C-14-Q083mediumsingle_mcq

In the proof of Theorem 35, $\dfrac{h}{p} = ?$

a $\dfrac{BC}{EF}$
b $\dfrac{EF}{BC}$
c $\dfrac{BC^{2}}{EF^{2}}$
d $\dfrac{1}{2}$

ব্যাখ্যা

The two similar triangles' corresponding heights $h$ and $p$ scale in the same ratio as their corresponding bases, so $\dfrac{h}{p} = \dfrac{BC}{EF}$ .

এই অধ্যায়ের আরও প্রশ্ন

C-14-Q084Two similar triangles with matching side ratio have area ratio —
C-14-Q085If areas of two similar triangles are and , the ratio of matching sides is —
C-14-Q086If lies between and with , then divides —
C-14-Q087Construction 12 divides a segment in a given internal ratio using a parallel line whose existence relies on —
C-14-Q088In Construction 12, on ray we cut and . Drawing gives —
C-14-Q089Example 1: a cm segment is divided in the ratio . The two parts measure —
C-14-Q090To divide a cm segment in ratio , the parts are —
C-14-Q091In , with , . Then and are —
C-14-Q092With in , the correct relation is —
C-14-Q093With , the area ratio is —

In the proof of Theorem 35, $\dfrac{h}{p} = ?$

\dfrac{BC}{EF}

\dfrac{EF}{BC}

\dfrac{BC^{2}}{EF^{2}}

\dfrac{1}{2}

ব্যাখ্যা

The two similar triangles' corresponding heights $h$ and $p$ scale in the same ratio as their corresponding bases, so $\dfrac{h}{p} = \dfrac{BC}{EF}$ .

In the proof of Theorem 35, hp=?\dfrac{h}{p} = ?ph​=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof of Theorem 35, hp=?\dfrac{h}{p} = ?ph​=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof of Theorem 35, $\dfrac{h}{p} = ?$

In the proof of Theorem 35, $\dfrac{h}{p} = ?$