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Heron’s Formula

Class 9th Mathematics NCERT Exemplar Solution
Exercise 12.1
  1. An isosceles right triangle has area 8 cm2. The length of its hypotenuse isA.…
  2. The perimeter of an equilateral triangle is 60 m. The area isA. 10√3m^2 B.…
  3. The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the…
  4. The area of an equilateral triangle with side 2√3 cm isA. 5.196 cm^2 B. 0.866…
  5. The length of each side of an equilateral triangle having an area of 9√3 cm2…
  6. If the area of an equilateral triangle is 16√3 cm2, then the perimeter of the…
  7. The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of…
  8. The area of an isosceles triangle having base 2 cm and the length of one of the…
  9. The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting…
Exercise 12.2
  1. The area of a triangle with base 4 cm and height 6 cm is 24cm^2
  2. The area of delta ABC is 8 cm2 in which AB = AC = 4 cm and ∠A = 90°.…
  3. The area of the isosceles triangle is 5/4 root 11 cm^2 if the perimeter is 11…
  4. The area of the equilateral triangle is 20√3cm^2 whose each side is 8 cm.…
  5. If the side of a rhombus is 10 cm and one diagonal is 16 cm, the area of the…
  6. The base and the corresponding altitude of a parallelogram are 10 cm and 3.5…
  7. The area of a regular hexagon of side ‘a’ is the sum of the areas of the five…
  8. The cost of leveling the ground in the form of a triangle having the sides 51m,…
  9. In a triangle, the sides are given as 11 cm, 12 cm and 13 cm. The length of the…
Exercise 12.3
  1. Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65…
  2. The triangular side walls of a flyover have been used for advertisements. The…
  3. From a point in the interior of an equilateral triangle, perpendiculars are…
  4. The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to…
  5. Find the area of a parallelogram given in Fig. 12.2. Also find the length of…
  6. A field in the form of a parallelogram has sides 60m and 40 m and one of its…
  7. The perimeter of a triangular field is 420 m and its sides are in the ratio…
  8. The sides of a quadrilateral ABCD are 6 cm, 8cm, 12 cm and 14 cm (taken in…
  9. A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted…
  10. Find the area of the trapezium PQRS with height PQ given in Fig. 12.3…
Exercise 12.4
  1. How much paper of each shade is needed to make a kite given in Fig. 12.4, in…
  2. The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer…
  3. The area of a trapezium is 475 cm^2 and the height is 19 cm. Find the lengths…
  4. A rectangular plot is given for constructing a house, having a measurement of…
  5. A field is in the shape of a trapezium having parallel sides 90 m and 30 m.…
  6. In Fig. 12.5, ΔABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC…
  7. The dimensions of a rectangle ABCD are 51 cm × 25 cm. A trapezium PQCD with its…
  8. A design is made on a rectangular tile of dimensions 50 cm × 70 cm as shown in…
Exercise 12.1
Question 1.

An isosceles right triangle has area 8 cm2. The length of its hypotenuse is
A. √32cm

B. √16cm

C. √48cm

D. √24cm
Answer:

Let height of triangle = h

As the triangle is isosceles, base = height = h


Area of triangle = 8cm2 (given)


⇒ 1/2 × Base × Height = 8


⇒ 1/2 × h × h = 8


⇒ h2 = 16


⇒ h = 4cm


Base = Height = 4cm


As the triangle is right angled,


Hypotenuse2 = Base2 + Height2


⇒ Hypotenuse2 = 42 + 42


⇒ Hypotenuse2 = 32


⇒ Hypotenuse = √ 32


Hence, the answer is Options A.
Question 2.

The perimeter of an equilateral triangle is 60 m. The area is
A. 10√3m2

B. 15√3m2

C. 20√3m2

D. 100√3m2
Answer:

Area of an equilateral triangle of side a is given by

Perimeter of triangle = 60m (given)


⇒ a + a + a = 60


⇒ 3a = 60


⇒ a = 20m


Area(Δ) =


Hence the correct option is (D).
Question 3.

The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is
A. 1322 cm2

B. 1311 cm2

C. 1344 cm2

D. 1392 cm2
Answer:

a = 56, b = 60, c = 52

s = (a + b + c)/2


⇒ s = (56 + 60 + 52)/2 = 168/2 = 84.


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √84(84-56)(84-60)(84-52)


⇒ Area(Δ) = √84×28×24×32


⇒ Area(Δ) = 1344cm2


Hence, the correct option is (C).
Question 4.

