RD Sharma Solutions Class 9 Herons Formula Ex 12.1 RD Sharma Solutions Class 9 Chapter 12 Ex 12.1 Q1. Find the area o f a triangle whose sides are respectively 150 cm, 120 cm and200 cm. Solution: Let the sides of the given triangle be a, b, c respectively. So given, a = 150 cm b = 120 cm c = 200 cm By using Heron's Formula The area of the triangle = y/s x (s — a) x (s — b) x (s — c) Semi perimeter of a triangle = s 2s = a + b + c (a+5+c) S 2 (150+200+120) S 2 s = 235 cm Therefore, area of the triangle = y/s x (s — a) x (s — b) x (s — c) = -^235 x (235 - 150) x (235 - 200) x (235 - 120) = 8966.56cm2 Q2. Find the area o f a diangle whose sides are respectively 9 cm, 12 cm and 15 cm. Solution: Let the sides of the given triangle be a, b, c respectively. So given, a = 9 cm b = 12 cm c = 15 cm By using Heron's Formula The area of the triangle = -\/s x (s — a) x (s — b) x (s — c) Semi perimeter of a triangle = s 2s = a + b + c (o+5+c) S = ^ — (9+12+15) S = --------2-------- s = 18 cm Therefore, area of the triangle = y/sx (s — a) x (s — b) x (s — c) = y/18 x (18 - 9) x (18 - 12) x (18 - 15) = 54cm2 Q3. Find the area o f a diangle two sides o f which are 18 cm and 10 cm and the perimeter is 42cm. Solution: Whenever we are given the measurements of all sides of a triangle, we basically look for Heron's formula to find out the area of the triangle. If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by: A = i^/s x (s — a) x (s — b) x (s — c) Where, s = (o+5+c) 2
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RD Sharma Solutions Class 9 Herons Formula Ex 12.1
RD Sharma Solutions Class 9 Chapter 12 Ex 12.1
Q 1. Find the area o f a triangle whose sides are respectively 150 cm, 120 cm and200 cm.
Solution:
Let the sides of the given triangle be a, b, c respectively.
So given,
a = 150 cm
b = 120 cm
c = 200 cm
By using Heron's Formula
The area of the triangle = y /s x (s — a) x (s — b) x (s — c)
Semi perimeter of a triangle = s
2s = a + b + c
(a+5+c)S 2
(150+200+120)S 2
s = 235 cm
Therefore, area of the triangle = y /s x (s — a) x (s — b) x (s — c)
= -^235 x (235 - 150) x (235 - 200) x (235 - 120)= 8966.56cm2
Q2. Find the area o f a diangle whose sides are respectively 9 cm, 12 cm and 15 cm.
Solution:
Let the sides of the given triangle be a, b, c respectively.
So given,
a = 9 cm
b = 12 cm
c = 15 cm
By using Heron's Formula
The area of the triangle = -\/s x (s — a) x (s — b) x (s — c)
Semi perimeter of a triangle = s
2s = a + b + c
(o+5+c)S = ^ —
(9+12+15)S = --------2--------
s = 18 cm
Therefore, area of the triangle = y / s x (s — a) x (s — b) x (s — c)
= y/18 x (18 - 9) x (18 - 12) x (18 - 15)= 54cm2
Q3. Find the area o f a diangle two sides o f which are 18 cm and 10 cm and the perim eter is 42cm .
Solution:
Whenever we are given the measurements of all sides of a triangle, we basically look for Heron's formula to find out the area of the triangle.
If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by:
A = i^/s x (s — a) x (s — b) x (s — c)
Where, s =(o+5+c)
2
We are given:
a = 18 cm
b = 10 cm, and perimeter = 42 cm
We know that perimeter = 2s,
So, 2s = 42
Therefore, s = 21 cm
(a+b+c)We know that, s = — —̂
_ (18+10+c)Z ” 2
42 = 28 + c
c = 14 cm
So the area of the triangle is:
a=V2T ir(2r^n^r(2nn^r(2r^i4)A = ^ /2 1 x (3) x (11) x (7)
A = -v/4851
A = 21 -v /ll cm0
Q4. In a triangle ABC, AB = 15cm, BC = 13cm and AC = 14cm. Find the area o f triangle ABC and hence Its altitude on AC.
