Question

plz.. answer the questions​

Answer 1

FORMULA TO USE :-

$\dfrac{a+b+c}{2} = s$ Where s is the semi perimeter

$\sqrt{s(s-a)(s-b)(s-c)}$ Where a, b and c are the sides of the triangle

Question 1 :-

Find the area of a triangle whose two sides are 8cm and 11cm and perimeter is 32cm.

Given :-

• Two sides of the triangle are 8cm and 11cm
• Perimeter of the triangle is 32cm

To find :-

Area of the triangle

We first have to find the third side of the triangle.

PERIMETER = Sum of sides

Hence let the third side be x.

32 = 8 + 11 + x

13 = x

As we already know the perimeter of the triangle, the semi perimeter :-

$\dfrac{32}{2} = 16$

Now, by using the formula to find area :-

$\sqrt{16(16-8)(16-11)(16-13)}$

$\sqrt{16\times8\times5\times3}$

$\sqrt{1920}$

Area = 43.908cm² (approximately)

Question 2 :-

The sides of a triangular plot is 3:5:7 and the perimeter is 300m. Find the area of the triangle

Given :-

• The sides of a triangular plot are in the ratio 3:5:7
• Perimeter of the triangle is 300m

To find :-

Area of the plot

Let the sides be 3y, 5y and 7y respectively.

As we know perimeter is the sum of sides,

300 = 3y + 5y + 7y

300 = 15y

$\dfrac{300}{15} = y$

20 = y.

Hence the sides are :-

3y => 3(20) = 60m

5y => 5(20) = 100m

7y => 7(20) = 140m

Now, by using the heron's formula,

$\dfrac{300}{2} = s$

$150 = s$

By substituting the values,

$\sqrt{150(150-60)(150-100)(150-10)}$

$\sqrt{150\times90\times50\times10}$

$\sqrt{6750000}$

Area = 2598.08m² (approximately)

Question 3 :-

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

Given :-

• Two sides of the triangle are 18cm and 10cm
• Perimeter of the triangle is 42cm

To find :-

Area of the triangle

Let the third side be z.

z + 18 + 10 = 42

z = 14

Semi perimeter = $\dfrac{42}{2} = 21cm$

By using heron's formula,

$\sqrt{21(21-18)(21-10)(21-14)}$

$\sqrt{21\times3\times11\times7}$

$\sqrt{4851}$

Area = 69.65cm² (approximately)

Question 4 :-

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

Given :-

• The triangle sides are in the ratio 12:17:25
• Perimeter of the triangle is 540cm

To find :-

Area of the triangle

Let the sides be 12w, 17w and 25w respectively.

12w + 17w + 25w = 540

54w = 540

$\dfrac{540}{54} = w => 10$

Hence the sides are :-

12w => 12(10) = 120cm

17w => 17(10) = 170cm

25w => 25(10) = 250cm

Using heron's formula to calculate the area,

$\dfrac{540}{2} = s$

$270 = s$

$\sqrt{270(270-120)(270-170)(270-250)}$

$\sqrt{270\times150\times100\times20}$

$\sqrt{81,000,000}$

Area = 9000cm²

Question 5 :-

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

Given :-

• The two equal sides of the triangle are 12cm
• Perimeter of the triangle is 30cm

To find :-

Area of the triangle

Let the third side be c

c + 12 + 12 = 30

c = 6cm

By using heron's formula, Area :-

$\dfrac{30}{2} = s => 15$

$\sqrt{15(15-12)(15-12)(15-6)}$

$\sqrt{15\times3\times3\times9}$

$\sqrt{1215}$

Area = 34.86cm² (approximately)

Some more formulas :-

Area of square = (side)²

Area of rectangle = length × breadth

Area of parallelogram = base × height

Area of rhombus = 1/2 × Diagonal 1 × Diagonal 2