Question

the altitude drawn to the base of the isosceles triangle is 8 cm and perimeter of the triangle is 32 . find the area of the triangle
​

Answer 1

(Diagram given in attachment)

In an Isosceles Triangle, 2 sides are equal.

Let

• AB = AC = x

Altitude divides the base of Isosceles Triangle into equal sides.

• BD = DC = y

Perimeter = Sum of all sides

⇒ Perimeter = AB + BC + CA

⇒ Perimeter = x + 2y + x

⇒ 32 = 2x + 2y

⇒ 2x + 2y = 32

⇒ 2(x + y) = 32

⇒ x + y = 32 ÷ 2

⇒ x + y = 16    

⇒ x = 16 - y     → → → [Equation 1]

Also, In ΔABD;

By Pythagoras Theorem,

(AB)² = (AD)² + (BD)²

⇒ x² = 8² + y²

⇒ x² = 64 + y²

Now let's substitute value of x from Equation 1

x² = 64 + y²

⇒ (16 - y)² = 64 + y²

Split using (a - b)² = a² + b² - 2ab

256 + y² - 2(16)y = 64 + y²

⇒ 256 + y² - 32y = 64 + y²

⇒ 256 - 32y = 64

⇒ 256 - 64 = 32y

⇒ 32y = 192

⇒ y = 192 ÷ 32

⇒ y = 6 cm

Substitute value of y in Equation 1 to find value of x

x = 16 - y

⇒ x = 16 - 6

⇒ x = 10 cm

Area = ¹/₂ × Base × Height

⇒ Area = ¹/₂ × 2y × 8

⇒ Area = ¹/₂ × 12 × 8

★ Area = 48 cm² ★

Answer 2

Solving in simple method :-

Refer attachment for Image,

Altitude of the isosceles triangle = 8

Let equal side be a.

∴ Perimeter of the triangle = 32

$2a + b = 32$

$a + \frac{b}{2} = 16.........(1)$

${a}^{2} - \frac{ {b}^{2} }{4} = {8}^{2}$

$(a + \frac{b}{2} )(a - \frac{b}{2}) = 64$

$16(a - \frac{b}{2} ) = 64$

$a - \frac{b}{2} = 4........(2)$

$from \: equation(1) \: - (2) \\ = > b = 12$

∴ Base of the triangle is = 12

$∴area \: of \: triangle = \\ \frac{1}{2} \times base \times height$

$= > \frac{1}{2} \times 12 \times 8$

$= > 48{cm}^{2}$

$hence \: verified$