Question

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help me out in q no 9​

Answer 1

(8)In Given Quadrilateral ABCD.

Consider Triangle BCD.

➡️In which MC is Altitude Of Triangle.

➡️BD is base of Triangle.

$\boxed{\mathtt{Ar\:of\: Triangle=\frac{1}{2}\times\:Base\:\times\:Altitude}}$

$\boxed{\mathtt{Ar\:of\: Triangle(BCD)=\frac{1}{2}\times\:BD\:\times\:MC}}$

Consider Triangle ABD

➡️In which LA is Altitude of Triangle.

➡️BD is base of Triangle.

$\boxed{\mathtt{Ar\:of\: Triangle(ABD)=\frac{1}{2}\times\:LA\:\times\:BD}}$

Area of Given Figure.

Area of Triangle BCD+Area of Triangle ABD

$( \frac{1}{2} \times mc \times bd )+( \frac{1}{2} \times la \times bd)$

Common can be Done

$\frac{1}{2} \times bd$

Now,

$\frac{1}{2} \times bd(mc + la)$

Hence Its Proven.

$\boxed{\mathtt{Ar.\:of\:ABCD=\frac{1}{2}\times\:BD\:\times(CM+LA)}}$

(9) ➡️Area of AMCD=24 cm²
➡️AC Diagonal Divide Figure and Area in 2 Equal Parts.
➡️Area of Triangle ADC+ABC +Area of ABCD.
➡️Area of ADC=1/2 of ABCD
➡️Area of AMCD=1/2 of ABCD.

Consider To Attachment.

$\boxed{\mathbf{Ar.of\:\triangle\:ABC={16\:cm}^{2}}}$