Question

7.
The base of a triangle is 4cm longer than its altitude. If the area of the triangle is 48 sq.cm
then find its base and altitude,​

Answer 1

Given :-

• The base of a triangle is 4cm longer than its altitude. If the area of the triangle is 48 sq.cm

To find :-

• Base and Altitude of triangle

Solution :-

Let the Altitude be x then base be (x + 4)

• Area of triangle = 48cm²

As we know that

→ Area of triangle = ½ × b × h

Where " b " is base and " h " is height of triangle

According to question

→ Area of triangle = 48 cm²

→ ½ × b × h = 48cm²

→ ½ × (x + 4) × x = 48

→ x(x + 4) = 48 × 2

→ x² + 4x = 96

→ x² + 4x - 96 = 0

Splitting middle term

→ x² - 8x + 12x - 96 = 0

→ x(x - 8) + 12(x - 8) = 0

→ (x - 8)(x + 12) = 0

Either

→ x - 8 = 0

→ x = 8

Or

→ x + 12 = 0

→ x = - 12

• Height and Altitude never in negative

Hence,

• Altitude of triangle = x = 8cm

• Height of triangle = x + 4 = 12cm

Answer 2

AnswEr :-

• Altitude of the triangle = 8cm
• Base of the triangle = 12cm

Given :-

• The base of a triangle is 4cm longer than it's altitude. Area of the triangle is 48cm².

To Find :-

• Base and altitude of the triangle.

SoluTion :-

Let,

• Altitude of the triangle be x cm
• Base of the triangle will be (x + 4)cm

It is given that the area of the triangle is 48cm².

According to question :-

Area of the triangle = ½ base × height

→ ½ × b × h = 48

→ ½ × (x+4) × x = 48

→ x (x + 4) = 48 × 2

→ x² + 4x = 96

→ x² + 4x - 96 = 0

Split the middle term

→ x² - 8x + 12x - 96 = 0

→ x (x - 8) + 12 (x - 8) = 0

→ (x + 12) (x - 8) = 0

• x + 12 = 0

→ x = 0 - 12

→ x = -12

• x - 8 = 0

→ x = 0 + 8

→ x = 8

Value of x is 8 [ ignore negative value ]

• Altitude of the triangle = x = 8cm
• Base of the triangle = x + 4 = 8 + 4 = 12cm

Hence, the altitude and base of the triangle are 8cm and 12cm respectively.

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