The base of a triangle is $15$ dm and its height is $12$ dm. The height of another triangle of double the area, the base of which is $20$ dm, is given by

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Solution

The correct option is B $18$ dm
Let $A_{1}$ be the area of 1st triangle and $A_{2}$ be the area of 2nd triangle.
$∴ A_{1} = \frac{1}{2} \times b \times h$
$A_{1} = \frac{1}{2} \times 15 \times 12$
$A_{1} = 90 d m^{2}$
$∴ A_{2} = 2 A_{1} = 180 d m^{2}$
$A_{2} = \frac{1}{2} \times b \times h$
$\Rightarrow b = 18 d m$

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The base of a triangle is 15 dm and its height is 12 dm. The height of another triangle of double the area, the base of which is 20 dm, is given by

The correct option is B 18 dmLet A1 be the area of 1st triangle and A2 be the area of 2nd triangle.∴A1=12×b×hA1=12×15×12A1=90dm2∴A2=2A1=180dm2A2=12×b×h⇒b=18dm

The base of a triangle is $15$ dm and its height is $12$ dm. The height of another triangle of double the area, the base of which is $20$ dm, is given by