RS Aggarwal Solutions for Class 10 Maths Chapter 17 Exercise 17.2

RS Aggarwal Solutions for Class 10 Maths Chapter 17 Exercise 17.2: RS Aggarwal Solutions for Class 10 Maths Chapter 17 Exercise 17.2 provide detailed guidance on solving problems related to the perimeter and areas of plane figures. This exercise focuses on more complex problems, including the calculation of areas and perimeters of various geometric shapes such as triangles, parallelograms, trapeziums, and circles.

RS Aggarwal Maths Solutions Chapter 17 Exercise 17.2 

RS Aggarwal Solutions for Class 10 Maths Chapter 17 Exercise 17.2 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 17 Exercise 17.2 PDF

RS Aggarwal Solutions Chapter 17 Perimeter and Areas of Plane Figures Exercise 17.2

Q. 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.

 

perimeter=80 $2(l+b)=80$ $l+b=40$ $l+16=40$ $l=40-16=24$ $Area=l\times b=24\times 16=\mathrm{384m}$

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

Solution:

 

Let $l=2b$ $2(l+b)=840$ $2(2b+b)=840$ $6b=840$ $b=\frac{840}{6}=140$ so $l=140\times 2=280$ $Area=l\times b=140\times 280=39200m^{}$

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

Solution:

 

Using Pythagoras theorem, diagonal is the hypotenuse Other side $=\sqrt{(37\times 37)-(12\times 12)}=\sqrt{(1369-144)}=35cm$ $Area=12\times 35=420m^{}$

Q. The area of a rectangle plot is 462 $m^{}$ and its length is 28 m. Find its perimeter.

Solution:

 

Area of rectangle = $length\times breadth=462m^{}$ Length = 28 m $28\times b=462$ $b=\frac{462}{28}$ b= 16.5m Perimeter = 2(l+b) = 2(28+16.5) =2 $\times$ 44.5 =89m

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

Solution:

 

area of room = length $\times$ breadth=16 $\times$ 13.5 = 216 $m^{}$ width of carpet = 75 m length of carpet required = $\frac{216}{75}$ =2.88 m cost = 60 $\times$ 2.88= 172.8 rupees

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

Solution:

 

area of path = area of whole field - area of rec. plot = 131 $\times$ 84 - 125 $\times$ 78 =11004 - 9750 = 1254 $m^2$ cost of gravelling = 1254 $\times$ 75 =Rs. 94050 Q. (i) A footpath of uniform width runs all around the inside of a rectangular field 54 m long and 35 m wide. If the area of the path is 420 $m^{}$ , find the width of the path. (ii) A carpet is laid on the floor of a room 8 m by 5m. There is a border of constant width all around the carpet. If the area of the border is 12 $m^{}$ , find its width.

Solution:

 

(i)

Let width = x m Area of path = ar(EFGH) - ar(ABCD) $420=54\times 35-(54-2x)(35-2x)420=1890-[54\times 34-108x-70x+4x^2]420=1890-1890+178x+4x^{2}4x^2-178x+420=02x^2-89x+210=0D=b^2-4ac=(-89{)}^2-4(2\left)\right(210)=7921-1680=6241x=\frac{-b\pm \sqrt{D}}{2a}=\frac{-(-89)\pm \sqrt{6241}}{2\times 2}=\frac{89\pm 79}{4}=\frac{89+79}{4}or\frac{89-79}{4}=\frac{168}{4}or\frac{10}{4}=42or2.5$ Width of path cannot be 42, So width of path = 2.5 m

(ii)

Let width of carpet be 'x'm Therefore, inclusive of border: Length of room = 8m Breadth of room = 5m Length of carpet = 8-x-x = (8-2x) m Breadth of carpet = 5-x-x = (5-2x) m Area of room (carpet + border) = 8 $\times$ 5 = 40 sq.m Area of just carpet \)= (8-2x)(5-2x) \ = (40-16x-10x+4x^2)\= (4x^2-26x+40)~m^2\) Area of just border = 12sq.m (given) Therefore, Area of ground = Area of just border + Area of just carpet $\Rightarrow 40=12+(4x^2-26x+40)\Rightarrow 0=4x^2-26x+12\Rightarrow 0=2x^2-13x+6\text{(dividing equation by 2)}\Rightarrow 0=2x^2-12x-x+6\Rightarrow 0=2x(x-6)-1(x-6)\Rightarrow 0=(2x-1)(x-6)\Rightarrow x=\frac{1}{2}=0.5mor6m$ discarding 6m since it is longer than room then Therefore, width of border = 0.5m

