RS Aggarwal Solutions: Areas of Triangles and Quadrilaterals | Mathematics (Maths) Class 9 PDF Download

 Page 1


Q u e s t i o n : 1
Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.
S o l u t i o n :
We have:
Base = 24 cm
Height = 14.5 cm
Now,
Area of triangle =
1
2
×Base ×Height =
1
2
×24 ×14. 5 = 174 cm
2
Q u e s t i o n : 2
The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height.
S o l u t i o n :
Let the height of the triangle be h m.
? Base = 3h m
Now,
Area of the triangle = 
Total Cost
Rate
=
783
58
= 13. 5 ha = 135000 m
2
We have:
Area of triangle = 135000 m
2
?
1
2
×Base ×Height = 135000 ?
1
2
×3h ×h = 135000 ? h
2
=
135000×2
3
? h
2
= 90000 ? h = 300 m
Thus, we have:
Height = h = 300 m
Base = 3h = 900 m
Q u e s t i o n : 3
Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.
S o l u t i o n :
Let: a = 42 cm, b = 34 cm and c = 20 cm ? s = 
a+b+c
2
=
42+34+20
2
= 48 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 48(48 -42)(48 -34)(48 -20) = v 48 ×6 ×14 ×28
We know that the longest side is 42 cm.
Thus, we can find out the height of the triangle corresponding to 42 cm.
We have:
Area of triangle = 336 cm
2
?
1
2
×Base ×Height = 336 ? Height = 
336×2
42
= 16 cm
Q u e s t i o n : 4
Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.
S o l u t i o n :
Let: a = 18 cm, b = 24 cm and c = 30 cm ? s = 
a+b+c
2
=
18+24+30
2
= 36 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 36(36 -18)(36 -24)(36 -30) = v 36 ×18 ×12 ×6
We know that the smallest side is 18 cm.
Thus, we can find out the altitude of the triangle corresponding to 18 cm.
We have:
Area of triangle = 216 cm
2
?
1
2
×Base ×Height = 216 ? Height = 
216×2
18
= 24 cm
Q u e s t i o n : 5
Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length. Find the height corresponding to the longest side.
S o l u t i o n :
Let: a = 91 m, b = 98 m and c = 105 m ? s = 
a+b+c
2
=
91+98+105
2
= 147 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 147(147 -91)(147 -98)(147 -105) = v 147 ×56 ×
We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.
Area of triangle = 4116 m
2
?
1
2
×Base ×Height = 4116 ? Height = 
4116×2
105
= 78. 4 m
Q u e s t i o n : 6
The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 m. Find the area of the triangle.
S o l u t i o n :
Let the sides of the triangle be 5x m, 12x m and 13x m.
We know:
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, we obtain the sides of the triangle.
5 ×
5 = 25 m
12 ×
5 = 60 m
13 ×
5 = 65 m
Now,
Let: a = 25 m, b = 60 m and c = 65 m ? s = 
150
2
= 75 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 75(75 -25)(75 -60)(75 -65) = v 75 ×50 ×15 ×10 = v 15 ×5 ×5 ×10
Page 2


Q u e s t i o n : 1
Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.
S o l u t i o n :
We have:
Base = 24 cm
Height = 14.5 cm
Now,
Area of triangle =
1
2
×Base ×Height =
1
2
×24 ×14. 5 = 174 cm
2
Q u e s t i o n : 2
The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height.
S o l u t i o n :
Let the height of the triangle be h m.
? Base = 3h m
Now,
Area of the triangle = 
Total Cost
Rate
=
783
58
= 13. 5 ha = 135000 m
2
We have:
Area of triangle = 135000 m
2
?
1
2
×Base ×Height = 135000 ?
1
2
×3h ×h = 135000 ? h
2
=
135000×2
3
? h
2
= 90000 ? h = 300 m
Thus, we have:
Height = h = 300 m
Base = 3h = 900 m
Q u e s t i o n : 3
Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.
S o l u t i o n :
Let: a = 42 cm, b = 34 cm and c = 20 cm ? s = 
a+b+c
2
=
42+34+20
2
= 48 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 48(48 -42)(48 -34)(48 -20) = v 48 ×6 ×14 ×28
We know that the longest side is 42 cm.
Thus, we can find out the height of the triangle corresponding to 42 cm.
We have:
Area of triangle = 336 cm
2
?
1
2
×Base ×Height = 336 ? Height = 
336×2
42
= 16 cm
Q u e s t i o n : 4
Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.
S o l u t i o n :
Let: a = 18 cm, b = 24 cm and c = 30 cm ? s = 
a+b+c
2
=
18+24+30
2
= 36 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 36(36 -18)(36 -24)(36 -30) = v 36 ×18 ×12 ×6
We know that the smallest side is 18 cm.
Thus, we can find out the altitude of the triangle corresponding to 18 cm.
We have:
Area of triangle = 216 cm
2
?
1
2
×Base ×Height = 216 ? Height = 
216×2
18
= 24 cm
Q u e s t i o n : 5
Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length. Find the height corresponding to the longest side.
S o l u t i o n :
Let: a = 91 m, b = 98 m and c = 105 m ? s = 
a+b+c
2
=
91+98+105
2
= 147 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 147(147 -91)(147 -98)(147 -105) = v 147 ×56 ×
We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.
Area of triangle = 4116 m
2
?
