RS Aggarwal Solutions for Class 10 Maths Chapter 17 Exercise 17.1

RS Aggarwal Solutions for Class 10 Maths Chapter 17 Exercise 17.1: RS Aggarwal Solutions for Class 10 Maths Chapter 17, Exercise 17.1 provide a detailed approach to solving problems related to the perimeter and area of plane figures. This exercise includes a range of problems designed to help students apply the formulas for various shapes such as rectangles, squares, and triangles.

RS Aggarwal Solutions for Class 10 Maths Chapter 17 Exercise 17.1 Overview

RS Aggarwal Solutions for Class 10 Maths Chapter 17 Exercise 17.1 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 17 Exercise 17.1 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 17 Exercise 17.1

Q. 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.

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

 

Given a = 18 cm, b = 24 cm, c = 30 cm. Semi perimeter $s=\frac{a+b+c}{2}=18+24+30/2=72/2=36$ Using heron's formula, Area $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{36(36-18)(36-24)(36-30)}$ $=\sqrt{36\times 18\times 12\times 6}$ $=\sqrt{2\times 2\times 3\times 3\times 3\times 3\times 2\times 2\times 2\times 3\times 2\times 3}$ $=2\times 2\times 2\times 3\times 3\times 3=216c{m}^2$

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

Solution:

 

Let the sides of the triangle be $5x,12x,13x$ where $x$ is a positive rational number. $\therefore$ Perimeter $=5x+12x+13x=150$ $\Rightarrow 30x=150$ $\Rightarrow x=\frac{150}{30}$ $\Rightarrow x=5$ Hence, the sides of the triangle are $25m,60m,65m$ Also, these sides form a Pythagoras Triplet, as ${25}^2+{60}^2=625+3600=4225={65}^2$ Hence, the triangle is a right angles triangle whose hypotenuse is $65m$ . So, area of triangle $=\frac{1}{2}\times \text{base}\times \text{height}$ $=\frac{1}{2}\times 25\times 60$ $=750sq.units$

Q. 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 Rs. 40 per 100 $m^2$ .

Solution:

 

The perimeter of a triangular field = 540 m Let the sides are 25x, 17x, 12 x The perimeter of a $△$ = sum of three sides 25x + 17x + 12x = 540 54x = 540 x = 10 1st side (a) - 25x = 25 $\times$ 10= 250 m 2nd side(b)= 17x = 17 $\times$ 10= 170 m 3rd side (c)= 12x = 12 $\times$ 10 =120 m Semi - perimeter (S) = $\frac{a+b+c}{2}=\frac{(250+170+120)}{2}=\frac{540}{2}=270$ m Area of the $\Delta =\sqrt{s(s-a)(s-b)(s-c)}$ [By Heron’s Formula] $=\sqrt{S(S-250)(S-170)(S-120)}=\sqrt{270(270-250)(270-170)(270-120)}=\sqrt{270\times 20\times 100\times 150}=\sqrt{81000000}=9000m^2$ Area of the $\Delta =9000m^2$ Cost of ploughing the field at Rs. 40 per 100 $m^2=\frac{9000\times 40}{100}=Rs.3600$

Q. The perimeter of a right triangle is 40 cm and its hypotenuse measures 17 cm. Find the area of the triangle.

Solution:

 

Let the base and height of right triangle is x and y respectively Given: hypotenuse = 17 cm So, $x+y+17=40cmx+y=23cmx=23-y.......\left(1\right)and{x}^2+y^2=z^2......\left(2\right)$ $\Rightarrow (23-y{)}^2+y^2={17}^2$ $\Rightarrow 529-46y+y^2+y^2=289$ $\Rightarrow 2y^2-46y+240=0\Rightarrow y^2-23y+120=0\Rightarrow (y-8)(y-15)=0\Rightarrow y=15or8$ $\Rightarrow x=8or15,So,Area=\frac{1}{2}\times b\times h=\frac{1}{2}\times 8\times 15=60c{m}^{}$

Q. The difference between the sides at right angles in a right-angled triangle is 7 cm, The area of the triangle is 60 $c{m}^2$ . Find its perimeter.

