Question

Find the altitude of a triangle, whose area is 48 sq.cm and base is 3.6 cm.

Answer 1

★Given:-

• Area of triangle = 48sq.cm.
• Base = 3.6cm

★To find:-

• Altitude of triangle.

★Solution:-

We know,

✦Area of triangle = 1/2(Base×height)

Substituting values in the formula,

→Area = 1/2 ( 3.6 ×h)

→48 = 1/2 ( 3.6 ×h)

→96 = 3.6×h

→h = 96/3.6

→h = 26.67cm.

Hence,the altitude of the triangle is 26.67cm.

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Know more:-

Important area formula's:

• Area of square = (side)²
• Area of rectangle = length×breadth.
• Ar.Triangle  A=√s(s−a)(s−b)(s−c),where  s=(a + b + c)/2
• Parallelogram  A = bh
• Circle  A=πr²

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Answer 2

Given :-

• Area of traingle = 48sq.cm.
• Base = 3.6cm

To find :-

• Altitude of triangle.

Solution :-

★ We know that ★

$\boxed{\rm{Area \: of \: triangle = \frac{1}{2} (Base \: \times \: Height)}}$

Substituting values of the formula,

● Area = ${\bf{\frac{1}{2} (3.6 \: \times \: h)}}$

● 48 = ${\bf{\frac{1}{2} (3.6 \: \times \: h)}}$

● 96 = ${\bf{3.6 × h }}$

● h = ${\bf{\frac{96}{3.6}}}$

● h = 26.67 cm.

Hence,

• The altitude of the triangle is $\boxed{\sf{26.67 cm}}$

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Know More :

➝ Important area formula's :

• Area of square = (side)²

• Area of reactangle = length × breath

• Ar.Triangle A = √s(s-a)(s-b)(s-c), where s = ${\bf{\frac{a \: + \: b \: + \: c}{2} }}$

• Parellogram A = bh

• Circle A = 2πr²

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