Solution:
Let the sides of the given triangle be AB = a, BC = b, AC = c respectively.
So given,
a = 15 cm
b = 13 cm
c = 14 cm
By using Heron's Formula
The area of the triangle = ^ s x (s — a) x (s — b) x (s — c)
Semi perimeter of a triangle^ 2s
2s = a + b + c
(a+b+c)s = Hr^(15+13+14)
S= 2
s = 21 cm
Therefore, area of the triangle = y /s x (s — a) x (s — b) x (s — c)
= -^21 x (21 - 15) x (21 - 13) x (21 - 14)
= 84cm2
BE is a perpendicular on AC
Now, area of triangle = 84cm2
\ x BE x AC = 84cm2
BE = 12cm
The length of BE is 12 cm
Q 5. The perimeter o f a triangular field Is 540 m and Its sides are In die ratio 25:17:12. Find the area o f Wangle.
Solution:
Let the sides of the given triangle be a = 25x, b = 17x, c = 12x respectively.
So,
a = 25x cm
b = 17x cm
c = 12x cm
Given Perimeter = 540 cm
2s = a + b + c
a + b + c = 540 cm
25x + 17x + 12x = 540 cm
54x = 540 cm
x = 10 cm
Therefore, the sides of a triangle are
a = 250 cm
b = 170 cm
c = 120 cm
, , „ . . (o+6+c)Now, Semi perimeter s = — -—
_ 540 ” 2
= 270 cm
By using Heron's Formula
The area of the triangle = y /s x (s — a) x (s — b) x (s — c)
= y/270 x (270 - 250) x (270 - 170) x (270 - 120)
= 9000cm2
Therefore, the area of the triangle is 9000cm2
Q6. The perimeter o f a triangle Is 300m. I f its sides are In the ratio o f 3 :5 :7 . Find die area o f the triangle.
Solution:
Given the perimeter of a triangle is 300 m and the sides are in a ratio of 3: 5:7
Let the sides a, b, c of a triangle be 3x, 5x, 7x respectively
So, the perimeter = 2s = a + b + c
200 = a + b + c
300 = 3x + 5x + 7x
300 = 15x
Therefore, x = 20 m
So, the respective sides are
a = 60 m
b = 100 m
c = 140 m
Now, semi perimeter s = °+ +̂c
_ 60+100+140 - 2
= 150 m
By using Heron's Formula
The area of a triangle = y /s x (s — a) x (s — b) x (s — c)
= y/150 x (150 - 60) x (150 - 100) x (150 - 140)
= 1500V3m2
Thus, the area of a triangle is 1500-\/3m2
Q7. The perimeter o f a triangular field is 240 dm. I f two o f its sides are 78 dm and 50 dm, find the length o f the perpendicular on the side o f length 50 dm from the opposite vertex.
Solution:
Given,
In a triangle ABC, a = 78 dm = AB, b = 50 dm = BC
Now, Perimeter = 240 dm
Then, AB + BC + AC = 240 dm
78 + 50 + AC = 240
AC = 240-(78+50)
AC = 112 dm = c
Now, 2s = a + b + c
2s = 78+ 50+ 112
s = 120 dm
Area of a triangle ABC = y/s x (s — a) x (s — b) x (s — c)
= ^120 x (120 - 78) x (120 - 50) x (120 - 112)= 1680dm2Let AD be a perpendicular on BC
Area of the triangle ABC = ^ x AD x BC
} x AD x B C = 1680dm2
AD = 67.2 dm
Q8. A triangle has sides 35 cm, 54 cm, 61 cm long. Find Its area. Also, find die smallest o f Its altitudes?
Solution:
Given,
The sides of the triangle are
a = 35 cm
b = 54 cm
c = 61 cm
Perimeter 2s = a + b + c
2s = 35 + 54 + 61 cm
Semi perimeter s = 75 cm
By using Heron's Formula,
Area of the triangle = y /s x (s — a) x (s — b) x (s — c)
= y/75 x (75 - 35) x (75 - 54) x (75 - 61)= 939.14cm2
The altitude will be smallest provided the side corresponding to this altitude is longest.
