Question 1

Question 1A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

Accepted Answer

We know that,Each side of the equilateral triangle is equal.Given,Length of each side of an equilateral triangle = a cm.Perimeter of traffic signal board (equilateral triangle) = sum of all the sides = a + a + a = 3a cm.By formula,Semi perimeter (s) = Perimeter of triangle2=3a2\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{3a}{2}2Perimeter of triangle​=23a​ cm.By Heron's formula,Area of triangle (A) = s(s−a)(s−b)(s−c)\sqrt{s(s - a)(s - b)(s - c)}s(s−a)(s−b)(s−c)​ sq.units, where a, b and c are sides of triangle.Substituting values we get :A=3a2×(3a2−a)×(3a2−a)×(3a2−a)=3a2×(3a−2a2)×(3a−2a2)×(3a−2a2)=3a2×a2×a2×a2=3a416=34a2.A = \sqrt{\dfrac{3a}{2} \times \Big(\dfrac{3a}{2} - a\Big) \times \Big(\dfrac{3a}{2} - a\Big) \times \Big(\dfrac{3a}{2} - a\Big)} \[1em] = \sqrt{\dfrac{3a}{2} \times \Big(\dfrac{3a - 2a}{2}\Big) \times \Big(\dfrac{3a - 2a}{2}\Big) \times \Big(\dfrac{3a - 2a}{2}\Big)} \[1em] = \sqrt{\dfrac{3a}{2} \times \dfrac{a}{2} \times \dfrac{a}{2} \times \dfrac{a}{2}} \[1em] = \sqrt{\dfrac{3a^4}{16}} \[1em] = \dfrac{\sqrt{3}}{4}a^2.A=23a​×(23a​−a)×(23a​−a)×(23a​−a)​=23a​×(23a−2a​)×(23a−2a​)×(23a−2a​)​=23a​×2a​×2a​×2a​​=163a4​​=43​​a2.Given,Perimeter = 180 cm∴ 3a = 180⇒ a = 1803\dfrac{180}{3}3180​ = 60 cm.Substituting value of a, we get :Area of triangle (A)=34a2=34×602=34×3600=9003 cm2.\text{Area of triangle (A)} = \dfrac{\sqrt{3}}{4}a^2 \[1em] = \dfrac{\sqrt{3}}{4} \times 60^2 \[1em] = \dfrac{\sqrt{3}}{4} \times 3600 \[1em] = 900\sqrt{3}\text{ cm}^2.Area of triangle (A)=43​​a2=43​​×602=43​​×3600=9003​ cm2.Hence, the area of the signal board is 9003900\sqrt{3}9003​ cm2.

Question 2

Question 2The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m. The advertisements yield an earning of ₹ 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?

Accepted Answer

Here a, b and c are the sides of the triangle.Let a = 122 m, b = 22 m and c = 120 mBy formula,Semi Perimeter (s) = Perimeter of triangle2\dfrac{\text{Perimeter of triangle}}{2}2Perimeter of triangle​s = a+b+c2=122+22+1202=2642\dfrac{a + b + c}{2} = \dfrac{122 + 22 + 120}{2} = \dfrac{264}{2}2a+b+c​=2122+22+120​=2264​ = 132 m.By Heron's formula,Area of triangle (A) = s(s−a)(s−b)(s−c)\sqrt{s(s - a)(s - b)(s - c)}s(s−a)(s−b)(s−c)​ sq.unitsSubstituting values we get :Area of one wall=132(132−122)(132−22)(132−120)=132×10×110×12=1742400=1320 m2.\text{Area of one wall} = \sqrt{132(132 - 122)(132 - 22)(132 - 120)} \[1em] = \sqrt{132 \times 10 \times 110 \times 12} \[1em] = \sqrt{1742400} \[1em] = 1320 \text{ m}^2.Area of one wall=132(132−122)(132−22)(132−120)​=132×10×110×12​=1742400​=1320 m2.We know that,The rent of advertising per year = ₹ 5000 per m2So,The rent of one complete triangular wall for 1 month= Rent per sq. unit× Area12\dfrac{\text{Rent per sq. unit} \times \text{ Area}}{12}12Rent per sq. unit× Area​= (5000×1320)12=11×5000\dfrac{(5000 \times 1320)}{12} = 11 \times 500012(5000×1320)​=11×5000 = ₹ 5,50,000.∴ The rent of one wall for 3 months = ₹ 5,50,000 x 3 = ₹ 16,50,000.Hence, the rent paid by the company = ₹ 16,50,000.

