Question

plz answer this 2 qstns

1. in the fig if, DE parallel to BC , DE /BC ,then calculate x


2. two circles touch each othe externally at P. AB is a common tangent to the circle touching them at A and B . find angle APB ..



plz answer....​

Answer 1

Ans.1)

use midpoint therome :-

BC = 2DE
DE=1/2BC
x=1/2(14)
x=7 cm

Ans.2)

Tangent always considered 60 degree because it became equaltriel triangle.




plz mark as brainliest answer.

Answer 2

Given:

X and Y are two circles touch each other externally at P. AB is the common tangent to the circles X and Y at point A and B respectively.

To find : ∠APB

Proof:

let ∠CAP = α and ∠CBP = β.

CA = CP

[lengths of the tangents from an external point C]

In a triangle PAC,

∠CAP = ∠APC = α

similarly CB = CP and ∠CPB = ∠PBC = β

now in the triangle APB,

∠PAB + ∠PBA + ∠APB = 180° [sum of the interior angles in a triangle]

α + β + (α + β) = 180°

2α + 2β = 180°

α + β = 90°

∴ ∠APB = α + β = 90°

1st question answer is in image