construct a right triangle whose hypotenuse measures 5.3 cm & the length of one. Of whose sides contaning the right angle measures 4.5
help!
my ans is looking like in the pic.
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Draw the base PQ as 4.5cm. Then draw a ray PM perpendicular to base PQ of a suitable length. Now take a length of 5.3cm in a rounder and put its point on 'Q' and draw an arc intersecting the ray PM. Name the point as R. PQR is the required triangle
kvnmurty:
right
Answered by
22
Your answer shown in the picture is wrong. A small mistake about point R.
The point R should be on the semicircular arc. Line PQ and semicircle drawn with PQ as the diameter are correct. It is the circumcircle of the triangle PQR. (we know angle in a semicircle ie., ∠PRQ = 90°).
The distance OR = OP = OQ = circumradius.
So you have to do:
1. Draw PQ = 5.3 cm Hypotenuse.
2. Bisect it by the standard procedure. O is the midpoint.
3. Draw a semicircular arc above PQ with OP = OQ as the radius.
4. From P draw a circular arc with radius = 4.5 cm and intersecting the previous semicircular arc.
5. The point of intersection of two arcs is R.
6. Join PR and QR.
There are other methods also to construct it.
The point R should be on the semicircular arc. Line PQ and semicircle drawn with PQ as the diameter are correct. It is the circumcircle of the triangle PQR. (we know angle in a semicircle ie., ∠PRQ = 90°).
The distance OR = OP = OQ = circumradius.
So you have to do:
1. Draw PQ = 5.3 cm Hypotenuse.
2. Bisect it by the standard procedure. O is the midpoint.
3. Draw a semicircular arc above PQ with OP = OQ as the radius.
4. From P draw a circular arc with radius = 4.5 cm and intersecting the previous semicircular arc.
5. The point of intersection of two arcs is R.
6. Join PR and QR.
There are other methods also to construct it.
:-)
;-)
thank a lot sir!
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