1
2
×Base ×Height = 4116 ? Height = 
4116×2
105
= 78. 4 m
Q u e s t i o n : 6
The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 m. Find the area of the triangle.
S o l u t i o n :
Let the sides of the triangle be 5x m, 12x m and 13x m.
We know:
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, we obtain the sides of the triangle.
5 ×
5 = 25 m
12 ×
5 = 60 m
13 ×
5 = 65 m
Now,
Let: a = 25 m, b = 60 m and c = 65 m ? s = 
150
2
= 75 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 75(75 -25)(75 -60)(75 -65) = v 75 ×50 ×15 ×10 = v 15 ×5 ×5 ×10
Q u e s t i o n : 7
The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the field. Also, find the cost of ploughing the field at  5 per m
2
.
S o l u t i o n :
Let the sides of the triangle be 25x m, 17x m and 12x m.
We know:
Perimeter = Sum of all sides
or, 540 = 25x + 17x + 12x
or, 54x = 540
or, x = 10
Thus, we obtain the sides of the triangle.
25 ×
10 = 250 m
17 ×
10 = 170 m
12 ×
10 = 120 m
Now,
Let: a = 250 m, b = 170 m and c = 120 m ? s = 
540
2
= 270 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 270(270 -250)(270 -170)(270 -120) = v 270 ×20 ×100 ×150
Cost of ploughing 1 m
2
 field = Rs 5
Cost of ploughing 9000 m
2
 field = 5 ×9000 = Rs 45000
.
Q u e s t i o n : 8
Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find i
the area of the field and ii
the length of the perpendicular from the opposite vertex on the side measuring 154 m.
S o l u t i o n :
i Let: a = 85 m and b = 154 m Given: Perimeter = 324 mor, a +b +c = 324 ? c = 324 -85 -154 = 85 m ? s = 
324
2
= 162 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(
ii We can find out the height of the triangle corresponding to 154 m in the following manner:
We have:
Area of triangle = 2772 m
2
?
1
2
×Base ×Height = 2772 ? Height = 
2772×2
154
= 36 m
Q u e s t i o n : 9
Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm.
S o l u t i o n :
We have: a = 13 cm and b = 20 cm ? Area of isosceles triangle =
b
4
v
4a
2
-b
2
                                         =
20
4
×
v
4 13)
2
-20
2
                                         = 5 × v 676 -400                                       
Q u e s t i o n : 1 0
The base of an isosceles triangle measures 80 cm and its area is 360 cm
2
. Find the perimeter of the triangle.
S o l u t i o n :
Let ?
PQR be an isosceles triangle and PX ?
QR.
Now,
Area of triangle = 360 cm
2
 ?
1
2
×QR ×PX = 360 ? h =
720
80
= 9 cmNow, QX = 
1
2
×80 = 40 cm and PX = 9 cm
Also,
PQ =
v
QX
2
+PX
2
a =
v
40
2
+9
2
= v 1600 +81 = v 1681 = 41 cm
? Perimeter = 80 + 41 + 41  = 162 cm
Q u e s t i o n : 1 1
The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find  the area of the triangle.
S o l u t i o n :
The ratio of the equal side to its base is 3 : 2.
?
Ratio of sides = 3 : 3 : 2.
Let the three sides of triangle be 3x, 3x, 2x.
The perimeter of isosceles triangle = 32 cm.
? 3x +3x +2x = 32 cm ? 8x = 32 ? x = 4 cm
( )
(
Page 3


Q u e s t i o n : 1
Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.
S o l u t i o n :
We have:
Base = 24 cm
Height = 14.5 cm
Now,
Area of triangle =
1
2
×Base ×Height =
1
2
×24 ×14. 5 = 174 cm
2
Q u e s t i o n : 2
The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height.
S o l u t i o n :
Let the height of the triangle be h m.
? Base = 3h m
Now,
Area of the triangle = 
Total Cost
Rate
=
783
58
= 13. 5 ha = 135000 m
2
We have:
Area of triangle = 135000 m
2
?
1
2
×Base ×Height = 135000 ?
1
2
×3h ×h = 135000 ? h
2
=
135000×2
3
? h
2
= 90000 ? h = 300 m
Thus, we have:
Height = h = 300 m
Base = 3h = 900 m
Q u e s t i o n : 3
Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.
S o l u t i o n :
Let: a = 42 cm, b = 34 cm and c = 20 cm ? s = 
a+b+c
2
=
42+34+20
2
= 48 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 48(48 -42)(48 -34)(48 -20) = v 48 ×6 ×14 ×28
We know that the longest side is 42 cm.
Thus, we can find out the height of the triangle corresponding to 42 cm.
We have:
Area of triangle = 336 cm
2
?
1
2
×Base ×Height = 336 ? Height = 
336×2
42
= 16 cm
Q u e s t i o n : 4
Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.
S o l u t i o n :
Let: a = 18 cm, b = 24 cm and c = 30 cm ? s = 
a+b+c
2
=
18+24+30
2
= 36 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 36(36 -18)(36 -24)(36 -30) = v 36 ×18 ×12 ×6
We know that the smallest side is 18 cm.
Thus, we can find out the altitude of the triangle corresponding to 18 cm.
We have:
Area of triangle = 216 cm
2
?
