In the figure,Delta ABC has sides AB = 7.5, c | vedclass

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(B) The sides of $\Delta ABC$ are $a = 6.5\, cm$,$b = 7\, cm$,and $c = 7.5\, cm$.The semi-perimeter $s$ of $\Delta ABC$ is:$s = \frac{6.5 + 7 + 7.5}{2

Vedclass · Accepted Answer

$3$

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(B) The sides of $\Delta ABC$ are $a = 6.5\, cm$,$b = 7\, cm$,and $c = 7.5\, cm$.The semi-perimeter $s$ of $\Delta ABC$ is:$s = \frac{6.5 + 7 + 7.5}{2} = \frac{21}{2} = 10.5\, cm$.Using Heron's formula,the area of $\Delta ABC$ is:$	ext{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$= \sqrt{10.5(10.5 - 6.5)(10.5 - 7)(10.5 - 7.5)}$$= \sqrt{10.5 	imes 4 	imes 3.5 	imes 3}$$= \sqrt{441} = 21\, cm^2$.Since the parallelogram $DBCE$ is constructed on the same base $BC$ and has the same area as $\Delta ABC$:$	ext{Area of parallelogram } DBCE = 	ext{Area of } \Delta ABC = 21\, cm^2$.The area of a parallelogram is given by $	ext{base} 	imes 	ext{height}$.$BC 	imes DF = 21\, cm^2$$7\, cm 	imes DF = 21\, cm^2$$DF = \frac{21}{7} = 3\, cm$.Thus,the height $DF$ of the parallelogram is $3\, cm$.

In the figure,$\Delta ABC$ has sides $AB = 7.5\, cm$,$AC = 6.5\, cm$,and $BC = 7\, cm$. On base $BC$,a parallelogram $DBCE$ of the same area as that of $\Delta ABC$ is constructed. Find the height $DF$ of the parallelogram (in $cm$).

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