The area of an equilateral triangle with side 2√3 cm is
A. 5.196 cm2

B. 0.866 cm2

C. 3.496 cm2

D. 1.732 cm
Answer:

Area of an equilateral triangle of side a is given by

Given that a = 2√3


Area(Δ) = = 5.196


Hence the correct option is A.
Question 5.

The length of each side of an equilateral triangle having an area of 9√3 cm2 is
A. 8 cm

B. 36 cm

C. 4 cm

D. 6 cm
Answer:

Area of an equilateral triangle of side a is given by

Given that Area(Δ) = 9√3 cm2


⇒


⇒a2 = 9× 4


⇒a2 = 36


⇒a = 6cm
Question 6.

If the area of an equilateral triangle is 16√3 cm2, then the perimeter of the triangle is
A. 48 cm

B. 24 cm

C. 12 cm

D. 36 cm
Answer:

Area of an equilateral triangle of side a is given by

Given that Area(Δ) = 16√3 cm2


⇒


⇒ a2 = 16× 4


⇒ a2 = 64


⇒ a = 8cm


Perimeter(Δ) = a + a + a = 3a = 3×8 = 24cm


Hence the correct option is (B).
Question 7.

The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude
A. 16√5 cm

B. 10√5 cm

C. 24√5 cm

D. 28 cm
Answer:

a = 35, b = 54, c = 61

s = (a + b + c)/2


⇒ s = (35 + 54 + 61)/2 = 150/2 = 75.


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √75(75-35)(75-54)(75-61)


⇒ Area(Δ) = √75×40×21×14


⇒ Area(Δ) = 420√5cm2


Area(Δ) = 1/2 × Base × Altitude


As the area of the triangle is fixed, for the longest altitude we need smallest base.


So, the length of base = 35cm


Area(Δ) = 1/2 × Base × Altitude


⇒ 420√5 = 1/2 × 35 × Altitude


⇒ 24√5 = Altitude.


Hence, the correct option is (C).
Question 8.

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is
A. √15 cm2

B. √15/2 cm2

C. 2√15 cm2

D. 4√15 cm2
Answer:

a = 2, b = 4, c = 4

s = (a + b + c)/2


⇒ s = (2 + 4 + 4)/2 = 10/2 = 5.


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √5(5-2)(5-4)(5-4)


⇒ Area(Δ) = √5×3×1×1


⇒ Area(Δ) = √15cm2


Hence, the correct option is A.
Question 9.

The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm2 is
A. Rs 2.00

B. Rs 2.16

C. Rs 2.48

D. Rs 3.00
Answer:

a = 6, b = 8, c = 10

s = (a + b + c)/2


⇒ s = (6 + 8 + 10)/2 = 24/2 = 12.


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √12(12-6)(12-8)(12-10)


⇒ Area(Δ) = √12×6×4×2


⇒ Area(Δ) = 24cm2


Cost of painting = Area(Δ)×Cost per cm2


⇒ Cost of painting = 24×9 = 216 paise = Rs 2.16


Hence, the correct option is (B).
Exercise 12.2
Question 1.

The area of a triangle with base 4 cm and height 6 cm is 24cm2
Answer:

Area(Δ) = 1/2 × Base × Altitude

⇒ Area(Δ) = 1/2 × 4 × 6 = 12cm2


Hence, the statement is False.
Question 2.

The area of ABC is 8 cm2 in which AB = AC = 4 cm and ∠A = 90°.
Answer:

Area(Δ) = 1/2 × Base × Altitude

⇒ Area(Δ) = 1/2 × 4 × 4 = 8cm2


Hence, the statement is True.
Question 3.

The area of the isosceles triangle is cm2 if the perimeter is 11 cm and the base is 5 cm.
Answer:

Given Perimeter = 11cm

And a = 5


As the triangle is isosceles, b = c


Perimeter = 11


⇒ a + b + c = 11


⇒ 5 + b + b = 11


⇒ 5 + 2b = 11


⇒ 2b = 6


⇒ b = 3


So, we have a = 5, b = 3, c = 3


s = (a + b + c)/2


⇒ s = (5 + 3 + 3)/2 = 11/2


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) =


⇒ Area(Δ) =


⇒ Area(Δ) = (5√11)/4cm2


Hence, the statement is True.
Question 4.

The area of the equilateral triangle is 20√3cm2 whose each side is 8 cm.
Answer:

Area of an equilateral triangle of side a is given by

Given that Area(Δ) = 20√3 cm2


⇒


⇒ a2 = 20× 4


⇒ a2 = 80


⇒ a = 4√5 cm


Hence, the statement is False.
Question 5.