Solution:

 

Let base = x and altitude be x + 7 Now area = 60 $c{m}^2$ i.e, $\frac{1}{2}\times base\times altitude=60\frac{1}{2}\times x\times (x+7)=60x^2+7x=120x^2+7x-120=0x^2-8x+15x-120=0x(x-8)+15(x-8)=0(x-8)(x+15)=0x=8or-15$ Length annot be negative. So base = 8 cm and altitude = 8 + 7 = 15 cm Hypotenuse $=\sqrt{8^2+{15}^2}=\sqrt{64+225}=\sqrt{289}=17cm$ Perimeter $=8+15+17=40cm$

Q. The lengths of the two sides of a right triangle containing the right angle differ by 2 cm. If the area of the triangle is 24 $c{m}^2$ , find the perimeter of the triangle.

Solution:

 

Let the base of the triangle be b cm and its height or perpendicular be p cm. Given, p - b = 2 ... (1) and area of triangle = 24 $c{m}^2$ ⇒ $\frac{1}{2}\times b\times p=24$ ​​​​​​​ ⇒ $b\times p=48$ ​​​​​​​​​​​​​​ ⇒ $b=\frac{48}{p}$ ​​​​​​​ On putting the value of b in (1), we get $p-\frac{48}{p}=2$ ​​​​​​​ ⇒ $p^2-2p-48=0$ ​​​​​​​ ⇒ $p^2-8p+6p-48=0$ ​​​​​​​ ⇒ $(p-8)(p+6)=0$ ​​​​​​​ ⇒ $p=8,-6$ ​​​​​​​ p ≠ -6. So, p = 8 cm On putting p in (1), we get ⇒ b = 6 cm So, Hypotenuse (h) = $\sqrt{p^2+b^2}=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10$ ​​​​​​​ ​​​​​​​ Hence, perimeter of triangle = p + b + h = 8 cm + 6 cm + 10 cm = 24 cm

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

Solution:

 

In the above triangle, the measurement of the side DC is 5 cm. In the triangle ADC , the side AC is called hypotenuse side whose length is 10 cm. So we can say $A{C}^2=A{D}^2+D{C}^2{10}^2=A{D}^2+5^{2}100=A{D}^2+25100-25=A{D}^{2}A{D}^2=75AD=\sqrt{75}AD=5\sqrt{3}$ Now we can apply the formula to find the area of triangle = $\frac{1}{2}\times base\times height$ Here base is 10 cm and height is $5\sqrt{3}$ cm Area of the given triangle = $\frac{1}{2}\times 10\times 5\sqrt{3}=5\times 5\sqrt{3}=25\sqrt{3}c{m}^2$

Q. The height of an equilateral triangle is 6 cm. Find its area. [Take $\sqrt{3}$ = 1.73.]

Solution:

 

Altitude = Height = 6 cm Let length of side be 'a' So , = 6 So a = So a = $4\sqrt{3}$ cm This 'a' is the length of the side Now, $Area=\frac{\sqrt{3}}{4}\times a^2$ $Area=\frac{\sqrt{3}}{4}\times (4\sqrt{3}{)}^2$ $Area=\frac{\sqrt{3}}{4}\times 16\times 3$ So, $Area=\sqrt{3}\times 4\times 3$ So, $Area=12\sqrt{3}$ So, $Area=12\times 1.732$ So, $Area=20.784c{m}^2$

Q. If the area of an equilateral triangle is 36 $\sqrt{3}c{m}^2$ , find its perimeter.

Solution:

 

Area of equilateral triangle of side a $=\frac{\sqrt{3}}{4}\times a^2$ $\frac{\sqrt{3}}{4}\times a^2=36\sqrt{3}$ $a^2=36\times 4$ a = 12 perimeter of equilateral triangle = 3a = 3 $\times$ 12 = 36 cm.

