Triangles 12 | CAT Preparation 2024 | Geometry

Triangles 12 | CAT Preparation 2024 | Geometry | Quantitative Aptitude

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5 жыл бұрын

Triangles by Ravi Prakash | Geometry Quantitative Aptitude for CAT 2024
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Пікірлер: 241

@mynotes9802 4 жыл бұрын

Last answer was 10/3✓66

@170_vermaamriteshsunilkuma9 4 жыл бұрын

@sabyasachichakraborty6637 3 жыл бұрын

yes

@devanshupadhyay6314 3 жыл бұрын

okk

@nirmalraj5602 2 жыл бұрын

No isn't it 5root 66

@kartikkhatri739 2 жыл бұрын

@@nirmalraj5602 no, it's 10/3 root 66

@hammadali4350 2 ай бұрын

At 18:19, AB:BC should be 3:5 because BE is parallel to DC therefore BC=DE that is 10.

@namankapoor3341 18 күн бұрын

It is not given that BE = CD So this cann't be applied

@gawadeaditi-2368 14 күн бұрын

@@namankapoor3341 as lines are parallel it should be equal to length of ED. whether given or not. isn't it?

@tanmaychandra9434 2 жыл бұрын

Sir I am following you lectures since 4 months. Covered almost 65 % of your lectures. You are amazing. I suggest you to publish a books for quants and LRDI comprising of all these concepts. I can bet it will be massive hit. Current books in market doesn't even covers 50% of these concepts and tricks.

@iramfatima2714 7 ай бұрын

Hey, did you find your CAT preparation beneficial from this channel?

@noamskyamom6105 2 жыл бұрын

Hands down to the best teacher for CAT!

@anyamehta8561 3 жыл бұрын

I joined iquanta this year for the cat preparation and their concept is no where near to yours, you are just excellent Like seriously my scores are improving day by day You should start your own company and all

@sonty6589 3 жыл бұрын

He is already a legendary person...he make mocks for ims times and all

@shubhamjaiswal1031 Жыл бұрын

@@sonty6589 nothing is more true than what you said

@ShivamVerma-cp4bk Жыл бұрын

Why can't we find the area of 🔼AED by 1/2 base height ??? we have DC as Height and 10 as base

@barshasaha1658 10 ай бұрын

@anyamehta Can you suggest how is iQuanta for CAT preparation I request?

@divyanshjaiswal6236 10 ай бұрын

bro even i also joined iquanta for this year...and what u saying is 100% true

@gautamsachdeva1358 3 жыл бұрын

Rodha>>> Time, CL, Ims, etc

@zoro3188 2 жыл бұрын

Agreed with time

@shivsahetya4323 2 жыл бұрын

I Agree 😃

@shorttrickking5119 2 жыл бұрын

i agree

@Vijay98Yadav 3 жыл бұрын

Such a beautiful teaching approach u have 🔥 We love it sirr🙏 Keep growing and your students too❤️

@sumit.7256 2 жыл бұрын

sir at 24:00 , it will be 10*sqrt(66)/3 .

@Animesthan Жыл бұрын

You are the bestest teacher i have known in my life sir Thankyou 🙏

@yashasharma9963 3 жыл бұрын

Best teacher ever.

@bhavyashandilya1864 10 ай бұрын

These questions and the explainations both are excellent. Thank You so much sir.

@aahistachirps 2 жыл бұрын

Sir, you are the best. 🙌🏼❤️

@fitasticansh1918 5 жыл бұрын

08:27 we can also find the side TN by Pythagorean triplet that is 6,8,10

@sshvibha 3 жыл бұрын

no words sir ji 👌👏

@theatomicstories 2 жыл бұрын

I am so blessed that i only purchased mocks and rest is made free and easy by Ravi Prakash sir in the best way.

@ayushraman1821 5 жыл бұрын

great sir 🙏🙏

@vasudhatapriya6315 Ай бұрын

YOU ARE AMAZING. Kudos !!

@sriramkusumanchi9335 Ай бұрын

great explaniation sir these videos are so benfical to us sir ..

@170_vermaamriteshsunilkuma9 4 жыл бұрын

Sir, your explanation is best . But sir can you pls.. Also paste the link of geometry related questions you explain in this video in your description.

