# RD Sharma Solutions Class 9 Herons Formula Exercise 12.2

## RD Sharma Solutions Class 9 Chapter 12 Ex 12.2

Q1.Find the area of the quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Solution:

From the figure, in  △ABC, by Pythagoras theorem,

$AC^{2} = BC^{2}+AB^{2}$

25 = 9 + 16

Hence, △ABC is a right-angled triangle which is right-angled at point B.

Area of △ABC = $\frac{1}{2}\times AB\times BC$

= $\frac{1}{2}\times 3\times 4$

= 6 $cm^{2}$

Perimeter of △CAD  = 2s = AC + CD + DA

2s = 5 cm+ 4 cm+ 5 cm

2s   = 14 cm

s = 7 cm

From Heron’s Formula,

Area of the △CAD = $\sqrt{s\times (s-a)\times(s-b) \times(s-c) }$

= $\sqrt{7\times (7-5)\times(7-4) \times(7-5) }$

= $9.16cm^{2}$

= (6+9.16) $cm^{2}$

= 15.16$cm^{2}$. (Answer)

Q2. The sides of a quadrilateral field, taken in order are 26 m, 27 m, 7 m, 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.

Solution:

Given here, the length of the sides of the quadrilateral;

AB = 26 m, BC = 27 m, CD = 7 m, DA = 24 m

AC is the diagonal joined at A to C point.

From Pythagoras theorem;

$AC^{2} = AD^{2}+CD^{2}$

$AC^{2} = 14^{2}+7^{2}$

AC = 25 m

Now, area of △ABC

Perimeter of △ABC= 2s = AB + BC + CA

2s = 26 m + 27 m + 25 m

s = 39 m

From Heron’s Formula

The area of a triangle = $\sqrt{s\times (s-a)\times(s-b) \times(s-c) }$

= $\sqrt{39\times (39-26)\times (39-27) \times (39-25)}$

= $291.84m^{2}$

Thus, the area of a triangle is $291.84m^{2}$

Now, for area of △ADC, we need to find the perimeter first.

= 25 m + 24 m + 7 m

s = 28 m

From Heron’s Formula, we know,

The area of a △ADC = $\sqrt{s\times (s-a)\times(s-b) \times(s-c) }$

= $\sqrt{28\times (28-24)\times (28-7) \times (28-25)}$

= $84m^{2}$

Thus, the area of a △ADC is $84m^{2}$

Therefore, Area of rectangular field ABCD

= Area of △ABC + Area of △ADC

= 291.84 + 84

Q3. The sides of a quadrilateral, taken in order as 5 m, 12 m, 14 m, 15 m respectively, and the angle contained by first two sides is a right angle. Find its area.

Solution:

Given here, the sides of the quadrilateral are;

AB = 5 m, BC = 12 m, CD =14 m and DA = 15 m

Join the diagonal AC.

Now, area of △ABC = $\frac{1}{2}\times AB\times BC$

= $\frac{1}{2}\times 5\times 12$

= 30 m2

In △ABC, from Pythagoras theorem, we know

$AC^{2} = AB^{2}+BC^{2}$

AC = $\sqrt{5^{2}+12^{2}}$

AC = 13 m

2s = 15 m +14 m +13 m

s = 21 m

From Heron’s Formula,

Area of the △PSR = $\sqrt{s\times (s-a)\times(s-b) \times(s-c) }$

= $\sqrt{21\times (21-15)\times(21-14) \times(21-13) }$

= $84m^{2}$

Area of quadrilateral ABCD = Area of △ABC + Area of △ADC

= (30 + 84) $m^{2}$

= 114$m^{2}$ (Answer)

Q4. A park in the shape of a quadrilateral ABCD, has angle C = 900, AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m. How much area does it occupy?

Solution:

Given here, sides of a quadrilateral are AB = 9 m, BC = 12 m, CD = 5 m, DA = 8 m.

Join the points BD, forming diagonal to ABCD.

In △BCD , from Pythagoras theorem;

$BD^{2} = BC^{2}+CD^{2}$

$BD^{2} = 12^{2}+5^{2}$

BD = 13 m

Area of △BCD = $\frac{1}{2}\times BC\times CD$

= $\frac{1}{2}\times 12\times 5$

= 30 m2

Now, in △ABD,

Perimeter of △ABD = 2s = 9 m + 8m + 13m

s = 15 m

From Heron’s Formula,

Area of the △ABD = $\sqrt{s\times (s-a)\times(s-b) \times(s-c) }$

= $\sqrt{15\times (15-9)\times(15-8) \times(15-13) }$

= $35.49m^{2}$

Area of quad.ABCD = Area of △ABD + Area of △BCD

= (35.496 + 30) $m^{2}$

= 65.5$m^{2}$(Answer)

Q5. Two parallel sides of a trapezium are 60 m and 77 m and the other sides are 25 m and 26 m. Find the area of the trapezium?

