Find in square metres the area of a triangle whose
(a) base $ =8 \mathrm{~cm}, $ altitude $ =6 \mathrm{~cm} $
(b) base $ =38000 \mathrm{~mm} $, altitude $ =4500 \mathrm{~mm} $

Given:

The measures of a triangle are:

(a) base \( =8 \mathrm{~cm}, \) altitude \( =6 \mathrm{~cm} \)
(b) base \( =38000 \mathrm{~mm} \), altitude \( =4500 \mathrm{~mm} \)

To do:

We have to find the areas of the triangles in square metres.
Solution:

We know that,

Area of a triangle of base $b$ and altitude $h$ is $\frac{1}{2}bh$.
$1\ m=100\ cm$

$1\ cm=10\ mm$

Therefore,

(a) base \( =8 \mathrm{~cm} \)

$=\frac{8}{100}\ m$

altitude \( =6 \mathrm{~cm} \)

$=\frac{6}{100}\ m$

Area of the triangle $=\frac{1}{2}\times\frac{8}{100}\times\frac{6}{100}\ m^2$

$=\frac{24}{10000}\ m^2$

$=0.0024\ m^2$

(b) base \( =38000 \mathrm{~mm} \)

$=\frac{38000}{10}\ cm$

$=3800\ cm$

$=\frac{3800}{100}\ m$

$=38\ m$

altitude \( =4500 \mathrm{~mm} \)

$=\frac{4500}{10}\ cm$

$=450\ cm$

$=\frac{450}{100}\ m$

$=4.5\ m$

Area of the triangle $=\frac{1}{2}\times38\times4.5\ m^2$

$=19\times4.5\ m^2$

$=85.5\ m^2$

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Updated on: 10-Oct-2022

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