Advertisements
Advertisements
प्रश्न
Construct an isosceles ∆ PQR such that PQ = PR = 6.5 cm and ∠PQR = 75°. Using a ruler and compasses only constructs a circumcircle to this triangle.
Advertisements
उत्तर
Steps of Construction :
(i) Draw a line segment PQ = 6.5 cm
(ii) At Q, draw a ray making an angle of 75°.
(iii) Through P, with a radius of 6.5 cm, draw an arc which intersects the angle ray at R.
(iv) Join PR,
Δ PQR is the required triangle.
(v) Draw the perpendicular bisectors of sides PQ and PR intersecting each other at O.
(vi) Join OP, OQ and OR.
(vii) With centre O and radius equal to OP or OQ or OR draw a circle which passes through P, Q and R. This is the required circumcircle of Δ PQR
APPEARS IN
संबंधित प्रश्न
Construct a ∆ABC such that:
AB = 6 cm, BC = 4 cm and CA = 5.5 cm
Construct a ∆ PQR such that :
PQ = 6 cm, ∠Q = 60° and ∠P = 45°. Measure ∠R.
Construct an isosceles ∆ ABC such that:
AB = AC = 6.5 cm and ∠A = 60°
Construct an isosceles ∆ ABC such that:
One of the equal sides = 6 cm and vertex angle = 45°. Measure the base angles.
Construct an isosceles ∆ ABC such that:
BC = AB = 5.8 cm and ZB = 30°. Measure ∠A and ∠C.
Construct an equilateral Δ ABC such that:
AB = 5 cm. Draw the perpendicular bisectors of BC and AC. Let P be the point of intersection of these two bisectors. Measure PA, PB, and PC.
Construct an equilateral Δ ABC such that:
Each side is 6 cm.
Construct a ∆ ABC such that AB = 6 cm, BC = 5.6 cm and CA = 6.5 cm. Inscribe a circle to this triangle and measure its radius.
Construct an isosceles ∆ MNP such that base MN = 5.8 cm, base angle MNP = 30°. Construct an incircle to this triangle and measure its radius.
Construct an equilateral ∆ DEF whose one side is 5.5 cm. Construct an incircle to this triangle.