Q. The length an breadth of a rectangular garden are in the ratio 9:5.A path 3.5 m wide, running all around inside it has an area of 1911 ${\mathrm{m.}}^{}$ Find the dimensions of the garden.

Solution:

 

Given: Length and breadth of the rectangular garden are in the ratio 9: 5 and the width of the path is 3.5 m. Let the length and breadth of the garden be 9xand 5x respectively. The length and breadth of the inner rectangle are $9x–2\times 3.5$ and $5x–2\times 3.5$ ie, $9x–7$ and $5x–7$ Now area of garden $=9x\times 5x=45x^2$ and area of the inner rectangle $=(9x–7)(5x–7)$ $=45x^2–63x–35x+49$ $=45x^2–98x+49$ also the area of path = 1911 $m^{}$ ⇒ $45x^2–(45x^2–98x+49)=1911$ ⇒ $98x-49)=1911$ ⇒ $49(2x–1)=1911$ ⇒ $2x–1=39$ ⇒ $2x=40\Rightarrow x=20$ Length $=9x=9\times 20=180$ Breadth $=5x=5\times 20=100$

Q. A room 4.9 m long and 3.5 m broad is covered with carpet, leaving an uncovered margin of 25 cm all around the room. If the breadth of the carpet is 80 cm, find its cost at Rs.80 per metre.

Solution:

 

A room is covered with carpet of length = 4.9 m and breadth = 3.5 m. leaving an uncovered margin of 25cm or 0.25 m all around the room. so, length = original length - 2 $\times$ uncovered margin = 4.9 m - 2 $\times$ 0.25 m = 4.4 m Breadth = original breadth -2 $\times$ uncovered margin = 3.5 m - 2 $\times$ 0.25 m = 3 m given, breadth of the carpet is 0.8 m Then, length of carpet = $\frac{\text{area of carpet}}{\text{breadth of carpet}}$ = 4.4 m $\times \frac{3m}{0.8m}$ = 16.5 m So, length of carpet = 16.5 m Now, cost = length of carpet $\times$ rate = 16.5 $\times$ 80 = ₹ 1320 Hence, the total cost = ₹ 1320

Q. An 80 m by 64 m rectangular lawn has two roads, each 5 m wide, running through its middle, one parallel to its length and the other parallel to its breadth. Find the cost of gravelling the roads at Rs.40 per ${\mathrm{m.}}^{}$

Solution:

 

$\text{Area of the road}=ar\left(KLMN\right)+ar\left(PQRS\right)-ar\left(EFGH\right)$ $=(64\times 5)+(80\times 5)-(5\times 5)$ $=320+400-25$ $=695m^{}$ $\text{Cost of gravelling the roads per}{\mathrm{m\; =\; 40}}^{}$ $\text{Total cost of gravelling the roads}$ $=695\times 40=Rs.27800$

Q. The cost of painting the four walls of a room 12 m long ar Rs.30 per $m^2$ is Rs.7560 and the cost of covering the floor with mat at Rs.25 per $m^2$ is Rs.2700. Find the dimensions of the room.

