Question

the base of triangle is 4cm longer than its attitude. the area of base is 48sq.cm then find the base and attitude​

Answer 1

Que. the base of triangle is 4cm longer than its attitude. the area of triangle is 48sq.cm then find the base and attitude.

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We know that the area of a triangle is

= 1/2 × base × altitude

Let the triangle be ∆ABC , In which

AB = Altitude = y ( let )

BC = Base = y + 4 ( as per given in question )

Now we will apply the formula of area of triangle

Ar ( ∆ABC ) = 1/2 × AB × BC

48 = 1/2 × y × ( y + 4 )

96 = y( y +4 )

96 = y² + 4y

y² + 4y - 96 = 0

Now by splitting the term method , we will factorise the following equation.

y² + (12 - 8)y - 96 = 0

y² + 12y - 8y - 96 = 0

y(y + 12) - 8 (y + 12) = 0

(y + 12) (y - 8) = 0

Now , two cases will form.

CASE 1

if

y + 12 = 0

y = -12

Which is not possible.

CASE 2

if

y - 8 = 0

y = 8

So y = 8 is the correct answer.

AB = Altitude = y = 8 cm

BC = Base = y + 4 = 8 + 4 = 12 cm

Answer 2

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⏩The base of triangle is 4cm longer than its attitude. the area of base is 48sq.cm. Then, find the base and attitude.??

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⏩Altitude= 8cm,

⏩Base=12cm.

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➡So It is given that the base is 4 cm longer than altitude.

➡So let altitude be x.

➡Now put it in equation

➡That is, 1/2×(4+x)×x=48

➡So, 4x × 2 + x × 2 - 96

➡So by factorising it we get two values of x ,

➡One is 8 and other is - 12.

Case-1

if

y + 12 = 0

y = -12

Which is not possible.

Case-2

if

y - 8 = 0

y = 8

So y = 8 is the correct answer.

➡So neglecting - 12 .

➡We take the value of x as 8 .

⏩ Now altitude = 8 cm and base =8 +4 = 12 cm.

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