In the figure, ABCD is a rectangle. G is a point on CD. The area of
the triangle ∆ AGH is 7 square centimetres and the area of the
triangle ∆ BGH is 4 square centimetres.
a) What is the area of the triangle ∆ AGD?
b) What is the area of the triangle ∆ BGC?
c) What is the area of the triangle ∆ ABG?
d) What is the area of the rectangle ABCD?
e) What is the relation between the area of the triangle ABG
and that of the rectangle ABCD
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Answer:
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Using Pythagoras theorem in right triangle ABC, we obtain
AC
2
=AB
2
+BC
2
⇒(5+2)
2
+(2−t)
2
={(5−2)
2
+(2+2)
2
}+{(2+2)
2
+(−2−t)
2
}
⇒49+(4−4t+t
2
)=(9+16)+(16+4+4t+t
2
)
⇒t
2
−4t+53=t
2
+4t×45
⇒−8t=−8
⇒t=1
Step-by-step explanation:
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Answer:
a) question answer please
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