Question

19. In a AABC, ZA = 25, B = 350 and AB = 16 units. In APQR, P = 35°, ZQ = 120
°= =
PR = 4 units. Then, ar(AABC) =
fo) Zor APQR) (b) 4ar APQR)
(c) Bar(APQR) (d) 16ar(APQR)
​

Answer 1

Answer:

(i) ∠PRB=∠BAP=35

o

(Angles in the same segment of the circle)

(ii) ∠BPA=90

o

(angle in a semicircle)

∴∠BPQ=90

o

∴∠PBR=∠BQP+∠BPQ=25

o

+90

o

=115

o

(exterior angle in △BPQ)

(iii) ∠ABP=90

o

−∠BAP=90

o

−35

o

=55

o

∴∠ABR=∠PBR=∠ABP=115

o

−55

o

=60

o

∴∠APR=∠ABR=60

o

(Angle in the same segment of circle)

Hence, ∠BPR=90

o

−∠APR=90

o

−60

o

=30

o

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Step-by-step explanation:

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Answer 2

$\huge\colorbox{white}{ \colorbox{white}{\colorbox{white}{Correct\:Question}}}$

In a AABC, $\angle$A = 25°, $\angle$ B = 35°and AB = 16 units. In ∆PQR, $\angle$P = 35°, $\angle$Q = 120° & PR = 4 units. Then, ar(∆ABC) =

(i) 2 ar(∆PQR)

(ii) 4ar (∆PQR)

(iii) 8ar (∆PQR)

(iv) 16ar (∆PQR)

$\huge\colorbox{white}{ \colorbox{white}{\colorbox{white}{Answer}}}$

Given :

• $\tt \:\color{green} {In ∆ABC ,}$

$\rightarrow$ ZA = 25

$\rightarrow$ B = 350

$\rightarrow$ AB = 16 units

• $\tt \: \color{green}{In ∆PQR}$

$\rightarrow$ P = 35°

$\rightarrow$ ZQ = 120

$\rightarrow$ PR = 4 units

To Find :-

• $\tt \: \color{green}{ar (∆ABC = ? )}$

Solution :-

$\rightarrow$ $\tt \: \color{green}{In ∆ABC and ∆PQR}$

$\rightarrow$ $\tt \: \angle A = \angle R = 25°$

$\rightarrow$ $\tt \: \angle{B} = \angle{P} = 35°$

$\therefore$ $\tt \: ∆ABC ~ ∆RPQ....[ By AA similarity]$

$\color{green} {So,}$

$\sf\huge\implies$$\frac{ar(∆ABC)}{ar(RPQ)}$

$\sf\huge\implies$ $\frac{AB²}{PR²}$

$\sf\huge\implies$ $\frac{16²}{4²}$

$\sf\huge\implies$ $\frac{256}{16}$

$\sf\huge\implies$ $\frac{16}{1}$

$\sf\huge\implies$ $\tt \: ar(∆ABC) = 16 ar (∆PQR)$

$\color{Red} {Hence,\: Correct\:Answer\:is \: option\: (d)}$