Q. If the area of an equilateral triangle is 81 $\sqrt{3}c{m}^2$ , find its height.

Solution:

 

Given that we know that $Area=\frac{\sqrt{3}}{4}\times a^2$ area of the equilateral triangle = 81 $\sqrt{3}c{m}^2$ $\frac{\sqrt{3}}{4}\times a^2=81\sqrt{3}$ $\frac{1}{4}{a}^2=81$ $a^2=81\left(4\right)$ $a^2=324$ a = 18cm Then the height of the equilateral triangle $=\frac{\sqrt{3}a}{2}$ $=\frac{\sqrt{3}}{2}\times 18=9\sqrt{3}cm$

Q. The base of a right-angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle.

Solution:

 

base = 48 cm hypotenuse= 50 cm . By Pythagoras theorem, we get $h^2=b^2+p^{2}So,p^2=h^2-b^2$ $\Rightarrow p^2={50}^2-{48}^2$ $\Rightarrow p^2=2500-2304$ $\Rightarrow p^2=196$ $\Rightarrow p=\sqrt{196}$ $\Rightarrow p=14So,Perpendicular=14cm.Nowareaoftriangle=\frac{1}{2}\times base\times height=\frac{1}{2}\times 48\times 14=48\times 7=336c{m}^2$

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

Solution:

 

Hypotenuse = 65 cm Base = 60 cm According to Pythagoras theorem, $(\text{Hypotenuse}{)}^2=(\text{Base}{)}^2+(\text{Altitude}{)}^2{65}^2={60}^2+(\text{Altitude}{)}^2(\text{Altitude}{)}^2={65}^2-{60}^2\text{Altitude}=\sqrt{4225-3600}=\sqrt{625}=25cm$ Area of triangle = $\frac{1}{2}\times base\times height=\frac{1}{2}\times 60\times 25=750c{m}^2$

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

Solution:

 

Let ABC be a right-angled triangle inside a circle having centre 'O' Given, OA = OB = CC = 8 cm [OA, 0B and OC be the radius of circle] Let AD be the altitude drawn from the opposite vertex to the hypotenuse AD = 6 cm Since, $△$ ABC is a right triangle right angled at A A circle can be drawn which passes the vertices of $△$ ABC (Angle in a semi-circle is 90 ${}^{}$ ) Now, the Area of $△$ ABC = $\frac{1}{2}\times base\times height=\frac{1}{2}\times BC\times AD=\frac{1}{2}\times (OB+OC)\times AD=\frac{1}{2}\times (8+8)\times 6=\frac{1}{2}\times 16\times 6=48c{m}^2$

Q. Find the lenght of the hypotenuse of an isosceles right-angled triangle whose area is 200 $c{m}^2$ . Also, find its perimeter. [Given, $\sqrt{2}$ = 1.41.]

Solution:

 

In an isosceles triangle,any two sides are equal.So,in this question, base=height or (b=h) Area of ∆ = $\frac{1}{2}\times b\times h$ $200=\frac{1}{2}\times b^2$ $400=b^2$ $b=\sqrt{400}=20$ So, b=h=20 Hypotenuse $=\sqrt{400+400}=20\sqrt{2}=20\times 1.414=28.28$ Perimeter = 20 + 20 + 28.28 = 68.28 cm

Q. The base of an isosceles triangle measures 80 cm and its area is 360 $c{m}^2$ Find the perimeter of the triangle.

Solution:

Q. Each side of an equilateral triangle is $8cm$ . Its area is (a) $24c{m}^2$ (b) $24\sqrt{3}c{m}^2$ (c) $16\sqrt{3}c{m}^2$ (d) $8\sqrt{3}c{m}^2$

Solution:

 

Area of equilateral triangle $=\frac{\sqrt{3}}{4}{a}^2$ sq. units Given: Each side of an equilateral triangle is $\left(a\right)=8cm$ $\Rightarrow$ Area of equilateral triangle $=\frac{\sqrt{3}}{4}(8cm{)}^2$ $=\frac{\sqrt{3}}{4}\times 8cm\times 8cm$ $=\sqrt{3}\times 2cm\times 8cm$ $=16\sqrt{3}c{m}^2$ Hence, Option C is correct.

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 17 Exercise 17.1

• Detailed Explanations: Each solution is broken down step-by-step, making it easy for students to follow and understand the process of calculating perimeters and areas.
• Expert Guidance: The solutions are prepared by subject experts ensuring that the methods and explanations are accurate and reliable.
• Improved Understanding: Working through these solutions helps students grasp important geometric concepts and apply the correct formulas to various problems.
• Confidence Building: By practicing with these solutions, students can build confidence in their ability to solve geometry problems, leading to better performance in their exams.