Chapter – 13 : Construction | Chapter Solution Class 9

Let us Do – 13

Question 1

We draw a line segment PQ of length 5cm. We take an external point A of the line segment Let us draw a parallel line through point A to line segment PQ.

Solution

Method of Construction:

1. Draw a fixed straight line PQ (length 5 cm) using a scale. Mark a point R outside the line PQ.
2. Mark a point S on the straight line PQ. Join points R and S using a scale. This forms the angle ∠RSQ.
3. Using a scale and compass, draw ∠VRS at point R such that its measurement is equal to ∠RSQ, but on the opposite side of RS. Join points V and R, and extend the line VR in both directions to obtain the straight line TU.

Since ∠VRS = ∠RSQ (alternate angles), ∴ PQ ∥ TU.

Question 2

We draw a triangle with sides 5 cm, 8 cm, and 11 cm, and then construct a parallelogram equal in area to that triangle with one angle measuring 60°.

[ Let us write instruction process and proof]

Solution

Method of Construction:

1. Draw triangle ABC with sides AB = 5 cm, AC = 8 cm, and BC = 11 cm using a scale and compass.
2. Bisect side BC at point D using a compass and scale.
3. Through point A, draw a line PR parallel to BC using a scale and compass.
4. At point D on BC, construct an angle GDC = 60° so that the line meets PR at point E.
5. On line ER, measure and cut segment EF equal to DC using a compass.
6. Join C to F. The quadrilateral CDEF is the required parallelogram.

Question 3

We draw a triangle with AB = 6 cm, BC = 9 cm, and ∠ABC = 55°. We then construct a parallelogram equal in area to that triangle, having one angle of 60° and one side equal to half of AC.

Solution

Method of Construction:

1. Draw triangle ABC with AB = 6 cm, BC = 9 cm, and ∠ABC = 55°.
2. Bisect side AC at point D using a pencil, compass, and scale.
3. Through point C, draw a line PR parallel to AC using a scale, pencil, and compass.
4. At point D on AC, construct ∠GDC = 60° so that the line meets PR at point E.
5. On ER, mark point F such that EF = DC.
6. Join C to F. Quadrilateral EDCF is the required parallelogram.

Question 4

In triangle PQR, ∠PQR = 30°, ∠PRQ = 75°, and QR = 8 cm. Let us draw a rectangle equal in area to that triangle.

Solution

Method of Construction:

1. First, draw the perpendicular bisector AB of side QR at the point T.
2. Draw the straight line CD parallel to QR through the point P of triangle PQR, which intersects the perpendicular bisector AB at E.
3. Cut EF from ED equal to TR. Joining points F and R, we get parallelogram ETRF whose area is equal to the area of triangle PQR and whose angle ∠ETR = 90°. Thus, we draw a rectangle ETRF equal in area to triangle PQR

Question 5

Draw an equilateral triangle with side length 6.5 cm, and let us construct a parallelogram equal in area to that triangle and having one angle of 45°.

Solution

Method of Construction:

1. First, draw the given triangle ABC with each side measuring 6.5 cm.
2. Bisect side BC of triangle ABC at point D using a pencil, compass, and scale.
3. Through point A, draw a line PR parallel to BC using a scale, pencil, and compass.
4. At point D on BC, construct ∠GDC = 45° so that the line meets PR at point E.
5. On ER, mark point F such that EF = DC using a scale and compass. Join points C and F to form parallelogram EDCF.

Question 6

Length of each equal side of an isosceles triangle is 8 cm and the length of the base is 5 cm. Let us draw a parallelogram equal in area to that triangle, having one angle equal to one of the equal angles of the isosceles triangle, and one side equal to half of an equal side.

Solution

Method of Construction:

1. First, draw the given triangle ABC with base BC = 5 cm and equal sides AB = AC = 8 cm.
2. Bisect side BC at point D using a pencil, compass, and scale.
3. Through point A, draw a line PR parallel to BC using a scale, pencil, and compass.
4. At point D on BC, construct ∠GDC equal to one of the equal angles of triangle ABC so that the line meets PR at point E.
5. On ER, mark point F such that EF = DC using a scale and compass. Join points C and F to form parallelogram EDCF.