Question

In the given figure : ZABD = ZEBC, BD = BC and ZACB = ZEDB
Prove that AB = BE.
[Hint. ZEBD - ZABE + ZABD and ZABC = ZABE + ZEBC.
So, ZEBD = ZABC. .
Now, show that AACB = AEDB.]​

Answer 1

Answer:

$\small{\fcolorbox{red}{indigo}{\small{\fcolorbox{red}{violet}{\small{\fcolorbox{red}{pink} {\small{\fcolorbox{red}{red}{\small{\fcolorbox{red}{springgreen}{\small{\fcolorbox{red}{blue}{\small{\fcolorbox{red}{yellow} {\small{\fcolorbox{red}{azure} {\small{\fcolorbox{red}{blue}{\small{\fcolorbox{red}{orange}{\huge{\fcolorbox{red}{green}{\large{\fcolorbox{blue}{red}{{\fcolorbox{orange}{aqua}{ANSWER}}}}}}}}}}}}}}}}}}}}}}}}}}$

$\mathfrak{ \large{\blue{ \underline{ \purple{ Given: }}}}}$

Radius of circle = 20 cm

Value of π = 3.14

$\mathfrak{ \large{\blue{ \underline{ \purple{ To \: Find: }}}}}$

Area and perimeter of circle

$\mathfrak{ \large{\blue{ \underline{ \purple{ Solution:}}}}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\: \mathfrak{ \underline{ \green{Formula \: to \: calculate \: the \: circumference \: of \: circle}}}$

$\boxed{ \mathfrak{ \star \: \: \: { \red{ \large{circumference = 2 \times \pi \times radius}}}}}$

According to question,

$\large{ \rm \longmapsto \: \: \: \: \: \: \: circumference = 2 \times 3.14 \times 20}$

$\large{ \rm \longmapsto \: \: \: \: \: \: \: circumference = \boxed{ \orange{ \mathfrak 125.6 \: cm}}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \mathfrak{ \underline{ \green{Formula \: to \: calculate \: the \: circumference \: of \: circle}}}$

$\boxed{ \mathfrak{ \star \: \: \: \: \: \: { \red{ \large{area = \pi \times {radius}^{2} }}}}}$

According to question,

$\large{ \rm \longmapsto \: \: \: \: \: \: \: area = 3.14 \times {20}^{2}}$

$\large{ \rm \longmapsto \: \: \: \: \: \: \: area = 3.14 \times 20 \times 20}$

$\large{ \rm \longmapsto \: \: \: \: \: \: \: area = \boxed{ \orange{ \mathfrak{1256 \: {cm}^{2} }}}}$

$\mathfrak{ \large{\blue{ \underline{ \purple{ Hence:}}}}}$

The area and circumference of circle is 125.6 cm and 1256 cm²

Step-by-step explanation:

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Answer 2

In, ∆ACB & ∆EDB ,we have,

$•BD = BC$

$•\angle{ACB} = \angle{EDB}$

$•\angle{EBD} = \angle{EBA}+\angle{ABD}\\ \angle{EBD} = \angle{ABE}+\angle{EBC}\\ \angle{EBD }= \angle{ABC}$

By ASA property ∆ACB $\cong$ ∆EDB

By Congruence, AB = BE