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#10.5

Q 3. Construct an isosceles right-angled triangle ABC, where $m\angle ACB=90^{\circ}$ 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

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Pankaj Sanodiya

#10.5

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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Posted by

Pankaj Sanodiya

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#10.5

Q 1. Construct the right angled $\Delta PQR$ , where $\angle Q=90^{\circ}$ , QR = 8cm and PR = 10 cm.

the right-angled $\Delta PQR$ , where $\angle Q=90^{\circ}$ , 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

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Posted by

Pankaj Sanodiya

#10.5

Q: 11 ABC and ADC are two right triangles with common hypotenuse AC. Prove that
$\small \angle CAD =\angle CBD$ .

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

To prove : $\small \angle CAD =\angle CBD$

Proof :

Triangle ABC and ADC are on common base BC and $\angle$ BAC = $\angle$ 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.)

$\angle$ CAD = $\angle$ CBD (Angles in same segment are equal)

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seema garhwal

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#10.5

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. $\small 10.40$ ). Prove that $\small \angle ACP=\angle QCD$ .

$\angle ABP=\angle QBD$ ................1(vertically opposite angles)

$\angle ACP=\angle ABP$ ..................2(Angles in the same segment are equal)

$\angle QBD=\angle QCD$ .................3(angles in the same segment are equal)

From 1,2,3 ,we get

$\angle ACP=\angle QCD$

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Posted by

mansi

#10.5

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, $\angle AOC = 90 \degree$ (diagonals of a rhombus bisect each other at $90 \degree$ )

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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Posted by

seema garhwal

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#10.5

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.

$\angle A + \angle C = 180 \degree$ .......................1(sum of either pair of opposite angles of a cyclic quadrilateral)

$\angle A = \angle C$ ........................................2(opposite angles of a parallelogram are equal )

From 1 and 2,

$\angle A + \angle A = 180 \degree$

$\Rightarrow 2\angle A = 180 \degree$

$\Rightarrow \angle A = 90 \degree$

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

Hence, ABCD is a rectangle.

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Posted by

mansi

#10.5

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.

Construction: Join AD.

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

$\angle$ ADB = $90 \degree$ ........................1(angle in a semi-circle)

Similarly,

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

$\angle$ ADC = $90 \degree$ ........................2(angle in a semi-circle)

From 1 and 2, we have

$\angle$ ADB+ $\angle$ ADC= $90 \degree$ + $90 \degree$ = $180 \degree$

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

Hence, point D lies on this side.

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mansi

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#10.5

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 $\triangle$ EBC,

BE = BC (Proved above )

Thus, $\angle C = \angle 2$ ...........1(angles opposite to equal sides )

$\angle A= \angle 1$ ...............2(Opposite angles of the parallelogram )

From 1 and 2, we get

$\angle 1+\angle 2=180 \degree$ (linear pair)

$\Rightarrow \angle A+\angle C=180 \degree$

Thus, ABED is a cyclic quadrilateral.

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Posted by

seema garhwal

#10.5

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, $\angle ADC=90 \degree$ and $\angle ABC=90 \degree$ ............................1(Angle in a semi-circle is right angle)

Similarly, BD is the diameter of the circle.

Thus, $\angle BAD=90 \degree$ and $\angle BCD=90 \degree$ ............................2(Angle in a semi-circle is right angle)

From 1 and 2, we get

$\angle BCD=\angle ADC=\angle ABC=\angle BAD =90 \degree$

Hence, ABCD is a rectangle.

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Posted by

mansi

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Get Answers to all your Questions

All Questions

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

Posted by

Pankaj Sanodiya

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

Posted by

Pankaj Sanodiya

Crack CUET with india's "Best Teachers"

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

Posted by

Pankaj Sanodiya

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

Posted by

seema garhwal

JEE Main high-scoring chapters and topics

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 .

Q 3. Construct an isosceles right-angled triangle ABC, where $m\angle ACB=90^{\circ}$ and AC = 6 cm.

Q 1. Construct the right angled $\Delta PQR$ , where $\angle Q=90^{\circ}$ , QR = 8cm and PR = 10 cm.

Q: 11 ABC and ADC are two right triangles with common hypotenuse AC. Prove that
$\small \angle CAD =\angle CBD$ .

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. $\small 10.40$ ). Prove that $\small \angle ACP=\angle QCD$ .