Q.1 It is given that in triangles 𝐴𝐵𝐶 and 𝐷𝐸𝐹, $\frac{AB}{DE} = \frac{BC}{FD}$ , they will be similar if

1. ∠𝐵 = ∠𝐸
2. ∠𝐴 = ∠𝐷
3. ∠𝐵 = ∠𝐷
4. ∠𝐴 = ∠𝐹

 

(OR)

 

Q.2 The distance of a point 𝑄 from the centre of a circle is 25 𝑐𝑚, if the length of the tangent to a circle from point Q is 24 𝑐𝑚, find the radius of the circle.

1. 7 𝑐𝑚
2. 12 cm
3. 15 cm
4. 22 cm

Answer : 3

We are given that in triangles 𝐴𝐵𝐶 and 𝐷𝐸𝐹, $\frac{AB}{DE} = \frac{BC}{FD}$
So, we can say that 𝐴𝐵 and 𝐷𝐸 are similar and 𝐵𝐶 and 𝐹𝐷 are similar.
To make the triangles 𝐴𝐵𝐶 and 𝐷𝐸𝐹 similar by 𝑆𝐴𝑆 criterion, the angle between 𝐴𝐵 and 𝐵𝐶
& 𝐷𝐸 and 𝐹𝐷 must be equal.
The angle between 𝐴𝐵 and 𝐵𝐶 is ∠𝐵, and between 𝐷𝐸 and 𝐹𝐷 is ∠𝐷.
To make the two triangles similar, these two angles must be similar, that is, ∠𝐵 = ∠𝐷.
Hence, the correct option is (c) ∠𝐵 = ∠𝐷.

(OR)

Answer : 1

We are given the length of the tangent and the distance of point 𝑄 from the centre, and we have to find the circle’s radius.

Length of the tangent, 𝑃𝑄 = 24 𝑐𝑚

Distance of 𝑄 from the centre, 𝑂𝑄 = 25 𝑐𝑚

𝑂𝑃 denotes the radius of the circle.

We know that the tangent and the radius of a circle are perpendicular to each other; that is, the angle between 𝑃𝑄 and 𝑂𝑃 is 90$°$.

So, ∆𝑂𝑃𝑄 is a right-angled triangle in which two sides are known, and we have to find the third side.

We can use Pythagoras’ theorem for this, which can be written as

ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒² = 𝑏𝑎𝑠𝑒² + ℎ𝑒𝑖𝑔ℎ𝑡²

Here 𝑂𝑄 is the hypotenuse, 𝑃𝑄 is the base, and 𝑂𝑃 is the height. So, 𝑂𝑄² = 𝑃𝑄² + 𝑂𝑃²

Putting the value of the sides in the above equation, we get

252 = 242 +𝑂𝑃²

⇒ 625 = 576 + 𝑂𝑃²

⇒ 𝑂𝑃² = 625 − 576

⇒ 𝑂𝑃² = 49

⇒ 𝑂𝑃 =    $\sqrt{49}$ 

⇒ 𝑂𝑃 = 7 𝑐𝑚