RS Aggarwal Solutions Class 10 Perimeter And Areas Of Plane Figures

Q.1: Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.

Sol:

Area of given triangle = \(\frac{1}{2}\times Base\times Height\) = \((\frac{1}{2}\times 24\times 14.5)cm^{2}=174cm^{2}\)

Q.2: Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm. Also, find the height corresponding to the longest side.

Sol: Let a = 42cm, b = 34cm and c = 20cm

Then, s = \(\frac{1}{2}(42+34+20)cm\) = 48cm

(s – a) = 6cm, (s – b) = 14cm and (s – c) = 28cm

(i) Area of triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) = \(\sqrt{48\times 6\times 14\times 28}\;cm^{2}=336\;cm^{2}\)

(ii) Let base = 42 cm and corresponding Height = h cm

Then area of triangle = \((\frac{1}{2}\times 42\times h)cm^{2}=(21h)cm^{2}\)

21h = 336 i.e. h = \(\frac{336}{21}=16cm\)

Hence, the height corresponding to the longest side = 16cm

Q.3: Find the area of the triangle whose sides are 18 cm, 24 cm and 30 cm. Also, find the height of the corresponding to the smallest side.

Sol: Let a = 18cm, b = 24cm, c = 30cm

Then, 2s = (18 + 24 + 30) cm = 72 cm

s = 36cm

(s – a) = 18cm, (s – b) = 12cm and (s – c) = 6cm

(i) Area of triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) = \(\sqrt{36\times 18\times 12\times 6}\;cm^{2}\) = \(216\;cm^{2}\)

(ii) Let base = 18 cm and altitude = x cm

Then, area of triangle = \((\frac{1}{2}\times 18\times x)=9x\;cm^{2}\)

9 x = 216 x = \(\frac{216}{9}\) = 24

Hence, altitude corresponding to the smallest side = 24 cm

Q.5: The sides of the triangle are in the ratio 5 : 12 : 13 , and it perimeter is 150 m. Find the area of the triangle.

Sol: On dividing 150m in the ratio 5 : 12 : 13, we get

Length of one side = \((150\times \frac{5}{30})m=25m\)

Length of the second side = \((150\times \frac{12}{30})m=60m\)

Length of third side = \((150\times \frac{13}{30})m=65m\)

Let a = 25m, b = 60m, c = 65m

Then, s = \(\frac{1}{2}(25+60+65)m=75m\)

(s – a) = 50cm, (s – b) = 15cm, and (s – c) = 10cm

Area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) = \(\sqrt{75\times 50\times 15\times 10}\;m^{2}\) = \(750\;m^{2}\)

Hence, area of the triangle = \(750\;m^{2}\)

Q.6: The perimeter of the triangular field is 540m, and its sides are in the ratio 25 : 17 : 12. Find the area of the field. Also find the cost of plugging the field at Rs. 40 per 100 \(m^{2}\).

Sol: On dividing 540m in ratio 25 : 17 : 12, we get

Length of one side = \((540\times \frac{25}{54})m=250m\)

Length of second side = \((540\times \frac{17}{54})m=170m\)

Length of third side = \((540\times \frac{12}{54})m=120m\)

Let a = 250m, b = 170m and c = 120m

Then, s = \(\frac{1}{2}(250+170+120)m=270m\)

Then, (s – a) = 29m, (s – b) = 100m, and (s – c) = 150m

Area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{270\times 29\times 100\times 150}\;m^{2}\) = \(9000\;m^{2}\)

The cost of plugging 100 area is = Rs. 18.80

The cost of plugging 1 is Rs. \(\frac{18.80}{100}\)

The cost of plugging 9000 area = Rs. \((\frac{18.80}{100}\times 9000)\) = Rs. 1692

Hence, area of the triangle = \(60cm^{2}\)

Q.7: The difference between the sides at right angles in right angled triangle is 7 cm. The area of the triangle is 60 \(cm^{2}\).Find its perimeter.

