1.

A DeltaPQR , given that QR = 3 cm, anglePQR = 45^(@) and QP - PR = 2 cm.

Answer»

Solution :Given, in`DeltaPQR`, `QR = 3 cm`
`anglePQR = 45^(@)`
and `QP - PR` = `2 cm `
To construct `DeltaPQR` , USE the FOLLOWING STEPS.
(i) Draw the base QR of length 3cm.
(ii) Make an angle XQR = `45^(@)` at point Q of base QR.
(iii) Cut the line segment QS = QP - PR = 2cm from the ray QX.
(iv) Join SR and draw the perpendicular bisector of SR say AB.
(v) Let bisector AB intersect QX at P.Join PR Thus, `DeltaPQR` is the required triangle.
Justification
Base QR and `anglePQR` are drawn as given. Since, the point P lies on the perpendicular bisector of SR.
`:.` PS = PR
Now,QS = PQ-PS
= PQ - PR
Thus , our construction is justified.


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