Area of a triangle having base $6 c m$ and altitude $8 c m$ is:

48 c m^{2}

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24 c m^{2}

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64 c m^{2}

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36 c m^{2}

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Solution

The correct option is B $24 c m^{2}$
Area of $△ = \frac{1}{2} \times b a s e \times a l t i t u d e$
$= \frac{1}{2} \times 6 \times 8 = 24 {c m}^{2}$

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Similar questions

Q. Two triangles ALM and BLM have same base LM = 6cm. These two triangles have an equal area of

24 c m^{2}

. The distance between their altitudes is PQ. If PQ is equal to the length of one of the altitudes then the area of the quadrilateral formed by the altitudes and the base PQ is equal to:

Two triangles have equal area of $24 c m^{2}$ and stand on the same base of 6 cm, if the distance PQ (on base) between their altitudes is equal to the length of one of the altitude. Then the area of quadrilateral formed by the altitudes and the base PQ is equal to:

Q. Two triangles have equal area of

24 c m^{2}

and stand on the same base of 6 cm. If the distance PQ between their altitudes is equal to the length of one of the altitude, then the area of quadrilateral formed by the altitudes and the base PQ is equal to:___.

If the base BC = 6 cm and the altitude AP = 4 cm, then the area of the triangle ABC = .

Here AB $∥$ CD. If the altitude of $△$ ACD is 6 cm and CD is 8 cm then the area of $△$ BCD is:

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Area of a triangle having base 6cm and altitude 8cm is:

The correct option is B 24cm2Area of △=12×base×altitude=12×6×8=24 cm2

Area of a triangle having base $6 c m$ and altitude $8 c m$ is:

The correct option is B $24 c m^{2}$
Area of $△ = \frac{1}{2} \times b a s e \times a l t i t u d e$
$= \frac{1}{2} \times 6 \times 8 = 24 {c m}^{2}$