If the side of a rhombus is 10 cm and one diagonal is 16 cm, the area of the rhombus is 96 cm2
Answer:

To find the area of rhombus, we divide it into two triangles.

As all the sides of a rhombus are equal, we have for a triangle


a = 10, b = 10, c = 16


s = (a + b + c)/2


⇒ s = (10 + 10 + 16)/2 = 36/2 = 18.


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √18(18-10)(18-10)(18-16)


⇒ Area(Δ) = √18×8×8×2


⇒ Area(Δ) = 48cm2


As the sides of the other triangle are also same, so their areas will also be equal.


Area(Rhombus) = Area(Δ) + Area(Δ)


⇒ Area(Rhombus) = 48 + 48 = 96cm2


Hence, the statement is True.
Question 6.

The base and the corresponding altitude of a parallelogram are 10 cm and 3.5 cm, respectively. The area of the parallelogram is 30 cm2.
Answer:

Area of parallelogram = Base × Altitude = 10 × 3.5 = 35cm2

Hence, the statement is False.
Question 7.

The area of a regular hexagon of side ‘a’ is the sum of the areas of the five equilateral triangles with side a.
Answer:


The area of a regular hexagon of side ‘a’ is the sum of the areas of the six equilateral triangles with side a.

Hence, the statement is False.
Question 8.

The cost of leveling the ground in the form of a triangle having the sides 51m, 37m and 20m at the rate of Rs 3 per m2 is Rs 918.
Answer:

a = 51, b = 37, c = 20

s = (a + b + c)/2


⇒ s = (51 + 37 + 20)/2 = 108/2 = 54.


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √54(54-51)(54-37)(54-20)


⇒ Area(Δ) = √54×3×17×34


⇒ Area(Δ) = 306m2


Cost of painting = Area(Δ)×Cost per m2


⇒ Cost of painting = 306×3 = Rs. 918


Hence, the statement is True.
Question 9.

In a triangle, the sides are given as 11 cm, 12 cm and 13 cm. The length of the altitude is 10.25 cm corresponding to the side having length 12 cm.
Answer:

a = 11, b = 12, c = 13

s = (a + b + c)/2


⇒ s = (11 + 12 + 13)/2 = 36/2 = 18.


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √18(18-11)(18-12)(18-13)


⇒ Area(Δ) = √18×7×6×5


⇒ Area(Δ) = 6√105 cm2 = 61.48 ∼ 61.5 cm2


Area(Δ) = 1/2 × Base × Altitude


⇒ Area(Δ) = 1/2 × 12 × 10.25


⇒ Area(Δ) = 61.5 cm2


Hence the statement is True.
Exercise 12.3
Question 1.

Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs 7 per m2
Answer:

a = 50, b = 65, c = 65

s = (a + b + c)/2


⇒ s = (50 + 65 + 65)/2 = 180/2 = 90.


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √90(90-50)(90-65)(90-65)


⇒ Area(Δ) = √90×40×25×25


⇒ Area(Δ) = 1500m2


Cost of laying grass = Area(Δ)×Cost per m2


⇒ Cost of laying grass = 1500×7 = Rs.10500
Question 2.

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m and 15 m. The advertisements yield and earning of Rs. 2000 per m2 a year. A company hired one of its walls for 6 months. How much rent did it pay?
Answer:

a = 13, b = 14, c = 15

s = (a + b + c)/2


⇒ s = (13 + 14 + 15)/2 = 42/2 = 21.


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √21(21-13)(21-14)(21-15)


⇒ Area(Δ) = √21×8×7×6


⇒ Area(Δ) = 84m2


Cost of advertisements for a year = Area(Δ) × Cost per m2


⇒ Cost of advertisements for a year = 84×2000 = Rs. 168000


As the company only rented the board for 6 months:


Cost of advertisements for 6 months =



Question 3.

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 14 cm, 10 cm and 6 cm. Find the area of the triangle.
Answer:

Area of equilateral side of side ‘a’ =

We divide the triangle in three triangles,


Area of triangle = (1/2 × a × 14) + (1/2 × a × 10) + (1/2 × a × 6)


Area of triangle = 1/2 × a × (14 + 10 + 6)


= 1/2 × a × 30


⇒ cm


Area (Δ) =


⇒ Area (Δ) =


⇒ Area(Δ) = 300√3 cm2
Question 4.