Sol: Let the sides containing the right angle be x cm and (x – 7) cm

Then, its area = [1/2 * x * (x-7)] \(cm^{2}\)

But, area = 60 \(cm^{2}\)

Therefore, 1/2 * (x-7) = 60

\(x^{2}-7x-120=0\)

i.e \(x^{2}-15x+8x-120=0\)

x(x – 15) + 8(x – 15) = 0

(x – 15)(x + 8) = 0

x = 15[Neglecting x = -8]

One side = 15cm and other = (15 * 7) cm = 8 cm

Hypotenuse = \(\sqrt{(15)^{2}+(8)^{2}}cm\) = \(\sqrt{225+64}cm\)

\(\sqrt{289}cm\) = 17 cm

Perimeter of triangle (15 + 8 + 17) cm = 40 cm

Q.8: The length of the two sides of a right triangle containing the right angle differ by 2 cm. If the area of the triangle is 24 \(cm^{2}\) , find the perimeter of the triangle.

Sol: let the sides containing the right angle be x and (x – 2) cm

Then, its area = (1/2 * x * (x – 2)) \(cm^{2}\)

But area = 24\(cm^{2}\)

Therefore, 1/2 * (x – 2) = 24

\(x^{2}-2x-48=0\)

i.e. \(x^{2}-8x+6x-48=0\)

x(x – 8) + 6(x – 8) =0

(x – 8)(x + 6) = 0

x = 8 [Neglecting x = -6]

One side = 8cm, and other (8 * 2) cm = 6cm

Hypotenuse = \(\sqrt{(8)^{2}+(6)^{2}}cm\) = \(\sqrt{64+36}cm\) = \(\sqrt{100}cm\)

Therefore, perimeter of the triangle = 8 + 6 +10 = 24 cm

Q.9: Each side of the equilateral triangle is 10cm. Find (i) the area of the triangle and (ii) the height of the triangle.

Sol: Here a = 10 cm

Area of the triangle = \((\frac{\sqrt{3}}{4}\times a^{2})\) SQ.. unit

= \((\frac{\sqrt{3}}{4}\times 10\times 10)=43.3cm^{2}\)

(ii) Height of the triangle = \((\frac{\sqrt{3}}{4}\times a)\) SQ.. units

= \((\frac{\sqrt{3}}{4}\times 10)=4.3cm\)

Hence, area = 43.3cm2 and height = 4.3 cm

Q.10: The height of an equilateral triangle is 6 cm. Find its area. Take \( \sqrt{3} = 1.73 \).

Sol: Let the sides of the equilateral triangle be x cm.

As, the area of an equilateral triangle = \(\frac{\sqrt{3}}{4} \; (side)^{2} = \frac{x^{2}\sqrt{3}}{4}\)

Also, the area of the triangle = \(\frac{1}{2}\times base\times height = \frac{1}{2}\times x\times 6 = 3x\)

\(\frac{x^{2}\sqrt{3}}{4}= 3x\)

\(\frac{x \sqrt{3}}{4}= 3 \)

\(x= \frac{12}{\sqrt{3}}\)

\(x= \frac{12}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\)

\(x= \frac{12 \sqrt{3}}{3} = 4\sqrt{3}\) cm

Now, area of the equilateral triangle = 3x

= \(x= 3 \times 4\sqrt{3}\) cm

\(= 12\sqrt{3}\) cm

\(= 12 \times 1.73\) = \(20.76 cm^{2}\)

Q.11: If the area of an equilateral triangle is \(36\sqrt{3}\) , find its height.

Sol: Let each side of the equilateral triangle be a cm

Then, its area = \((\frac{\sqrt{3}}{4}\times a^{2})\) cm

Therefore, \((\frac{\sqrt{3}}{4}\times a^{2})\) = \(36\sqrt{3}\Rightarrow a^{2}=(\frac{36\sqrt{3}\times 4}{\sqrt{3}})=144\)

a = 12 cm

Perimeter of equilateral triangle = 3a = (3 * 12) cm = 36 cm.

Q.12: If the area of an equilateral triangle is \(81\sqrt{3}\), find its height.

