Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Tanuj Agarwal 5 years, 2 months ago
- 7 answers
Anshuman Singh 5 years, 2 months ago
Posted by Rakesh Shakya 5 years, 2 months ago
- 2 answers
Golu Golu Raunak Raj 5 years, 2 months ago
Posted by Aashish Shahi 5 years, 2 months ago
- 0 answers
Posted by Piyush Kumar 5 years, 2 months ago
- 1 answers
Sia ? 5 years, 2 months ago
Given: In {tex}\triangle{/tex}ABC, BC = 7 cm, {tex}\angle{/tex}B = 75o and AB + AC = 13 cm.
Required: To construct the triangle ABC.
Steps of construction :
- Draw the base BC = 7 cm.
- At the point B construct an angle YBC = 75o.
- Cut an arc from B as centre and radius equal to AB + AC = 13 cm. on the ray BY. Name it D.
- Join DC.
- Draw the perpendicular bisector of line segment DC which intersects BD at some point name it A.
- Join AC.
ABC is the required triangle.
Posted by Gourav Gupta 5 years, 2 months ago
- 1 answers
Posted by Rajesh Kumar 5 years, 2 months ago
- 3 answers
Posted by Khushi Sahuji 5 years, 2 months ago
- 1 answers
Posted by Sahil Raj 5 years, 2 months ago
- 1 answers
Sia ? 5 years, 2 months ago
SAS congruence criterion: If two sides and included an angle of one triangle are equal to the corresponding two sides and included angle of another triangle then the two triangles are said to be congruent.
Consider {tex}\triangle{/tex}ABC, {tex}\triangle{/tex}PQR
here AB = PQ, AC = PR and {tex}\angle{/tex}A = {tex}\angle{/tex}P
Hence {tex}\triangle{/tex}ABC {tex}\cong{/tex} {tex}\triangle{/tex}PQR by SAS congruence criterion.
Posted by Swati Chaturvedi 5 years, 2 months ago
- 0 answers
Posted by Priya Seth 5 years, 2 months ago
- 2 answers
Posted by Rupali Singh 5 years, 2 months ago
- 1 answers
Sia ? 5 years, 2 months ago
Given: Diagonals of quadrilateral intersect each other at right angles.
To Prove: Quadrilateral is a rhombus.
Proof : In {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}AOD,
AO = AO . . . [Common]
OB = OD . . . [Given]
{tex}\angle{/tex}AOB = {tex}\angle{/tex}AOD . . .[Each 90o]
{tex}\therefore{/tex} {tex}\angle{/tex}AOB {tex}\cong{/tex} {tex}\triangle{/tex}AOD . . . [By SAS property]
{tex}\therefore{/tex} AB = AD . . . [c.p.c.t.] . . . . (1)
Similarly, we can prove that
AB = BC . . . . (2)
BC = CD . . . . (3)
CD = AD . . . . (4)
From (1), (2), (3) and (4)
AB = BC = CD = DA
Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a rhombus.
Posted by Bindass Jockar 5 years, 2 months ago
- 0 answers
Posted by Raunak Jain 5 years, 2 months ago
- 6 answers
Posted by Vinayak Gupta 5 years, 2 months ago
- 1 answers
Sia ? 5 years, 2 months ago
Given: Diagonals of quadrilateral intersect each other at right angles.
To Prove: Quadrilateral is a rhombus.
Proof : In {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}AOD,
AO = AO . . . [Common]
OB = OD . . . [Given]
{tex}\angle{/tex}AOB = {tex}\angle{/tex}AOD . . .[Each 90o]
{tex}\therefore{/tex} {tex}\angle{/tex}AOB {tex}\cong{/tex} {tex}\triangle{/tex}AOD . . . [By SAS property]
{tex}\therefore{/tex} AB = AD . . . [c.p.c.t.] . . . . (1)
Similarly, we can prove that
AB = BC . . . . (2)
BC = CD . . . . (3)
CD = AD . . . . (4)
From (1), (2), (3) and (4)
AB = BC = CD = DA
Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a rhombus.
Posted by Niraj Singh Kushwaha 5 years, 2 months ago
- 1 answers
Dpsangwan Choudhary 5 years, 2 months ago
Posted by Ayushi Maheshwari 5 years, 2 months ago
- 1 answers
Posted by Ranvijay Kumar 5 years, 2 months ago
- 0 answers
Posted by Yuvraj Singh 5 years, 2 months ago
- 4 answers
Posted by Hunting Man Crazy 5 years, 2 months ago
- 0 answers
Posted by Rajeev Rajan 5 years, 2 months ago
- 2 answers
Mukul Kumar 5 years, 2 months ago
Posted by Golve Sree Shashank 5 years, 2 months ago
- 1 answers
Posted by Reshika Navy 5 years, 2 months ago
- 1 answers
Parul Goenka 5 years, 2 months ago
Posted by Govind Ram Saini 5 years, 2 months ago
- 1 answers
Posted by Sneha Singh 5 years, 2 months ago
- 0 answers
Posted by Abhilash Abhilash 5 years, 2 months ago
- 2 answers
Gaurav Seth 5 years, 2 months ago
Tangent to a circle is a line which intersects the circle in exactly one point.
At a point of a circle there is one and only one tangent.
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
The lengths of tangents drawn from an external point to a circle are equal.
Centre of the circle lies on the bisector of the angle between the two tangents.
Posted by Aman Jaiswal 5 years, 2 months ago
- 0 answers
Posted by Vaishnavi Pawar 5 years, 2 months ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Sia ? 5 years, 2 months ago
Egypt
0Thank You