C-07-Q019mediumsingle_mcq

In the constructed triangle $\triangle ABC$ , $BA + AC$ equals—

a $a$
b $s$
c $a+s$
d $s-a$

ব্যাখ্যা

Because $AC = AD$ , the path $BA + AC$ equals $BA + AD = BD$ , which was constructed equal to $s$ . Hence $BA + AC = s$ , satisfying the required sum of the two sides.

এই অধ্যায়ের আরও প্রশ্ন

C-07-Q020In the alternate method of Construction 1, line segment is bisected by—
C-07-Q021In and , the included angle equals—
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—

Practical Geometry — সব প্রশ্ন দেখুন (150টি) →

In the constructed triangle △ABC\triangle ABC△ABC, BA+ACBA + ACBA+AC equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the constructed triangle △ABC\triangle ABC△ABC, BA+ACBA + ACBA+AC equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the constructed triangle $\triangle ABC$ , $BA + AC$ equals—

In the constructed triangle $\triangle ABC$ , $BA + AC$ equals—