Question

5. Find the area of an isosceles triangle whose one of the
equal sides measures 10 cm and the third side is 8 cm.​

Answer 1

Answer:

• Area of an isosceles triangle = 18.33 cm²

Step-by-step explanation:

Given that:

• Let us suppose that ABC is an isosceles triangle.
• D is the point between BC.

Whose,

• Equal sides = AB = AC = 10 cm
• Third side = BC = 8 cm

To Find:

• Area of an isosceles triangle.

Formula used:

Area of isosceles triangle = ½ (height × base) sq. unit

Here,

• Base = BC/2 = 8/2 = 4 cm
• Height = AD

First finding the length of height:

• By using pythagoras theorem.

⇒ (AD)² + (BD)² = (AB)²

• Substituting the values.

⇒ (AD)² + (4 cm)² = (10 cm)²

⇒ (AD)² + 16 cm² = 100 cm²

⇒ (AD)² = (100 - 16) cm²

⇒ (AD)² = 84 cm²

⇒ AD = √84 cm

Now finding the area of isosceles triangle:

⇒ Area = ½ (height × base) sq. unit

⇒ Area = ½(√84 × 4) cm²

⇒ Area = 2√84 cm²

⇒ Area = 18.33 cm² (approx.)

Answer 2

Given,

• Side a = 10cm
• Side b = 10cm
• Side c = 8cm

To Find,

• The Area Of Isosceles Triangle.

Solution,

a = 10cm

b = 10cm

c = 8cm

P = a + b + c

= 10cm + 10cm + 8cm

= 28cm

S = P/2

= 28cm/2

= 14cm

$:\implies Area = \sqrt{{s} \times (s - a) \times (s - b) \times (s - c)} \\ \\ :\implies Area = \sqrt{{14cm} \times (14cm - 10cm) \times (14cm - 10cm) \times (14cm - 8cm)} \\ \\ :\implies Area = \sqrt{14cm \times 4cm \times 4cm \times 6cm} \\ \\ :\implies Area = 4 {cm}^{2} \sqrt{14 \times 6} \\ \\ :\implies Area = 4 {cm}^{2} \times 2 \sqrt{21} \\ \\ :\implies Area = 8 \sqrt{21} {cm}^{2}$

Required Answer,

$Area \: \: \: of \: \: \: Isosceles \: \: \: Triangle = 8 \sqrt{21} {cm}^{2}$