Mathematics Part I Solutions Solutions for Class 9 Maths Chapter 5 Area are provided here with simple step-by-step explanations. These solutions for Area are extremely popular among Class 9 students for Maths 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 Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 74:

#### 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*.

Δ*ABC* 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 Δ*ABC*.

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*.

Δ*ABD* is the required isosceles triangle with *DA* equal to *DB* and area of Δ*ABC* equal to area of Δ*ABD*.

#### Page No 74:

#### 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*.

Δ*ABC* 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 Δ*ABC*.

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*.

Δ*BCD* is the required triangle with *BC* = 6 cm and *BD* = 7 cm and area of Δ*ABC* equal to the area of Δ*ADC*.

#### Page No 74:

#### 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*.

Δ*ABC* 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 Δ*ABC*.

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*.

Δ*ABD* is the required triangle whose one side *AB *= 5 cm, angle at *B *= 30° and area of Δ*ABC* is equal to the area of Δ*ABD*.

#### Page No 74:

#### 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 Δ*OAB*.

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*.

Δ*ABC* is the required triangle whose area is equal to the area of Δ*OAB*.

#### Page No 77:

#### 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*.

Δ*ADE* 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.

#### Page No 77:

#### 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*.

Δ*ADE* 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.

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