C-07-Q021mediumsingle_mcq

In $\triangle ACR$ and $\triangle ADR$ , the included angle $\angle ARC = \angle ARD$ equals—

aan acute angle
ban obtuse angle
ca right angle
da straight angle

ব্যাখ্যা

$R$ lies on the perpendicular bisector, so $AR \perp CD$ , making both $\angle ARC$ and $\angle ARD$ right angles ( $90^\circ$ ). These equal right angles serve as the included angle for proving the two triangles congruent.

এই অধ্যায়ের আরও প্রশ্ন

C-07-Q022The congruence is established by—
C-07-Q023After joining and in the alternate method, equals—
C-07-Q024In Construction 2 the difference is cut on ray as line segment—
C-07-Q025The angle constructed at in Construction 2 equals—
C-07-Q026In the resulting triangle , the difference equals—
C-07-Q027If the given base-adjacent angle is NOT acute, Construction 2 is—
C-07-Q028In Construction 2, because in —
C-07-Q029Construction 2 essentially differs from Construction 1 in that—
C-07-Q030In Construction 3, on ray the part is cut equal to—
C-07-Q031At points and the angles and are made equal respectively to—

Practical Geometry — সব প্রশ্ন দেখুন (150টি) →

In $\triangle ACR$ and $\triangle ADR$ , the included angle $\angle ARC = \angle ARD$ equals—

aan acute angle

ban obtuse angle

ca right angle

da straight angle

In △ACR\triangle ACR△ACR and △ADR\triangle ADR△ADR, the included angle ∠ARC=∠ARD\angle ARC = \angle ARD∠ARC=∠ARD equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In △ACR\triangle ACR△ACR and △ADR\triangle ADR△ADR, the included angle ∠ARC=∠ARD\angle ARC = \angle ARD∠ARC=∠ARD equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In $\triangle ACR$ and $\triangle ADR$ , the included angle $\angle ARC = \angle ARD$ equals—

In $\triangle ACR$ and $\triangle ADR$ , the included angle $\angle ARC = \angle ARD$ equals—