Question

find the area of triangle ABC and the value of x for the following Triangle ​

Answer 1

find the area of triangle ABC and the value of x for the following Triangle

• Area of triangle ABC = 3√7 cm^2 and the value of x = √7 cm
• Given,
• As, ∠BPC = 90°
• ∴ Δ BPC is a right angled triangle.
• ⇒ BC^2 = BP^2 + PC^2
• ⇒ BC^2 = (AB/2)^2 + PC^2
• ⇒ 4^2 = (6/2)^2 + x^2
• ⇒ x^2 = 16 - 9 = 7
• ∴ x = √7
• As, ∠APC = 90°
• ∴ Δ APC is a right angled triangle.
• ⇒ AC^2 = AP^2 + PC^2
• ⇒ AC^2 = 3^2 + (√7)^2
• ⇒ AC^2 = 9 + 7
• ⇒ AC^2 = 16
• ∴ AC = 4 cm
• Here, BC = AC = 4cm  and AB = 6 cm
• We can notice that, since the triangle has 2 sides equal, therefore it should be an isosceles triangle.
• Area of isosceles triangle
• A = bh/2
• = (6 * √7) / 2
• = 3√7 cm^2

Answer 2

Answer:

$area = 3 \sqrt{7 }$

$x = 4cm$

Step-by-step explanation:

Area of triangle ABC = 3√7 cm^2 and the value of x = √7 cm

Given,

As, ∠BPC = 90°

∴ Δ BPC is a right angled triangle.

⇒ BC^2 = BP^2 + PC^2

⇒ BC^2 = (AB/2)^2 + PC^2

⇒ 4^2 = (6/2)^2 + x^2

⇒ x^2 = 16 - 9 = 7

∴ x = √7

As, ∠APC = 90°

∴ Δ APC is a right angled triangle.

⇒ AC^2 = AP^2 + PC^2

⇒ AC^2 = 3^2 + (√7)^2

⇒ AC^2 = 9 + 7

⇒ AC^2 = 16

∴ AC = 4 cm

Here, BC = AC = 4cm and AB = 6 cm

We can notice that, since the triangle has 2 sides equal, therefore it should be an isosceles triangle.

Area of isosceles triangle

A = bh/2

= (6 * √7) / 2

= 3√7 cm^2