@ganjiganesh7706 3 жыл бұрын

Mind blowing question. At 16.00

@ManuSrivastava_ 2 жыл бұрын

Thank you so much sir.❤❤

@corn-pq6ui 3 ай бұрын

The converse of the midpoint theorem states that ” if a line is drawn through the midpoint of one side of a triangle, and parallel to the other side, it bisects the third side”.

@rishiagrahari4212 10 ай бұрын

In last question If we consider ∆ ABF similar ∆ACD Then, BF comes out to be 7 and hence FE = 5 Then applying 1/2x 5x10 formula in ∆ EFD comes out to be 10

@amarjeetkashyap3375 4 жыл бұрын

Awesome lectures

@abhijithnambiar2115 2 жыл бұрын

Sir, in the last question.. Lets say we extend the line AE to meet line CD at point G, then AB/BC=AF/FD=AE/EG= 7:5. Which means EG has to be 10, but ED is already 10. Kindly correct me if I am wrong.

@samyakjain512 Жыл бұрын

bro this figure is not correct actually. if you draw it you will find EG and ED will be the same line. HOPE IT HELPS...

@ShivamVerma-cp4bk Жыл бұрын

Why can't we find the area of 🔼AED by 1/2 base height ??? we have DC as Height and 10 as base 18:00

@AryanKhannaX Жыл бұрын

had the same doubt, I guess the figure is roughly drawn and no additional information is given so it might be a far fetched inference to make@@ShivamVerma-cp4bk

@ShivamVerma-cp4bk Жыл бұрын

@@AryanKhannaX Yeah 👍

@mayapunj9257 2 жыл бұрын

Thankyou sir:)

@syedahmad4466 4 жыл бұрын

Sir , we can directly say area of ANC will 60 because AN is median and it divide the traingle two equal part😊🙏😊

@Kapilsharmaashow Ай бұрын

In the last ques ration should be 3:5 bcz BC ll ED and ED is 10 therefore BC should be 10 and 16 can't be divided into 7:5 that's why use 3:5 and then the final ans we get is 5√66

@bhuvendrasingh2686 2 жыл бұрын

Thank you sir

@hanshikagupta5658 Жыл бұрын

I can conclude that I HAVE GAINED SO MUCH KNOWLEGE FROM THE BEST TEACHER.

@tejpratapsingh5480 Жыл бұрын

are you taking coaching? if not what is your source for doubt solving?

@hanshikagupta5658 Жыл бұрын

@@tejpratapsingh5480 I am not taking any coaching, Ravi sir covers most of the concepts required to solve maximum sums. In case I have doubts I reread the questions and concepts again and again, plus I refer to Arun Sharma's books and give some time to understand it myself. This has worked for me. However, if this also doesn't lead to any conclusion, the telegram groups that I have joined may help. There are many CAT telegram groups.

@monishmoolya8919 Жыл бұрын

@@hanshikagupta5658can u provide some links of telegram groups or name them it would be help full for me

@ShivamVerma-cp4bk Жыл бұрын

Why can't we find the area of 🔼AED by 1/2 base height ??? we have DC as Height and 10 as base 18:00

@sanjayk1286 3 жыл бұрын

Great sir

@Shivam_General_Fitness Жыл бұрын

Thanku sir

@chandlervikramsingh5985 5 ай бұрын

Sir we can solve these que easily with the help of mass point geometry 😊 but i know u want to make our concepts stronger

@thebeerhub.8969 3 жыл бұрын

Sir charan kaha h aapke.

@tarunchaudhary9215 3 жыл бұрын

Ghutne ke neeche

@mitochondria4537 2 жыл бұрын

Chapal main

@anirmanjoshi309 4 жыл бұрын

Amazing

@sadafshaikh1250 3 ай бұрын

Uhm, in second question. We knew that those lines AN and BQ are medians actually...so we could have assumed answer already in the ratio 2:1 ... ultimately *centroid property*

@venugarg6991 4 жыл бұрын

since triangle is bw two parallel lines hence height should be 16*5/12

@avijitdhar6887 3 жыл бұрын

Why

@kunalmjain 4 жыл бұрын

Sir last question answer is 10/3✓66

@user-of9br5xk6b 11 ай бұрын

In the last question of this video the final answer is 10/3 of root over 66... not 8/3 of root over 66.

