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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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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.

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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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

Q 11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that $\angle C A D=\angle C B D$..

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

To prove : $\angle C A D=\angle C B D$

Proof :

Triangle ABC and ADC are on a common base BC and $\angle \mathrm{BAC}=\angle \mathrm{BDC}$.

Thus, points A, B, C, and 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 the line containing the line segment, four points lie on the circle.)

$\angle C A D=\angle C B D$ (Angles in the same segment are equal)

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seema garhwal

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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. \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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mansi

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 a rhombus.

Thus, $\angle A O B=90^{\circ} \quad$ (diagonals of a rhombus bisect each other at $90^{\circ}$ )
So, a circle drawn AB as diameter will pass through point O .

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

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seema garhwal

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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.
$\angle A+\angle C=180^{\circ}$------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,
$\angle A+\angle A=180^{\circ}$
$\Rightarrow 2 \angle A=180^{\circ}$
$\Rightarrow \angle A=90^{\circ}$
We know that a parallelogram with one angle right angle is a rectangle.
Hence, $A B C D$ is a rectangle.

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Posted by

mansi

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

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

Construction: Join AD.

Proof: $A B$ is the diameter of the circle and $\angle A D B$ is formed in a semi-circle.
$\angle A D B=90^{\circ}$-------1 (angle in a semi-circle)
Similarly,
$A C$ is the diameter of the circle and $\angle A D C$ is formed in a semi-circle.
$\angle A D C=90^{\circ}$------2(angle in a semi-circle)
From 1 and 2, we have
$\angle \mathrm{ADB}+\angle \mathrm{ADC}=90^{\circ}+90^{\circ}=180^{\circ}$

$\angle A D B$ and $\angle A D C$ are forming a linear pair.

Hence, point D lies on this side.

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mansi

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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 the 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
$\angle 1+\angle 2=180^{\circ}$
$\Rightarrow \angle A+\angle C=180^{\circ}$
Thus, $ABED$ is a cyclic quadrilateral.

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Posted by

seema garhwal

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, $\angle A D C=90^{\circ}$ and $\angle A B C=90^{\circ}$--------1(Angle in a semi-circle is right angle)
Similarly, BD is the diameter of the circle.
Thus, $\angle B A D=90^{\circ}$ and $\angle B C D=90^{\circ}$--------2(Angle in a semi-circle is right angle)
From 1 and 2, we get
$\angle B C D=\angle A D C=\angle A B C=\angle B A D=90^{\circ}$
Hence, $A B C D$ is a rectangle.

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Posted by

mansi

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