Construct a right angled triangle PQR with hypotenuse PQ = 9 cm and RQ = 5 cm.
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We know,
In right angle triangle ∠R = 90°,shown in the figure attached.
For constructing Δ
1.Draw a line from point R to Q(RQ = 5 cm.) using scale.
2.Now Make ∠MRQ=90° using protractor.
3.Now keep needle of the bow compass at Q and extend equal to 9 cm. draw an arc N and chop it with.
4.Give point of intersection name P.
5.By joining point PR and PQ we got right angle ΔPQR.
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It is given that ,
In right angled triangle PQR ,
Hypotenuse PQ = 9 cm,
RQ = 5 cm
< PRQ = 90°
Construction steps :
1 ) Draw RQ = 5 cm line segment .
2 ) Draw <XRQ = 90° .
3 ) Take Q as center with radius 9 cm
draw an arc which intersects RX at P .
4 ) Join P to Q .
Required right angled ∆ PRQ formed.
I hope this helps you.
: )
In right angled triangle PQR ,
Hypotenuse PQ = 9 cm,
RQ = 5 cm
< PRQ = 90°
Construction steps :
1 ) Draw RQ = 5 cm line segment .
2 ) Draw <XRQ = 90° .
3 ) Take Q as center with radius 9 cm
draw an arc which intersects RX at P .
4 ) Join P to Q .
Required right angled ∆ PRQ formed.
I hope this helps you.
: )
Attachments:
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