Area of the triangle $M N O$ of Fig is

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

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

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

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

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Solution

The correct option is D $\frac{1}{2} N O \times O Q$
As we know that,
Area of triangle $= \frac{1}{2} \times b a s e \times h e i g h t$
$= \frac{1}{2} \times N O \times O Q$ [as $M P$ and $O N$ are parallel]

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Similar questions

Q. Question 9

Area of

Δ M N O

in the figure is

(a)

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

(b)

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

(c)

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

(d)

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

The area of triangle MNO in figure is (a) $ \frac{1}{2}MN\times NO$ (b) $ \frac{1}{2}NO\times MO$ (c)$ \frac{1}{2}MN\times OQ$ (d) $ \frac{1}{2}NO\times OQ$

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

Q. In the given figure OQ is the perpendicular bisector of RS and UP is perpendicular to OQ. To prove

\frac{1}{O P} + \frac{1}{O Q} = ?

Find the sum of the series
$\sum_{n}^{r = 0} (- 1)^{r}^{n} C_{r} [\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \frac{15^{r}}{2^{4 r}} \dots u p t o m t e r m s]$

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Area of the triangle MNO of Fig is

The correct option is D 12NO×OQAs we know that, Area of triangle =12×base×height=12×NO×OQ [as MP and ON are parallel]

Area of the triangle $M N O$ of Fig is

The correct option is D $\frac{1}{2} N O \times O Q$
As we know that,
Area of triangle $= \frac{1}{2} \times b a s e \times h e i g h t$
$= \frac{1}{2} \times N O \times O Q$ [as $M P$ and $O N$ are parallel]