@bhagirathsharma7505 3 жыл бұрын

Nice video 👍🏼

@jaisibisivakumar3157 Жыл бұрын

Nice one sir

@sukyuit 7 ай бұрын

Last question mein FBCD ka area nikalein fir rectangle BCDE se subtract karein toh fir answer alag aa rha hai

@rehanjeelani3534 Жыл бұрын

Yaaaar😂 phaadu concept i literally tried another lengthy method to solve the first question then sir did the mid point thing, solving it easily 😂 Awesome sir 👌👌

@restyletime4918 2 ай бұрын

Sir no way i can think like this and solve these many sums in exam , no way

@khusi843 Жыл бұрын

I think we can solve last que directly through pythagoras theorem .by taking sides 10,24,26 of triangle def and area of right angle traingle is 1/2× multiplycationn of 2 side except longest side i .e 1/2 ×24×10 = 120.

@ShivamVerma-cp4bk Жыл бұрын

Why can't we find the area of 🔼AED by 1/2 base height ??? we have DC as Height and 10 as base 18:00

@nisheshojha3539 Жыл бұрын

@@ShivamVerma-cp4bk that is because DC is not the height of Triangle AED. You cant assume that. You must verify first. Check in Triangle AEB, By Pythagoras th., you get EB = 10.xx not equal to CD =12. => ED is not parallel to AC. => 12 cant be the height of triangle AED. Method 2, given that ED = 10 and by ratio BC = 20/3 => ED not equal to BC. Hence EBCD is not a parallelogram.

@saumyaverma7593 3 жыл бұрын

Thank you sir 🙏

@vibhanshurathaur20 Ай бұрын

What's your percentile?👀

@santasak 9 ай бұрын

At 16:00, when the area of ANC is 60, why aren't we connecting N to Q? since the midpoint of AC is Q as given, then connecting N to Q would divide ANC further into two equal triangles ANQ and NQC of area 30 and 30 each. Further in ANQ Since P is the midpoint of AN and drawn towards Q opposite vertex, it shall divide ANQ further into two equal triangles APQ and PNQ of area 15 each. Thus, ans being 15. Can someone help understand?

@revatigokhale3877 Жыл бұрын

amazinggggg

@keshavbubna3942 5 жыл бұрын

Sir congruency is not covered?

@manas905 Жыл бұрын

In the last question, for triangle AED, can we not consider DC to be the height of the triangle AED (since it is perpendicular from ED till AC) and calculate the area as half*base*height or 0.5 * ED * DC or 0.5*10*12 or 60? Am I missing something?

@muneelay2814 Жыл бұрын

AC is not parallel to ED and don't assume that FED =90° they haven't mentioned those points and draw the diagram with ED is slightly upwards or downwards then you can easily understand the mistake.if the angle is 90° then the concept which you are mentioned is correct but it is not mentioned in question.

@user-yk1yi1nb8v Жыл бұрын

6:20 how can we say lines are parallel. As only one corresponding angle is equal

@himalayanthink4421 Жыл бұрын

23:45 what is the purpose of giving DE parallel to CD. i didint get it instead confused

@intothestreets8099 Жыл бұрын

🙏🙏

@vasudhatapriya6315 Ай бұрын

I joined IMS this years, paid 50k and that compares nothing to you! I'm soooooo thankful

@Special1min Ай бұрын

Sir if BE parallel to DC then BC and ED should be of same length because the gap between the parallel line are same everywhere.. sir please reply 🙏🙏🙏🙏

@amoljain3297 10 ай бұрын

Last Question 23:30 answer is 10/3 √66

@melonsucks1563 2 жыл бұрын

at 23:41 it will be 10/3 root(66)

@alokkumarjha6133 4 жыл бұрын

Sir in last question can't we find area of triangle AED by 1/2b×h method?

@shubhammaheshwari1191 3 жыл бұрын

You don't have the hight for the triangle

@ksaikiran1781 4 жыл бұрын

in the last question,the answer should be 25 as by considering the triangle ABF and ACD,we can apply the proportionality theorem to find the ratio of AF/AD = AB/AC=BF/CD=7:12....so BF=7 and FE=5....therefore the area must be 0.5 x 10 x 5=25

@ksaikiran1781 4 жыл бұрын

it is true if i take DE as parallel to BC...my bad...i assumed it in a wrong way....

