Question

- In a maths class, the teacher draws two circles that touch each other externally at point K with centres A and B and radii 5cm and 4cm respectively, as shown in the figure....
Based on the above information answer the following...


1. the value of PA
1. 12cm
2. 5cm
3. 13cm
4. can't be determined


2. the value of BQ...
1. 4cm
2. 5cm
3. 6cm
4. none of these

3. the value of PK...
1. 13cm
2. 15cm
3. 16cm
4. 18cm

4. the value of QY...
1. 2cm
2. 5cm
3. 1cm
4. 3cm


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Answer 1

Answer:

1-c,2-b,3-d,4-c

Step-by-step explanation:

Used pythogoras to solve these because tangent point join the center at 90°

Answer 2

Given figure answer the following questions

Explanation:

1. PS is a tangent to the circle with center A (radius $r_1=5\ cm$)
2. hence, PS makes an angle of $90$° with the radius AS. Then in ΔAPS,                              $PS=12\ cm \ \ \ \ \ and\ \ \ \ \ AS=r_1=5\ cm \\PA^2= PS^2+AS^2 \\PA=\sqrt{12^2+5^2} \\PA= 13\ cm$-> Q.1 has answer option (3)
3. TQ is tangent to the circle with center B (radius $r_2=4\ cm$)          
4. hence, TQ makes an angle of $90$° with radius BT. Then in ΔBQT,               $TQ=3\ cm \ \ \ \ \ and\ \ \ \ \ BT=r_1=4\ cm \\BQ^2= TQ^2+BT^2 \\BQ=\sqrt{3^2+4^2} \\BQ=5 \ cm$ -> Q.2 has answer option (2)            
5. from point 2. we have, $PA=13\ cm$ and $AK= r_1=5\ cm$  hence we have,                                                                                     $PK=PA+AK\\PK=18\ cm$                                    -> Q.3 has answer option (4)    
6. from point 4. we have $BQ=5\ cm$  and  $BY=r_2=4\ cm$ hence we have,                                                                                                                              $QY=BQ-BY\\QY=1\ cm$                                    ->Q.4 has answer option (3)