Question

The altitude of a triangle is two thirds the length of its corresponding base if the altitude is decreased by 1cm and the base is increased by 4cm , the area of the new triangle remains the same . find the base and the altitude of the original triangle . ( Hint : the answer Base = 2.4 , Altitude =1.6 ) I know the answer but dont know to to explain it pleze help meeee...​

Answer 1

$\bigstar \underline{\underline{\sf Given:-}}$

• The altitude of a triangle is two thirds the length of its corresponding base.
• If the altitude is decreased by 1cm and the base is increased by 4cm , the area of the new triangle remains the same.

$\bigstar \underline{\underline{\sf To\ find:-}}$

• The base & altitude of the original triangle.

$\bigstar \underline{\underline{\sf Solution:-}}$

↪Let the base = 'b'

Then as the altitude of a triangle is two thirds the length of its corresponding base,

↪Altitude = 2/3b

We know:

❥Area of a triangle = 1/2 (base)(height)

Hence,

$\mapsto \sf Area=\frac{1}{2} (b)(\frac{2}{3} b)$

            = b²/3 .

Given that altitude is decreased by 1cm,

↪New Altitude = 2/3 b-1 = 2b -3 /3

Also the base is increased by 4cm,

↪New base = b+4

Now,

↪Area of new triangle = 1/2 (b+4)(2b -3/3)

According to question :-

❥ Area of new triangle = Area of original triangle.

$:\implies \sf \frac{b^2}{\cancel{3}} =\frac{1}{2} (b+4)(\frac{2b-3}{\cancel{3}})\\\\:\implies \sf \cancel{2b^2} = \cancel{2b^2}-3b+8b-12\\\\:\implies \sf 5b -12 = 0 \\\\:\implies \sf 5b = 12\\\\:\implies \sf b=\frac{12}{5} \\\\:\implies \boxed{\boxed{\sf b=2.4cm}}$

Therefore, the value of base is 2.4cm.

And,

Altitude of a triangle is two thirds the length of its corresponding base.

$:\implies \sf Original\ Altitude = \frac{2}{3} \times base\\\\:\implies \sf Altitude = \frac{2}{3} \times 2.4 = 1.6\\\\:\implies \boxed{\boxed{\sf Altitude =1.6cm}}$

Hence,

❥ Altitude = 1.6cm

❥ Base = 2.4cm

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Answer 2

Answer:

altitude 1.6cm

base 2.4 cm