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Area of Triangle Between Two Parallels

Area of trian...

Question

Area of triangle MNO in the given figure is______.

½ MN × NO

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½ NO × MO

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½ MN × OQ

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½ NO × OQ

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Solution

The correct option is D
½ NO × OQ

Area of triangle = $\frac{1}{2}$ $\times$ Base $\times$ Height

= $\frac{1}{2}$ $\times$ OQ (Height) $\times$ NO (Base)

The figure is a parallelogram, the height of the parallelogram is same throughout, therefore both the triangles will have the same height.

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Similar questions

What is the area of the right-angled triangle MNO? (right angled at N)

Δ M N O

\frac{1}{2} M N \times N O

\frac{1}{2} N O \times M O

\frac{1}{2} M N \times O Q

\frac{1}{2} N O \times O Q

△ M N O

N O = 6.7 c m

∠ M N O = 45^{\circ}

M O - M N = 2.8 c m .

△ M N O

N O = 6.2 c m

∠ N = 50^{\circ}

M O - M N = 2.4 c m .

In the reaction $3 M n O_{2} + 4 A l \to 3 M n + 2 A l_{2} O_{3}$ ,the oxidising agent is :

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