \[=\text{ }60/13\text{ }=\text{ }4.6\]

(to one decimal place)

i

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