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.
Answer:

Ratio of equal side to base = 3 : 2

Let the equal side = 3x


So, base = 2x


Perimeter(Δ) = 32


⇒ 3x + 3x + 2x = 32


⇒ 8x = 32


⇒ x = 4.


Equal side = 3x = 3×4 = 12


Base = 2x = 2×4 = 8


The sides of the triangle are 12cm, 12cm and 8cm.


a = 12, b = 12, c = 8


s = (a + b + c)/2


⇒ s = (12 + 12 + 8)/2 = 32/2 = 16.


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √16(16-12)(16-12)(16-8)


⇒ Area(Δ) = √16×4×4×8


⇒ Area(Δ) = 32√2 cm2
Question 5.

Find the area of a parallelogram given in Fig. 12.2. Also find the length of the altitude from vertex A on the side DC.
Answer:

Area of parallelogram(ABCD) = Area(ΔBCD) + Area(ΔABD)

For Area(ΔBCD),


a = 12, b = 17, c = 25


s = (a + b + c)/2


⇒ s = (12 + 17 + 25)/2 = 54/2 = 27.


Area(ΔBCD) = √s(s-a)(s-b)(s-c)


⇒ Area(ΔBCD) = √27(27-12)(27-17)(27-25)


⇒ Area(ΔBCD) = √27×15×10×2


⇒ Area(ΔBCD) = 90 cm2


As ABCD is a parallelogram, Area(ΔBCD) = Area(ΔABD)


⇒ Area of parallelogram(ABCD) = Area(ΔBCD) + Area(ΔABD)


⇒ Area of parallelogram(ABCD) = 90 + 90


⇒ Area of parallelogram(ABCD) = 180 cm2


Also, Area of parallelogram(ABCD) = CD × (Altitude from A)


⇒ 180 = 12 × (Altitude from A)


⇒ Altitude from A = 15 cm
Question 6.

A field in the form of a parallelogram has sides 60m and 40 m and one of its diagonals is 80 m long. Find the area of the parallelogram.
Answer:


For Area(ΔBCD),


a = 60, b = 40, c = 80


s = (a + b + c)/2


⇒ s = (60 + 40 + 80)/2 = 180/2 = 90.


Area(ΔBCD) = √s(s-a)(s-b)(s-c)


⇒ Area(ΔBCD) = √90(90-60)(90-40)(90-80)


⇒ Area(ΔBCD) = √90×30×50×10


⇒ Area(ΔBCD) = 300√15 cm2


As ABCD is a parallelogram, Area(ΔBCD) = Area(ΔABD)


⇒ Area of parallelogram(ABCD) = Area(ΔBCD) + Area(ΔABD)


⇒ Area of parallelogram(ABCD) = 300√15 + 300√15


⇒ Area of parallelogram(ABCD) = 600√15 = 2323.79 cm2
Question 7.

The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field.
Answer:

Given that the sides are in ratio of 6:7:8

Let the sides of the triangle be 6x, 7x and 8x.


Perimeter(Δ) = 420


⇒ 6x + 7x + 8x = 420


⇒ 21x = 420


⇒ x = 20


Therefore the sides are 120m, 140m and 160m.


a = 120, b = 140, c = 160


s = (a + b + c)/2


⇒ s = (120 + 140 + 160)/2 = 420/2 = 210.


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √210(210-120)(210-140)(210-160)


⇒ Area(Δ) = √210×90×70×50


⇒ Area(Δ) = 2100√15 m2
Question 8.

The sides of a quadrilateral ABCD are 6 cm, 8cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. Find its area.
Answer:


We join A to C to complete the Δ ABC,


In Δ ABC,


AC2 = AB2 + BC2


⇒ AC2 = 62 + 82


⇒ AC2 = 36 + 64


⇒ AC2 = 100


⇒ AC = 10cm


Area(ΔABC) = 1/2 × AB × BC


⇒ Area(ΔABC) = 1/2 × 6 × 8


⇒ Area(ΔABC) = 24cm2


For area of ΔACD,


a = 12, b = 14, c = 10


s = (a + b + c)/2


⇒ s = (12 + 14 + 10)/2 = 36/2 = 18.


Area(ΔACD) = √s(s-a)(s-b)(s-c)


⇒ Area(ΔACD) = √18(18-12)(18-14)(18-10)


⇒ Area(ΔACD) = √18×6×4×8


⇒ Area(ΔACD) = 24√6 = 58.78cm2


Area(ABCD) = Area(ΔABC) + Area(ΔACD)


⇒ Area(ABCD) = 24 + 58.78 = 82.78 cm2
Question 9.

