Question

17. An altitude of a triangle is five-thirds the length of its corresponding base. If the altitude be
increased by 4 cm and the base decreased by 2 cm, the area of the triangle remains the same. find the base and the altitude of the triangle ​

Answer 1

$\huge\boxed{Answer}$

Let the length of corresponding base be x.

Thus, altitude will be 35x.

Now, it is given that altitude is increased by 4 cm i.e., 35x+4 and base is decreased by 2 cm i.e., x−2.

The area is same for both.

Therefore, 21×x×35x=21×(35x+4)×(x−2)

⇒35x2=35x2−310x+4x−8

⇒8=32x

⇒x=12

Thus, length of base is 12 cm

and length of altitude is 35×12=20 cm

$\huge\mathfrak\red{itz\:Jyotsana☺}$

Answer 2

» To Find :

The Corresponding Base and Altitude of the triangle .

» Given :

For Case 1 :

Let the Corresponding Base be x.

So According to the question , the Altitude is $\dfrac{5}{3}x$

• Altitude $\rightarrow H_{1} = \dfrac{5}{3}x$

• Corresponding Base $\rightarrow B_{1} = x$

For Case 2 :

Taken the Corresponding Base is x ,so Base is (x - 2)

Taken Altitude is $\dfrac{5}{3}x$ so ,new Altitude is $\left(\dfrac{5}{3}x + 4\right)$

• Altitude $\rightarrow H_{2} = \left(\dfrac{5}{3}x + 4\right)$

• Corresponding Base $\rightarrow B_{2} = (x - 2)$

» We Know :

Area of a Triangle :

$\sf{\underline{\boxed{A = \dfrac{1}{2} \times base \times height}}}$

» Concept :

According to the question , the Area of the Triangle remains same ,even after the Altitude and Corresponding Base is changed .i.e,

$\therefore$ Area of Orginal Triangle = Area of New triangle.

Area of Orginal Triangle :

• Altitude $\rightarrow H_{1} = \dfrac{5}{3}x$

• Corresponding Base $\rightarrow B_{1} = x$

Formula :

$\sf{A = \dfrac{1}{2} \times base \times height}$

Substituting the values in it ,we get :

$\sf{\Rightarrow A = \dfrac{1}{2} \times B_{1} \times H_{1}}$

$\sf{\Rightarrow A = \dfrac{1}{2} \times x \times \dfrac{5}{3}x}$

Hence , the original area is $\sf{\dfrac{1}{2} \times x \times \dfrac{5}{3}x}$

Area of New Triangle :

• Altitude $\rightarrow H_{2} = \left(\dfrac{5}{3}x + 4\right)$

• Corresponding Base $\rightarrow B_{2} = (x - 2)$

Formula :

$\sf{A = \dfrac{1}{2} \times base \times height}$

Substituting the values in it ,we get :

$\sf{\Rightarrow A = \dfrac{1}{2} \times B_{2} \times H_{2}}$

$\sf{\Rightarrow A = \dfrac{1}{2} \times (x -2) \times \left(\dfrac{5}{3}x + 4\right)}$

Hence ,the area of new triangle is $\sf{A = \dfrac{1}{2} \times (x -2) \times \left(\dfrac{5}{3}x + 4\right)}$

Now According to the question ,the area are same ,so we get the equation as :

$\sf{\dfrac{1}{2} \times x \times \dfrac{5}{3}x = \dfrac{1}{2} \times (x -2) \times \left(\dfrac{5}{3}x + 4\right)}$

By solving this Equation ,we will be able to find the value of x ,and the by putting the value of x ,we can find the corresponding base and Altitude of the triangle .

» Calculation :

Given Equation :

$\sf{\dfrac{1}{2} \times x \times \dfrac{5}{3}x = \dfrac{1}{2} \times (x - 2) \times \left(\dfrac{5}{3}x + 4\right)}$

By solving it ,we get :

$\sf{\Rightarrow \dfrac{1x}{2} \times \dfrac{5}{3}x = \dfrac{1}{2} \times (x - 2) \times \dfrac{5x + 12}{3}}$

$\sf{\Rightarrow \dfrac{5x}{3 \times 2}x = \dfrac{(x - 2)}{2} \times \dfrac{5x + 12}{3}}$

$\sf{\Rightarrow \dfrac{5x^{2}}{6} = \dfrac{(x - 2) \times (5x + 12)}{2 \times 3}}$

$\sf{\Rightarrow \dfrac{5x^{2}}{6} = \dfrac{5x^{2} + 12x - 10 x - 24}{6}}$

$\sf{\Rightarrow \dfrac{5x^{2}}{6} = \dfrac{5x^{2} + 2x - 24}{6}}$

$\sf{\Rightarrow \dfrac{5x^{2}}{\cancel{6}} = \dfrac{5x^{2} + 2x - 24}{\cancel{6}}}$

$\sf{\Rightarrow 5x^{2} = 5x^{2} + 2x - 24}$

$\sf{\Rightarrow 5x^{2} - 5x^{2} = 2x - 24}$

$\sf{\Rightarrow \cancel{5x^{2}} - \cancel{5x^{2}} = 2x - 24}$

$\sf{\Rightarrow 0 = 2x - 24}$

$\sf{\Rightarrow - 2x = - 24}$

$\sf{\Rightarrow \cancel{-} 2x = \cancel{-} 24}$

$\sf{\Rightarrow 2x = 24}$

$\sf{\Rightarrow x = \dfrac{24}{2}}$

$\sf{\Rightarrow x = \cancel{\dfrac{24}{2}}}$

$\sf{\Rightarrow x = 12}$

Hence ,the value of x is 12.

Base :

Since , we have taken base as x , so the Corresponding Base of the Triangle is 12 cm.

Altitude :

Since , the altitude is $\dfrac{5}{3}x$ ,we can put the value of x ,in the Equation to find out the Altitude of the triangle.

$\sf{\Rightarrow Altitude = \dfrac{5}{3}x}$

$\sf{\Rightarrow Altitude = \dfrac{5}{3} \times 12}$

$\sf{\Rightarrow Altitude = \dfrac{5}{\cancel{3}} \times \cancel{12}}$

$\Rightarrow Altitude = 5 \times 4$

$\Rightarrow Altitude = 20 cm$

So , Altitude is 20 cm

Hence , the corresponding base is 12 cm and the Altitude is 20 cm.

» Additional Information :

• Area of a Circle = πr²

• Circumference of a Circle = 2πr

• Volume of a Cylinder = πr²h