Question

in triangle ABC,AB=10cm,AC=8cm,triangleA=45 a) find perpendicular distance from C to AB b) find the area of the triangle​

Answer 1

Draw a ∆ABC ,where AB=10cm,AC=8cm and ∠A=45.∠A=45. Draw a perpendicular from vertex C which meets the line AB at D. And ∠D=90°∠D=90°

Let AD=x cm

In ∆ADC

Using Pythagoras theorem

82=x2+x282=x2+x2

x=42–√x=42

CD=42–√cmCD=42cm

In ∆ BDC

BC2=(10−42–√)2+(42–√)2BC2=(10−42)2+(42)2

BC=4(41−802–√)cmBC=4(41−802)cm

So all sides are different. So the ∆ABC is a scelence triangle.

Step-by-step explanation:

hopes its helps