EXERCISE - 8.3
-
1. Equilateral triangles are drawn on the three sides of a right angled triangle. Show that the
area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the
other two sides.
2. Prove that the area of the equilateral triangle described on the side of a square is half the
area of the equilateral triangles described on its diagonal.
3. D, E, F are mid points of sides BC, CA, AB of AABC. Find the ratio of areas of ADER
and AABC
4. In AABC, XY || AC and XY divides the triangle into two parts of equal area. Find the
AX
ratio of
XB
5. Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of
their corresponding medians.
6. AABC-ADEF. BC=3cm EF = 4cm and area of AAB
Answers
Answer:
it is true that the equilateral triangle drawn on the hypotenuse is equal to the sum of areas of equilateral triangles drawn on the other two sides.
Let us go a step further.
What happens if its semicircles instead of triangles?
Or any similar figure?
Beauty is that Pythogoras theorem still holds good!
Step-by-step explanation:
ANSWER
Given A right angled triangle ABC with right angle at B. Equilateral triangles PAB, QBC and RAC are described on sides AB, BC and Ca respectively.
To prove Area(△PAB)+Area(△QBC)=Area(△RAC).
Proof Since triangles PAB, QBC and RAC are equilateral. Therefore, they are equiangular and hence similar.
∴
Area(△RAc)
Area(△PAB)
+
Area(△RAC)
Area(△QBC)
=
AC 2
AB
2
+
AC
2
BC
2
⇒
Area(△RAC)
Area(△PAB)
+
Area(△RAC)
Area(△QBC)
=
AC
2
AB
2
+BC
2
⇒
Area(△RAC)
Area(△PAB)
+
Area(△RAC)
Area(△QBC)
=
AC
2
AC
2
=1
[∵△ is a right angled triangle with ∠B=90
0
∴AC
2
=AB
2
+BC
2
]
⇒
Area(△RAC)
Area(△PAB)+Area(△QBC)
=1
⇒ Area(△PAB)+Area(△QBC)=Area(△RAC) [Hence proved]