Sol: Let each side of the equilateral triangle be a cm

Area of equilateral triangle = \((\frac{\sqrt{3}}{4}\times a^{2})\) cm

\((\frac{\sqrt{3}}{4}\times a^{2})\) = \(81\sqrt{3}\Rightarrow a^{2}(\frac{81\sqrt{3}\times 4}{\sqrt{3}})=324\)

a = \(\sqrt{324}\) = 18cm

Height of equilateral triangle = \(\frac{\sqrt{3}}{2}a=(\frac{\sqrt{3}}{2}\times 18)cm=9\sqrt{3}cm\)

Q.13: The base of a right angled triangle measures 48 cm and its hypotenuse measures 50 cm. find the area of the triangle.

Sol: Base of right angled triangle = 48 cm

Height of the right angled triangle = \(\sqrt{(hypotenuse)^{2}-(base)^{2}}\)

Height = \(\sqrt{(50)^{2}-(48)^{2}}cm\) = \(\sqrt{2500-2304}cm\) = \(\sqrt{196}cm\) = 14cm

Area of triangle = \((\frac{1}{2}\times 48\times 14)cm^{2}\) = \(336\;cm^{2}\)

Q.14: The hypotenuse of a right-angled triangle is 65 cm and its base is 60cm. Find the length of perpendicular and the area of the tringle.

Sol:

Hypotenuse = 65 cm

Base = 60 cm

In a right-angled triangle,

\((Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2}\)

\((65)^{2} = (60)^{2} + (Perpendicular)^{2}\)

\((Perpendicular)^{2} = (65)^{2} – (60)^{2}\)

\((Perpendicular)^{2} = (65 – 60 )(65 + 60)\)

\((Perpendicular)^{2} = 5 \times 125\)

\((Perpendicular)^{2} = 625\)

\((Perpendicular) = 25\)

Area of triangle = \(\frac{1}{2} \times base \times perpendicular\)

\(\frac{1}{2} \times 60 \times 25\)

\(= 750 cm^{2}\)

Q.15: Find the area of a right-angled triangle, the radius of whose circumcircle measures 8cm and the altitude drawn to the hypotenuse measures 6cm.

Sol: The circumference of a right angled triangle is the midpoint of the hypotenuse

https://lh4.googleusercontent.com/xLiMO77mL0ms_m97WLeoZ-UZ7KAO_sbgCDiLKNZb5bIMNVc8LAR7Y8eCvh7UJALaUI5qn0qBTkigelqDVKUkr0QAeQAmRYK226tZhgzjyXm6eZrWbSue4UOZUd5F1w2TCp0hrA15zXPRtWDbsw

Hypotenuse = 2 * (radius of circumcircle) = (2 * 8) cm = 16 cm

Base = 16 cm, height = 6 cm

Area of right angled triangle = (1/2 * base * height) = (1/2 * 16 * 6) cm = \(48\;cm^{2}\)

Q.16: Find the length of the hypotenuse of an isosceles right angled triangle whose area is \(200\;cm^{2}\) . Also find its perimeter.

Sol: Let each eQ.ual side be a cm in length.

Then,

1/2 * a * a =200 → a = 20 cm

Hypotenuse (h) = \(\sqrt{a^{2}+a^{2}}\;cm\)

= \(a\sqrt{2}cm=20\sqrt{2}cm\) = (20 * 1.414) cm = 28.28 cm

Therefore, Perimeter of the triangle = (2a = h) cm = (2 * 20 + 28.28) cm =68.28 cm

Hence, hypotenuse =28.28 cm and perimeter = 68.28 cm

Q.17: The base of an isosceles triangle measures 80 cm and its area is \(360\;cm^{2}\).Find the perimeter of the triangle.

Sol: Let each eQ.ual side be ‘a’ cm and base = 80 cm

Area = \(\frac{1}{4}b\times \sqrt{4a^{2}-b^{2}}\) square units

=\(\frac{1}{4}\times 80\times \sqrt{4a^{2}-6400}\;cm^{2}\) = \(20\times \sqrt{4a^{2}-6400}\;cm^{2}\)

But area = \(360\;cm^{2}\)

Therefore, \(20\sqrt{4a^{2}-6400}=360\)

\(20\times 2\sqrt{a^{2}-1600}=360\)

\(\sqrt{a^{2}-1600}=9\)

\(a^{2}-1600=81\)

\(a^{2}=1681\)

Therefore, a = 41 cm

Perimeter of triangle = (2a + b) cm

= (2 * 41 + 80) cm = (82 + 80) cm = 162 cm

Hence, perimeter of the triangle = 162 cm

Q.18: Each of the eQ.ual sides isosceles triangle measures 2 cm more than its height, and the base of the triangle measures 12 cm. Find the area of the triangle?

