Triangles – Exercise 7.5 – (MATHEMATICS) – 9th Class

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Exercise 7.5

1. ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.


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Sol. We know that any point on the perpendicular bisector of the line segment is equidistant from the end points of the line segment. Hence,
(i) Draw the perpendicular bisector of line segment BC.
(ii) Draw the perpendicular bisector of line segment AC. Let these perpendicular bisectors meet at O.
Then we have OB = OC = OA, i.e., 0 is a point equidistant from the vertices of a triangle ABC.

2. In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

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Sol. We know that any point on the bisector of an angle is equidistant from the arms of an angle.
(i) Draw the bisector of LB.
(ii) Draw the bisector of LC. Let these bisectors meet at I.


IL, IM and IN are perpendiculars drawn on the sides BC, AB and AC B respectively.

As I lies on the bisector of ∠ABC ⇒ IL = IM

and I lies on the bisector of ∠ACB ⇒ IL = IN.

Then IL = IM = IN.
Hence, I is equidistant from BC, AC and AB.


3. In a huge park, people are concentrated at three points (see figure):
A: where there are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exit.
Where should an ice cream parlour be set up so that maximum number of persons can approach it?
[Hint: The parlour should be equidistant from A, B and C.]


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Sol. Position should be equidistant from A, B and C.
We know that any point on the perpendicular bisector of the line segment is equidistant from the end points of the line segment. Hence,


(i)  Draw the perpendicular bisector of line segment BC.
(ii) Draw the perpendicular bisector of line segment AC.
Let these perpendicular bisectors meet at P.

Then we have PB = PC = PA, i.e., P is a point equidistant from the vertices of a triangle ABC. Hence parlour should be situated at P.


4. Complete the hexagonal and star shaped Rangolies [see figure (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?


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Sol.   Try yourself .
e.g.,

Side of bigger equilateral triangle is 5 cm and side of each smaller equilateral triangle is 1 cm.


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