The longest side = 61 cm
Area of the triangle = | x A x 61
} x h x 61 = 939.14cm2
h = 30.79cm
Hence the length of the smallest altitude is 30.79cm
Q9. The lengths o f the sides o f a triangle are In a ratio o f 3 : 4 : 5 and Its perimeter Is 144 on. Find the area o f the triangle and the height corresponding to die longest side?
Solution:
Given the perimeter of a triangle is 160m and the sides are in a ratio of 3 : 4 : 5
Let the sides a, b, c of a triangle be 3x, 4x, 5x respectively
So, the perimeter = 2s = a + b + c
144 = a + b + c
144 = 3x+ 4x+ 5x
Therefore, x = 12cm
So, the respective sides are
a = 36cm
b = 48cm
c = 60cm
Now, semi perimeter s = — -—
_ 36+48+60 " 2
= 72 cm
By using Heron's Formula
The area of a triangle = ^ /s x (s — a) x (s — b) x (s — c)
= x (72 - 36) x (72 - 48) x (72 - 60)
= 864cm2
Thus, the area of a triangle is 864cm2
The altitude will be smallest provided the side corresponding to this altitude is longest.
The longest side = 60 cm
Area of the triangle = x h x 60
j x h x 60 = 864cm2
h = 28.8 cm
Hence the length of the smallest altitude is 28.8 cm
Q10. The perimeter o f an isosceles triangle is 42 cm and its base is | times each o f the equal side. Find the length o f each o f the triangle, area
o f the triangle and the height o f the triangle.
Solution:
Let 'x' be the length of two equal sides.
Therefore the base = x x
Let the sides a, b, c of a triangle be j x i , x and x respectively
So, the perimeter = 2s = a + b + c
42 = a + b + c
42 = | x i t x t x
Therefore, x = 12 cm
So, the respective sides are
a = 12 cm
b = 12 cm
c = 18 cm
Now, semi perimeter s = °+ +̂c
_ 12+12+18 " 2
= 21 cm
By using Heron's Formula
The area of a triangle = ■s/s x (s — a) x (s — b) x (s — c)
= s j 2 \ x (21 - 12) x (21 - 12) x (21 - 18)
= 71.42cm2
Thus, the area of a triangle is 71.42cm2
The altitude will be smallest provided the side corresponding to this altitude is longest.
The longest side = 18 cm
Area of the triangle = x h x 18
} x h x 18 = 71.42cm2
h = 7.93 cm
Hence the length of the smallest altitude is 7.93 cm
Q11. Find die area o f the shaded region in fig. below
Solution:
Area of the shaded region = Area o f A ABC — Area o f AADB Now in triangle ADB
AB2 = AD2 + BD2....... (i)
Given, AD = 12 cm, BD =16 cm
Substituting the value of AD and BD in eq (i), we get
AB2 = 122 + 162= 400cm2
AB = 20 cm
Now, area of a triangle = x AD x BD
= 96cm2Now in triangle ABC,
s = | x (AB + BC + CA)
= \ x (52 + 48 + 20)
= 60 cm
By using Heron's Formula
The area of a triangle = x (s — a) x (s — b) x (s — c)
= ^60 x (60 - 20) x (60 - 48) x (60 - 52)= 480cm2Thus, the area of a triangle is 480cm2Area of shaded region = Area of triangle ABC - Area of triangle ADB
= (480 - 96) cm2
= 384 cm2
Area of shaded region = 384 cm2
RD Sharma Solutions Class 9 Herons Formula Ex 12.2
RD Sharma Solutions Class 9 Chapter 12 Ex 12.2
Q1 .Find the area o f the quadrilateral ABCD in which AB = 3 cm, BC = 4cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Solution:
For triangle ABC
AC2 = BC2 + AB2 25 = 9 + 16
So, triangle ABC is a right angle triangle right angled at point R
Area of triangle ABC = x AB x BC= \ x 3 x 4
= 6 cm 2
From triangle CAD
Perimeter = 2s = AC + CD + DA
2s = 5 cm+ 4 cm+ 5 cm
2s = 14 cm
s = 7 cm
By using Fleron's Formula
Area of the triangle CAD = -\/s x (s — a) x (s — b) x (s — c)
= y /7 x (7 - 5) x (7 - 4) x (7 - 5)
= 9.16cm2
Area of ABCD = Area of ABC + Area of CAD
= (6+9.16) cm 2
= 15.16 cm 2
Q2. The sides o f 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:
D 7 m C
Here the length of the sides of the quadrilateral is given as
AB = 26 m, BC = 27 m, CD = 7 m, DA = 24 m
Diagonal AC is joined.