Question 3

Question 3There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Accepted Answer

Let a, b and c be the sides of the triangle.Let a = 11 m, b = 6 m and c = 15 m.By formula,Semi Perimeter (s) = Perimeter of triangle2=a+b+c2\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{a + b + c}{2}2Perimeter of triangle​=2a+b+c​= 11+6+152=322\dfrac{11 + 6 + 15}{2} = \dfrac{32}{2}211+6+15​=232​ = 16 m.By Heron's formula,Area of triangle (A) = s(s−a)(s−b)(s−c)\sqrt{s(s - a)(s - b)(s - c)}s(s−a)(s−b)(s−c)​ sq.unitsSubstituting values we get :A=16(16−11)(16−6)(16−15)=16×5×10×1=800=202 m2.A = \sqrt{16(16 - 11)(16 - 6)(16 - 15)} \[1em] = \sqrt{16 \times 5 \times 10 \times 1} \[1em] = \sqrt{800} \[1em] = 20\sqrt{2} \text{ m}^2.A=16(16−11)(16−6)(16−15)​=16×5×10×1​=800​=202​ m2.Hence, area of the wall painted in colour = 20220\sqrt{2}202​ m2.

Question 4

Question 4Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Accepted Answer

Let a, b and c be the sides of the triangle.Let a = 18 cm, b = 10 cm.Given,Perimeter = 42 cm∴ a + b + c = 42⇒ 18 + 10 + c = 42⇒ 28 + c = 42⇒ c = 42 - 28 = 14 cm.By formula,Semi Perimeter (s) = Perimeter of triangle2=422\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{42}{2}2Perimeter of triangle​=242​ = 21 cm.By Heron's formula,Area of triangle (A) = s(s−a)(s−b)(s−c)\sqrt{s(s - a)(s - b)(s - c)}s(s−a)(s−b)(s−c)​ sq.unitsSubstituting values we get :A=21(21−18)(21−10)(21−14)=21×3×11×7=4851=2111 cm2A = \sqrt{21(21 - 18)(21 - 10)(21 - 14)} \[1em] = \sqrt{21 \times 3 \times 11 \times 7} \[1em] = \sqrt{4851} \[1em] = 21\sqrt{11} \text{ cm}^2A=21(21−18)(21−10)(21−14)​=21×3×11×7​=4851​=2111​ cm2Hence, area of triangle = 211121\sqrt{11}2111​ cm2.

Question 5

Question 5Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.

Accepted Answer

Given,Sides of a triangle are in the ratio of 12 : 17 : 25.Let sides of the triangle are :a = 12x, b = 17x and c = 25x.Given,Perimeter = 540 cm∴ 12x + 17x + 25x = 540 cm⇒ 54x = 540 cm⇒ x = 54054\dfrac{540}{54}54540​⇒ x = 10 cm⇒ a = 12 × 10 = 120 cm,⇒ b = 17 × 10 = 170 cm,⇒ c = 25 × 10 = 250 cm.Semi perimeter (s) = Perimeter of triangle2=5402\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{540}{2}2Perimeter of triangle​=2540​ = 270 cm.By Heron's formula,Area of triangle (A) = s(s−a)(s−b)(s−c)\sqrt{s(s - a)(s - b)(s - c)}s(s−a)(s−b)(s−c)​ sq.unitsSubstituting values we get :A=270(270−120)(270−170)(270−250)=270×150×100×20=81000000=9000 cm2.A = \sqrt{270(270 - 120)(270 - 170)(270 - 250)} \[1em] = \sqrt{270 \times 150 \times 100 \times 20} \[1em] = \sqrt{81000000} \[1em] = 9000 \text{ cm}^2.A=270(270−120)(270−170)(270−250)​=270×150×100×20​=81000000​=9000 cm2.Hence, area of triangle = 9000 cm2.

Question 6

Question 6An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Accepted Answer

Length of equal sides (a and b) = 12 cmLet third side of triangle (c) = x cmGiven,Perimeter of triangle = 30∴ 12 + 12 + x = 30⇒ 24 + x = 30⇒ x = 30 - 24⇒ x = 6 cm.∴ c = 6 cm.By formula,Semi perimeter (s) = Perimeter of triangle2=302\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{30}{2}2Perimeter of triangle​=230​ = 15 cm.By Heron's formula,Area of triangle (A) = s(s−a)(s−b)(s−c)\sqrt{s(s - a)(s - b)(s - c)}s(s−a)(s−b)(s−c)​ sq.unitsSubstituting values we get :A=15(15−12)(15−12)(15−6)=15×3×3×9=1215=915 cm2.A = \sqrt{15(15 - 12)(15 - 12)(15 - 6)} \[1em] = \sqrt{15 \times 3 \times 3 \times 9} \[1em] = \sqrt{1215} \[1em] = 9\sqrt{15} \text{ cm}^2.A=15(15−12)(15−12)(15−6)​=15×3×3×9​=1215​=915​ cm2.Hence, area of triangle = 9159\sqrt{15}915​ cm2.

Heron's Formula