Question

$\mathfrak \red{Question \: in \: the \: attachment}$​

Answer 1

Answer:

Given :

The base and height of a triangle are in the ratio 4:3.

If the base is increased by 4 cm and the height is decreased by 2 cm, the area of the triangle remains the same.

To Find :

Base and Height of the Triangle .

Solution :

$\longmapsto\tt{Let\:base\:be(b)=4x}$

$\longmapsto\tt{Let\:height\:be(h)=3x}$

Using Formula :

$\longmapsto\tt\boxed{Area\:of\:Triangle=\dfrac{1}{2}\times{b}\times{h}}$

Putting Values :

$\longmapsto\tt{\dfrac{1}{{\cancel{2}}}\times{{\cancel{4x}}}\times{3x}}$

$\longmapsto\tt{2x\times{3x}}$

$\longmapsto\tt\bf{6x}^{2}$

Now ,

If the base is increased by 4 cm and the height is decreased by 2 cm, the area of the triangle remains the same.

$\longmapsto\tt{Base=4x+4}$

$\longmapsto\tt{Height=3x-2}$

Using Formula :

$\longmapsto\tt\boxed{Area\:of\:Triangle=\dfrac{1}{2}\times{b}\times{h}}$

Putting Values :

$\longmapsto\tt{{6x}^{2}=\dfrac{1}{2}\times{(4x+4)}\times{(3x-2)}}$

$\longmapsto\tt{6x}^{2}\times{2}=(4x+4)\:\:(3x-2)$

$\longmapsto\tt{{{\cancel{12x}^{2}}}={\cancel{{12x}^{2}}}-8x+12x-8}$

$\longmapsto\tt{-8x+12x-8=0}$

$\longmapsto\tt{4x-8=0}$

$\longmapsto\tt{4x=8}$

$\longmapsto\tt{x=\cancel\dfrac{8}{4}}$

$\longmapsto\tt\bf{x=2}$

Value of x is 2 .

Therefore :

$\longmapsto\tt{Base\:of\:Triangle=4(2)}$

$\longmapsto\tt\bf{8\:cm}$

$\longmapsto\tt{Height\:of\:Triangle=3(2)}$

$\longmapsto\tt\bf{6\:cm}$

Answer 2

Answer:

Given :

The base and height of a triangle are in the ratio 4:3.

If the base is increased by 4 cm and the height is decreased by 2 cm, the area of the triangle remains the same.

To Find :

Base and Height of the Triangle .

Solution : Answer