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Two line segment AB and AC include an angle of 60^(@), where AB=5 cm and AC= 7 cm. Locate points P and Q on AB and AC, respectively such that AP=(3)/(4)AB and AQ=(1)/(4)AC. Join P and Q and measure the length PQ. |
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Answer» Solution :Given that ,AB= 5 cm and AC= 7 cm. Also,`AP=(3)/(4)AB` and `AQ=(1)/(4)AC` …..(i) From Eq. (i), `AP=(3)/(4).AB=(3)/(4)xx5=(15)/(4)cm` Then, `PB=AB-AP=5-(15)/(4)=(20-15)/(4)=(5)/(4) cm` [ `:.` P is any point on the AB] `:. AP:PB=(15)/(4):(5)/(4)implies AP:PB=3:1` i.e, scale factor of line segment AB is `(3)/(1)`. Again from Eq. (i), `AQ=(1)/(4)AC=(1)/(4)xx7=(7)/(4)cm` Then, `QC=AC-AQ=7-(1)/(4)` `=(28-7)/(4)=(21)/(4) cm` [`:.` Q is any point on the AC] `:. AQ:QC=(7)/(4):(21)/(4)=1:3` `implies AQ:QC =1:3` i.e., scale factor of line segment AQ is `(1)/(3)`. Steps of construction 1. Draw a line segment AB=5 cm. 2. Now draw a ray AZ making an acute `angleBAZ=60^(@)`. 3. Whit A as centre and radius EQUAL to 7 cm draw an arc CUTTING the line AZ at C. 4. Draw a ray AX, making an acute ` angle BAX`. 5. Along AX , mark 1+3=4 points `A_(1),A_(2),A_(3), " and " A_(4)`. Such that `A A_(1)=A_(1)A_(2)=A_(2)A_(3)=A_(3)A_(4)` 6. Join `A_(4)B` ltbr. 7. From `A_(3)` draw `A_(3)P||A_(4)B` meeting AB at P. [ by making an angle equal to `angleAA_(4)B`] Then, P is the point on AB which DIVIDED it in the ratio 3:1. So, AP:PB=3:1 ,brgt 8. Draw a ray AY, making an acute `angleCAY`. ![]() 9. Along AY, mark 3+1=4 points `B_(1),B_(2),B_(3) " and " B_(4)` Such that `AB_(1)=B_(1)B_(2)=B_(2)B_(3)=B_(3)B_(4)` 10. Join `B_(4)C`. 11. From `B_(1)` draw `B_(1)Q||B_(4)C` meeting AC at Q. [by making an angle equal to `angleAB_(4)C`] Then, Q is the point on AC which divides in the ratio 1:3. So, AQ:QC=1:3 ltbtgt 12. Finally, join PQ and its MEASUREMENT is 3.25 cm. |
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