What is the area of the right-angled triangle - MNO- right angled at N- 1-2- MO - NO1-2- MN nbsp- NO1-2- MN nbsp- MOMN nbsp- NO

The correct option is B 1/2× MN nbsp;× NOArea of a triangle nbsp; = = 1/2× base × height = 1/2× MN × NO. ...

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Byju's Answer

Standard X

Mathematics

Area of triangle given its vertices

What is the a...

Question

What is the area of the right-angled triangle $Δ M N O$ , right angled at N.

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

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

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

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M N \times N O

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Solution

The correct option is A $\frac{1}{2} \times M N \times N O$
Area of a triangle = $= \frac{1}{2} \times b a s e \times h e i g h t = \frac{1}{2} \times M N \times N O$ .

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Similar questions

Q. In ∆LMN, m∠M = 40°, m∠N = 50°. Which of the following statements is true?

(1) [l (MN)]² = [l (LM)]² + [l (LN)]²
(2) [l (LM)]² = [l (MN)]² + [l (LN)]²
(3) [l (LN)]² = [l (LM)]² + [l (MN)]²

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

Q. What is the area of the right-angled triangle

Δ M N O

, right angled at N.

Q. What is the angle between the straight lines

(m^{2} - m n) y = (m n + n^{2}) x + n^{3}

and

(m n + m^{2}) y = (m n - n^{2}) x + m^{3}

, where

m > n

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$

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What is the area of the right-angled triangle ΔMNO, right angled at N.

The correct option is A 12×MN×NOArea of a triangle = =12×base×height=12×MN×NO.

What is the area of the right-angled triangle $Δ M N O$ , right angled at N.

The correct option is A $\frac{1}{2} \times M N \times N O$
Area of a triangle = $= \frac{1}{2} \times b a s e \times h e i g h t = \frac{1}{2} \times M N \times N O$ .