Question

2.) Find the area of triangle whose sides are:
(i) 9 cm, 10 cm and 17 cm
(ii) 13 cm, 5 cm and 12 cm
(iii) 35 m, 45 m and 50 m
]​

Answer 1

Given -

• Sides of three triangles.

To find -

• Area of triangles.

Formula used -

• Heron's formula

Solution -

In the question, we are provided with the sides of 3 triangles, and we need to find it's area. For that, we will apply heron's formula and will find every triangle's area. For that, first we will add up side a , b and c and then, semi - perimeter and then applying heron's formula, we will find their area

Let's do it!

Heron's formula says -

• $\sf \: \sqrt{s \:(s \: - a)(s \: - b)(s \: - c)}$

For 1st triangle -

Side a = 9 cm

Side b = 10 cm

Side c = 17 cm

Semi - Perimeter -

$\sf \: s \: = \dfrac{9 \: + \: 10 \: + \: 17}{2} \\ \\ \sf\: s \: = \cancel\frac{36}{2} \\ \\ \sf\: s \: = 18$

Area of 1st triangle -

$\sf \sqrt{18(18 \: - \: 9)(18 \: - \: 10)( 18\: - \: 17)} \\ \\ \sf \: \sqrt{18 \: \times \: 9 \: \times \: 8 \: \times \: 1} \\ \\ \sf \: \sqrt{1296} \\ \\ \sf \: 36 { \: cm}^{2}$

For 2nd triangle -

Side a = 13 cm

Side b = 5 cm

Side c = 12 cm

Semi - Perimeter -

$\sf \: s \: = \dfrac{13 \: + \: 5 \: + \: 12}{2} \\ \\ \sf \: s \: = \cancel \dfrac{30}{2} \\ \\ \sf \: s \: = 15$

Area of 2nd triangle -

$\sf \: \sqrt{15(15 \: - \: 13)(15 \: - \: 5)(15 \: - \: 12)} \\ \\ \sf \: \sqrt{15 \: \times 2 \: \times \: 10 \: \times \: 3} \\ \\ \sf \: \sqrt{900} \\ \\ \sf \: 30 { \: cm}^{2}$

For 3rd triangle -

Side a = 35 m

Side b = 45 m

Side c = 50 m

Semi - perimeter -

$\sf \: s \: = \dfrac{35 \: + \: 45 \: + \: 50}{2} \\ \\ \sf \: s \: = \cancel\frac{130}{2} \\ \\ \sf \: s \: = 65$

Area of 3rd triangle -

$\sf \: \sqrt{65(65 \: - \: 35)(65 \: - \: 45)(65 \: - \: 50)} \\ \\ \sf \: \sqrt{65 \: \times \: 30 \: \times \: 20 \: \times \: 15} \\ \\ \sf \: \sqrt{585000} \\ \\ \sf \: 764.8 \: m^{2} \: (appox)$

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Answer 2

Answer:

Given :-

1. 9 cm , 10 cm and 17 cm
2. 13 cm , 5 cm and 12 cm
3. 35 m , 45 m and 50 m

To Find :-

• What is the area of triangle.

Formula Used :-

★ Semi-perimeter of triangle :

$\sf\boxed{\bold{Semi-perimeter (s) =\: \dfrac{Sum\: of\: all\: sides}{2}}}$

★ Area of triangle by using heron's formula :

$\sf\boxed{\bold{Area\: of\: triangle =\: \sqrt{(s(s - a)(s - b)(s - c)}}}$

Solution :-

❶ 9 cm , 10 cm and 17 cm.

Given :

• a = 9 cm
• b = 10 cm
• c = 17 cm

First, we have to find the semi-perimeter of a triangle,

As we know that,

$\leadsto$ $\sf\bold{\pink{Semi-perimeter (s) =\: \dfrac{a + b + c}{2}}}$

Then,

↦ $\sf Semi-perimeter =\: \dfrac{9 + 10 + 17}{2}$

↦ $\sf Semi-perimeter =\: \dfrac{19 + 17}{2}$

↦ $\sf Semi-perimeter =\: \dfrac{\cancel{36}}{\cancel{2}}$

➦ $\sf\bold{\green{Semi-perimeter =\: 18\: cm}}$

Now, we have to find the area of triangle,

⇒ $\sf Area =\: \sqrt{18(18 - 9)(18 - 10)(18 - 17)}$

⇒ $\sf Area =\: \sqrt{18(9)(8)(1)}$

⇒ $\sf Area =\: \sqrt{18 \times 72}$

⇒ $\sf Area =\: \sqrt{1296}$

➠ $\sf\bold{\red{Area =\: 36\: {cm}^{2}}}$

$\therefore$ The area of a triangle is 36 cm².

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❷ 13 cm , 5 cm and 12 cm.

Given :

• a = 13 cm
• b = 5 cm
• c = 12 cm

As we know that,

$\leadsto$ $\sf\bold{\pink{Semi-perimeter (s) =\: \dfrac{a + b + c}{2}}}$

Then,

↦ $\sf Semi-perimeter =\: \dfrac{13 + 5 + 12}{2}$

↦ $\sf Semi-perimeter =\: \dfrac{\cancel{30}}{\cancel{2}}$

➦ $\sf\bold{\green{Semi-perimeter =\: 15\: cm}}$

Now, we have to find the area of a triangle,

⇒ $\sf Area =\: \sqrt{15(15 - 13)(15 - 5)(15 - 12)}$

⇒ $\sf Area =\: \sqrt{15(2)(10)(3)}$

⇒ $\sf Area =\: \sqrt{15 \times 60}$

⇒ $\sf Area =\: \sqrt{900}$

➠ $\sf\bold{\red{Area =\: 30\: {cm}^{2}}}$

$\therefore$ The area of a triangle is 30 cm².

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❸ 35 m , 45 m and 50 m.

Given :

• a = 35 m
• b = 45 m
• c = 50

As we know that,

$\leadsto$ $\sf\bold{\pink{Semi-perimeter (s) =\: \dfrac{a + b + c}{2}}}$

Then,

↦ $\sf Semi-perimeter =\: \dfrac{35 + 45 + 50}{2}$

↦ $\sf Semi-perimeter =\: \dfrac{\cancel{130}}{\cancel{2}}$

➦ $\sf\bold{\green{Semi-perimeter =\: 65\: cm}}$

Now, we have to find the area of a triangle,

⇒ $\sf Area =\: \sqrt{65(65 - 35)(65 - 45)(65 - 50)}$

⇒ $\sf Area =\: \sqrt{65(30)(20)(15)}$

⇒ $\sf Area =\: \sqrt{65 \times 9000}$

⇒ $\sf Area =\: \sqrt{585000}$

➠ $\sf\bold{\red{Area =\: 764.8\: {cm}^{2}}}$

$\therefore$ The area of a triangle is 764.8 cm².

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