| 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