Sol: Let the height be h cm, then a = (h + 2) cm and b = 12 cm

1/2 *12*h = 1/4 * 12 * \(\sqrt{4(h+2)^{2}-144}\)

6h = \(6\sqrt{(h+2)^{2}-36}\)

h = \(\sqrt{(h+2)^{2}-36}\)

On squaring both sides:

\(h^{2}=(h+2)^{2}-36\)

\(h^{2}=h^{2}+4+4h-36\)

-4h = -32 i.e. h = 8cm

Therefore, a = h+2 = (8+2)cm = 10 cm

Area of isosceles triangle = \(\frac{1}{4}b\times \sqrt{4a^{2}-b^{2}}\)

\(\frac{1}{4}\times 12\times \sqrt{4\times (10)^{2}-(12)^{2}}\)

= \(3\sqrt{400-144}=3\times \sqrt{256}\) = 3 * 16 = 48 \(cm^{2}\)

Hence, area of the triangle = 48 \(cm^{2}\)

Q.19: Find the area and perimeter of an isosceles right triangle, each of whose eQ.ual sides measures 10cm.

Sol: Let \(\bigtriangleup ABC\) is a isosceles triangle. Let AC, BC be the equal sides

Then AC = BC = 10cm. Let AB be the base of \(\bigtriangleup ABC\) right angle at C.

AB = \(\sqrt{AC^{2}+BC^{2}}=\sqrt{(10)^{2}+(10)^{2}}\;cm^{2}\)

= \(\sqrt{20}cm=10\sqrt{2}cm\)

Perimeter = (2a + b) square units

= (2*10 + \(10\sqrt{2}\))cm = (20 +10 *1.414)cm = (20 + 14.14) cm = 34.14 cm

Area of isosceles triangle ABC

= 1/2 * 10 *10 \(cm^{2}\) = 50 \(cm^{2}\)

Hence, the area = 50 \(cm^{2}\) and perimeter = 34.14 cm

Q.20: In the given figure, \(\bigtriangleup ABC\) is an equilateral triangle the length of whose side is eQ.ual to 10 cm, and \(\bigtriangleup DBC\) is the right angled triangle at D and BD = 8cm. Find the area of the shaded region.

https://lh6.googleusercontent.com/8Rb7KARgDwAwxvolCdoQdgh3wSnqxXyr8S_T8jtOwO1vIIeKhJeQ2EAlLkXYufL1Wz7OChxZVrK6JMrRQdONA9n_aQUfMTuttOo6Wl2OYQK1VYLnE8JQ7F8v3387J6Z-8J1t_yWNstYR9VgvBw

Sol: Area of shaded region = Area of \(\bigtriangleup ABC\) – Area of \(\bigtriangleup DBC\)

First we find area of \(\bigtriangleup ABC\)

Therefore, Area = \(\frac{\sqrt{3}}{4}a^{2}=(\frac{\sqrt{3}}{4}\times 10\times 10)cm^{2}\) = 43.30 \(cm^{2}\)

Second we find area of \(\bigtriangleup DBC\) which is right angled

Therefore area of triangle DBC = 1/2 * Base * Height

Height = \(\sqrt{BC^{2}-DB^{2}}=\sqrt{10^{2}-8^{2}}\) = \(\sqrt{100-64}=\sqrt{36}=6\;cm\)

Therefore, Area = 1/2  *DB *DC = (1/2  * 8 * 6)  \(cm^{2}\) = 24 \(cm^{2}\)

Area of shaded region = Area of \(\bigtriangleup ABC\) – Area of \(\bigtriangleup DBC\) = (43.30 – 24) = 19.30

Area of shaded region = 19.3

Exercise – 17B

Q.1: The perimeter of a rectangular plot of land is 80 m and its breadth is 16 m. find the length and area of the plot.

