Question

In the adjoining figure, AC=10cm, AB=6cm and BC=8cm.Is the ∆ ABC right angled ? Give reason.​

Answer 1

Step-by-step explanation:

Area ABC =√( 12*2*4*6) = 24 (cm)^2 using Heron's formula.

Area ABC is also = (1/2) 10*8 * sin B= 24, and so sin B = 0.6., B = 36.87 °

The line perpendicular to AB at X cuts BC at D. Let DX = h & BX = x.

Area BXD =(1/2) x h = 12 (cm)^2. ( given condition)

h = x tan B = 0.6 x.

Area BXD =(1/2) x ( 0.6x) = 12

x = BX = 6.33 cm

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Verify the result.

h = 6.33* sin B = 6.33* 0.6 = 3.798

Area (1/2) 6.33 *3.798 = 12 cm ^2

Answer 2

AC²=AB²+BC²

10²=6²+8²

100=36+64

100=100

By this we can say that by the converse of Pythagoras theorem,ABC is a right angled triangle.