Solution:

 

Q. Find the length of the diagonal of a square whose area is 128 $c{m}^{}$ . Also, find its perimeter.

Solution:

 

 

Q. The cost of fencing a square lawn at Rs.14 per metre is Rs.28000. Find the cost of moving the lawn at Rs.54 per 100 ${\mathrm{m.}}^{}$

 

Solution:

 

Cost of fencing the lawn = Rs 28000 Let l be the length of each side of the lawn. Then, the perimeter is 4l. We know: Cost=Rate $\times$ Perimeter ⇒28000=14 $\times$ 4l ⇒28000 = 56l Or, Area of the square lawn = 500 $\times$ 500 = 250000 $m^2$ Cost of mowing 100 $m^2$ of the lawn = Rs 54 Cost of mowing 1 $m^2$ ​​​​​​​ of the lawn=Rs ∴ Cost of mowing 250000 $m^2$ ​​​​​​​ of the lawn= =Rs 135000

Q. Find the area of the quadrilateral ABCD in which AD = 24 cm, $\angle$ BAD = ${90}^o$ and $\Delta$ BCD is an equilateral triangle having each side equal to 26 cm. Also, find the perimeter of the quadrilateral. [Given, $\sqrt{3}$ = 1.73.]

Solution:

 

ΔBDC is an equilateral triangle with side a = 26 cm Area of ΔBDC $=\frac{\sqrt{3}{a}^2}{4}=\frac{\sqrt{3}\times {26}^2}{4}=\frac{\sqrt{3}\times 676}{4}=292.37c{m}^2$ By using Pythagoras theorem in the right-angled triangle ΔDAB, we get: $A{D}^2+A{B}^2=B{D}^2{24}^2+A{B}^2={26}^{2}A{B}^2={26}^2-{24}^2=676-{576}^2=100AB=10cm$ Area of ΔABD $=\frac{1}{2}\times base\times height\frac{1}{2}\times 10\times 24=120c{m}^{}$ Area of the quadrilateral = Area of ΔBDC + Area of ΔABD $=292.37+120=412.37c{m}^2$ Perimeter of the quadrilateral $=AB+AC+CD+AD=24+10+26+26=86cm$

Q. The area of a parallelogram is 392 $m^{}$ , If its altitude is twice the corresponding base, determine the base and the altitude.

Solution:

 

Let the base(b) be 'x' Given altitude is twice the base It means,altitude(h) = 2x Area of a parallelogram = bh Given area of the parallelogram = 392 m² ⇒(bh)=392m² ⇒(x)(2x)=392m² ⇒2x²=392m² ⇒x²=392m²/2 ⇒x²=196m² ⇒x=√196m² ⇒x=14m. Therefore, the base of the parallelogram(x)=14m and the altitude of the parallelogram(2x)=2(14)=28m

 

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

Solution:

 

Let ABCD is a parallelogram; AD =20cm CD =34 cm and AC = 42 cm in ΔADC; s= 34+20+42 = 48; As per Heron's formula area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}$ Area of Δ ADC = $\sqrt{48(48-42)(48-34)(48-20)}$ Area of Δ ADC = $\sqrt{48\times 6\times 14\times 28}$ Area of Δ ADC =336 cm² Area of parallelogram = 2 $\times$ area of Δ ADC = 672 cm²

Q. If the circumferrence of a circle and the perimeter of a square are equal then

(a) area of the circle = area of the square (b) (area of the circle) > (area of the square) (c) (area of the circle) < (area of the square) (d) none of these

Solution:

 

Q. The radius of a wheel is 0.25 m. How many revolutions will it make in covering 11 km ? (a) 2800 (b) 4000 (c) 5500 (d) 7000

Solution:

 

Radius $=0.25m$ Circumference $=2\pi r=2\times \frac{22}{7}\times 0.25m$ No: of revolutions $=\frac{\text{total distance}}{\text{circumference}}$ $=\frac{11000\times 7}{2\times 22\times 0.25}=7000$ revolutions

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 17 Exercise 17.2

• Clear Explanations: These solutions provide step-by-step explanations for each problem, making it easy for students to understand complex concepts.
• Expert Guidance: Prepared by subject experts form Physics Wallah the solutions ensure accuracy and thorough coverage of all topics.
• Improved Problem-Solving Skills: By practicing these solutions, students can enhance their ability to solve problems related to the perimeter and areas of plane figures.
• Exam Preparation: The solutions cover a wide range of problems that are likely to appear in exams helping students prepare effectively.
• Conceptual Understanding: Detailed explanations help students build a strong foundation in geometry, particularly in calculating perimeters and areas.
• Time Management: With clear methods and strategies students can learn to solve problems more efficiently saving time during exams.