Question

the altitude of triangle is three-fifth of the length of the corresponding base. if the altitude is decreased by, 4 CM and the base is decrease by 2 CM, the area of the triangle remains same. find the base of the triangle​

Answer 1

Step-by-step explanation:

Base of triangle is 20 cm and altitude is 12 cm.

Step-by-step explanation:

1. First consider old triangle, which base and altitude are

Let

Base = X (cm) ...a)

\mathbf{Altitude= \frac{3}{5}X=\frac{3X}{5}}Altitude=

5

3

X=

5

3X

...b) (Given)

Then

\mathbf{Area of old triangle= \frac{1}{2}\times Base\times Altitude=\frac{1}{2}\times X\times \frac{3X}{5}}Areaofoldtriangle=

2

1

×Base×Altitude=

2

1

×X×

5

3X

Which can be write as

\mathbf{Area of old triangle=\frac{1}{2}\times \frac{3X^{2}}{5}}Areaofoldtriangle=

2

1

×

5

3X

2

...1)

2. Now consider new triangle, which base and altitude are

Base = (X+10) (cm) (Given)

\mathbf{Altitude= ( \frac{3X}{5}-4 )}Altitude=(

5

3X

−4) (Given)

Then

\mathbf{Area of new triangle= \frac{1}{2}\times Base\times Altitude=\frac{1}{2}\times (X+10)\times ( \frac{3X}{5}-4 )}Areaofnewtriangle=

2

1

×Base×Altitude=

2

1

×(X+10)×(

5

3X

−4)

Which can be write as

\mathbf{Area of new triangle=\frac{1}{2}\times (X+10)\times ( \frac{3X}{5}-4 )}Areaofnewtriangle=

2

1

×(X+10)×(

5

3X

−4) ...2)

3. It is given that

Area of new triangle = Area of old triangle ...3)

So form equation 1) and equation 2)

\mathbf{\frac{1}{2}\times (X+10)\times ( \frac{3X}{5}-4 )=\frac{1}{2}\times \frac{3X^{2}}{5}}

2

1

×(X+10)×(

5

3X

−4)=

2

1

×

5

3X

2

\mathbf{(X+10)\times ( \frac{3X}{5}-4 )=\frac{3X^{2}}{5}}(X+10)×(

5

3X

−4)=

5

3X

2

\mathbf{\frac{3X^{2}}{5}-4X+10\times \frac{3X}{5}-40=\frac{3X^{2}}{5}}

5

3X

2

−4X+10×

5

3X

−40=

5

3X

2

\mathbf{-4X+6X-40=0}−4X+6X−40=0

So on solving

X = 20

4. So from equation a) and equation b)

Base = X =20 (cm) ...a)

\mathbf{Altitude=\frac{3X}{5}=12(cm)}Altitude=

5

3X

=12(cm)

Hope it helps you

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