What is the area of the right-angled triangle MNO- right angled at N A- MN-NOB- NO-MNC- Area -1-2 - MO - NM D- Area - I -2 - MO - NO

The correct options are A MN,NO B NO,MN Area of triangle MNO = 1/2 × base × height = 1/2 × MN × NO. The order of the vertices does not matter while naming sid ...

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Standard X

Mathematics

Area of a Triangle

What is the a...

Question

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

MN,NO

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NO,MN

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Area = (

\frac{1}{2}

)*(MO)*(NM)

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Area = (

\frac{1}{2}

)*(MO)*(NO)

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Solution

The correct options are
A
MN,NO

B
NO,MN

Area of triangle MNO = $\frac{1}{2}$ $\times$ base $\times$ height = $\frac{1}{2}$ $\times$ MN $\times$ NO. The order of the vertices does not matter while naming sides here hence the area can also be = $\frac{1}{2} \times N M \times O N$

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Similar questions

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△ M N O

such that

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∠ N = 50^{\circ}

and

M O - M N = 2.4 c m .

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where base

N O = 6.7 c m

∠ M N O = 45^{\circ}

and

M O - M N = 2.8 c m .

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

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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

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) of

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

The correct options are A MN,NO B NO,MN Area of triangle MNO = 12 × base × height = 12 × MN × NO. The order of the vertices does not matter while naming sides here hence the area can also be =12×NM×ON

The correct options are
A
MN,NO

B
NO,MN

Area of triangle MNO = $\frac{1}{2}$ $\times$ base $\times$ height = $\frac{1}{2}$ $\times$ MN $\times$ NO. The order of the vertices does not matter while naming sides here hence the area can also be = $\frac{1}{2} \times N M \times O N$