Question

In ∆abc, if ∠B=90, AB=4√5 units, BD is perpendicular AC and AD=4 units, then area of triangle ABC is equal to

Answer 1

Step-by-step explanation:

Hey here's the answer.

Answer 2

Given:

• ∠B = 90°
• AB = $4\sqrt{5}$ units
• BD is perpendicular to AC
• AD = 4 units

To Find:

• Area of ΔABC

Solution:

• From the given data we come to know that ΔABC is a right triangle.
• We can apply the Pythagoras theorem, $BD^2+AD^2=AB^2$
• substituting the values we get, $BD^2 + 4^2=(4\sqrt{5})^2$
• $BD^2 +16 = 80$
• BD = $\sqrt{80-16}$ = $\sqrt{64}$ = 8 cm
• BD = 8cm
• Area of ΔABC = $\frac{1}{2}(AB)(BC)$
• To find the value of BC consider, ΔBDC
• Let CD = x, BC = y, and AC = x + 4
• In ΔBDC,
• $x^2+8^2=y^2$
• ⇒ $y^2 = x^2+64$  → (1)
• In ΔABC,
• $y^2+80=(x+4)^2$  
• Substitute (1) in the above equation,
• $x^2+64+80=x^2+16-8x$  
• 144 - 16 = 8x
• 8x = 128
• x = 128/8 = 16
• x = 16, substitue x value in (1)
• $y^2 = 16^2 + 64 = 256+64 = 320$
• y = $\sqrt{320} = 8\sqrt{5}$
• BC = $8\sqrt{5}$
• Area of triangle ABC = $\frac{1}{2}(AB)(BC) = \frac{1}{2}(4\sqrt{5})(8\sqrt{5})$
• Area of ΔABC = $\frac{1}{2}(160) = \frac{160}{2} = 80$

Area of ΔABC = 80 sq.units.