Mix Examples - Areas of Parallelograms and Triangles Questions in English

(A) In $\Delta ABC$,since $\angle B = 90^{\circ}$,by the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Substituting the values,$AC^2 = 12^2 + 16^2 = 144 + 256 = 400$.
Thus,$AC = \sqrt{400} = 20 \, cm$.
The area of $\Delta ABC$ can be calculated as $\frac{1}{2} \times \text{base} \times \text{height}$.
Using $AB$ and $BC$ as base and height,Area $= \frac{1}{2} \times 12 \times 16 = 96 \, cm^2$.
Alternatively,using $AC$ as the base and $BM$ as the altitude,Area $= \frac{1}{2} \times AC \times BM$.
Equating the two expressions for area: $\frac{1}{2} \times 20 \times BM = 96$.
$10 \times BM = 96$.
$BM = 9.6 \, cm$.

(N/A) In the given figure,we have a parallelogram $ABCD$ and a trapezium $ABST$.
For figures to lie on the same base and between the same parallels,they must share a common side as a base,and their remaining vertices must lie on a line parallel to that base.
Here,the parallelogram $ABCD$ and the trapezium $ABST$ share the common base $AB$.
However,the vertices $D$ and $C$ of the parallelogram lie on the line $DC$,while the vertices $T$ and $S$ of the trapezium lie on the line $TS$.
Since the line $DC$ is not the same as the line $TS$,the two figures do not lie between the same parallels.
Therefore,the answer is negative.

(N/A) Yes,parallelogram $ABCD$ and $\Delta MBC$ lie on the same base and between the same parallels.
Common base: $BC$.
Parallel lines: $AD$ and $BC$.

(N/A) No,the parallelogram $ABCD$ and the triangle $QBC$ do not lie on the same base and between the same parallels. For figures to lie on the same base and between the same parallels,they must share a common side as a base,and their vertices opposite to the base must lie on a line parallel to the base. In this figure,the parallelogram $ABCD$ has base $AB$ (or $CD$),while the triangle $QBC$ has base $BC$. Since they do not share a common base,the condition is not satisfied.