@ShivamVerma-cp4bk Жыл бұрын

Why can't we find the area of 🔼AED by 1/2 base height ??? we have DC as Height and 10 as base 18:00

@MaxwellNaradd Жыл бұрын

I used the same method as you and got 25. Can someone please help??

@vatsalparikh3574 8 ай бұрын

High level sets

@kvivek3474 2 жыл бұрын

@ RODHA classes sir IN LAST QUESTION BE|| TO CD SO BC MUST BE 10 AND ACCORDINGLY AB WILL BE 14 THEN HOW AC =16 ??

@user-ex9cd3eo3l 3 ай бұрын

Why cant we find the side EF by proving triangle EDF similar to ACD and then find the area by considering EF as height and ED as base,then using A=1/2 *B*H.

@abhishekrajora9224 4 жыл бұрын

sir 15th class of geometry is missing

@408uchilriddhi4 Ай бұрын

Sir which author's book do you recommend for CAT preparation

@KhushiKumari-ll2ru 3 жыл бұрын

Can someone elaborate the last ans how he calculated???

@bullstreamlive Жыл бұрын

Sir, In the last question as ED = 10 and BE || CD, Shouldn't BC be equal to 10 as well? and then by this AB becomes 6 and ratio comes out to be 3:5 and not 7:5. Do reply and also pls correct me if I am wrong '

@arpitsingla2002 Жыл бұрын

yes right

@sarrah3944 10 ай бұрын

BE is parallel to CD , but CD can still be longer BE... unless the question says BE=CD I think we shdnt assume otherwise.

@ShivamVerma-cp4bk Жыл бұрын

Why can't we find the area of 🔼AED by 1/2 base height ??? we have DC as Height and 10 as base 18:00

@zeleoz463 4 ай бұрын

DC is not the altitude for ∆AED . So we cant

@unavailable_0 2 жыл бұрын

In last question it should be BF//DC not BE//DC

@prateekthakur1848 11 ай бұрын

A doubt in the last section which I couldn't find in the comments, AE : ED = 7 : 5 (from the given values). Suppose we extend AE and CD to meet, at a point X, then the ratio AE : EX must be 7:5 as well, which means that if ED is 10, then AED must be a straight line, and instead of meeting at pt D, a quadrilateral should be formed. For the question to be correct, ED has to be a particular value less than 10. Can anyone confirm?

@lipsamarandi6192 10 ай бұрын

AEX will form a straight line . But why AED ? when it’s already given in the question as a diff point…. Ratios are same length might be same that need not mean the lines to be overlapping 🙄

@prateekthakur1848 10 ай бұрын

Well, because the length of AE is given as 14. If there is a line EX, after extending, then it must be of the length 10 units (7:5). Any other line, as we make it not straight, and bring EX towards D, will be shorter than EX (which must be 10 units). The length of AE is fixed as 14. Yes, ratio of two non straight lines can be the same, but since BE and CD are parallel, it is intuitive EX has to be longer than ED. So the ratio also will be less.

@manisharaj5143 11 ай бұрын

If BE||DC and DE=10 Then BC should also be equal to 10 bcz distance between two parallel lines is always same

@yashchhimwal7913 10 ай бұрын

yes many failed to acknowledge that which means ac should be equal 24 not 16

@parkyoung0302 Жыл бұрын

sir in the last question, EBllDC, so wouldn't ED be equal to BC? according to the calculations, BC= 5/12*16, which is obv not 10. Please guide me where i'm wrong.

@ShivamVerma-cp4bk Жыл бұрын

Why can't we find the area of 🔼AED by 1/2 base height ??? we have DC as Height and 10 as base 18:00

@sachinkumarsinha2381 Жыл бұрын

I have a dought in the last question ...... can any one decode this question

@Aman3 Жыл бұрын

if we calculate area of triangle AED by 1/2*base *height= 1/2*10*12= 60 and in the video area by heron's formula is sqrt.66 why this difference?

@nisheshojha3539 Жыл бұрын

that is because DC is not the height of Triangle AED. You cant assume that. You must verify first. Check in Triangle AEB, By Pythagoras th., you get EB = 10.xx not equal to CD =12. => ED is not parallel to AC. => 12 cant be the height of triangle AED. Method 2, given that ED = 10 and by ratio BC = 20/3 => ED not equal to BC. Hence EBCD is not a parallelogram.

@karshpersonal9 3 жыл бұрын

Sir, where are trigonometry lectures ?