Solution:

Given,

Two parallel sides of trapezium are AB = 77 m and CD = 60 m

The other two parallel sides of trapezium are BC = 26 m, AD = 25m

Join diagonals, AE and CF

DE is perpendicular to AB and also, CF is perpendicular to AB

Therefore, we get, DC = EF = 60 m

Say, AE = x

So, BF = 77 – 60 – x

BF = 17 – x

$DE^{2} = AD^{2}-AE^{2}$

$DE^{2} = 25^{2}-x^{2}$

In △BCF, from Pythagoras theorem,

$CF^{2} = BC^{2}-BF^{2}$

$CF^{2} = 26^{2}-(17-x)^{2}$

Here, DE = CF

So, $DE^{2} = CF^{2}$

$25^{2}-x^{2} = 26^{2}-(17-x)^{2}$

$25^{2}-x^{2} = 26^{2}-(17^{2}\;-\;34x\;+x^{2})$

$25^{2}-x^{2} = 26^{2}-17^{2}\;+\;34x\;+x^{2}$

$25^{2} = 26^{2}-17^{2}\;+\;34x$

x = 7

$DE^{2} = 25^{2}-x^{2}$

DE = $\sqrt{625-49}$

DE = 24 m

Area of trapezium = $\frac{1}{2}\;\times(60+77) \;\times24$

Area of trapezium = $1644\;m^{2}$ (Answer)

Q6. Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.

Solution:

Given here,

Perimeter of a rhombus = 80 m

Perimeter of a rhombus = $4 \times side$ = $4 \times a$

$4 \times a$ = 80 m

a = 20 m

Say, AC = 24 m

Therefore OA = $\frac{1}{2}\times AC$

OA = 12 m

In △AOB,

$OB^{2} = AB^{2}-OA^{2}$

$OB^{2} = 20^{2}-12^{2}$

OB = 16 m

Also, OB = OD because diagonal of rhombus bisect each other at 90 degrees.

Therefore, BD = 2 OB = 2 x 16 = 32 m

Area of rhombus = $\frac{1}{2}\times BD\times AC$

Area of rhombus = $\frac{1}{2}\times 32\times 24$

Area of rhombus = 384 $m^{2}$ (Answer)

Q7. A rhombus sheet, whose perimeter is 32 m and whose diagonal is 10 m long, is painted on both the sides at the rate of Rs 5 per meter square. Find the cost of painting.

Solution:

Given here,

Perimeter of a rhombus = 32 m

Perimeter of a rhombus = $4 \times side$

$4 \times side$ = 32 m

$4 \times a$ = 32 m

a = 8 m

Say, AC = 10 m

OA = $\frac{1}{2}\times AC$

OA = $\frac{1}{2}\times 10$

OA = 5 m

From Pythagoras theorem;

$OB^{2} = AB^{2} – OA^{2}$

$OB^{2} = 8^{2} – 5^{2}$

OB =$\sqrt{39}$ m

BD = 2 x OB

BD = 2$\sqrt{39}$ m

Area of the sheet = $\frac{1}{2}\times BD\times AC$

Area of the sheet = $\frac{1}{2}\times 2\sqrt{39}\times 10$

Therefore, cost of printing on both sides of the sheet, at the rate of Rs. 5 per m2

= Rs 2 x $10\sqrt{39}$x5

Q8. Find the area of the quadrilateral ABCD in which AD = 24 cm, angle BAD = $90^{\circ}$ and BCD forms an equilateral  triangle whose each side is equal to 26 cm. [Take $\sqrt{3}$ = 1.73]

Solution:

BCD is an equilateral triangle and the sides BC = CD = BD = 26 cm

In △BAD, by using Pythagoras theorem,

$BA^{2} = BD^{2}-AD^{2}$

$BA^{2} = 26^{2}+24^{2}$

BA = $\sqrt{100}$

BA = 10 cm

Area of the △BAD = $\frac{1}{2}\times BA\times AD$

Area of the △BAD = $\frac{1}{2}\times 10\times 24$

Area of the △BAD = 120 $cm^{2}$

Area of the equilateral triangle = $\frac{\sqrt{3}}{4}\times side$

Since, △QRS and △BCD are equilateral triangle, then,

Area of the △QRS = $\frac{\sqrt{3}}{4}\times 26$

Area of the △BCD = 292.37 $cm^{2}$

Therefore, the area of quadrilateral ABCD = area of △BAD + area of the △BCD

The area of quadrilateral ABCD = 120 + 292.37

= 412.37 $cm^{2}$ (Answer)

Q9.  Find the area of quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and the diagonal BD = 20 cm.

Solution:

Given here;

AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm, and

the diagonal, BD = 20 cm.

