Q.1 In ∆𝐴𝐵𝐶, 𝐷𝐸||𝐵𝐶, find the value of 𝑥.

 

(OR)

 

Q.2 In the given figure, 𝑃𝑄 and 𝑃𝑅 are tangents to the circle with center 𝑂, such that ∠𝑄𝑃𝑅 = 50𝑜, then find ∠𝑂𝑄𝑅.

As 𝐷𝐸||𝐵𝐶, the triangles 𝐴𝐵𝐶 and 𝐴𝐷𝐸 are similar in ∆𝐴𝐵𝐶. 

So, we can write 

𝐴𝐵/𝐴𝐷 = 𝐴𝐶/𝐴𝐸
⇒ 𝐴𝐷+𝐷𝐵/ 𝐴𝐷 = 𝐴𝐸+𝐸𝐶/𝐴𝐸
⇒ 1 + 𝐷𝐵/𝐴𝐷 = 1 + 𝐸𝐶/𝐴𝐸
⇒ 𝐷𝐵/𝐴𝐷 = 𝐸𝐶/𝐴𝐸
Here, 𝐴𝐷 = 𝑥
𝐷𝐵 = 𝑥 + 1
𝐴𝐸 = 𝑥 + 3
𝐸𝐶 = 𝑥 + 5

Putting these values in the above equation, we get

𝑥+1/𝑥 = 𝑥+5/𝑥+3
⇒ 𝑥(𝑥 + 5) = (𝑥 + 3)(𝑥 + 1)
⇒ 𝑥2 + 5𝑥 = 𝑥2 + 𝑥 + 3𝑥 + 3
⇒ 5𝑥 = 4𝑥 + 3
⇒ 5𝑥 − 4𝑥 = 3
⇒ 𝑥 = 3
Hence, the value of 𝑥 is 3.

(OR)

We are given two tangents, 𝑃𝑄 and 𝑃𝑅, and the angle                                     𝑜. 

We have to find the value of ∠𝑂𝑄𝑅.

As we know that in a cyclic quadrilateral, that is, the quadrilateral inside a circle, the sum of opposite angles is 180$°$

In the figure, we are given a cyclic quadrilateral 𝑂𝑄𝑃𝑅, in which the opposite angles are
∠𝑄𝑂𝑅 and ∠𝑄𝑃𝑅.
So, ∠𝑄𝑂𝑅 + ∠𝑄𝑃𝑅 = 180$°$ (as mentioned above) 𝑜
Putting the value of ∠𝑄𝑃𝑅 in the above equation, we get
∠𝑄𝑂𝑅 + 50$°$ = 180$°$
⇒ ∠𝑄𝑂𝑅 = 180$°$ − 50$°$
⇒ ∠𝑄𝑂𝑅 = 130$°$ … eq(1) 
A triangle 𝑂𝑄𝑅 will form if we join the points 𝑄 and 𝑅.
In triangle 𝑂𝑄𝑅, 𝑂𝑄 and 𝑂𝑅 are the radii of the circle, so, 𝑂𝑄 = 𝑂𝑅
As two sides of the Δ𝑂𝑄𝑅 are equal, this is an isosceles triangle.
In an isosceles triangle, the angles opposite to the equal sides are also equal, i.e.,
∠𝑂𝑄𝑅 = ∠𝑂𝑅𝑄
We know that the sum of the angles of a triangle is 180$°$
So, for Δ𝑂𝑄𝑅, ∠𝑄𝑂𝑅 + ∠𝑂𝑄𝑅 + ∠𝑂𝑅𝑄 = 180$°$
Putting the value of ∠𝑄𝑂𝑅 from eq(1), we get
130$°$  + ∠𝑂𝑄𝑅 + ∠𝑂𝑄𝑅 = 180$°$ ∠𝑂𝑄𝑅 = ∠𝑂𝑅𝑄
⇒ 2∠𝑂𝑄𝑅 = 180$°$ − 130$°$

⇒ 2∠𝑂𝑄𝑅 = 50$°$

∠𝑂𝑄𝑅 = 50$°$/2

⇒ ∠𝑂𝑄𝑅 = 25$°$

Hence, the ∠𝑂𝑄𝑅 is 25$°$