RD Sharma 2020 solution class 9 chapter 17 Heron's Formula Exercise 17.2

Exercise 17.2

Page-17.19

Question 1:

Find the area of a quadrilateral ABCD is which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Answer 1:

The quadrilateral ABCD having sides AB, BC, CD, DA and diagonal AC=5 cm is given, where AC divides quadrilateralABCD into two triangles ΔABC and ΔADC. We will find the area of the two triangles separately and them to find the area of quadrilateral ABCD.

In triangle ΔABC, observe that,

So the triangle ΔABC is right angled triangle.

Area of right angled triangle ΔABC, is given by

In ΔACD all the sides are known, so just use Heron’s formula to find out Area of triangle ΔACD, 

s=AD+DC+AC2=5+4+52=7 cm

The area of the ΔACD is:

Area of quadrilateral ABCD will be,

Area = Area of triangle ABC + Area of triangle ADC

Question 2:

The sides of a quadrangular field, taken in order are 26 m, 27 m, 7 m are 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.

Answer 2:

We assume quadrilateral ABCD be the quadrangular field having sides AB, BC, CD, DA and.

We take a diagonal AC, where AC divides quadrilateral ABCD into two triangles ΔABC and ΔADC

We will find the area of two triangles ΔABC and ΔADC separately and add them to find the area of the quadrangular ABCD.

In triangle ΔADC, we have

AD = 24 m; DC = 7 m

We use Pythagoras theorem to find side AC, 

AC2 = AD2 + DC2

 

Area of right angled triangle ΔADC, say A1 is given by

Where, Base = DA = 24 m; Height = DC = 7 m

Area of triangle ΔABC, say A2 having sides a, b, c and s as semi-perimeter is given by

Where, a = AC = 25 m; b = AB = 26 m; c = BC = 27 m 

Area of quadrilateral ABCD, say A

A = Area of triangle ΔADC + Area of triangle ΔABC

Question 3:

The sides of a quadrilateral, taken in order are 5, 12, 14 and 15 metres respectively, and the angle contained by the first two sides is a right angle. Find its area.

Answer 3:

We assume ABCD be the quadrilateral having sides AB, BC, CD, DA and angle.

We take a diagonal AC, where AC divides quadrilateral ABCD into two triangles ΔABC and ΔADC. We will find the area of these two triangles and add them to find the area of the quadrilateral ABCD

In triangle ΔABC, we have

AB = 5 m; BC = 12 m

We will use Pythagoras theorem to calculate AC 

AC2 = AB2 BC2

 

Area of right angled triangle ΔABC, say A1 is given by

Where, Base = AB = 5 m; Height = BC = 12 m

Area of triangle ΔADC, say Ahaving sides a, bc and s as semi-perimeter is given by

Where, = AC = 13 m; b = DC = 14 m; c = AD = 15 m 

Area of quadrilateral ABCD, say A

A = Area of triangle ΔABC + Area of triangle ΔADC

Question 4:

A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD= 5 m and AD = 8 m. How much area does it occupy?

Answer 4:

We assume ABCD be the quadrilateral having sides AB, BC, CD, DA and.

We take a diagonal DB, where DB divides ABCD into two triangles ΔBCD and ΔABD

In ΔBCD, we have

DC = 5 m; BC = 12 m

Use Pythagoras theorem 

BD2 = DC2 + BC2

 

Area of right angled triangle ΔBCD, say A1 is given by

Where, Base = DC = 5 m; Height = BC = 12 m

Area of triangle ΔABD, say A2 having sides a, b, c and s as semi-perimeter is given by

Where, a = AD = 8 m; b = AB = 9 m; c = BD = 13 m 

Area of quadrilateral ABCD, say A

A = Area of triangle DCB + Area of triangle ABD

Question 5:

Two parallel side of a trapezium are 60 cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of the trapezium.

Answer 5:

We assume ABCD be the given trapezium where AB is parallel to DC.

We draw CE parallel to AD from point C.

Therefore, a parallelogram ADCE is formed having AD parallel to CE and DC parallel to AE.

AE = 60 cm; CE = 25 cm; BE = 

Basically we will find the area of the triangle BCE and area of the parallelogram AECD and add them to find the area of the trapezium ABCD.

Area of triangle ECB, say A1 having sides a, bc and s as semi-perimeter is given by

Where, = EB = 17 cm; b = EC = 25 cm; c = BC = 26 cm 

Here we need to find the height of the parallelogram AECD which is CM to calculate area of AECD.

Where, BE = Base = 17 cm ; Height = CM = h



Thus area of parallelogram will be,
A2b×h
=60×24=1440 cm2

Total area of the trapezium will be
A = A+ A2
= 204 + 1440
= 1644 cm2

Question 6:

A rhombus, sheet, whose perimeter is 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of Rs 5 per m2. Find the cost of painting.

Answer 6:

We assume ABCD be the given rhombus having 

AB = BC = CD = DA 

BD and AC be the diagonals of rhombus

We need to find cost of painting both sides

Perimeter of rhombus ABCD, say P is 32 m

We know that BD and AC diagonals of rhombus and.So,

(Diagonals in rhombus intersect at right angle)

BD = 24 m; AC = h; side = AB = 8m

Taking square of both sides, we get

Area of rhombus, say A1

Area of both sides of rhombus;

Page-17.20

Question 7:

Find the area of a quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90° and BCD forms an equilateral triangle whose each side is equal to 26 cm.