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs 5 per m2. Find the cost of painting.
Answer:

In a rhombus, all sides are equal.

Perimeter of rhombus = 40cm


Side of rhombus = 40/4 = 10cm


a = 10, b = 10, c = 12


s = (a + b + c)/2


⇒ s = (10 + 10 + 12)/2 = 32/2 = 16.


Area(ΔBCD) = √s(s-a)(s-b)(s-c)


⇒ Area(ΔBCD) = √16(16-10)(16-10)(16-12)


⇒ Area(ΔBCD) = √16×6×6×4


⇒ Area(ΔBCD) = 48 cm2


As ABCD is a rhombus, Area(ΔBCD) = Area(ΔABD)


⇒ Area of rhombus(ABCD) = Area(ΔBCD) + Area(ΔABD)


⇒ Area of rhombus(ABCD) = 48 + 48


⇒ Area of rhombus(ABCD) = 96cm2
Question 10.

Find the area of the trapezium PQRS with height PQ given in Fig. 12.3
Answer:


We draw perpendicular from R on PS, which is also parallel to PQ.


PQRT is a rectangle.


ST = PS – TP


ST = 12 – 7 = 5m (TP = RQ = 7m)


In Δ RTS,


RS2 = ST2 + RT2


⇒ 132 = 122 + RT2


⇒ 169 = 144 + RT2


⇒ RT2 = 169-144


⇒ RT2 = 25


⇒ RT = 5m


Area (PQRS) =


Area(PQRS) =


Area(PQRS) = 9.5 × 5 = 47.5 m2
Exercise 12.4
Question 1.

How much paper of each shade is needed to make a kite given in Fig. 12.4, in which ABCD is a square with diagonal 44 cm.
Answer:

AC = BD = 44cm

AO = 44/2 = 22cm


BO = 44/2 = 22cm


In ΔAOB,


AB2 = AO2 + BO2


⇒ AB2 = 222 + 222


⇒ AB2 = 2 × 222


⇒ AB = 22√ 2 cm


Area of square = (Side)2 = (22√2)2 = 968 cm2


Area of each triangle (I, II, III, IV) = Area of square /4 = 968 /4 = 242 cm2


Area of lower triangle,


a = 28, b = 28, c = 14


s = (a + b + c)/2


⇒ s = (28 + 28 + 14)/2 = 70/2 = 35.


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √35(35-28)(35-28)(35-14)


⇒ Area(Δ) = √35×7×7×21


⇒ Area(Δ) = 49√15 = 189.77cm2


Area of Red = Area of IV = 242 cm2


Area of Yellow = Area of I + Area of II = 242 + 242 = 484 cm2


Area of Green = Area of III + Area of lower triangle = 242 + 189.77 = 431.77 cm2
Question 2.

The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. Find the area of the triangle.
Answer:

Let the smaller side be x cm

Larger side = (x + 4) cm


Third side = (2x-6) cm


Perimeter = 50 cm


⇒ x + (x + 4) + (2x-6) = 50


⇒ 4x-2 = 50


⇒ 4x = 52


⇒ x = 13


Therefore, smaller side = 13cm


Larger side = x + 4 = 13 + 4 = 17cm


Third side = 2x-6 = 2×13 – 6 = 26-6 = 20cm


a = 13, b = 17, c = 20


s = (a + b + c)/2


⇒ s = (13 + 17 + 20)/2 = 50/2 = 25.


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √25(25-13)(25-17)(25-20)


⇒ Area(Δ) = √25×12×8×5


⇒ Area(Δ) = 20√30 cm2
Question 3.

The area of a trapezium is 475 cm2 and the height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.
Answer:


Let PQRS be the given trapezium,


Given PQ = 19cm


Let RQ be x cm


Then, PS = (x + 4)cm


We draw Perpendicular from R on PS which will also be parallel to PQ.


Now, PQRT is a rectangle,


Area(PQRT) = PQ × QR


⇒ Area(PQRT) = 19x


Now,


PS = PT + TS


⇒ (x + 4) = x + TS (As PT = QR = x cm)


⇒ TS = 4cm


Area(ΔRST) = 1/2 × RT × ST


⇒ Area(ΔRST) = 1/2 × 19 × 4 = 38cm2 (As RT = PQ = 19cm)


Area(PQRS) = Area(PQRT) + Area(ΔRST)


⇒ 475 = 19x + 38


⇒ 19x = 475 -38


⇒ 19x = 437


⇒ x = 23 cm


(x + 4) = 23 + 4 = 27cm


∴ Lengths of parallel sides is 23cm and 27cm.
Question 4.

