Question

In fig., Triangle ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC, a parallelogram DBCE is constructed. Find the height of DF of the parallelogram.​

Answer 1

Answer:

Now, first determine the area of

triangle

ABC. <br> The sides of a triangle are <br> AB=a=7.5 cm, BC=b= 7 cm and CA=c=6.5 cm <br> Now, semi-perimeter of a triangle, <br>

s=(a+b+c)/(2)=(7.5+7+6.5)/(2)=(21)/(2)=10.5 cm

<br>

therefore" ""Area of "triangleABC=sqrt(s(s-a)(s-b)(s-c))" "["by Heron's formula"]

<br>

=sqrt(10.5(10.5-7.5)(10.5-7)(10.5-6.5))

<br>

=sqrt(10.5xx3xx35xx4)=sqrt441=21cm^(2)

<br> Now, area of parallelogram

BCED="Base"xx"Height"

<br>

=BCxxDF=7xxDF

<br> According to the question, <br> Area of

triangleABC

= Area of parallelogram BCED <br>

rArr " "21=7xxDF" "["from Eqs.(i) and (ii)"]

<br>

rArr " "DF=(21)/(4)=3cm

<br> Hence, the height of parallelogram is 3 cm.

Step-by-step explanation:

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