In the given figure AB parallel to PQ, if PQ=1.5 cms, QC= 2 cm and BQ= 8 cm, then the measure of AB?
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Asked on October 15, 2019 by
Nitin Ameen
In the given figure, AB∥PQ. If PQ=1.5cm,QC=2cm and BQ=8cm, then the measure of AB is:
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ANSWER
Given : AB∥PQ, PQ=1.5cm,QC=2cm and BQ=8cm
Now, △PQC is 90
o
.
∴ by Pythagoras theorem,
PC
2
=PQ
2
+QC
2
=1.5
2
+2
2
=2.25+4=6.25
∴ PC=2.5cm
Since, AB∥PQ
⟹
QC
BQ
=
PC
AP
⟹
2
8
=
PC
AP
⟹ AP=4PC
⟹ AP=4×2.5cm=10cm
∴ AP=10cm
∴ AC=AP+PC=10+2.5=12.5cm
∴ AC=12.5cm and BC=BQ+QC=8+2=10cm
Since, △ABC is 90
o
then by Pythagoras theorem,
AC
2
=AB
2
+BC
2
⟹ AB
2
=AC
2
−BC
2
⟹ AB
2
=12.5
2
−10
2
⟹ AB
2
=156.25−100
⟹ AB
2
=56.25
⟹ AB=7.5cm
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Given : AB∥PQ, PQ=1.5cm,QC=2cm and BQ=8cm
Now, △PQC is 90°
∴ by Pythagoras theorem,
PC²=PQ²+QC²
=1.5²+2²
=2.25+4
=6.25
∴ PC=2.5cm
Since, AB∥PQ
⟹ BQ/QC = AP/PC
⟹ 8/2 = AP/PC
⟹ AP=4PC
⟹ AP=4×2.5cm=10cm
∴ AP=10cm
∴ AC=AP+PC=10+2.5=12.5cm
∴ AC=12.5cm and BC=BQ+QC=8+2=10cm
Since, △ABC is 90°
then by Pythagoras theorem,
AC ²=AB²+BC²
⟹ AB²=AC²−BC²
⟹ AB²=12.5²−10²
⟹ AB²=156.25−100
⟹ AB²=56.25
⟹ AB=7.5cm