Answer 7:

We assume ABCD be the quadrilateral having sides AB, BC, CD, DA, diagonal BD and angle where BCD forms an equilateral triangle having equal sides.

We need to find area of ABCD

In triangle BAD, we have

BD2 = BA2 + AD2 .So

Area of right angled triangle ABD, say A1 is given by

Where,

Base = BA = 10 cm; Height = AD = 24 cm

Area of equilateral triangle BCD, say A2 having sides a, b, c is given by

, where

a = BC = CD = BD = 26 cm

Area of quadrilateral ABCD, say A

A = Area of triangle BAD + Area of triangle BCD

Question 8:

Find the area of a quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.

Answer 8:

The quadrilateral ABCD having sides AB,BC,CD,DA and diagonal BD is given, where BD divides ABCD into two triangles 

ΔDBC and ΔDAB

In triangle DBC, we can observe that

DC2 = DB2 + BC2

Therefore, it is a right angled triangle. 

Area of right angled triangle DBC, say A1 is given by

Where,

Base = BC = 21 cm; Height = BD = 20 cm

Area of triangle DAB, say A2 having sides a, b, c and s as semi-perimeter is given by

, where

a = DB = 20 cm; b = AD = 34 cm; c = AB = 42 cm

Area of quadrilateral ABCD, say A

A = Area of triangle DBC + Area of triangle DAB

Question 9:

The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.

Answer 9:

We are given the measure of adjacent sides of a parallelogram AB and BC that is the sides having same point of origin and the diagonal AC which divides parallelogram ABCD into two congruent triangles ABC and ADC.

Area of triangle ABC is equal to Area of triangle ADC as they are congruent triangles.

Area of parallelogram ABCD, say A is given by

A =

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

 

Therefore the area of a triangle, say A1 having sides 20 cm, 34 cm and 42 cm is given by

a = 20 cm ; b = 34 cm ; c = 42 cm

Area of parallelogram ABCD, say A is given by

Question 10:

Find the area of the blades of the magnetic compass shown in the given figure. (Take 11 = 3.32)

Answer 10:


The blades of the magnetic compass are forming a rhombus having all equal sides measuring 5 cm each. A diagonal measuring 1 cm is given which is forming the triangular shape of the blades of the magnetic compass and diving the rhombus into two congruent triangles, say triangle ABC and triangle DBC having equal dimensions.

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

 


Therefore the area of a triangle ABC, say A1 having sides 5 cm, 5 cm and 1 cm is given by:

a = 5 cm ; b = 5 cm ; c = 1 cm


 

 

 

 

Area of blades of magnetic compass, say A is given by

A = Area of one of triangle ABC

Question 11:

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.

Answer 11:

It is given that the area of triangle and parallelogram are equal.

We will calculate the area of triangle with the given values and it will also give us the area of parallelogram as both are equal.

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

Therefore the area of a triangle; say A, having sides 15 cm, 13 cm and 14 cm is given by

a = 15 cm ; b = 13 cm ; c = 14 cm

We have to find the height of the parallelogram, say h

Area of parallelogram AECD say A1 is given by

Base = 60 cm; Height = h cm; A =A1 = 84 cm2

Question 12:

Two parallel side of a trapezium are 60 cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of the trapezium.

Answer 12:

Consider a trapezium ABCD as shown below.

Draw a line BF parallel to AD such that ABFD becomes a parallelogram. Also, draw a perpendicular BE on CD as shown in the figure.


In ΔBCF, the three sides are given as a=BF=25 cm, b=BC=26 cm, c=CF=CD-FD=17cm.
The semi-perimeter of ΔBCF = 25+26+172=34.
The area of ΔBCF can be calculated using Heron's formula as
ss-as-bs-c=3434-2534-2634-17=34×9×8×17=204 cm2

Also, the area of ΔBCF = 12×base×height
204=12×17×BE204=12×17×BEBE=24 cm

Therefore, Area of Trapezium =12AB+CD×BE=1260+77×24=1644 cm2.

Question 13:

Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, ∠ACB = 90° and AC = 15 cm.

Answer 13:

We assume ABCD be the quadrilateral having sides AB, BC, CD, DA and ∠ACB = 90.

We take a diagonal AC, where AC divides ABCD into two triangles ΔACB and ΔADC

Since ACB is right angled at C, we have

AC = 15 cm; AB = 17 cm

AB2 = AC2 + BC2

 

Area of right angled triangle ABC, say A1 is given by

, where,

Base = BC = 8 cm; Height = AC = 15 cm

Area of triangle ADC, say A2 having sides a, b, c and s as semi-perimeter is given by

, where

a = AD = 9 cm; b = DC = 12 cm; c = AC = 15 cm 

Area of quadrilateral ABCD, say A

A = Area of ACB + Area of ∆ADC

Perimeter of quadrilateral ABCD, say P

Question 14:

A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown in Fig. 12.28. The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each type of paper needed to make the hand fan.

Answer 14:

We have to find the area of each type of triangular strips needed for the fan.

There are 5 strips of each type having equal dimensions, so we will calculate the area of a single strip and then multiply it by 5 to ascertain the area of each type of strip needed. 

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

 

Therefore the area of a triangular strip, say A1 having sides 25 cm, 25 cm and 14 cm is given by: 

a = 25 cm ; b = 25 cm ; c = 14 cm

 

 

 

 

Area of each type of strip needed, say A.

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