Now, in triangle ADC
By applying Pythagoras theorem
AC2 = AD2 + CD2 AC2 = 142 + 72
AC = 25 m
Now area of triangle ABC
Perimeter = 2s = AB + BC + CA
2s = 26 m + 27 m + 25 m
s = 39 m
By using Heron's Formula
The area of a triangle = \ / s x (a — a) x (a — b) x (a — c)
= ^/39 x (39 - 26) x (39 - 27) x (39 - 25)
= 291.84m2
Thus, the area of a triangle is 291.84m2
Now for area of triangle ADC
Perimeter = 2s = AD + CD + AC
= 25m + 24m + 7m
s = 28 m
By using Heron's Formula
The area of a triangle = ^ a x (a — a) x (a — b) x (a — c)
= x/28x“(28^r 24y^ 2 8 ^ 7 7 x 1 2 8 ^ 2 5 )
= 84m2
Thus, the area of a triangle is 84m2
Therefore, Area of rectangular field ABCD
= Area of triangle ABC + Area of triangle ADC
= 291.84 + 84
= 375.8 m2
Q3. The sides o f a quadrilateral, taken in order as 5 m, 12m, 14m, 15 m respectively, and the angle contained by first two sides is a right angle. Find its area.
Given that the sides of the quadrilateral are
AB = 5 m, BC = 12 m, CD =14 m and DA = 15 m
Join AC
Now area of triangle ABC = | x AB x BC= \ x 5 x 12
= 30 m2
In triangle ABC, By applying Pythagoras theorem
AC2 = AB2 + BC2
AC = \ / 5 2 + 122
AC = 13 m
Now in triangle ADC,
Perimeter = 2s = AD + DC + AC
2s = 15 m +14 m +13 m
s = 21 m
By using Heron's Formula,
Area of the triangle PSR = ^ /s x (s — a) x (s — b) x (s — c)
= y/21 x (21 - 15) x (21 - 14) x (21 - 13)
= 84m2
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC
= (30 + 84) m 2
= 114 m 2Q4. A park in the shape o f a quadrilateral ABCD, has angle C = 9(fi, AB = 9 m,BC= 12 m,CD = 5 m, AD = 8m. How much area does it occupy?
Solution:
Given sides of a quadrilateral are AB = 9 m, BC = 12 m, CD = 5 m, DA = 8 m.
Let us join BD
In triangle BCD, apply Pythagoras theorem
B D 2 = B C 2 + C D 2
B D 2 = 122 + 52
BD = 13 m
Area of triangle BCD = ^ x B C x C D
= \ x 12 x 5
= 30 m2
Now, in triangle ABD
Perimeter = 2s = 9m + 8m + 13m
s = 15 m
By using Heron's Formula,
Area of the triangle ABD = y /s x (s — a) x (s — b) x (s — c)
= ^/15 x (15 - 9) x (15 - 8) x (15 - 13)
= 35.49m2
Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD
= (35.496 + 30) m 2
= 65.5m2.
Q5. Two parallel sides o f a trapezium are 60 m and 77 m and the other sides are 25 m and 26 m. Find the area o f the trapezium?
