Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Praduman Bhatia 6 years ago
- 1 answers
Posted by Rashika Jangra 6 years ago
- 2 answers
Lucifer ? Morningstar? 6 years ago
Posted by Sushil Thua 6 years ago
- 1 answers
Unnati Agarwal 6 years ago
Posted by Alan Xavier 6 years ago
- 4 answers
Posted by Sushma Sushma.G 6 years ago
- 0 answers
Posted by Parveen Tomar 6 years ago
- 0 answers
Posted by Parveen Boora 6 years ago
- 1 answers
Yogita Ingle 6 years ago
Let the two similar triangles ABC and DEF in which AP and DQ are the medians respectively.
we have to prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
If ΔABC~ΔDEF
{tex}⇒ \frac{ar(BCA)}{ar(EFD)}=\frac{AB^2}{DE^2} {/tex} → (1)
As ΔABC~ΔDEF
{tex}\frac{AB}{DE}=\frac{BC}{EF}=\frac{2BP}{2EQ}{/tex}
Hence,{tex} \frac{AB}{DE}=\frac{BP}{EQ}{/tex}
In ΔABP and ΔDEQ
{tex}\frac{AB}{DE}=\frac{BP}{EQ}{/tex}
∠B=∠E (∵ΔABC~ΔDEF)
By SAS rule, ΔABP~ΔDEQ
{tex}⇒ \frac{AB}{DE}=\frac{AP}{DQ}{/tex}
Squaring, we get
{tex}\frac{AB^2}{DE^2}=\frac{AP^2}{DQ^2} {/tex} → (2)
Comparing (1) and (2), we get
{tex}\frac{ar(BCA)}{ar(EFD)}=\frac{AP^2}{DQ^2} {/tex}
Hence, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Posted by Rajendra Rajendra P G 6 years ago
- 2 answers
Posted by Sehjog Pannu 6 years ago
- 0 answers
Posted by Ayushi Tiwari 6 years ago
- 1 answers
Posted by Mercy Johny 6 years ago
- 2 answers
Aadya Singh ? 6 years ago
Smallest composite number = 4.
So, HCF = 2.
Posted by Vikas Jha 6 years ago
- 1 answers
Yogita Ingle 6 years ago
Let the speed of the stream be x km/hr
Then,
Speed downstream = (15 + x) km/hr
Speed upstream = (15 - x) km/hr
∴ 30/(15+x) + 30/ (15−x) = 4 1/2
⇒ 900/(225−x2) =9/ 2
⇒ 9x2 = 225
⇒x2 = 25
⇒x = 5km/hr
Posted by Punit Darak 6 years ago
- 0 answers
Posted by Satendra Sahu 6 years ago
- 0 answers
Posted by Sanjeev Singh 6 years ago
- 2 answers
Sneha Tyagi 6 years ago
Yogita Ingle 6 years ago
let root 5 be rational
then it must in the form of p/q [q is not equal to 0][p and q are co-prime]
root 5=p/q
=> root 5 × q = p
squaring on both sides
=> 5 ×q ×q = p ×p ------> 1
p ×p is divisible by 5
p is divisible by 5
p = 5c [c is a positive integer] [squaring on both sides ]
p ×p = 25c ×c --------- > 2
sub p ×p in 1
5 ×q ×q = 25 ×c ×c
q ×q = 5 ×c ×c
=> q is divisble by 5
thus q and p have a common factor 5
there is a contradiction
as our assumsion p &q are co prime but it has a common factor
so √5 is an irrational
Posted by Sam Raheja 6 years ago
- 5 answers
Posted by Sanjay Gupta 6 years ago
- 1 answers
Posted by Vaibhav Kumar Rai 6 years ago
- 0 answers
Posted by Vikas Saini 6 years ago
- 2 answers
Yogita Ingle 6 years ago
Integral coordinates are coordinates that are whole numbers. Integral coordinates cannot be fractional or have decimals.
Posted by Ankesh .. ??Now I Am Fully Ready? 6 years ago
- 19 answers
Ankesh .. ??Now I Am Fully Ready? 6 years ago
Kashvi ? 6 years ago
Kashvi ? 6 years ago
Ankesh .. ??Now I Am Fully Ready? 6 years ago
Kashvi ? 6 years ago
Posted by Samaya Rastogi 6 years ago
- 3 answers
Posted by Rachna Vashist 6 years ago
- 1 answers
Posted by Abhay Chauhan 6 years ago
- 2 answers
Posted by Samiksha Dhamdhere 6 years ago
- 7 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
myCBSEguide
Krishna Sharma 5 years, 11 months ago
0Thank You