Question

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Answer 1

$\sf\large\underline\purple{Given:-}$

In triangle ∆QAP => AP = 6cm, QA = 8cm , PQ=?

In triangle ∆QDR => QD = 4cm , DR=3cm , QR=?

In triangle ∆ PCB => BC= AQ + QD => 8+4=12cm

$\sf\large\underline\purple{To\: Find:-}$

$\sf{\implies PQ\:\:,QR\:\:,PC=?}$

$\sf\large\underline\purple{Solution:-}$

To calculate PQ, QR and PC at first we have to notice on the given figure, by observing we get that 3 right angle triangle is given and two sides are given each of the triangle so, here we have to Pythagoras theorem to calculate it's hypotenuse. We know that in Pythagoras theorem a formula is applied to calculate the hypotenuse of the right angle triangle. h²=p²+b²:-]

$\sf\large\underline\purple{Formula\:used:}$

$\tt{\implies hypotenuse^2=perpendicular^2+base^2}$

$\sf\small\underline\green{Calculation\:for\triangle\:QAP:-}$

$\tt{\implies AP^2+QA^2=PQ^2}$

$\tt{\implies 6^2+8^2=PQ^2}$

$\tt{\implies PQ^2=36+64}$

$\tt{\implies PQ=\sqrt{100}}$

$\tt\pink{\implies PQ=10cm}$

$\sf\small\underline\green{Calculation\:for\triangle\:QDR:-}$

$\tt{\implies QD^2+DR^2=QR^2}$

$\tt{\implies 4^2+3^2=QR^2}$

$\tt{\implies QR^2=16+9}$

$\tt{\implies QR=\sqrt{25}}$

$\tt\pink{\implies QR=5cm}$

$\sf\small\underline\green{Calculation\:for\triangle\:PCB:-}$

$\tt{\implies BC^2+PB^2=PC^2}$

$\tt{\implies 12^2+5^2=PC^2}$

$\tt{\implies PC^2=144+25}$

$\tt{\implies PC=\sqrt{169}}$

$\tt\pink{\implies PC=13cm}$

$\sf\large{Hence,}$

$\tt{\implies PC=13cm\:\:,QR=5cm\:\:,PQ=10cm}$

Answer 2

Given:−

In triangle ∆QAP => AP = 6cm, QA = 8cm , PQ=?

In triangle ∆QDR => QD = 4cm , DR=3cm , QR=?

In triangle ∆ PCB => BC= AQ + QD => 8+4=12cm

\sf\large\underline\purple{To\: Find:-}

ToFind:−

\sf{\implies PQ\:\:,QR\:\:,PC=?}⟹PQ,QR,PC=?

\sf\large\underline\purple{Solution:-}

Solution:−

To calculate PQ, QR and PC at first we have to notice on the given figure, by observing we get that 3 right angle triangle is given and two sides are given each of the triangle so, here we have to Pythagoras theorem to calculate it's hypotenuse. We know that in Pythagoras theorem a formula is applied to calculate the hypotenuse of the right angle triangle. h²=p²+b²:-]

\sf\large\underline\purple{Formula\:used:}

Formulaused:

\tt{\implies hypotenuse^2=perpendicular^2+base^2}⟹hypotenuse

2

=perpendicular

2

+base

2

\sf\small\underline\green{Calculation\:for\triangle\:QAP:-}

Calculationfor△QAP:−

\tt{\implies AP^2+QA^2=PQ^2}⟹AP

2

+QA

2

=PQ

2

\tt{\implies 6^2+8^2=PQ^2}⟹6

2

+8

2

=PQ

2

\tt{\implies PQ^2=36+64}⟹PQ

2

=36+64

\tt{\implies PQ=\sqrt{100}}⟹PQ=

100

\tt\pink{\implies PQ=10cm}⟹PQ=10cm

\sf\small\underline\green{Calculation\:for\triangle\:QDR:-}

Calculationfor△QDR:−

\tt{\implies QD^2+DR^2=QR^2}⟹QD

2

+DR

2

=QR

2

\tt{\implies 4^2+3^2=QR^2}⟹4

2

+3

2

=QR

2

\tt{\implies QR^2=16+9}⟹QR

2

=16+9

\tt{\implies QR=\sqrt{25}}⟹QR=

25

\tt\pink{\implies QR=5cm}⟹QR=5cm

\sf\small\underline\green{Calculation\:for\triangle\:PCB:-}

Calculationfor△PCB:−

\tt{\implies BC^2+PB^2=PC^2}⟹BC

2

+PB

2

=PC

2

\tt{\implies 12^2+5^2=PC^2}⟹12

2

+5

2

=PC

2

\tt{\implies PC^2=144+25}⟹PC

2

=144+25

\tt{\implies PC=\sqrt{169}}⟹PC=

169

\tt\pink{\implies PC=13cm}⟹PC=13cm

\sf\large{Hence,}Hence,

\tt{\implies PC=13cm\:\:,QR=5cm\:\:,PQ=10cm}⟹PC=13cm,QR=5cm,PQ=10cm