C-07-Q028mediumsingle_mcq

In Construction 2, $AC = AD$ because in $\triangle ACD$ —

a $\angle ACD = \angle ADC$
b $\angle CAD = \angle ACD$
c $AD = CD$
dit is equilateral

ব্যাখ্যা

In $\triangle ACD$ the base angles $\angle ACD$ and $\angle ADC$ are equal, and equal base angles make the triangle isosceles. Hence the sides opposite them, $AC$ and $AD$ , are equal.

এই অধ্যায়ের আরও প্রশ্ন

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—
C-07-Q032The rays and are—
C-07-Q033At the point where the bisectors meet, is constructed equal to—
C-07-Q034In , because—
C-07-Q035In , because—
C-07-Q036The perimeter in the constructed triangle equals—
C-07-Q037In Construction 3, equals—
C-07-Q038In Example 1, is drawn equal to—

Practical Geometry — সব প্রশ্ন দেখুন (150টি) →

In Construction 2, $AC = AD$ because in $\triangle ACD$ —

\angle ACD = \angle ADC

\angle CAD = \angle ACD

AD = CD

dit is equilateral

ব্যাখ্যা

In $\triangle ACD$ the base angles $\angle ACD$ and $\angle ADC$ are equal, and equal base angles make the triangle isosceles. Hence the sides opposite them, $AC$ and $AD$ , are equal.

In Construction 2, AC=ADAC = ADAC=AD because in △ACD\triangle ACD△ACD—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In Construction 2, AC=ADAC = ADAC=AD because in △ACD\triangle ACD△ACD—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In Construction 2, $AC = AD$ because in $\triangle ACD$ —

In Construction 2, $AC = AD$ because in $\triangle ACD$ —