Question

Hello
please solve this ​

Answer 1

Step-by-step explanation:

✞︎ Given

• In ThE Attachment

✞︎We Know,

➪∆DC = 6

➪∆BC= 15

✞︎To Find

A(ABD) : A (ABC)

Solution

∆ABC, ∆ABD, ∆ADC all have same height-h

• Buse length of ∆ABC = 15

Buse length of ∆ABC = 15}

• Buse length of ∆ADC = 6

$\tt\large\underline{\longmapsto \: Buse \: length \: of \: ∆ADC = 6}$

.

$\tt\large\underline{ \longmapsto\: Base \: Length \: Of \: ∆ABD =9}$

$\tt\large\underline{\longmapsto \: Area \: Cy \: Of ∆ABC = \frac{1}{2} \times h \times 15 }$

$\tt\large\underline\red{\longmapsto \: Area \: of \: ∆ABD = \frac{1}{2} \times h \times 9 }$

$\tt\large\underline{\longmapsto \: Area \: of \: ∆ADC = \frac{1}{2} \times h \times 6}$

A(∆ABD): A(ABC)=

$\tt\large\underline{\longmapsto( \frac{1}{2} \times h \times 9)}$

$\tt\large\underline{ \longmapsto(\frac{1}{2} \times h \times 15) }$

$\tt\large\underline\red{\longmapsto=3:5}$

Hence AnS is 3:5

$\:$

Hope It Helps!

ᗷᗴ ᗷᖇᗩIᑎᒪY

ᗩᗪᗪITIOᑎᗩᒪ IᑎᖴOᖇᗰᗩTIOᑎ-

$\tt\large\underline\pink{ \longmapsto\: Perimeter \: of \: Equilateral \: Triangle \: : P = 3a.}$

$\large{✰} \bf \underline\color{brown}{\longmapsto \: Area \: of \: Equilateral \: Triangle: K = ( \frac{1}{4} \times \sqrt{3} \times a. }$

$\tt\large\underline\purple{\longmapsto \: In \: an \: equilateral \: triangle \: ABC, AB = BC = CA.}$

Kindly Scrool Right For Further Sol.

Answer 2

✞︎We Know,

➪∆DC = 6

➪∆BC= 15

✞︎To Find

A(ABD) : A (ABC)

Solution

∆ABC, ∆ABD, ∆ADC all have same height-h

$\tt\large\underline{\longmapsto \: Buse \: length \: of \: ∆AbC = 15}$$\tt\large\underline{\longmapsto \: Buse \: length \: of \: ∆ADC = 6}$

$\tt\large\underline{ \longmapsto\: Base \: Length \: Of \: ∆ABD =9}$$\tt\large\underline{\longmapsto \: Area \: Cy \: Of ∆ABC = \frac{1}{2} \times h \times 15 }$$\tt\large\red{\longmapsto \: Area \: of \: ∆ABD = \frac{1}{2} \times h \times 9 }$$\tt\large\underline{\longmapsto \: Area \: of \: ∆ADC = \frac{1}{2} \times h \times 6}$

A(∆ABD) : A(ABC) =

$\tt\large{\longmapsto( \frac{1}{2} \times h \times 9)}$$\tt\large\underline{ \longmapsto(\frac{1}{2} \times h \times 15) }$$\tt\large\red{\longmapsto=3:5}$

Hence AnS is 3:5 .