@jignashasanchala5305 2 жыл бұрын

Sir if BE||DC & AC is 16. then if ED is 10 BC is also 10.ratio of AB:BC 3:5.

@vanshgoel7458 2 жыл бұрын

I am also facing the same issue

@paritoshjain5451 Жыл бұрын

Question is maybe flawed

@saketvaidya1019 Жыл бұрын

@RODHA. In last question since BE II DC and angle BCD= 90 degree, would'nt BC= 10 ??

@dhroovindesai4047 Жыл бұрын

Same doubt

@_surajbhandari_ Жыл бұрын

But bc is not = EC. It could be a trapezium

@soham1277 Жыл бұрын

The line ED can be at a slope not perpendicular to EB or CD

@mansisharma767 3 жыл бұрын

If in the last question shouldn't BC = ED = 10 as BE is parallel to CD ???

@tarlikapatel7673 3 жыл бұрын

BE and DC parallel, there height is not equal

@anuragkr7892 4 жыл бұрын

If AB:BC is 7:5 Then length of BC = 20/3 But ED=BC => BC = 10 So change the ratio to 3:5 only ...

@vinodluhana9776 4 жыл бұрын

can it be 5;7?

@anuragkr7892 4 жыл бұрын

No... Even then u'll get BC = 28/3 But BC has to be 10 only

@vinodluhana9776 4 жыл бұрын

@@anuragkr7892 yeah thank you so much

@vinodluhana9776 4 жыл бұрын

Do you have any link of whatsapp group which can help to solve the doubt, it will be very helpful.

@anuragkr7892 4 жыл бұрын

@@vinodluhana9776 no ... #aatmanirbhar 😅

@DevendraSingh-hs1gj 4 жыл бұрын

ED =10 is not possible since both line are parallel and perpendicular (BE & DC) then DEF is also 90 but ya good question for practice

@syedahmad4466 4 жыл бұрын

So what's the problem bro

@DevendraSingh-hs1gj 4 жыл бұрын

@@syedahmad4466 it can't be 10

@syedahmad4466 4 жыл бұрын

Kyu Bhai smjh ni aa rha really 😂

@DevendraSingh-hs1gj 4 жыл бұрын

@@syedahmad4466 syed ahmad arre bhai ED = 10 possible nhi h Bcz BE DC are pendicular or if ED=10 then BC=10 also

@anushkatanwar8121 4 жыл бұрын

Its just a figure. Its not compulsory that angle DEF is 90

@alokj2099 2 жыл бұрын

since BE||DC and ED=10 , also BC is perpendicular to CD BC has to be 10 , making AB = 6 which is in the ratio 3:5 and 7:5 is not possible

@Rodha 2 жыл бұрын

think again

@mahimanigam2845 Жыл бұрын

Bro BE||DC so BEDC is a qudrilateral with 2 sides parallel so it can be a part of a trapezium (with it's height as BC) or any random quadrilateral with 2 sides parallel. So if BEDC is a trapezium/quadrilateral then ED is NOT equal to DC.

@alokj2099 Жыл бұрын

@@mahimanigam2845 ed = dc ro bola hi nhi mene dhyaan se dekho eb||dc given h or 90 at c doesnt that make eb perpendicular to ac making bcde a reactangle

@alokj2099 Жыл бұрын

@@Rodha sir can you please check again isnt BCDE a rectangle as BE||CD and DC is perpendicular to AC

@ziyadc8542 3 жыл бұрын

sir , in the last question ,it is given that ab:bc is 7:5 but if BE parallel to DC and angle BCD is 90,then BCDE would be rectangle which means that BC=ED=10 an therefore AB=6.which means AB:BC =3:5????????

@chandansinghpatni 2 жыл бұрын

DE is not parallel to bc right so cant be rectangle

@roshanshahid9262 Жыл бұрын

@@chandansinghpatni just see in 18.04secs Ravi sir said be is parallel to dc

@aadityasinha 3 жыл бұрын

Sir, last answer (10√66)/3 aa raha mera

@prayaspatra2673 3 жыл бұрын

Sir in the last qns how is BE parallel to CD?

