Mathematics Part I Solutions Solutions for Class 9 Math Chapter 5 Area are provided here with simple step-by-step explanations. These solutions for Area are extremely popular among Class 9 students for Math Area Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 9 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Question 1:

Draw ΔABC with AB = 4 cm, BC = 5 cm and CA = 6 cm. Draw an isosceles triangle of the same area with one side as AB itself.

#### Answer:

The dimensions of the triangle to be constructed are AB = 4 cm, BC = 5 cm and CA = 6 cm.

The steps of construction are as follows:

1) Draw a line segment AB of length 4 cm.

2) Taking A as the centre and 6 cm as the radius, draw an arc on the upper side of line segment AB.

3) Taking B as the centre and 5 cm as the radius, draw an arc cutting the previously drawn arc at point C.

4) Join AC and BC. ΔABC is the required triangle.

We know that triangles lying on the same base and between the same parallels are equal in area.

We also know that any point lying on the perpendicular bisector of a line segment is equidistant from the end points of that line segment.

Here, we need to draw an isosceles triangle whose base is AB and whose area is equal to the area of ΔABC.

So, we will use both these properties to construct the required triangle.

The steps of construction are as follows:

1) Draw a line l that is parallel to line segment AB and passes through point C. 2) Draw the perpendicular bisector of line segment AB by taking A and B as the centres. Draw arcs of radius more than half the length of AB. Let the perpendicular bisector intersect line l at point D. 3) Join AD and BD. ΔABD is the required isosceles triangle with DA equal to DB and area of ΔABC equal to area of ΔABD.

#### Question 2:

Draw ΔABC as in the first problem. Draw a triangle of the same area with two of its sides 6 centimetres and 7 centimetres.

#### Answer:

The dimensions of the triangle to be constructed are AB = 4 cm, BC = 5 cm and CA = 6 cm.

The steps of construction are as follows:

1) Draw a line segment AB of length 4 cm.

2) Taking A as the centre and 6 cm as the radius, draw an arc on the upper side of line segment AB.

3) Taking B as the centre and 5 cm as the radius, draw an arc cutting the previously drawn arc at point C.

4) Join AC and BC. ΔABC is the required triangle.

We know that triangles lying on the same base and between the same parallels are equal in area.

Here, we need to construct a triangle whose base is BC of length 6 cm and one other side of length 7 cm. The area of this triangle should be equal to the area of ΔABC.

The steps of construction are as follows:

1) Draw a line l that is parallel to line segment BC and passes through point A. 2) Taking B as the centre and 7 cm as the radius, draw an arc which intersects line l at a point, say D. 3) Join BD and CD. ΔBCD is the required triangle with BC = 6 cm and BD = 7 cm and area of ΔABC equal to the area of ΔADC.

#### Question 3:

Draw ΔABC, with AB = 5 cm and BC = 7 cm and ABC = 60°. Draw a triangle of the same area with one side as AB and the angle at B equal to 30°

#### Answer:

The measures of the sides and the angle of the triangle to be constructed are AB = 5 cm, BC = 7 cm and ABC = 60°

The steps of construction are as follows:

1) Draw a line segment BC of length 7 cm.

2) Draw XBC of measure 60° at point B.

3) Taking B as the centre and 5 cm as the radius, draw an arc which intersects ray BX at point A.

4) Join AC. ΔABC is the required triangle.

We know that triangles lying on the same base and between the same parallels are equal in area.

Here, we need to construct a triangle whose one side AB = 5 cm, angle at B = 30° and whose area is equal to the area of ΔABC.

The steps of construction are as follows:

1) Draw a line l that is parallel to line segment AB and passes through point C. 2) Draw DBC of measure 30° such that point D is lying on line l. 3) Join AD. ΔABD is the required triangle whose one side AB = 5 cm, angle at B = 30° and area of ΔABC is equal to the area of ΔABD.

#### Question 4:

Draw a circle and a triangle as in the figure below. One vertex of the triangle is at the centre of the circle and the other two vertices are on the circle. #### Answer:

The given figure shows a circle and an isosceles triangle, with two of its sides equal to the radius of the circle.

The steps of construction are as follows:

1) Draw a circle with centre O and of radius, say r units. 2) Join O with two points, say A and B on the circle. Join A and B also. This is the required figure.

We know that triangles lying on the same base and between the same parallels are equal in area.

Here, we need to construct a triangle whose all three vertices lie on the circle and whose area is equal to the area of ΔOAB.

The steps of construction are as follows:

1) Draw a line l that is parallel to line segment AB and passes through point O. 2) Join AC and BC. ΔABC is the required triangle whose area is equal to the area of ΔOAB.

#### Question 1:

Draw a quadrilateral with measures as given below: Draw a triangle of equal are and compute its area.

#### Answer:

The given quadrilateral can be drawn using the following steps of construction:

1) Draw a line segment AB of length 5 cm.

2) Draw XAB of measure 80° at point A.

3) Taking A as the centre and 3 cm as the radius, draw an arc which intersects ray AX at point D

4) Draw YDA of measure 120° at point D.

5) Taking D as the centre and 4 cm as the radius, draw an arc which intersects ray DY at point C.

6) Join BC. Quadrilateral ABCD is the required quadrilateral.

We know that triangles lying on the same base and between the same parallels are equal in area.

Here, we need to construct a triangle whose area is equal to the area of the quadrilateral ABCD.

The steps of construction are as follows:

1) Join diagonal BD of the quadrilateral ABCD. Draw a line l that is parallel to diagonal BD and passes through point C 2) Extend line segment AB such that it touches line l at point E. 3) Join DE ΔADE is the required triangle whose area is equal to the area of the quadrilateral ABCD.

Disclaimer: The given dimensions are not sufficient to compute the area of the given quadrilateral by using the methods of finding area taught till your grade.

#### Question 2:

Draw the quadrilateral ABCD with AB = 8.5 cm, BC = 4.5 cm, CD = 5 cm, DA = 6 cm, BD = 7.5 cm. Compute its area by drawing a triangle of equal area.

#### Answer:

The measures of the sides of the quadrilateral to be constructed are AB = 8.5 cm, BC = 4.5 cm, CD = 5 cm, DA = 6 cm and BD = 7.5 cm.

The steps of construction are as follows:

1) Draw a line segment AB of length 8.5 cm.

2) Taking A as the centre and 6 cm as the radius, draw an arc on the upper side of line segment AB

3) Taking B as the centre and 7.5 cm as the radius, draw an arc intersecting the previously drawn arc at point D

4) Join AD and BD.

5) Taking D as the centre and 5 cm as the radius, draw an arc on the right side of point D.

6) Taking B as the centre and 4.5 cm as the radius, draw an arc intersecting the previously drawn arc at point C.

7) Join DC and BC. Quadrilateral ABCD is the required quadrilateral.

We know that triangles lying on the same base and between the same parallels are equal in area.

Here, we need to construct a triangle whose area is equal to the area of the quadrilateral ABCD.

The steps of construction are as follows:

1) Draw a line l that is parallel to diagonal BD and passes through point C 2) Extend line segment AB such that it touches line l at point E. 3) Join DE ΔADE is the required triangle whose area is equal to the area of the quadrilateral ABCD.

Disclaimer: The given dimensions are not sufficient to compute the area of the given quadrilateral by using the methods of finding area taught till your grade.

View NCERT Solutions for all chapters of Class 9