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.**

*Method of Construction:*

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

(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 \mathrm{ABC} with the help of scale, pencil and compass.

(iv) We draw an angle \angle G D C at the point D of side BC of \triangle A B C with equal measure of an angle 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 get parallelogram EDCF.