1
2
×Base ×Height = 216 ? Height = 
216×2
18
= 24 cm
Q u e s t i o n : 5
Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length. Find the height corresponding to the longest side.
S o l u t i o n :
Let: a = 91 m, b = 98 m and c = 105 m ? s = 
a+b+c
2
=
91+98+105
2
= 147 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 147(147 -91)(147 -98)(147 -105) = v 147 ×56 ×
We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.
Area of triangle = 4116 m
2
?
1
2
×Base ×Height = 4116 ? Height = 
4116×2
105
= 78. 4 m
Q u e s t i o n : 6
The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 m. Find the area of the triangle.
S o l u t i o n :
Let the sides of the triangle be 5x m, 12x m and 13x m.
We know:
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, we obtain the sides of the triangle.
5 ×
5 = 25 m
12 ×
5 = 60 m
13 ×
5 = 65 m
Now,
Let: a = 25 m, b = 60 m and c = 65 m ? s = 
150
2
= 75 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 75(75 -25)(75 -60)(75 -65) = v 75 ×50 ×15 ×10 = v 15 ×5 ×5 ×10
Q u e s t i o n : 7
The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the field. Also, find the cost of ploughing the field at  5 per m
2
.
S o l u t i o n :
Let the sides of the triangle be 25x m, 17x m and 12x m.
We know:
Perimeter = Sum of all sides
or, 540 = 25x + 17x + 12x
or, 54x = 540
or, x = 10
Thus, we obtain the sides of the triangle.
25 ×
10 = 250 m
17 ×
10 = 170 m
12 ×
10 = 120 m
Now,
Let: a = 250 m, b = 170 m and c = 120 m ? s = 
540
2
= 270 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 270(270 -250)(270 -170)(270 -120) = v 270 ×20 ×100 ×150
Cost of ploughing 1 m
2
 field = Rs 5
Cost of ploughing 9000 m
2
 field = 5 ×9000 = Rs 45000
.
Q u e s t i o n : 8
Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find i
the area of the field and ii
the length of the perpendicular from the opposite vertex on the side measuring 154 m.
S o l u t i o n :
i Let: a = 85 m and b = 154 m Given: Perimeter = 324 mor, a +b +c = 324 ? c = 324 -85 -154 = 85 m ? s = 
324
2
= 162 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(
ii We can find out the height of the triangle corresponding to 154 m in the following manner:
We have:
Area of triangle = 2772 m
2
?
1
2
×Base ×Height = 2772 ? Height = 
2772×2
154
= 36 m
Q u e s t i o n : 9
Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm.
S o l u t i o n :
We have: a = 13 cm and b = 20 cm ? Area of isosceles triangle =
b
4
v
4a
2
-b
2
                                         =
20
4
×
v
4 13)
2
-20
2
                                         = 5 × v 676 -400                                       
Q u e s t i o n : 1 0
The base of an isosceles triangle measures 80 cm and its area is 360 cm
2
. Find the perimeter of the triangle.
S o l u t i o n :
Let ?
PQR be an isosceles triangle and PX ?
QR.
Now,
Area of triangle = 360 cm
2
 ?
1
2
×QR ×PX = 360 ? h =
720
80
= 9 cmNow, QX = 
1
2
×80 = 40 cm and PX = 9 cm
Also,
PQ =
v
QX
2
+PX
2
a =
v
40
2
+9
2
= v 1600 +81 = v 1681 = 41 cm
? Perimeter = 80 + 41 + 41  = 162 cm
Q u e s t i o n : 1 1
The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find  the area of the triangle.
S o l u t i o n :
The ratio of the equal side to its base is 3 : 2.
?
Ratio of sides = 3 : 3 : 2.
Let the three sides of triangle be 3x, 3x, 2x.
The perimeter of isosceles triangle = 32 cm.
? 3x +3x +2x = 32 cm ? 8x = 32 ? x = 4 cm
( )
(
Therefore, the three side of triangle are 3x, 3x, 2x = 12 cm, 12 cm, 8 cm.
Let S be the semi-perimeter of the triangle. Then, S =
1
2
(12 +12 +8) =
32
2
= 16
Area of the triangle will be
= v S(S -a)(S -b)(S -c) = v 16(16 -12)(16 -12)(16 -8) = v 16 ×4 ×4 ×8 = 4 ×4v 8 = 4 ×4 ×2v 2 = 32v 2 cm
2
D i s c l a i m e r : The answer does not match with the answer given in the book.
Q u e s t i o n : 1 2
The perimeter of triangle is 50 cm. One side of the triangle is 4 cm longer than the smallest side and the third side is 6 cm less than twice the smallest side. Find the area of the triangle.
S o l u t i o n :
Let ABC be any triangle with perimeter 50 cm.
Let the smallest side of the triangle be x.
Then the other sides be x + 4 and 2x - 6.
Now,
x + x + 4 + 2x - 6 = 50        ? perimeteris50cm
? 4x - 2 = 50
? 4x = 50 + 2
? 4x = 52
? x = 13
? The sides of the triangle are of length 13 cm, 17 cm and 20 cm.
? Semi-perimeter of the triangle is
s =
13+17+20
2
=
50
2
= 25 cm
? By Heron's formula,
Area of ? ABC = v s(s -a)(s -b)(s -c)                           = v 25(25 -13)(25 -17)(25 -20)                           = v 25(12)(8)(5)                           = 20v 30 cm
2
Hence, the area of the triangle is 20v 30 cm
2
.
