C-07-Q060mediumsingle_mcq

In $\triangle AOB$ and $\triangle COD$ , $OA = OC$ equals—

a $\frac{a}{2}$
b $\frac{b}{2}$
c $a$
d $b$

ব্যাখ্যা

In the parallelogram construction, $O$ is the intersection of the diagonals, which bisect each other. So $OA$ equals half of diagonal $a$ , giving $OA = OC = \frac{a}{2}$ .

এই অধ্যায়ের আরও প্রশ্ন

C-07-Q061The congruence uses—
C-07-Q062From being alternate angles, it follows that—
C-07-Q063In Construction 4, the included angle between the diagonals is—
C-07-Q064In Construction 5, on ray the segment is cut equal to—
C-07-Q065The arcs at and are drawn with radii respectively—
C-07-Q066The point of intersection of these arcs gives—
C-07-Q067To complete the parallelogram, is cut from equal to—
C-07-Q068In Construction 5, is justified by—
C-07-Q069In Example 3 the parallel sides satisfy—
C-07-Q070From ray , the line segment is cut equal to—

Practical Geometry — সব প্রশ্ন দেখুন (150টি) →

In △AOB\triangle AOB△AOB and △COD\triangle COD△COD, OA=OCOA = OCOA=OC equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In △AOB\triangle AOB△AOB and △COD\triangle COD△COD, OA=OCOA = OCOA=OC equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In $\triangle AOB$ and $\triangle COD$ , $OA = OC$ equals—

In $\triangle AOB$ and $\triangle COD$ , $OA = OC$ equals—