# Chapter – 14 : Construction | Chapter Solution Class 9

 Book Name : Ganit Prakash Subject : Mathematics (Maths) Class : 9 (Madhyamik/WB) Publisher : Prof. Nabanita Chatterjee Chapter Name : Construction (14th Chapter)

## Let us Do – 14

1. Pritam drew a quadrilateral ABCD of which AB = 5cm BC = 6cm, CD = 4cm, DA = 3cm and \angle \mathrm{ABC}=60^{\circ} draw a triangle with equal area of that quadrilateral.

Solution :

Method of Construction:

(i) I draw a quadrilateral ABCD.

(ii) Now, I draw diagonal AC of quadrilateral ABCD.

(iii) I draw a parallel line through the point D to the diagonal AC of quadrilateral ABCD which intersects at E of product BC with scale and compass.

(iv) Joining D and E we got triangle ABE.

2. Sahana drew a quadrilateral ABCD of which AB = 4cm, CD = 4.8cm, DA = 4.2cm and diagonal AC = 6cm. I draw a triangle with equal area of that quadrilateral.

Solution :

Method of Construction:

(i) I draw a given quadrilateral ABCD.

(ii) Now, I draw diagonal BD of quadrilateral ABCD.

(iii) I draw a parallel line through the point C to diagonal BD of quadrilateral ABCD which intersects at E of produced AB with scale and compass.

(iv) Joining D and E we got triangle ADE

3. Sahana drew a rectangle ABCD of which AB = 4cm, BC = 6cm. I draw a triangle with equal area of that rectangle.

Solution :

Method of Construction:

(i) I draw a given rectangle ABCD.

(ii) Now, I draw diagonal AC of rectangle ABCD.

(iii) I draw a parallel line through the point D to diagonal AC of rectangle \triangle B C D which intersect at E of produced BC with scale and compass.

(iv) Joining D and E we got triangle ABE.

4. I draw a quadrilateral ABCD of which is BC = 6cm, AB= 4cm, CD = 3cm, \angle ABC = 60^{\circ}, \angle BCD = 55^{\circ}, I draw a triangle with equal area of that quadrilateral of which one side is alongside AB and other side is alongside BC.

Solution

Method of Construction:

(i) I draw a quadrilateral ABCD.

(ii) Now, I draw diagonal AC of quadrilateral ABCD.

(iii) I draw a parallel line through the point D to diagonal AC of quadrilateral ABCD which intersects at E of produced BC with scale and compass.

(iv) Joining D and E we got triangle ABE.

5. I draw a square with side 5cm. I draw a parallelogram with equal area of square of which one angle is 60^{\circ}.

Solution :

Method of Construction:

(i) We draw a given square ABCD

(ii) I draw a triangle ABE equal in area of square ABCD draw.

(iii) with an equal area of triangle ABE, I draw a parallelogram FCEG of which one angle \angle F C E=60^{\circ}.

\therefore We got a shape of parallelogram field FCEG of which \angle F C E=60^{\circ} equal in area to shape of given square field ABCD.

6. I draw a square with side 6cm, and I draw triangle with equal area of that square.

Solution :

Method of Construction:

(i) I draw a square ABCD.

(ii) Now, I draw diagonal AC of square ABCD.

(iii) I draw a parallel line through the point D to the diagonal AC of square ABCD  which intersects at E of product BC with scale and compass.

(iv) Joining D and E we got triangle ABE.

7. I draw a quadrilateral ABCD  of which AD and BC are perpendicular on side AB and AB = 5cm, AD = 7cm and BC = 4 cm. I draw a triangle with equal area of that quadrilateral of which one angle is 30^{\circ}.

Solution :

Method of Construction:

(i) I draw a given quadrilateral ABCD.

(ii) Now, I draw diagonal BD  of quadrilateral ABCD.

(iii) I draw a parallel line through the point C to the diagonal BD  of quadrilateral ABCD  which intersects at E of product AB  with scale and compass.

(iv) Joining D and E we got triangle ADE.

(v) Now, \angle \mathrm{FAE}=30^{\circ} is constructed at the point A . The straight line DG is drawn parallel to AE to intersect AP at the point F. EF is drawn.

(vi) Thus \triangle \mathrm{AEF} is the required triangle.

8. I draw any pentagon ABCDE  and draw a triangle with equal area of it of which one vertex is C.

Solution :

Method of Construction:

(i) I draw a pentagon ABCDE.

(ii) I draw two diagonals AC and AD of pentagon ABCDE I draw two line segments BP and EQ parallel to AC and AD through the points B and E, which intersect extended both side CD  at the points P and Q respectively.

(iii) Join A, P,  and A, Q, we got \triangle A P Q

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