Q u e s t i o n : 1 3
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m, 15 m. The advertisements yield an earning of Rs 2000 per m
2
 a year. A
company hired one of its walls for 6 months. How much rent did it pay?
S o l u t i o n :
The sides of the triangle are of length 13 m, 14 m and 15 m.
? Semi-perimeter of the triangle is
s =
13+14+15
2
=
42
2
= 21 m
? By Heron's formula,
Area of ? = v s(s -a)(s -b)(s -c)                  = v 21(21 -13)(21 -14)(21 -15)                  = v 21(8)(7)(6)                  = 84 m
2
Now,
The rent of advertisements per m
2
 per year = Rs 2000
The rent of the wall with area 84 m
2
 per year = Rs 2000 × 84
                                                                         = Rs 168000
The rent of the wall with area 84 m
2
 for 6 months = Rs 
168000
2
                                                                                = Rs 84000
Hence, the rent paid by the company is Rs 84000.
Q u e s t i o n : 1 4
The perimeter of an isosceles triangle is 42 cm and its base is 1
1
2
times each of the equal sides. Find i
the length of each side of the triangle, ii
the area of the triangle, and iii
the height of the triangle.
S o l u t i o n :
Let the equal sides of the isosceles triangle be a cm each.
? Base of the triangle, b = 
3
2
a cm
i
Perimeter = 42 cm
or, a + a + 
3
2
a = 42
or, 2a +
3
2
a= 42
? 2a +
3
2
a = 42 ?
7a
2
= 42 ? a = 12 
 
So, equal sides of the triangle are 12 cm each.
Also,
Base = 
3
2
a = 
3
2
×12 = 18 cm
Page 4


Q u e s t i o n : 1
Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.
S o l u t i o n :
We have:
Base = 24 cm
Height = 14.5 cm
Now,
Area of triangle =
1
2
×Base ×Height =
1
2
×24 ×14. 5 = 174 cm
2
Q u e s t i o n : 2
The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height.
S o l u t i o n :
Let the height of the triangle be h m.
? Base = 3h m
Now,
Area of the triangle = 
Total Cost
Rate
=
783
58
= 13. 5 ha = 135000 m
2
We have:
Area of triangle = 135000 m
2
?
1
2
×Base ×Height = 135000 ?
1
2
×3h ×h = 135000 ? h
2
=
135000×2
3
? h
2
= 90000 ? h = 300 m
Thus, we have:
Height = h = 300 m
Base = 3h = 900 m
Q u e s t i o n : 3
Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.
S o l u t i o n :
Let: a = 42 cm, b = 34 cm and c = 20 cm ? s = 
a+b+c
2
=
42+34+20
2
= 48 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 48(48 -42)(48 -34)(48 -20) = v 48 ×6 ×14 ×28
We know that the longest side is 42 cm.
Thus, we can find out the height of the triangle corresponding to 42 cm.
We have:
Area of triangle = 336 cm
2
?
1
2
×Base ×Height = 336 ? Height = 
336×2
42
= 16 cm
Q u e s t i o n : 4
Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.
S o l u t i o n :
Let: a = 18 cm, b = 24 cm and c = 30 cm ? s = 
a+b+c
2
=
18+24+30
2
= 36 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 36(36 -18)(36 -24)(36 -30) = v 36 ×18 ×12 ×6
We know that the smallest side is 18 cm.
Thus, we can find out the altitude of the triangle corresponding to 18 cm.
We have:
Area of triangle = 216 cm
2
?
1
2
×Base ×Height = 216 ? Height = 
216×2
18
= 24 cm
Q u e s t i o n : 5
Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length. Find the height corresponding to the longest side.
S o l u t i o n :
Let: a = 91 m, b = 98 m and c = 105 m ? s = 
a+b+c
2
=
91+98+105
2
= 147 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 147(147 -91)(147 -98)(147 -105) = v 147 ×56 ×
We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.
Area of triangle = 4116 m
2
?
1
2
×Base ×Height = 4116 ? Height = 
4116×2
105
= 78. 4 m
Q u e s t i o n : 6
The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 m. Find the area of the triangle.
S o l u t i o n :
Let the sides of the triangle be 5x m, 12x m and 13x m.
We know:
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, we obtain the sides of the triangle.
5 ×
5 = 25 m
12 ×
5 = 60 m
13 ×
5 = 65 m
Now,
Let: a = 25 m, b = 60 m and c = 65 m ? s = 
150
2
= 75 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 75(75 -25)(75 -60)(75 -65) = v 75 ×50 ×15 ×10 = v 15 ×5 ×5 ×10
Q u e s t i o n : 7
The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the field. Also, find the cost of ploughing the field at  5 per m
2
.
S o l u t i o n :
Let the sides of the triangle be 25x m, 17x m and 12x m.
We know:
Perimeter = Sum of all sides
or, 540 = 25x + 17x + 12x
or, 54x = 540
or, x = 10
Thus, we obtain the sides of the triangle.
25 ×
10 = 250 m
17 ×
10 = 170 m
12 ×
10 = 120 m
Now,
Let: a = 250 m, b = 170 m and c = 120 m ? s = 
540
2
= 270 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 270(270 -250)(270 -170)(270 -120) = v 270 ×20 ×100 ×150
Cost of ploughing 1 m
2
 field = Rs 5
Cost of ploughing 9000 m
2
 field = 5 ×9000 = Rs 45000
.
