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Area and Perimeter

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Q 1. The sides of a triangle area in the ratio of {tex} \frac{1}{3}:\frac{1}{4}:\frac{1}{5} {/tex} and its perimeter is 94 Cm. Find the length of the smallest side of the triangle.

18 cm

22.5 cm

24 m

27 cm

Given ratio = {tex} \frac{1}{3}:\frac{1}{4}:\frac{1}{5} {/tex}

Let lengths of the sides be 20x,15x and 12x.

Then, according to the question,

20x + 15x + 12x = 94 =>47x = 94

x = {tex} \frac{94}{47} {/tex}=2

Smallest side =12x = {tex} 12\times 2 {/tex} = 24 cm

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Q 2. The ratio of the length of each equal side and the third side of an isosceles triangle is 3:4. If the area of the triangle is {tex} 18\sqrt{5} {/tex} sq units, the third side is

{tex} 8\sqrt{2} {/tex}units

12 units

16 units

{tex} 5\sqrt{10} {/tex}units

Let sides of isosceles triangle are 3x,3x and 4x.

Then, half-perimeter (s) = {tex} \frac{a+b+c}{2} {/tex}

= {tex} \frac{3x+3x+4x}{2} {/tex}=5x

Given, area of isosceles triangle

= {tex} 18\sqrt{5} {/tex}sq units

{tex} \sqrt{s(s-a)(s-b)(s-c)}=18\sqrt{5} {/tex}

{tex} \sqrt{5x(5x-3x)(5x-3x)(5x-4x)}=18\sqrt{5} {/tex}

{tex} \sqrt{5x\times 2x\times 2x\times x}=18\sqrt{5} {/tex}

=> {tex} 2\sqrt{5}x^{2}=18\sqrt{5} {/tex}

=> {tex} x^{2}=9 {/tex}

=> x=3

Third side of isosceles triangle

={tex} 4\times 3 {/tex}= 12units

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Q 3. Three sides of a triangular field are of length 15 m, 20 m and 25 m long, respectively. Find the cost of sowing seeds in the field at the rate of RS. 5 per Sq m.

750

150

300

600

Since,{tex} AC^{2}=AB^{2}+BC^{2} {/tex}

=> {tex} (25)^{2}=(15)^{2}+(20)^{2} {/tex}

=> 625=225+400

=> 625 = 625

So, the triangular field is right angled at B

Area of the field = {tex} \frac{1}{2}\times AB\times BC {/tex}

= {tex} \frac{1}{2}\times 15\times 20 {/tex}

={tex} 150m^{2} {/tex}

So, the cost of sowing seed is RS. 5 per sq m

Cost of sowing seed for

{tex} 150m^{2}=150\times 5 {/tex}

= RS. 750

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