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Q 3.      Construct an isosceles right-angled triangle ABC, where  and AC = 6 cm.

1. Draw CA = 6 cm

2. Draw a perpendicular CX to CA at C

3. From C and with length = 6 cm cut an arc on CX. The arc meet CX at B

4. Join BA

Triangle ABC is the required triangle

Q 2.     Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.

a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long:

1. Draw QR = 4 cm

2. Draw a perpendicular QX to QR at Q

3. From R and with length = 6 cm cut an arc on QX. The arc meet QX at P

4. Join PR

Triangle PQR is the required triangle

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Q 1.     Construct the right angled , where , QR = 8cm and PR = 10 cm.

the right-angled , where , QR = 8cm and PR = 10 cm:

1. Draw QR = 8 cm

2. Draw a perpendicular QX to QR at Q

3. From R and with length = 10 cm cut an arc on QX. The arc meet QX at P

4. Join PR

PQR is the required triangle

Q: 11         ABC and ADC are two right triangles with common hypotenuse AC. Prove that                    .

Given: ABC and ADC are two right triangles with common hypotenuse AC.

To prove :

Proof :

Triangle ABC and ADC are on common base BC and BAC = BDC.

Thus, point A,B,C,D lie in the same circle.

(If a line segment joining two points subtends equal angles at two other points lying on the same side of line containing line segment, four points lie on the circle.)

CAD = CBD  (Angles in same segment are equal)

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Q : 9     Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. ). Prove that .

................1(vertically opposite angles)

..................2(Angles in the same segment are equal)

.................3(angles in the same segment are equal)

From 1,2,3 ,we get

Q: 5     Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Given : ABCD is rhombus.

To prove : the circle drawn with AB  as diameter, passes through the  point O.

Proof :

ABCD is rhombus.

Thus,              (diagonals of a rhombus bisect each other at )

So, a circle drawn AB as diameter will pass through point O.

Thus,  the circle is drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

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Q: 12     Prove that a cyclic parallelogram is a rectangle.

Given: ABCD is a cyclic quadrilateral.

To prove: ABCD is a rectangle.

Proof :

In cyclic quadrilateral ABCD.

.......................1(sum of either pair of opposite angles of a cyclic quadrilateral)

........................................2(opposite angles of a parallelogram are equal )

From 1 and 2,

We know that a parallelogram with one angle right angle is a rectangle.

Hence, ABCD is a rectangle.

Q: 10    If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Given: circles are drawn taking two sides of a triangle as diameters.

Proof: AB is the diameter of the circle and ADB is formed in a semi-circle.

ADB = ........................1(angle in a semi-circle)

Similarly,

AC is the diameter of the circle and ADC is formed in a semi circle.

ADC = ........................2(angle in a semi-circle)

From 1 and 2, we have

ADB and ADC are forming a linear pair. So, BDC is a straight line.

Hence, point D lies on this side.

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Q: 8    If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Given: ABCD is a trapezium.

Construction: Draw AD || BE.

Proof: In quadrilateral ABED,

AB || DE    (Given )

AD || BE    ( By construction )

Thus, ABED is a parallelogram.

AD = BE     (Opposite sides of parallelogram )

AD = BC      (Given )

so,  BE = BC

In EBC,

BE = BC    (Proved above )

Thus, ...........1(angles opposite to equal sides )

...............2(Opposite angles of the parallelogram )

From 1 and 2, we get

(linear pair)

Thus, ABED is a cyclic quadrilateral.

Q: 7     If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

AC is the diameter of the circle.

Thus,   and    ............................1(Angle in a semi-circle is right angle)

Similarly, BD is the diameter of the circle.

Thus,   and    ............................2(Angle in a semi-circle is right angle)

From 1 and 2, we get

Hence, ABCD is a rectangle.