C-14-Q018mediumsingle_mcq

Triangles $ABC$ and $DEF$ have equal heights $h$ , and bases $BC=a$ , $EF=d$ . Then $\triangle ABC : \triangle DEF = ?$

a $a:d$
b $d:a$
c $h:h$
d $ah:dh$

ব্যাখ্যা

With equal heights $h$ , the areas are $\frac{1}{2}ah$ and $\frac{1}{2}dh$ ; the common factor $\frac{1}{2}h$ cancels in the ratio. This leaves $\triangle ABC:\triangle DEF = a:d$ .

এই অধ্যায়ের আরও প্রশ্ন

C-14-Q019Triangles with equal bases and heights , have area ratio —
C-14-Q020Two triangles have equal bases and heights in ratio . Their areas are in ratio —
C-14-Q021Two triangles have equal heights and bases cm and cm. The ratio of their areas is —
C-14-Q022A line parallel to one side of a triangle intersects the other two sides —
C-14-Q023In , with on and on . Then —
C-14-Q024The proof of Theorem 28 uses the fact that and have —
C-14-Q025Corollary 1 of Theorem 28 gives —
C-14-Q026Corollary 2 of Theorem 28 says: a line through the midpoint of one side parallel to another side —
C-14-Q027In , . If , , , then
C-14-Q028In , . If , , , then

Triangles $ABC$ and $DEF$ have equal heights $h$ , and bases $BC=a$ , $EF=d$ . Then $\triangle ABC : \triangle DEF = ?$

a:d

d:a

h:h

ah:dh

ব্যাখ্যা

With equal heights $h$ , the areas are $\frac{1}{2}ah$ and $\frac{1}{2}dh$ ; the common factor $\frac{1}{2}h$ cancels in the ratio. This leaves $\triangle ABC:\triangle DEF = a:d$ .

Triangles ABCABCABC and DEFDEFDEF have equal heights hhh, and bases BC=aBC=aBC=a, EF=dEF=dEF=d. Then △ABC:△DEF=?\triangle ABC : \triangle DEF = ?△ABC:△DEF=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Triangles ABCABCABC and DEFDEFDEF have equal heights hhh, and bases BC=aBC=aBC=a, EF=dEF=dEF=d. Then △ABC:△DEF=?\triangle ABC : \triangle DEF = ?△ABC:△DEF=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Triangles $ABC$ and $DEF$ have equal heights $h$ , and bases $BC=a$ , $EF=d$ . Then $\triangle ABC : \triangle DEF = ?$

Triangles $ABC$ and $DEF$ have equal heights $h$ , and bases $BC=a$ , $EF=d$ . Then $\triangle ABC : \triangle DEF = ?$