Q u e s t i o n : 8
Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find i
the area of the field and ii
the length of the perpendicular from the opposite vertex on the side measuring 154 m.
S o l u t i o n :
i Let: a = 85 m and b = 154 m Given: Perimeter = 324 mor, a +b +c = 324 ? c = 324 -85 -154 = 85 m ? s = 
324
2
= 162 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(
ii We can find out the height of the triangle corresponding to 154 m in the following manner:
We have:
Area of triangle = 2772 m
2
?
1
2
×Base ×Height = 2772 ? Height = 
2772×2
154
= 36 m
Q u e s t i o n : 9
Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm.
S o l u t i o n :
We have: a = 13 cm and b = 20 cm ? Area of isosceles triangle =
b
4
v
4a
2
-b
2
                                         =
20
4
×
v
4 13)
2
-20
2
                                         = 5 × v 676 -400                                       
Q u e s t i o n : 1 0
The base of an isosceles triangle measures 80 cm and its area is 360 cm
2
. Find the perimeter of the triangle.
S o l u t i o n :
Let ?
PQR be an isosceles triangle and PX ?
QR.
Now,
Area of triangle = 360 cm
2
 ?
1
2
×QR ×PX = 360 ? h =
720
80
= 9 cmNow, QX = 
1
2
×80 = 40 cm and PX = 9 cm
Also,
PQ =
v
QX
2
+PX
2
a =
v
40
2
+9
2
= v 1600 +81 = v 1681 = 41 cm
? Perimeter = 80 + 41 + 41  = 162 cm
Q u e s t i o n : 1 1
The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find  the area of the triangle.
S o l u t i o n :
The ratio of the equal side to its base is 3 : 2.
?
Ratio of sides = 3 : 3 : 2.
Let the three sides of triangle be 3x, 3x, 2x.
The perimeter of isosceles triangle = 32 cm.
? 3x +3x +2x = 32 cm ? 8x = 32 ? x = 4 cm
( )
(
Therefore, the three side of triangle are 3x, 3x, 2x = 12 cm, 12 cm, 8 cm.
Let S be the semi-perimeter of the triangle. Then, S =
1
2
(12 +12 +8) =
32
2
= 16
Area of the triangle will be
= v S(S -a)(S -b)(S -c) = v 16(16 -12)(16 -12)(16 -8) = v 16 ×4 ×4 ×8 = 4 ×4v 8 = 4 ×4 ×2v 2 = 32v 2 cm
2
D i s c l a i m e r : The answer does not match with the answer given in the book.
Q u e s t i o n : 1 2
The perimeter of triangle is 50 cm. One side of the triangle is 4 cm longer than the smallest side and the third side is 6 cm less than twice the smallest side. Find the area of the triangle.
S o l u t i o n :
Let ABC be any triangle with perimeter 50 cm.
Let the smallest side of the triangle be x.
Then the other sides be x + 4 and 2x - 6.
Now,
x + x + 4 + 2x - 6 = 50        ? perimeteris50cm
? 4x - 2 = 50
? 4x = 50 + 2
? 4x = 52
? x = 13
? The sides of the triangle are of length 13 cm, 17 cm and 20 cm.
? Semi-perimeter of the triangle is
s =
13+17+20
2
=
50
2
= 25 cm
? By Heron's formula,
Area of ? ABC = v s(s -a)(s -b)(s -c)                           = v 25(25 -13)(25 -17)(25 -20)                           = v 25(12)(8)(5)                           = 20v 30 cm
2
Hence, the area of the triangle is 20v 30 cm
2
.
Q u e s t i o n : 1 3
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m, 15 m. The advertisements yield an earning of Rs 2000 per m
2
 a year. A
company hired one of its walls for 6 months. How much rent did it pay?
S o l u t i o n :
The sides of the triangle are of length 13 m, 14 m and 15 m.
? Semi-perimeter of the triangle is
s =
13+14+15
2
=
42
2
= 21 m
? By Heron's formula,
Area of ? = v s(s -a)(s -b)(s -c)                  = v 21(21 -13)(21 -14)(21 -15)                  = v 21(8)(7)(6)                  = 84 m
2
Now,
The rent of advertisements per m
2
 per year = Rs 2000
The rent of the wall with area 84 m
2
 per year = Rs 2000 × 84
                                                                         = Rs 168000
The rent of the wall with area 84 m
2
 for 6 months = Rs 
168000
2
                                                                                = Rs 84000
Hence, the rent paid by the company is Rs 84000.
Q u e s t i o n : 1 4
The perimeter of an isosceles triangle is 42 cm and its base is 1
1
2
times each of the equal sides. Find i
the length of each side of the triangle, ii
the area of the triangle, and iii
the height of the triangle.
S o l u t i o n :
Let the equal sides of the isosceles triangle be a cm each.
? Base of the triangle, b = 
3
2
a cm
i
Perimeter = 42 cm
or, a + a + 
3
2
a = 42
or, 2a +
3
2
a= 42
? 2a +
3
2
a = 42 ?
7a
2
= 42 ? a = 12 
 
So, equal sides of the triangle are 12 cm each.
