Question

in ∆PQR, if <P+<Q = <R, PR=7 , QR= 24 then PQ=?
​

Answer 1

ANSWER

In △PQR

(PQ)

2

+(QR)

2

=(PR)

2

(24)

2

+(7)

2

=(PR)

2

⇒(PR)

2

=576+49=625

⇒PR=25

Now S is the mid point of PR

∴RS=

2

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PR

=

2

25

=12.5cm

Answer 2

GIVEN :-

• In ∆ PQR , ∠P + ∠Q = ∠R.
• PR = 7 cm , QR = 24 cm.

TO FIND :-

• The length of PQ.

SOLTUION :-

$\setlength{\unitlength}{1.9cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{P}}\put(7.7,1){\large\sf{R}}\put(10.6,1){\large\sf{Q}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(8.6,0.7){\sf{\large{24 cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\put(7.4,1.8){\sf 7 cm}\put(9.4,2.2){\sf ? cm}\end{picture}$

Here in the question it is given that , In ∆ PQR , ∠P + ∠Q = ∠R . So from the given outcomes we conclude that if the sum of two angles is equal to the other angle , then it must be a Right angled ∆ .

Now in ∆ PQR by Pythagoras theorem,

➳ (PQ)² = (PR)² + (QR)²

➳ (PQ)² = 7² + (24)²

➳ (PQ)² = 49 + 576

➳ (PQ)² = 625

➳ PQ = √625

➳ PQ = 25.

Hence the required length of PQ is 25 cm.