Solution:
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 AE and CF
DE is perpendicular to AB and also, CF is perpendicular to AB
Therefore, DC = EF = 60 m
Let AE = x
So, BF = 77 - 60 - x
BF = 17 - x
In triangle ADE,
By using Pythagoras theorem,
DE2 = AD2 - AE2 DE2 = 252 - a;2
In triangle BCF,
By using Pythagoras theorem,
CF2 = BC2 - BF2
CF2 = 262 - (17 - a;)2
Here, DE = CF
So, DE2 = CF2252 - a:2 = 262 - (17 - a;)2
252 - a:2 = 262 - (172 - 34a: + a:2)
252 - a:2 = 262 - 172 + 34a: + a:2
252 = 262 - 172 + 34a:
x = 7
DE2 = 252 - a:2
DE = -y/625^-49
DE = 24 m
Area of trapezium = ^ x (60 + 77) x 24
Area of trapezium = 1644 m 2
Q6. Find the area o f a rhombus whose perimeter is 80 m and one o f whose diagonal is 24 m.
Solution:
Given,
Perimeter of a rhombus = 80 m
As we know.
Perimeter of a rhombus = 4 x side = 4 x a
4 x a = 80 m
a = 20 m
There fore0A= | x AC OA = 12 m
In triangle AOB
OB2 = AB2 - OA2 OB2 = 202 - 122
OB = 16 m
Also, OB = OD because diagonal of rhombus bisect each other at 90°
Therefore, BD = 2 OB = 2x16 = 32 m
Area of rhombus = x BD x AC Area of rhombus = -| x 32 x 24
Area of rhombus = 384 m 2
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 o f Rs 5 per meter square. Find the cost o f painting.
Solution:
Given that.
Perimeter of a rhombus = 32 m
We know that.
Perimeter of a rhombus = 4 x side
4 x side = 32 m
4 x a = 32 m
a = 8 m
Let AC = 10 m
0A = } x ACOA = | x 10
OA = 5 m
By using Pythagoras theorem
OB2 = AB2- 0 A2 OB2 = 82-52
OB =v /39m
BD = 2 x OB
BD = 2-\/39 m
Area of the sheet = | x BD x AC Area of the sheet = j x 2y/39 x 10
Therefore, cost of printing on both sides at the rate of Rs. 5 per m2
= Rs 2x10^39x5
= Rs. 625
Q8. Find the area o f the quadrilateral ABCD in which AD = 24 cm, angle BAD = 90° and BCD forms an equilateral triangle whose each side is equal to 26 cm. [Take \ /3 = 1.73]
Solution:/p>
Given that, in a quadrilateral ABCD in which AD = 24 cm.
Angle BAD = 90°
BCD is an equilateral triangle and the sides BC = CD = BD = 26 cm
In triangle BAD, by applying Pythagoras theorem,
BA2 = BD2 - AD2 BA2 = 262 + 242
BA = •x/lOO
BA = 10 cm
Area of the triangle BAD = ^ x BA x AD Area of the triangle BAD = -| x 10 x 24
Area of the triangle BAD = 120 cm 2
Area of the equilateral triangle = — x side/n
Area of the equilateral triangle ORS = - j - x 26
Area of the equilateral triangle BCD = 292.37 cm 2
Therefore, the area of quadrilateral ABCD = area of triangle BAD + area of the triangle BCD
The area of quadrilateral ABCD = 120 + 292.37
= 412.37 cm2
Q9. Find the area o f quadrilateral ABCD in which AB = 42 cm, BC=21 cm, CD =29 cm, DA =34 cm and the diagonal BD =20 cm.
Solution:
Given
AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm, and the diagonal
BD = 20 cm.
Perimeter of triangle ABD 2s = AB + BD + DA
2s = 34 cm + 42 cm + 20 cm
s = 48 cm/p>
By using Heron's Formula,
Area of the triangle ABD = \ / s x (s — a) x (s — b) x (s — c)
= y^48 x (48 - 42) x (48 - 20) x (48 - 34)
= 336cm2
Now, for the area of triangle BCD
Perimeter of triangle BCD 2s = BC + CD + BD
2s = 29cm + 21 cm + 20cm
s = 35 cm
By using Heron's Formula,
Area of the triangle BCD = y /s x (s — a) x (s — b) x (s — c)
= ^/35 x (14) x (6) x (15)
= 210cm2Therefore, Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD
Area of quadrilateral ABCD = 336 + 210
Area of quadrilateral ABCD = 546 cm 2
Q10. Find the perimeter and the area o f the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, AC = 15 cm and angle ACB =90°.
