C-14-Q030mediumsingle_mcq

In $\triangle ABC$ , $DE \parallel BC$ , $AB = 9$ , $AD = 3$ , $AC = 12$ . Then $AE = ?$

a $3$
b $4$
c $6$
d $9$

ব্যাখ্যা

Since $DE \parallel BC$ , the Basic Proportionality Theorem gives $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ . Substituting, $\dfrac{3}{9} = \dfrac{AE}{12}$ , so $AE = \dfrac{12}{3} = 4$ .

এই অধ্যায়ের আরও প্রশ্ন

C-14-Q031The heights of and (with the common vertex pattern of the proof) are —
C-14-Q032If a line segment divides two sides of a triangle (or their produced sections) proportionally, it is —
C-14-Q033Theorem 29 is the converse of —
C-14-Q034In , if where , , then —
C-14-Q035In the proof of Theorem 29, because they are on the same base and —
C-14-Q036If on with , , and on with , , then is —
C-14-Q037The internal bisector of an angle of a triangle divides the opposite side in the ratio of —
C-14-Q038In , bisects and meets at . Then —
C-14-Q039In the proof of Theorem 30, the auxiliary line is drawn —
C-14-Q040In the proof, because they are —

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

In △ABC\triangle ABC△ABC, DE∥BCDE \parallel BCDE∥BC, AB=9AB = 9AB=9, AD=3AD = 3AD=3, AC=12AC = 12AC=12. Then AE=?AE = ?AE=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In △ABC\triangle ABC△ABC, DE∥BCDE \parallel BCDE∥BC, AB=9AB = 9AB=9, AD=3AD = 3AD=3, AC=12AC = 12AC=12. Then AE=?AE = ?AE=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In $\triangle ABC$ , $DE \parallel BC$ , $AB = 9$ , $AD = 3$ , $AC = 12$ . Then $AE = ?$

In $\triangle ABC$ , $DE \parallel BC$ , $AB = 9$ , $AD = 3$ , $AC = 12$ . Then $AE = ?$