If a triangle of a base $6 c m$ has the same area as that of a circle of radius $6 c m$ , then the altitude of the triangle is

6 π

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8 π

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10 π

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12 π

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Solution

The correct option is A $12 π$ cm
Area of $Δ =$ area of circle
$= \frac{1}{2} \times b a s e \times a l t i t u d e = π r^{2} = \frac{1}{2} \times 6 \times a l t i t u d e = π \times 6 \times 6$
altitude $= 12 π$ cm

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Similar questions

Q. A triangle has a base

6 c m

and has the same area as a circle of radius

6 c m

the altitude of the triangle in cm is .......

Q. The rate of change of the area of a circle with respect to its radius r at r = 6 cm is (A) 10π (B) 12π (C) 8π (D) 11π

The rate of change of the area of a circle with respect to its radius r at r = 6 cm is

(A) 10π (B) 12π (C) 8π (D) 11π

Q. Construct a triangle

A B C,

whose base

B C = 6 c m, ∠ B = 60^{o}

and altitude

A D = 4

cm.

The altitude of a triangle is three-fifth the length of its corresponding base. If the altitude is increased by 4 cm and the base is decreased by 2 cm, then the area of the triangle remains the same. Find the base of the triangle in cm.

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If a triangle of a base 6cm has the same area as that of a circle of radius 6cm, then the altitude of the triangle is

The correct option is A 12π cmArea of Δ=area of circle=12×base×altitude=πr2=12×6×altitude=π×6×6 altitude=12πcm

If a triangle of a base $6 c m$ has the same area as that of a circle of radius $6 c m$ , then the altitude of the triangle is

The correct option is A $12 π$ cm
Area of $Δ =$ area of circle
$= \frac{1}{2} \times b a s e \times a l t i t u d e = π r^{2} = \frac{1}{2} \times 6 \times a l t i t u d e = π \times 6 \times 6$
altitude $= 12 π$ cm