A rectangular plot is given for constructing a house, having a measurement of 40 m long and 15 m in the front. According to the laws, a minimum of 3 m, wide space should be left in the front and back each and 2 m wide space on each of other sides. Find the largest area where house can be constructed.
Answer:


Let PQRS be the given rectangle with PQ = 40m and QR = 15m


As 3m has be left in both front and back,


AB = PQ -3 -3


⇒ AB = 40 -6


⇒ AB = 34m


Also, as 2m has to be left at both the sides,


BC = QR -2 – 2


⇒ BC = 15 -4


⇒ BC = 11m


Area left for house construction is ABCD.


Area(ABCD) = AB × CD


Area(ABCD) = 34 × 11


Area(ABCD) = 374 m2
Question 5.

A field is in the shape of a trapezium having parallel sides 90 m and 30 m. These sides meet the third side at right angles. The length of the fourth side is 100 m. If it costs Rs 4 to plough 1m2 of the field, find the total cost of ploughing the field.
Answer:


PQ = 90m


RS = 30m


QR = 100m


PQ = PT + TQ = 90m


⇒ 30 + TQ = 90


⇒ TQ = 60


In ΔTRQ,


QR2 = TR2 + QT2


⇒ 1002 = TR2 + 602


⇒ TR2 = 10000-3600


⇒ TR2 = 6400


⇒ TR = 80m


Area (PQRS) =


Area(PQRS) =


Area(PQRS) = 60 × 80 = 4800 m2


Cost of ploughing = Area(PQRS) × Cost per m2


∴ Cost of ploughing = 4800 × 4 = Rs. 19200.
Question 6.

In Fig. 12.5, ΔABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC a parallelogram DBCE of same area as that of ΔABC is constructed. Find the height DF of the parallelogram.
Answer:

a = 6.5, b = 7, c = 7.5

s = (a + b + c)/2


⇒ s = (6.5 + 7 + 7.5)/2 = 21/2 = 10.5


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √10.5(10.5-6.5)(10.5-7)(10.5-7.5)


⇒ Area(Δ) = √10.5×4×3.5×3


⇒ Area(Δ) = 21 cm2


Area of parallelogram (BCED) = Area (Δ)


⇒ BC × DF = 21


⇒ 7 × DF = 21


⇒ DF = 3 cm
Question 7.

The dimensions of a rectangle ABCD are 51 cm × 25 cm. A trapezium PQCD with its parallel sides QC and PD in the ratio 9:8, is cut off from the rectangle as shown in the Fig. 12.6. If the area of the trapezium PQCD is th part of the area of the rectangle, find the lengths QC and PD.
Answer:

Area of rectangle (ABCD) = BC × CD

⇒ Area of rectangle (ABCD) = 51 × 25 = 1275 cm2


Area of trapezium PQCD = 5/6 × Area of rectangle (ABCD)


⇒ Area of trapezium PQCD = 5/6 × 1275 = 1062.5 cm2


Given that QC:PD = 9:8


Let QC = 9x and PD = 8x


Area (PQCD) =


⇒ Area (PQCD) =


⇒ 1062.5 =


⇒ 85 = 17x


⇒ x = 5cm


QC = 9x = 45cm


PD = 8x = 40 cm
Question 8.

A design is made on a rectangular tile of dimensions 50 cm × 70 cm as shown in Fig. 12.7. The design shows 8 triangles, each of sides 26 cm, 17 cm and 25 cm. Find the total area of the design and the remaining area of the tile.
Answer:

For area of one triangle,

a = 25, b = 17, c = 26


s = (a + b + c)/2


⇒ s = (25 + 17 + 26)/2 = 68/2 = 34


Area(Δ) = √s(s-a)(s-b)(s-c)


⇒ Area(Δ) = √34(34-25)(34-17)(34-26)


⇒ Area(Δ) = √34×9×17×8


⇒ Area(Δ) = 204 cm2


Area of design = 8 × Area(Δ)


∴ Area of design = 8 × 204 = 1632 cm2


Area of remaining tile = Area of rectangle – Area of design


⇒ Area of remaining tile = (50×70) – 1632


⇒ Area of remaining tile = 3500 – 1632 = 1868 cm2