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Practical Geometry

Question

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Draw ΔPQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of
triangle is this?

Solution

The steps of construction are as follows.

(i) Draw a line segment QR of length 3.5 cm.

(ii) Taking point Q as centre, draw an arc of 4 cm radius.

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(iii) Taking point R as centre, draw an arc of 4 cm radius to intersect the previous arc at point P.

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(iv) Join P to Q and R.

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ΔPQR is the required triangle. As the two sides of this triangle are of the same length (PQ = PR), therefore, ΔPQR is an isosceles triangle.

Some More Questions From Practical Geometry Chapter

Construct ΔABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.

Construct ΔABC, given m∠A = 60°, m∠B = 30° and AB = 5.8 cm.

Construct ΔPQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°.
(Hint: Recall angle sum property of a triangle).

Examine whether you can construct ΔDEF such that EF = 7.2 cm, m∠E =
110° and m∠F = 80°. Justify your answer.

Construct the right angled ΔPQR, where m∠Q = 90°, QR = 8 cm and PR = 10 cm

Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.

Construct an isosceles right-angled triangle ABC, where, m∠ACB = 90° and AC = 6 cm.

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