Construct a right angled Δ DEF such that the hypotenuse DF = 7.5 cm and EF = 5.5 cm
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We know,
In right angle triangle ∠DEF = 90°,shown in the figure attached.
For constructing Δ
1.Draw EF = 5.5 cm. using scale.
2.Now Make ∠DEF=90° using protractor.
3.Now using bow compass length extended equal to DF = 7.5 cm.chop it with.
4.By joing point P,Q,R we got right angle ΔPQR.
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It is given that ,
In ∆ DEF ,
Hypotenuse DF = 7.5 cm
EF = 5.5 cm
<DEF = 90°
Construction steps :
1 ) Draw EF = 5.5 cm line segment.
2 ) Draw <FEX = 90°.
3 ) Take F as center with radius 7.5 cm
draw an arc which intersects EX at D.
4 ) Join F to D .
We get required right angled triangle ∆DEF.
I hope this helps you.
: )
In ∆ DEF ,
Hypotenuse DF = 7.5 cm
EF = 5.5 cm
<DEF = 90°
Construction steps :
1 ) Draw EF = 5.5 cm line segment.
2 ) Draw <FEX = 90°.
3 ) Take F as center with radius 7.5 cm
draw an arc which intersects EX at D.
4 ) Join F to D .
We get required right angled triangle ∆DEF.
I hope this helps you.
: )
Attachments:
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