Now, for area of △ABD, we have to find the perimeter first;

Perimeter of △ABD, 2s = AB + BD + DA

2s = 34 cm + 42 cm + 20 cm

s = 48 cm

From Heron’s Formula,

Area of the △ABD = $\sqrt{s\times (s-a)\times(s-b) \times(s-c) }$

= $\sqrt{48\times (48-42)\times(48-20) \times(48-34)}$

= $336cm^{2}$

Now, in case of area of △BCD;

Perimeter of △BCD 2s = BC + CD + BD

2s = 29cm + 21cm + 20cm

s = 35 cm

From Heron’s Formula,

Area of the △BCD = $\sqrt{s\times (s-a)\times(s-b) \times(s-c) }$

= $\sqrt{35\times (14)\times (6)\times (15)}$

= $210cm^{2}$

Therefore, Area of quadrilateral ABCD = Area of △ABD + Area of △BCD

Area of quadrilateral ABCD = 336 + 210

Area of quadrilateral ABCD = 546 $cm^{2}$ (Answer)

Q10. Find the perimeter and the area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, AC = 15 cm and angle ACB = $90^{\circ}$.

Solution:

Given here, the sides of the quadrilateral ABCD are;

AB = 17 cm, AD = 9 cm, CD = 12 cm, AC = 15 cm and

∠ACB = 90°

From Pythagoras theorem, we know,

$BC^{2} = AB^{2}-AC^{2}$

$BC^{2} = 17^{2}-15^{2}$

BC = 8 cm

Now, area of △ABC = $\frac{1}{2}\times AC\times BC$

area of △BC = $\frac{1}{2}\times 8\times 15$

area of △ABC = 60 $cm^{2}$

Now, in △ACD,

Perimeter of △ACD 2s = AC + CD + AD

2s = 15 + 12 +9

s = 18 cm

From Heron’s Formula,

Area of the △ACD = $\sqrt{s\times (s-a)\times(s-b) \times(s-c) }$

= $\sqrt{18\times (3)\times(6) \times(9) }$

= $54cm^{2}$

Area of quadrilateral ABCD = Area of △ABC + Area of △ACD

Area of quadrilateral ABCD = 60 $cm^{2}$ + 54$cm^{2}$

Area of quadrilateral ABCD = 114 $cm^{2}$ (Answer)

Q11. The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of parallelogram.

Solution:

Given,

Given in the question, the adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm.

Area of the parallelogram = Area of △ADC + Area of △ABC

[Note: Diagonal of a parallelogram divides into two congruent triangles.]

Therefore,

Area of the parallelogram = 2 × (Area of △ABC)

Now, in △ABC;

Perimeter = 2s = AB + BC + CA

2s = 34 cm + 20 cm + 42 cm

s = 48 cm

From Heron’s Formula,

Area of the △ABC = $\sqrt{s\times (s-a)\times(s-b) \times(s-c) }$

= $\sqrt{48\times (14)\times (28) \times (6) }$

= 336 sq.cm.

Therefore, area of parallelogram ABCD = 2 × (Area of △ABC)

Area of parallelogram = 2 × 336 sq.cm.

Area of parallelogram ABCD = 672 sq.cm.  (Answer)

Q12. Find the area of the blades of the magnetic compass shown in the figure given below.

Solution:

Area of the blades of magnetic compass = Area of △ADB + Area of △CDB

Perimeter = 2s = AD + DB + BA

2s = 5 cm + 1 cm + 5 cm

s = 5.5 cm

From Heron’s Formula,

Area of the △DEF = $\sqrt{s\times (s-a)\times(s-b) \times(s-c) }$

= $\sqrt{5.5\times (0.5)\times (4.5) \times (0.5)}$

= $2.49cm^{2}$

Also, area of △ADB = Area of △CDB

Therefore, area of the blades of the magnetic compass = 2  × area of △ADB

= 2 × 2.49

Q13. A hand fan is made by sticking 10 equal size triangular strips of two different types of paper as shown in the figure. 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.

Solution: Given here, the sides of △AOB are;

AO = 25 cm

OB = 25 cm

BA = 14 cm

Area of each strip = Area of △AOB

Now, in △AOB;

Perimeter = AO + OB + BA

2s = 25 cm +25 cm + 14 cm

s = 32 cm

From Heron’s Formula,

Area of the △AOB = $\sqrt{s\times (s-a)\times(s-b) \times(s-c) }$

= $\sqrt{32\times (7)\times (4) \times(18)}$

= 168 sq.cm.

Therefore,

area of each type of paper needed to make a fan = 5 × Area of △AOB

= 5 × 168

Q14. 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 a parallelogram.

Solution: Given here; the sides of the △DCE are;

DC = 15 cm,

CE = 13 cm,

ED = 14 cm

Let the ‘h’ be the height of parallelogram ABCD.

Now, in △DCE;

Perimeter = DC + CE + ED

2s  = 15 cm + 13 cm + 14 cm

s = 21 cm

From Heron’s Formula,

Area of the △DCE = $\sqrt{s\times (s-a)\times(s-b) \times(s-c) }$

= $\sqrt{21\times (7)\times (8) \times (6)}$

= 84 sq.cm.

Also, area of △DCE = Area of parallelogram ABCD

Since, Area of //gmABCD = base × height

Therefore,

24 × h  = 84 sq.cm.