Question

Find the percentage increase
and in the area of a triangle
at its each side is doubled​

Answer 1

The area before :

area1=
$\frac{1}{2} \times h \times b$
Where h = height and b = base.

The area after increasing the sides,

area2=
$\frac{1}{2} (2h)(2b) \\ = 4( \frac{1}{2} \times h \times b)$

So the increase in area in % will be,

$\frac{4( area_{1} )- (area_{1}) }{ area_{2} } \times 100 \\ \frac{3area_{1}}{area_{2}} \times 100 \\ \frac{3}{4} \times 100 \\ = 75\%$

Answer 2

The area before :

area1=

\frac{1}{2} \times h \times b

2

1

×h×b

Where h = height and b = base.

The area after increasing the sides,

area2=

\begin{lgathered}\frac{1}{2} (2h)(2b) \\ = 4( \frac{1}{2} \times h \times b)\end{lgathered}

2

1

(2h)(2b)

=4(

2

1

×h×b)

So the increase in area in % will be,

\begin{lgathered}\frac{4( area_{1} )- (area_{1}) }{ area_{2} } \times 100 \\ \frac{3area_{1}}{area_{2}} \times 100 \\ \frac{3}{4} \times 100 \\ = 75\%\end{lgathered}

area

2

4(area

1

)−(area

1

)

×100

area

2

3area

1

×100

4

3

×100

=75%