Also,
Base = 
3
2
a = 
3
2
×12 = 18 cm
ii
Area of isosceles triangle = 
b
4
v
4a
2
-b
2
=
18
4
×
v
4 12)
2
-18
2
        a = 12 cm and b = 18 cm = 4. 5 × v 576 -324 = 4. 5 × v 252 = 4. 5 ×15. 87 = 71. 42 cm
2
iii
Area of triangle = 71. 42 cm
2
?
1
2
×Base ×Height = 71. 42 ? Height = 
71.42×2
18
= 7. 94 cm
Q u e s t i o n : 1 5
If the area of an equilateral triangle is 36v 3 cm
2
, find its perimeter.
S o l u t i o n :
Area of equilateral triangle = 
v
3
4
× Side)
2
?
v
3
4
× Side)
2
 = 36v 3
? Side)
2
= 144 ? Side = 12 cm
Thus, we have:
Perimeter = 3 × Side = 3 × 12 = 36 cm
Q u e s t i o n : 1 6
If the area of an equilateral triangle is 81v 3 cm
2
, find its height.
S o l u t i o n :
Area of equilateral triangle = 
v
3
4
× Side)
2
?
v
3
4
× Side)
2
 = 81v 3
? Side)
2
= 324 ? Side = 18 cm
Now, we have:
Height =
v
3
2
×Side =
v
3
2
×18 = 9v 3 cm
Q u e s t i o n : 1 7
Each side of an equilateral triangle measures 8 cm. Find i
the area of the triangle, correct to 2 places of decimal and ii
the height of the triangle, correct to 2 places of decimal. Take v 3 = 1. 732
.
S o l u t i o n :
Side of the equilateral triangle = 8 cm
i
Area of equilateral triangle = 
v
3
4
× Side)
2
=
v
3
4
× 8)
2
 =
1.732×64
4
= 27. 71 cm
2
ii
Height =
v
3
2
×Side =
v
3
2
×8 =
1.732×8
2
= 6. 93 cm
Q u e s t i o n : 1 8
The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 places of decimal. Take v 3 = 1. 732
.
S o l u t i o n :
Height of the equilateral triangle = 9 cm
Thus, we have:
Height =
v
3
2
×Side ? 9 =
v
3
2
×Side ? Side = 
18
v
3
=
18
v
3
×
v
3
v
3
= 6v 3 cm
Also,
Area of equilateral triangle = 
v
3
4
× Side)
2
=
v
3
4
× 6v 3)
2
 =
108
4
v 3 = 27v 3 = 46. 76 cm
2
Q u e s t i o n : 1 9
The base of a right-angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle.
S o l u t i o n :
(
( )
( (
(
( (
(
( (
( (
Page 5


Q u e s t i o n : 1
Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.
S o l u t i o n :
We have:
Base = 24 cm
Height = 14.5 cm
Now,
Area of triangle =
1
2
×Base ×Height =
1
2
×24 ×14. 5 = 174 cm
2
Q u e s t i o n : 2
The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height.
S o l u t i o n :
Let the height of the triangle be h m.
? Base = 3h m
Now,
Area of the triangle = 
Total Cost
Rate
=
783
58
= 13. 5 ha = 135000 m
2
We have:
Area of triangle = 135000 m
2
?
1
2
×Base ×Height = 135000 ?
1
2
×3h ×h = 135000 ? h
2
=
135000×2
3
? h
2
= 90000 ? h = 300 m
Thus, we have:
Height = h = 300 m
Base = 3h = 900 m
Q u e s t i o n : 3
Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.
S o l u t i o n :
Let: a = 42 cm, b = 34 cm and c = 20 cm ? s = 
a+b+c
2
=
42+34+20
2
= 48 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 48(48 -42)(48 -34)(48 -20) = v 48 ×6 ×14 ×28
We know that the longest side is 42 cm.
Thus, we can find out the height of the triangle corresponding to 42 cm.
We have:
Area of triangle = 336 cm
2
?
1
2
×Base ×Height = 336 ? Height = 
336×2
42
= 16 cm
Q u e s t i o n : 4
Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.
S o l u t i o n :
Let: a = 18 cm, b = 24 cm and c = 30 cm ? s = 
a+b+c
2
=
18+24+30
2
= 36 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 36(36 -18)(36 -24)(36 -30) = v 36 ×18 ×12 ×6
We know that the smallest side is 18 cm.
Thus, we can find out the altitude of the triangle corresponding to 18 cm.
We have:
Area of triangle = 216 cm
2
?
1
2
×Base ×Height = 216 ? Height = 
216×2
18
= 24 cm
Q u e s t i o n : 5
Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length. Find the height corresponding to the longest side.
S o l u t i o n :
Let: a = 91 m, b = 98 m and c = 105 m ? s = 
a+b+c
2
=
91+98+105
2
= 147 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 147(147 -91)(147 -98)(147 -105) = v 147 ×56 ×
We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.
Area of triangle = 4116 m
2
?
1
2
×Base ×Height = 4116 ? Height = 
4116×2
105
= 78. 4 m
Q u e s t i o n : 6
The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 m. Find the area of the triangle.
S o l u t i o n :
Let the sides of the triangle be 5x m, 12x m and 13x m.
We know:
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, we obtain the sides of the triangle.