Sol: Let the length of plot be x meters Its perimeter = 2 [length + breadth]

= 2(x + 16) = (2x +32) meters

Therefore, (2x + 32) = 75 i.e. 2x = 75 – 32

2x = 43 i.e. x = 43/2 = 21.5

Length of the rectangle is 21.5 meter

Area of the rectangular plot = length * breadth = (16 * 21.5) \(m^{2}\)

= 344 \(m^{2}\)

The length = 21.5 m and the area = 344 \(m^{2}\)

Q.2: The length of a rectangular park is twice its breadth and its perimeter is 840. Find the area of the park.

Sol: Let the breadth of a rectangular park be x meters

Then, its length = 2x meter

Therefore, perimeter = 2(length + breadth) = 2(2x + x) = 6x meters

Therefore, 6x = 840m [1km = 1000m]

i.e x = 140m

Then, breadth = 140m and length = 280m

Area of rectangular park = (length * breadth) = (140 * 280) \(m^{2}\)

= 39200 \(m^{2}\)

Hence, area of the park = 39200 \(m^{2}\)

Q.3: One side of a rectangle is 12cm long and its diagonal measures 37cm. Find the other side and the area of the rectangle.

Sol: Let ABCD be the rectangle in which in which AB =12cm and AC = 37cm

https://lh3.googleusercontent.com/zb3StagmwlQgmZF5BlscwMtI3RQtDz5iwJeSINmVA_-CHE4ElvTFY46MaCcuViT22O-eCBeg_YhcQq170L4xxsRuHToRJA-Qx9MEE39Cf9wTJ2LU0tmqSKVV7Np1RxGGYqTs-EH8AUfJdusicA

By Pythagoras theorem, we have BC = \(\sqrt{AC^{2}-AB^{2}}\) units

= \(\sqrt{(37)^{2}-(12)^{2}}\;cm^{2}\) = \(\sqrt{(37+12)(37-12)}\;cm\) = \(\sqrt{49\times 25}\;cm\) = \(\sqrt{1225}\;cm\) = 35 cm

Thus, length = 35 cm and breadth = 12cm

Area of rectangle = (12 * 35) \(cm^{2}\) = 420 \(cm^{2}\)

Q.4: The area of a rectangular plot is 462 \(m^{2}\) and its length is 28 m. Find its perimeter.

Sol: Let the breadth of the plot be x meter

Area = Length * Breadth = (28 * x) meter

= 28x \(m^{2}\)

Therefore, 28x = 462 i.e. x = 462/28 = 16.5 m

Breadth of plot is = 16.5 m

Perimeter of the plot is 2(length + breadth) = 2(28 + 16.5) m = 2 (44.5) m = 89 m

Q.5: A lawn is in the form of a rectangle whose sides are in the ratio 5 : 3. The area of the lawn is 3375 \(m^{2}\) . Find the cost of fencing the lawn at Rs. 65 per meter.

Sol: Let the length of lawn be 5x m and breadth of the lawn be 3x m

Area of rectangular lawn = (5x * 3x) \(m^{2}\) =

(\(15x^{2}\) ) \(m^{2}\)

Area of lawn = 3375 \(m^{2}\)

\(15x^{2}=3375\Rightarrow x^{2}=\frac{3375}{15}=225\)

\(x=\sqrt{225}\) =15 m

Length = 5 * 15 = 75

Breadth = (3 *15)m = 45 m

Perimeter of lawn = 2(length + breadth) = 2(75 +45)m =240 m

Cost of fencing the lawn per meter = Rs. 8.50 per meter

Cost of fencing the lawn = Rs. 8.50 * 240 = Rs. 2040

Q.6: A room is 16m long and 13.5m broad. Find the cost of covering its floor with 75m wide carpet at Rs. 60 per meter.

Sol: Length of the floor = 16 m

Breadth of the floor = 13.5 m

Area of floor = (16 * 13.5) \(m^{2}\)

Length of the carpet = \(\frac{Area\;of\;floor}{width\;of the\;carpet}\) =\(\frac{16\times 13.5}{0.75}m=288m\)

Cost of carpet = Rs. 15 per meter

Cost of 288 meters of carpet = Rs. (15 * 288) = Rs. 4320

Q.7: The floor of a rectangular hall is 24 m long 18m wide. How many carpets, each of the 2.5m and breadth 80cm, will be required to cover the floor of the hall?

