C-14-Q071mediumsingle_mcq

In the proof of Theorem 33, $\triangle APQ \cong \triangle DEF$ by —

aSSS
bSAS
cASA
dRHS

ব্যাখ্যা

$AP$ and $AQ$ were marked equal to $DE$ and $DF$ , and $PQ$ was shown equal to $EF$ , so all three pairs of sides match and the triangles are congruent by SSS.

এই অধ্যায়ের আরও প্রশ্ন

C-14-Q072In the proof, and are equiangular because —
C-14-Q073Theorem 33 is the converse of —
C-14-Q074If one angle of a triangle equals an angle of another and the sides adjacent to those angles are proportional,
C-14-Q075In the proof of Theorem 34, by —
C-14-Q076After establishing , we conclude —
C-14-Q077Theorem 34 is the analogue, for similarity, of —
C-14-Q078The ratio of the areas of two similar triangles equals the ratio of the squares of —
C-14-Q079If , then
C-14-Q080with , . The ratio of areas is —
C-14-Q081Two similar triangles have areas in ratio . The ratio of matching sides is —

Ratio, Similarity and Symmetry — সব প্রশ্ন দেখুন (160টি) →

In the proof of Theorem 33, △APQ≅△DEF\triangle APQ \cong \triangle DEF△APQ≅△DEF by —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof of Theorem 33, △APQ≅△DEF\triangle APQ \cong \triangle DEF△APQ≅△DEF by —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof of Theorem 33, $\triangle APQ \cong \triangle DEF$ by —

In the proof of Theorem 33, $\triangle APQ \cong \triangle DEF$ by —