Test: Problems on Similar Triangles - Grade 10 MCQ

Here, ∠ACB = c+180-(2c-b) = 180-(b+c)
So, We can say that Δ BCD and &delta ABC will be similar.
According to property of similarity,
AB/12 = 12/9 
Hence, 
AB = 16
AC/6 = 12/9 
AC = 8
Hence, AD = 7 and AC = 8
Now,
Perimeter of Δ ADC / Perimeter of ΔBDC,
= (6+7+8)/(9+6+12) 
= 21/27 = 7/9.

∆ ABC, in which sides are AB=BC= AC= a units and AD is perpendicular to BC ,

In ∆ADB ,

AB²= AD²+ BD²     (by Pythagoras theorem)

a² = AD² + (a/2)²   [BD= 1/2BC, since in an equilateral triangle altitude AD is  perpendicular bisector of BC ]

a²- a²/4 =AD²

⇒ ( 4a²-a²)/4 = AD²

⇒ 3a² /4 = AD²

⇒ 3AB²/4= AD²               [ AB= a]

⇒  3AB²= 4AD²

∠A = ∠R and ∠B = ∠Q
The names of triangles are written in such a manner in which the Angles are correspondingly equal which means that the vertices at which the angles of the two triangles are equal is placed accordingly. So since the the angle A and Angle R are equal and Angle B and Angle D are equal we have ΔABC - ΔRQP