5 ×
5 = 25 m
12 ×
5 = 60 m
13 ×
5 = 65 m
Now,
Let: a = 25 m, b = 60 m and c = 65 m ? s = 
150
2
= 75 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 75(75 -25)(75 -60)(75 -65) = v 75 ×50 ×15 ×10 = v 15 ×5 ×5 ×10
Q u e s t i o n : 7
The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the field. Also, find the cost of ploughing the field at  5 per m
2
.
S o l u t i o n :
Let the sides of the triangle be 25x m, 17x m and 12x m.
We know:
Perimeter = Sum of all sides
or, 540 = 25x + 17x + 12x
or, 54x = 540
or, x = 10
Thus, we obtain the sides of the triangle.
25 ×
10 = 250 m
17 ×
10 = 170 m
12 ×
10 = 120 m
Now,
Let: a = 250 m, b = 170 m and c = 120 m ? s = 
540
2
= 270 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 270(270 -250)(270 -170)(270 -120) = v 270 ×20 ×100 ×150
Cost of ploughing 1 m
2
 field = Rs 5
Cost of ploughing 9000 m
2
 field = 5 ×9000 = Rs 45000
.
Q u e s t i o n : 8
Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find i
the area of the field and ii
the length of the perpendicular from the opposite vertex on the side measuring 154 m.
S o l u t i o n :
i Let: a = 85 m and b = 154 m Given: Perimeter = 324 mor, a +b +c = 324 ? c = 324 -85 -154 = 85 m ? s = 
324
2
= 162 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(
ii We can find out the height of the triangle corresponding to 154 m in the following manner:
We have:
Area of triangle = 2772 m
2
?
1
2
×Base ×Height = 2772 ? Height = 
2772×2
154
= 36 m
Q u e s t i o n : 9
Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm.
S o l u t i o n :
We have: a = 13 cm and b = 20 cm ? Area of isosceles triangle =
b
4
v
4a
2
-b
2
                                         =
20
4
×
v
4 13)
2
-20
2
                                         = 5 × v 676 -400                                       
Q u e s t i o n : 1 0
The base of an isosceles triangle measures 80 cm and its area is 360 cm
2
. Find the perimeter of the triangle.
S o l u t i o n :
Let ?
PQR be an isosceles triangle and PX ?
QR.
Now,
Area of triangle = 360 cm
2
 ?
1
2
×QR ×PX = 360 ? h =
720
80
= 9 cmNow, QX = 
1
2
×80 = 40 cm and PX = 9 cm
Also,
PQ =
v
QX
2
+PX
2
a =
v
40
2
+9
2
= v 1600 +81 = v 1681 = 41 cm
? Perimeter = 80 + 41 + 41  = 162 cm
Q u e s t i o n : 1 1
The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find  the area of the triangle.
S o l u t i o n :
The ratio of the equal side to its base is 3 : 2.
?
Ratio of sides = 3 : 3 : 2.
Let the three sides of triangle be 3x, 3x, 2x.
The perimeter of isosceles triangle = 32 cm.
? 3x +3x +2x = 32 cm ? 8x = 32 ? x = 4 cm
( )
(
Therefore, the three side of triangle are 3x, 3x, 2x = 12 cm, 12 cm, 8 cm.
Let S be the semi-perimeter of the triangle. Then, S =
1
2
(12 +12 +8) =
32
2
= 16
Area of the triangle will be
= v S(S -a)(S -b)(S -c) = v 16(16 -12)(16 -12)(16 -8) = v 16 ×4 ×4 ×8 = 4 ×4v 8 = 4 ×4 ×2v 2 = 32v 2 cm
2
D i s c l a i m e r : The answer does not match with the answer given in the book.
Q u e s t i o n : 1 2
The perimeter of triangle is 50 cm. One side of the triangle is 4 cm longer than the smallest side and the third side is 6 cm less than twice the smallest side. Find the area of the triangle.
S o l u t i o n :
Let ABC be any triangle with perimeter 50 cm.
Let the smallest side of the triangle be x.
Then the other sides be x + 4 and 2x - 6.
Now,
x + x + 4 + 2x - 6 = 50        ? perimeteris50cm
? 4x - 2 = 50
? 4x = 50 + 2
? 4x = 52
? x = 13
? The sides of the triangle are of length 13 cm, 17 cm and 20 cm.
? Semi-perimeter of the triangle is
s =
13+17+20
2
=
50
2
= 25 cm
? By Heron's formula,
Area of ? ABC = v s(s -a)(s -b)(s -c)                           = v 25(25 -13)(25 -17)(25 -20)                           = v 25(12)(8)(5)                           = 20v 30 cm
2
Hence, the area of the triangle is 20v 30 cm
2
.
Q u e s t i o n : 1 3
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m, 15 m. The advertisements yield an earning of Rs 2000 per m
2
 a year. A
company hired one of its walls for 6 months. How much rent did it pay?
S o l u t i o n :
The sides of the triangle are of length 13 m, 14 m and 15 m.
? Semi-perimeter of the triangle is
s =
13+14+15
2
=
42
2
= 21 m
? By Heron's formula,
Area of ? = v s(s -a)(s -b)(s -c)                  = v 21(21 -13)(21 -14)(21 -15)                  = v 21(8)(7)(6)                  = 84 m
2
Now,
The rent of advertisements per m
2
 per year = Rs 2000
The rent of the wall with area 84 m
2
 per year = Rs 2000 × 84
                                                                         = Rs 168000
The rent of the wall with area 84 m
2
 for 6 months = Rs 
168000
2
                                                                                = Rs 84000
Hence, the rent paid by the company is Rs 84000.