Sol: Area of floor = length * breadth = (24 * 18) \(m^{2}\)

Area of carpet = Length * breadth = (2.5 * 0.8) \(m^{2}\)

Number of carpets = \(\frac{Area\;of\;floor}{Area\;of\;carpet}=\frac{(24\times 18)m^{2}}{(2.5\times 0.8)m^{2}}\) = 216

Hence the number of carpet pieces required = 216

Q.8: A 36m long, 15m broad verandah is to be paved with stones, each measuring 6dm by 5dm. How many stones will be required?

Sol: Area of verandah = (36 * 15) \(m^{2}\) = 540 \(m^{2}\)

Area of stone = (0.6 * 0.5) \(m^{2}\) [10dm = 1m]

Number of atones required = \(\frac{Area\;of\;verandah}{Area\;of\;stone}=\frac{540}{0.3}=1800\)

Hence, 1800 stones are required to pave the verandah.

Q.9: The area of a rectangle is 192\(cm^{2}\) and its perimeter is 56cm. Find the dimension of the rectangle.

Sol: Perimeter of rectangle = 2(l + b)

2(l + b) = 56 i.e. l + b = 28cm

b = (28 – l) cm

Area of rectangle = 192 \(cm^{2}\)

l * (28 – l) = 192

28l – \(l^{2}\) = 192

\(l^{2}-28l+192=0\)

\(l^{2}-16l-12l+192=0\)

l(l – 16) – 12(l – 16) = 0

(l – 16)(l – 12) = 0

l = 16 cm or l = 12

Therefore, length = 16cm and breadth = 12cm

Q.10: A rectangular park 35m long and 18m wide is to be covered with grass, leaving 2.5m uncovered all around it. Find the area to be laid with grass.

Sol: Length of the park = 35 m

Breadth of the park = 18 m

https://lh5.googleusercontent.com/u-Ee7IbvMbiZi2ItfDH9b4vbNv46bmOl36vOj0UQc8dHCssv1G5d1zFh20Df3Rr1_eoCAczVRPpsZPnFo36KiqAYACXZPgzggqekfWZQD-05OhqreXBYuvs2A-map-7Th8G-qpJ_ngEMhL1QWg

Area of the park = (35 * 18) \(m^{2}\) = 630 \(m^{2}\)

Length of the park with grass = (35 – 5) = 30 m

Breadth of the park with grass = (18 – 5) m = 13 m

Area of park with grass = (30 * 13) \(m^{2}\) =390 \(m^{2}\)

Area of path without grass = area of the whole park – area of park with grass = 630 – 390 = 240 \(m^{2}\)

Hence, area of the park to be laid with grass = 240 \(m^{2}\)

Q.11: A rectangular plot measures 125m by 78 m. It has a gravel path 3m wide all around on the outside. Find the area of the area of the path and the cost of gravelling it at Rs. 75 per \(m^{2}\).

Sol: Length of the plot = 125 m

Breadth of the plot = 78 m

https://lh6.googleusercontent.com/_UIX-Qz4us9bpafPxsZwNt-xyxVUnVblv3G7AbbfTdzEfjUIEeZOXu5O47G6CY0A1kpVPEqf-evqh5rDVNfASct5mi4xbvWJAa8QApZBxGC-j81jR8UBOHP5rASQMYlQ7GMSAMa_S6iryrelWQ

Area of plot ABCD = (125 * 78) \(m^{2}\) = 9750 \(m^{2}\)

Length of the plot including the path = (125 + 3 + 3) m = 131 m

Breadth of the plot including the path = (78 + 3 + 3) m = 84 m

Area of plot PQRS including the path = (131 * 84) \(m^{2}\) = 11004 \(m^{2}\)

Area of path = Area of plot PQRS – Area of plot ABCD

= (11004 – 9750) \(m^{2}\) = 1254 \(m^{2}\)

Cost of gravelling = Rs. 75 per \(m^{2}\)

Cost of gravelling the whole path = Rs. (1254 * 75) = Rs. 94050

Hence, cost of gravelling the path = Rs. 94050


Practise This Question

Floating generators are used in the sea to harness ______