In Delta ABC,angle B = 90^{circ},BC = 8 , tex | vedclass

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(A) $1$. In right-angled $\Delta ABC$,by Pythagoras theorem: $AB^2 + BC^2 = AC^2$.$2$. $AB^2 + 8^2 = 17^2 \implies AB^2 + 64 = 289 \implies AB^2 = 225

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) $1$. In right-angled $\Delta ABC$,by Pythagoras theorem: $AB^2 + BC^2 = AC^2$.$2$. $AB^2 + 8^2 = 17^2 \implies AB^2 + 64 = 289 \implies AB^2 = 225 \implies AB = 15 \, 	ext{cm}$.$3$. Area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 15 	imes 8 = 60 \, 	ext{cm}^2$.$4$. Since $BE$ is a median,it divides the triangle into two equal areas: $	ext{Area}(\Delta BCE) = \frac{1}{2} 	imes 	ext{Area}(\Delta ABC) = \frac{60}{2} = 30 \, 	ext{cm}^2$.$5$. In $\Delta BCE$,$M$ is the midpoint of $BE$,so $CM$ is a median of $\Delta BCE$.$6$. $A$ median divides a triangle into two equal areas,so $	ext{Area}(\Delta BMC) = \frac{1}{2} 	imes 	ext{Area}(\Delta BCE) = \frac{30}{2} = 15 \, 	ext{cm}^2$.

In $\Delta ABC$,$\angle B = 90^{\circ}$,$BC = 8 \, \text{cm}$,and $AC = 17 \, \text{cm}$. $BE$ is a median of the triangle and $M$ is the midpoint of $BE$. Find the area of $\Delta BMC$ in $\text{cm}^2$.

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