C-08-Q141mediumsingle_mcq

In a circle of radius $2.5$ cm with $AB = 3$ cm and $BD$ bisecting $\angle ADC$ , the length $AD$ equals—

a $3$ cm
b $4$ cm
c $5$ cm
d $2.5$ cm

ব্যাখ্যা

Because $BD$ bisects $\angle ADC$ , arcs $AB$ and $BC$ are equal so $AB = BC = 3$ ; with diameter $2\times2.5 = 5$ cm and $\angle ABD = 90^{\circ}$ in the semicircle, $AD = \sqrt{AB^2 + BD^2}$ works out to $4$ cm (a $3$ – $4$ – $5$ right triangle).

এই অধ্যায়ের আরও প্রশ্ন

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Circle — সব প্রশ্ন দেখুন (231টি) →

In a circle of radius $2.5$ cm with $AB = 3$ cm and $BD$ bisecting $\angle ADC$ , the length $AD$ equals—

3

4

5

2.5

In a circle of radius 2.52.52.5 cm with AB=3AB = 3AB=3 cm and BDBDBD bisecting ∠ADC\angle ADC∠ADC, the length ADADAD equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In a circle of radius 2.52.52.5 cm with AB=3AB = 3AB=3 cm and BDBDBD bisecting ∠ADC\angle ADC∠ADC, the length ADADAD equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In a circle of radius $2.5$ cm with $AB = 3$ cm and $BD$ bisecting $\angle ADC$ , the length $AD$ equals—

In a circle of radius $2.5$ cm with $AB = 3$ cm and $BD$ bisecting $\angle ADC$ , the length $AD$ equals—