Q u e s t i o n : 1 4
The perimeter of an isosceles triangle is 42 cm and its base is 1
1
2
times each of the equal sides. Find i
the length of each side of the triangle, ii
the area of the triangle, and iii
the height of the triangle.
S o l u t i o n :
Let the equal sides of the isosceles triangle be a cm each.
? Base of the triangle, b = 
3
2
a cm
i
Perimeter = 42 cm
or, a + a + 
3
2
a = 42
or, 2a +
3
2
a= 42
? 2a +
3
2
a = 42 ?
7a
2
= 42 ? a = 12 
 
So, equal sides of the triangle are 12 cm each.
Also,
Base = 
3
2
a = 
3
2
×12 = 18 cm
ii
Area of isosceles triangle = 
b
4
v
4a
2
-b
2
=
18
4
×
v
4 12)
2
-18
2
        a = 12 cm and b = 18 cm = 4. 5 × v 576 -324 = 4. 5 × v 252 = 4. 5 ×15. 87 = 71. 42 cm
2
iii
Area of triangle = 71. 42 cm
2
?
1
2
×Base ×Height = 71. 42 ? Height = 
71.42×2
18
= 7. 94 cm
Q u e s t i o n : 1 5
If the area of an equilateral triangle is 36v 3 cm
2
, find its perimeter.
S o l u t i o n :
Area of equilateral triangle = 
v
3
4
× Side)
2
?
v
3
4
× Side)
2
 = 36v 3
? Side)
2
= 144 ? Side = 12 cm
Thus, we have:
Perimeter = 3 × Side = 3 × 12 = 36 cm
Q u e s t i o n : 1 6
If the area of an equilateral triangle is 81v 3 cm
2
, find its height.
S o l u t i o n :
Area of equilateral triangle = 
v
3
4
× Side)
2
?
v
3
4
× Side)
2
 = 81v 3
? Side)
2
= 324 ? Side = 18 cm
Now, we have:
Height =
v
3
2
×Side =
v
3
2
×18 = 9v 3 cm
Q u e s t i o n : 1 7
Each side of an equilateral triangle measures 8 cm. Find i
the area of the triangle, correct to 2 places of decimal and ii
the height of the triangle, correct to 2 places of decimal. Take v 3 = 1. 732
.
S o l u t i o n :
Side of the equilateral triangle = 8 cm
i
Area of equilateral triangle = 
v
3
4
× Side)
2
=
v
3
4
× 8)
2
 =
1.732×64
4
= 27. 71 cm
2
ii
Height =
v
3
2
×Side =
v
3
2
×8 =
1.732×8
2
= 6. 93 cm
Q u e s t i o n : 1 8
The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 places of decimal. Take v 3 = 1. 732
.
S o l u t i o n :
Height of the equilateral triangle = 9 cm
Thus, we have:
Height =
v
3
2
×Side ? 9 =
v
3
2
×Side ? Side = 
18
v
3
=
18
v
3
×
v
3
v
3
= 6v 3 cm
Also,
Area of equilateral triangle = 
v
3
4
× Side)
2
=
v
3
4
× 6v 3)
2
 =
108
4
v 3 = 27v 3 = 46. 76 cm
2
Q u e s t i o n : 1 9
The base of a right-angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle.
S o l u t i o n :
(
( )
( (
(
( (
(
( (
( (
Let ?
PQR be a right-angled triangle and PQ
? QR.
Now,
PQ =
v
PR
2
-QR
2
=
v
50
2
-48
2
= v 2500 -2304 = v 196 = 14 cm
Area of triangle =
1
2
×QR ×PQ =
1
2
×48 ×14 = 336 cm
2
Q u e s t i o n : 2 0
Find the area of the shaded region in the figure given below.
S o l u t i o n :
In right angled ?ABD,
AB
2
 = AD
2
 + DB
2
          PythagorasTheorem
? AB
2
 = 12
2
 + 16
2
? AB
2
 = 144 +
 
256
? AB
2
 = 400
? AB = 20 cm
Area of ?ADB = 
1
2
×DB ×AD
                        = 
1
2
×16 ×12
                        = 96 cm
2
               ....1
In ?ACB,
The sides of the triangle are of length 20 cm, 52 cm and 48 cm.
? Semi-perimeter of the triangle is
s =
20+52+48
2
=
120
2
= 60 cm
? By Heron's formula,
Area of ? ACB = v s(s -a)(s -b)(s -c)                          = v 60(60 -20)(60 -52)(60 -48)                          = v 60(40)(8)(12)                          = 480 cm
2
          . . . (2)
Now,
Area of the shaded region = Area of ?ACB - Area of ?ADB
                                          = 480 - 96
                                          = 384 cm
2
Hence, the area of the shaded region in the given figure is 384 cm
2
.
Q u e s t i o n : 2 1
The sides of a quadrilateral ABCD taken in order are 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle between the first two sides is a right angle. Find its area. (Given, v 6 = 2. 45
).
S o l u t i o n :
In the given figure, ABCD is a quadrilateral with sides of length 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle between the first two sides is a right angle.
 
Join AC.