Easy geometry. For mathe-magicians(only). Please give the explanation,

1.	Easy geometry. For mathe-magicians(only). Please give the explanation, but don't give the answer. Last time I did not report when he gave answer with explanation, but this time if you don't give explanation, I'll report. Check the image carefully.
Answer» QUESTION: The QUADRILATERAL FGHE is inscribed inside a rectangle ABCD of length 100 cm and width 50 cm such that E is the midpoint of segment AB and G is the midpoint of segment DC. Find the area of the quadrilateral FEHG. ANSWER: Area of quadrilateral = 2500 cm² GIVEN: FGHE is a quadrilateral inscribed inside a rectangle ABCD Length of the rectangle = 100 cm Width of the rectangle = 50 cm E is the midpoint of segment AB G is the midpoint of segment DC TO FIND: Area of the quadrilateral FEHG. FORMULAE: AREA OF TRIANGLE = 1/2(base)(height) EXPLANATION: Draw a line from E to G and we will GET 2 triangles of height 25 cm and base 100 cm. Draw a perpendicular from E to X and from H to Y. These are the heights of the triangles respectively. $\large{\bf{\sf{Area \: \: of \: \: \triangle EFG = 1/2(EG)(FX)}}}$ $\large{\bf{Area \: \: of \: \: \triangle EFG = \dfrac{1}{2} (100)(25)}}} \\ \\ \\ \large{\bf{\sf{Area \: \: of \: \: \triangle EFG = 50 \times 25}}} \\ \\ \\ \large{\bf{\sf{Area \: \: of \: \: \triangle EFG = 1250 \: {cm}^{2}}}$ $\large{\sf{\bf{Area \: \: of \: \: \triangle EHG = 1/2(EG)(HY)}}}$ $\large{\bf{\sf{Area \: \: of \: \: \triangle EHG = \dfrac{1}{2} (100)(25)}}} \\ \\ \\ \large{\bf{\sf{Area \: \: of \: \: \triangle EHG = 50 \times 25}}} \\ \\ \\ \large{\bf{\sf{Area \: \: of \: \: \triangle EHG = 1250 \: {cm}^{2}}}$ $\sf{\bf{\large{Area\:\:of\:\:quadrilateral = Area \: \: of \: \: \triangle EFG+ Area \: \: of \: \: \triangle EHG}}$ Area of quadrilateral = 1250 + 1250 Area of quadrilateral = 2500 cm² Hence area of the quadrilateral FEHG = 2500 cm². NOTE : REFER ATTACHMENT FOR DIAGRAM.

Easy geometry. For mathe-magicians(only). Please give the explanation, but don't give the answer. Last time I did not report when he gave answer with explanation, but this time if you don't give explanation, I'll report. Check the image carefully.

Answer»

QUESTION:

The QUADRILATERAL FGHE is inscribed inside a rectangle ABCD of length 100 cm and width 50 cm such that E is the midpoint of segment AB and G is the midpoint of segment DC. Find the area of the quadrilateral FEHG.

ANSWER:

Area of quadrilateral = 2500 cm²

GIVEN:

FGHE is a quadrilateral inscribed inside a rectangle ABCD

Length of the rectangle = 100 cm

Width of the rectangle = 50 cm

E is the midpoint of segment AB

G is the midpoint of segment DC

TO FIND:

Area of the quadrilateral FEHG.

FORMULAE:

AREA OF TRIANGLE = 1/2(base)(height)

EXPLANATION:

Draw a line from E to G and we will GET 2 triangles of height 25 cm and base 100 cm.

Draw a perpendicular from E to X and from H to Y. These are the heights of the triangles respectively.

$\large{\bf{\sf{Area \: \: of \: \: \triangle EFG = 1/2(EG)(FX)}}}$

$\large{\bf{Area \: \: of \: \: \triangle EFG = \dfrac{1}{2} (100)(25)}}} \\ \\ \\ \large{\bf{\sf{Area \: \: of \: \: \triangle EFG = 50 \times 25}}} \\ \\ \\ \large{\bf{\sf{Area \: \: of \: \: \triangle EFG = 1250 \: {cm}^{2}}}$

$\large{\sf{\bf{Area \: \: of \: \: \triangle EHG = 1/2(EG)(HY)}}}$

$\large{\bf{\sf{Area \: \: of \: \: \triangle EHG = \dfrac{1}{2} (100)(25)}}} \\ \\ \\ \large{\bf{\sf{Area \: \: of \: \: \triangle EHG = 50 \times 25}}} \\ \\ \\ \large{\bf{\sf{Area \: \: of \: \: \triangle EHG = 1250 \: {cm}^{2}}}$

$\sf{\bf{\large{Area\:\:of\:\:quadrilateral = Area \: \: of \: \: \triangle EFG+ Area \: \: of \: \: \triangle EHG}}$