Chapter – 13 : Construction | Chapter Solution Class 9

Ganit Prakash Solution 9

Book Name	: Ganit Prakash
Subject	: Mathematics (Maths)
Class	: 9 (Madhyamik/WB)
Publisher	: Prof. Nabanita Chatterjee
Chapter Name	: Theorems On Area (13th Chapter)

Let us Do – 13

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.

Method of Construction:

(i) At first 1 draw a fixed straight line PQ by a scale and consider a point R out side the line PQ.

(ii) Let’s take a point S on the straight line PQ. Now I join two points R & S by a scale. As a result $<\operatorname{RSQ}$ is formed.

(iii) Now with a scale and compass we draw $<\operatorname{VRS}$ at R. On the line segment RS & its measurement is equal to the measurement of \angle RSQ and it is in the opposite side of $<\operatorname{RSQ}$ . We join the points V and R by a scale and produce line VR on both the sides. We get the straight line TU.

Now, $\angle V R S=\angle R S Q$ but they are alternate angle.

$\therefore \quad$ PQ and TU are parallel to each other, i,e. $PQ \|TU$

2. We draw a triangle with length of sides 5cm, 8cm and 11cm and draw a parallelogram equal in area to that triangle and having an angle $60^{\circ}$ .

[ Let us write instruction process and proof]

Method of Construction:

(i) At first we draw a given $\triangle \mathrm{ABC}$ with side 5cm, 8cm and 11cm.

(ii) Now, the side BC of $\triangle \mathrm{ABC}$ was bisected at the point D with the help of pencil, compass and scale.

(iii) We draw a parallel line PR to BC through the point A of $\triangle A B C$ with the help of scale and pencil compass.

(iv) We draw an angle $\angle G D C$ at the point D of side BC of $\triangle \mathrm{ABC}$ with equal measure of an angle $<60^{\circ}$ which intersect PR at the point E.

(v) We cut EF part from ER with measure of DC with the help of scale and compass and joining C, F we got parallelogram EDFC.

We see that from the figure,

$\because \quad \mathrm{DE}=\mathrm{CF}=4 \mathrm{~cm}.$

and $\mathrm{CD}=\mathrm{EF}=5.5 \mathrm{~cm} . \quad \therefore CDEF$ is a parallelogram

3. We draw a triangles which AB = 6cm, BC = 9cm, $\angle \mathrm{ABC} = 55^{\circ}$ , let us draw a parallelogram equal in area to that triangle having an angle $60^{\circ}$ and length of one side is 1 of AC.

Method of Construction:

(i) At first we draw a given $\triangle \mathrm{ABC}$ with AB = 6cm, BC = 9cm, $\angle \mathrm{ABC}=55^{\circ}.$

(ii) Now, the side AC of $\triangle A B C$ was bisected at the point D with the help of pencil, compass and scale.

(iii) We draw a parallel line PR to AC through the point C of $\triangle A B C$ with the help of scale, pencil and compass.

(iv) We draw an angle $\angle \mathrm{GDC}$ at the point D of side AC of $\triangle A B C$ with measure of an angle $<60^{\circ}$ which intersect PR at the point E.

(v) We cut EF part from ER with equal of measure of DC with the help of scale and compass and joining C, F we got parallelogram EDCF.

4. In $\triangle \mathrm{PQR}, \angle \mathrm{PQR}=30^{\circ}, \angle \mathrm{PRQ}=75^{\circ} \ and \ \mathrm{QR}=8 \mathrm{~cm}$ . Let us draw a rectangle equal in area to that triangle.

Method of Construction:

(i) First I draw Perpendicular bisector AB of side QR at the point T.

(ii) Now, I draw the straight line CD parallel to QR through the point P of $\triangle P Q R$ which intersects perpendicular bisector AB at E.

(iii) Cut EF part from ED equal to TR. Joining two points F and R we get parallelogram ETRF of which area is equal to area of $\triangle \mathrm{PQR}$ and of which one angle $\angle E T R=90^{\circ}.$ We draw a rectangle ETRF with equal area of $\triangle P Q R.$

5. Draw an equilateral triangle with length of side 6.5cm, and let us draw a parallelogram equal in area to that triangle and having and angle $45^{\circ}.$

Method of Construction:

(i) At first we draw a given $\triangle A B C$ with side 6.5cm,

(ii) Now, the side of $\triangle A B C$ was bisected at the point D with the help of pencil, compass and scale.

(iii) We draw a parallel line PR to BC through the point A of $\triangle A B C$ with the help of scale, pencil and compass.

(iv) We draw an angle $\angle \mathrm{GDC}$ at the point D of side BC of $\triangle A B C$ with equal measure of an angle $<45^{\circ}$ which intersect PR at the point E.

(v) We cut EF part frm ER with measure of DC with the help of scale and compass and joining C, F we gel parallelogram EDCF.

6. Length of each equal sides of an isosceles triangle is 8cm and length of base is 5cm. Let us draw a parallelogram equal in area to that circle and having one angle of parallelogram is equal to one of equal angle of isosceles triangle and one side is 1 / 2 of equal side.