Question

Answer the question in attachment

Answer 1

consider the figure attached :-

Let the length of diagonals be 2a and 2b respectively.

given, 2a + 2b = 17.

we know,

diagonals of a rhombus intersect at right angle .

thus,

a² + b² = 13³

=>now,

(a + b)² - 2ab = 169.

=>17² - 2ab = 169. ( since 2a + 2b = 34).

=>2ab = 120.

=>ab = 60.

Thus, product of half of diagonals = 60.

product of diagonals = 2a × 2b = 60 × 4 = 240 .

Hence, area =
$\frac{product \: of \: diagonals}{2} = \frac{240}{2} = 120 \: square \: unit$

Regards

KSHITIJ

Answer 2

Answer:

120

Step-by-step explanation:

Let the diagonals be d₁ and d₂.

(i)

Given that sum of its diagonals is 34.

d₁ + d₂ = 34.

d₁ = 34 - d₂

(ii)

Given that side (a) = 13.

⇒ (d₁/2)² + (d₂/2)² = a²

⇒ (d₁)² + (d₂)² = 13² * 4

⇒ (34 - d₂)² + (d₂)² = 676

⇒ 1156 + d₂² - 68d₂ + d₂² = 676

⇒ 2d₂² - 68d₂ + 480 = 0

⇒ d₂² - 34d₂ + 240 = 0

⇒ d₂² - 24d₂ - 10d₂ + 240 = 0

⇒ d₂(d₂ - 24) - 10(d₂ - 24) = 0

⇒ (d₂ - 10)(d₂ - 24) = 0

⇒ d₂ = 10,24.

When d₂ = 10:

d₁ = 34 - 10

d₁ = 24.

When d₂ = 24:

d₁ = 34 - 24

d₁ = 10.

∴ Hence, d₁ = 10,24 and d₂ = 24,10.

Area of rhombus:

⇒ (1/2) * [Product of diagonals]

⇒ (1/2)[d₁ * d₂]

⇒ (1/2) * 10 * 24

⇒ 240/2

⇒ 120.

Hope it helps!