Question

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Answer 1

$\large \bf \pink {៚ \: Question \: ✞}$

Find the base of a triangle whose area is $\sf 14cm^2$ and altitude $\sf 4cm$.

$\large \bf \pink {៚ \: Solution \: ✞}$

Given :-

• Area of triangle = $\sf 14cm^2$
• Altitude / Height = $\sf 4cm$

We have to find the base of triangle.

Let the unknown base of the triangle be h.

$\underline {\boxed {\tt \orange {Area \: of \: a \: triangle \: = \: \frac{1}{2} \times Base \times Height}}}$

$\orange {: \: \leadsto} \: \tt14cm^2=h\times 4cm$

$\orange {:~\leadsto}~\tt \cfrac{14cm^2}{4cm}=h$

$\underline {\boxed {\tt \orange {\therefore \: Height \: of \: the \: triangle \: = \: \cfrac{7}{2}cm} }}$

$\large \bf \pink {៚ \: Verification \: ✞}$

Substituting and verifying :-

$\orange {:~\leadsto}\:\tt 14cm^2=\cfrac{7}{2}cm\times4cm$

$\orange {:~\leadsto}~\tt 14cm^2= \cfrac{7}{\cancel2}cm \times {\cancel4} cm$

$\orange {:~\leadsto}\:\tt {14cm^2=14cm^2}$

$\large \underline {\boxed {\tt \orange {Hence, \: verified \: !!}}}$

$\large \bf \pink {៚ \: Know \: more\: ✞}$

$\orange {:~\leadsto}~\tt Area \: of \: square=Side\times Side$

$\orange {:~\leadsto}~\tt Area \: of \: rectangle=Length \times Breadth$

$\orange {:~\leadsto}~\tt Area \: of \: parallelogram=Base \times Height$

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Answer 2

DIAGRAM :

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S O L U T I O N :

$\: \:$

• Here, it is given that height of the triangle is 4cm and area of the triangle is 14cm². We've to find the base of the triangle. Let h be the height of the triangle and b be the base of the triangle.

$\: \:$

For finding the area of the triangle, formula is given as :

$\: \:$

$\sf:\implies \: \dfrac{1}{2} \times b \times h$

$\: \:$

Where, b denotes base of the triangle and h denotes height of the triangle.

$\: \:$

$\mathfrak {\underline{substituting \: the \: values \: :}}$

$\: \:$

$\sf : \implies \: Area_{(triangle)} = \dfrac{1}{2} \times b \times h \\ \\ \\ \sf :\implies \: 14 = \dfrac{1}{2} \times b \times 4 \\ \\ \\ \sf :\implies \:14 = \frac{4b}{2} \\ \\ \\ \sf :\implies \:b = \frac{14 \times 2}{4} \\ \\ \\ \sf :\implies \:b = \cancel\frac{28}{4} \\ \\ \\ \sf :\implies \:\underline{\boxed{\mathfrak\purple{b = 7cm}}} \: \bigstar$

$\:$

$\: \:$

$\sf\therefore{\underline{Hence, \: the \: base \: of \: the \: triangle \: is \: \bold{7 \: cm}.}}$