Solution:
Given are the sides of the quadrilateral ABCD in which
AB = 17 cm, AD = 9 cm, CD = 12 cm, AC = 15 cm and an angle ACB = 90°
By using Pythagoras theorem
BC2 = AB2 - AC2BC2 = 172 - 152
BC = 8 cm
Now, area of triangle ABC = x AC x BC area of triangle ABC = -| x 8 x 15
area of triangle ABC = 60 cm2
Now, for the area of triangle ACD
Perimeter of triangle ACD 2s = AC + CD + AD
2s = 15+ 12+9
s = 18 cm
By using Heron's Formula,
Area of the triangle ACD = y 's x (s — a) x (s — b) x (s — c)
= 0 8 x (3) x (6) x (9)
= 54cm2
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 60 cm2 + 54cm2
Area of quadrilateral ABCD = 114 cm 2
Q77. The adjacent sides o f a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal A C measures 42 cm. Find the area o f parallelogram.
Solution:
Given,
The adjacent sides of a parallelogram ABCD measures 34 cm and 20 cm, and the diagonal AC measures 42 cm.
Area of the parallelogram = Area of triangle ADC + Area of triangle ABC
Note: Diagonal of a parallelogram divides into two congruent triangles
Therefore,
Area of the parallelogram = 2 x (Area of triangle ABC)
Now, for area of triangle ABC
Perimeter = 2s = AB + BC + CA
2s = 34 cm + 20 cm + 42 cm
s = 48 cm
By using Heron's Formula,
Area of the triangle ABC = ^ /s x (s — a) x (s — b) x (s — c)
= ^/48 x (14) x (28) x (6)
= 336cm2
Therefore, area of parallelogram ABCD = 2 x (Area of triangle ABC)
Area of parallelogram = 2 x 336cm2
Area of parallelogram ABCD = 672 cm2
Q12. Find the area o f the blades o f the magnetic compass shown in figure given below.
Area of the blades of magnetic compass = Area of triangle ADB + Area of triangle CDB
Now, for the area of triangle ADB
Perimeter = 2s = AD + DB + BA
2s = 5 cm + 1 cm + 5 cm
s = 5.5 cm
By using Heron's Formula,
Area of the triangle DEF = y 's x (s — a) x (s — b) x (s — c)
= V 5 .5 x (0.5) x (4.5) x (0.5)
= 2.49cm2
Also, area of triangle ADB = Area of triangle CDB
Therefore area of the blades of the magnetic compass = 2 x area of triangle ADB
Area of the blades of the magnetic compass = 2 X2.49
Area of the blades of the magnetic compass = 4.98 cm 2
Q13.A hand fan is made by sticking 10 equal size triangular strips o f two different types o f paper as shown in the figure. The dimensions o f equal strips are 25 cm, 25 cm and 14 cm. Find the area o f each type o f paper needed to make the hand fan.
Given that.
The sides of AOB
AO = 25 cm
OB = 25 cm
BA = 14 cm
Area of each strip = Area of triangle AOB
Now, for the area of triangle AOB
Perimeter = AO + OB + BA
2s = 25 cm +25 cm + 14 cm
s = 32 cm
By using Heron's Formula,
Area of the triangle AOB = y /s x (s — a) x (s — b) x (s — c)
= v/32V(7yiT(4y^r(i8)
= 168cm2
Also, area of each type of paper needed to make a fan = 5 x Area of triangle AOB
Area of each type of paper needed to make a fan = 5 x 168 cm 2
Area of each type of paper needed to make a fan = 840cm2
Q14. A triangle and a parallelogram have the same base and the same area. I f the sides o f the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height o f a parallelogram.
Solution:
The sides of the triangle DCE are
DC = 15 cm,
CE = 13 cm,
ED = 14 cm
Let the h be the height of parallelogram ABCD
Now, for the area of triangle DCE
Perimeter = DC + CE + ED
2s = 15 cm + 13 cm + 14 cm
s = 21 cm
By using Heron's Formula,
Area of the triangle AOB = ^ s x (s — a) x (s — b) x (s — c)= V/2TlT(7)V(8)^r{6)= 84cm2
Also, area of triangle DCE = Area of parallelogram ABCD=>84cm2