@ziyadc8542 3 жыл бұрын

i have the same doubt since it is given that ab:bc is 7:5 but if BE parallel to DC and angle BCD is 90,then BCDE would be rectangle which means that BC=ED=10 an therefore AB=6.which means AB:BC =3:5

@udeepandit751 3 жыл бұрын

@@ziyadc8542 same doubt ... as then ans would be 30

@ziyadc8542 3 жыл бұрын

@@udeepandit751 from the figure it seems like a rectangle,but its actually not.it is not given that ED and BC are parallel

@Hemant_Gupta Жыл бұрын

Sir, In the last question if ed is 10, then bc is also supposed to be 10, as the distance between parallel lines is the same. If bc is 10, then ab should be 14 as ab:bc is 7:5. Therefore, shouldn't ac be 14+ 10 = 24?? why it's given as 16??

@nirajpatil1656 Жыл бұрын

You are that Ed is parallel to bc which is not the case

@aritraghosh7678 3 жыл бұрын

How can be is parallel to CD Bc is (5/12 *16) =20/3, but ed is 10 And still if I say it's parallel then Ab/(ab+bc) =bf/dc=af/ad??????????

@fleabag3458 3 жыл бұрын

I think ED is not PARALLEL to the BC. ED is a little slanted. That's why AB/BC is NOT equal to 6/10

@GAURAVSINGH-ww7ku 3 жыл бұрын

Sir I have a doubt in the triangle ADE how can we asssume that it will divide the triangle AFE and triangle EFD into 7:5. Please explain it.

@alphaallrounder4841 3 жыл бұрын

Using proportionality theorem...since BE is parallel to CD and if we draw a transversal from a single point i.e A then their ratio is same means AB/BC = AF/FD

@hussainpunawala6113 2 жыл бұрын

@@alphaallrounder4841 but according to this only AD should be divide in 7:5 area of ADE is why divided in 7:5

@nirmalraj5602 2 жыл бұрын

He did by mistake..it should be 3 is to 5

@vanditshrivastava3571 5 жыл бұрын

Sir you made a small mistake in the last sum... Ans will be 10/3 ×√66

@sakshiagarwal7578 4 жыл бұрын

Yes it's 10/3 √66

@sachinkumarsingh2518 3 жыл бұрын

Koi nhi bade bade shehero me choti😁😅 choti baat hoti rehti h bro..

@Aman3 Жыл бұрын

if AB:BC=7:5, then AB = 12x, that is length of AB should multiple of 12 then howi is AB 16?

@starkrobo 11 ай бұрын

yeah same thing i am wondering. If we calculate now then AB comes out to be 28/3. Using pythagoras theorem in ABE => BE comes out to be root(52). But thats not possible as BE is also eqaul to 12

@ArceusBlaze 3 жыл бұрын

As it is in the Ques 18:00 EB // CD and CD perpendicular to AC which makes Quad BCDE a Rectangle No if BCDE is a Rec then BC is 10 then AB:BC will be 3:5 not 7:5 So the Ques is wrong

@bewithpallabi3120 3 жыл бұрын

Yes u r right

@prasanthreddy5969 3 жыл бұрын

Even I thought same at first, but if u think ED need not be perpendicular to CD or BE, so there is no guarantee that BCDE is a rectangle

@ArceusBlaze 3 жыл бұрын

@prasanth reddy Ya u r right . ThankYou 😅😅

@agrim8863 2 жыл бұрын

@@prasanthreddy5969 No you are wrong bcoz if BE || CD then EDCB will be a rectangle if n only if it is given that angle C is 90° which is given so by corresponding angles and interior angle on the same side of the transversal using both these concepts ANGLE E,D,B in quad EDCB will turn out to be 90°. RAVI Sir pls correct me if I said anything wrong. Please clarify why EDCB is not a rectangle bcoz it is satisfying all conditions of it.

@farhakhan5113 2 жыл бұрын

same...I tried subtracting the area of bfdc and edbc by assuming and proving them as rectangle and trapezium respectively, but got diff answers

@Lalit_Sharma_20 11 ай бұрын

sir thank you always. Regarding last Question QUESTION : BE // DC so distance between 2 parallel lines will always remain same right ? ED = BC = 10 isn't it ? is there something that i am missing here ? coz from ratio it is coming as 5/12*16 = 20/3 which is not equal to 10. HOW IS THIS POSSIBLE ? Can Anyone else help me out here

@wwenabnas 11 ай бұрын

Not necessarily. For ease of understanding, consider a right angled triangle. Now draw a line such that it is parallel to the base of the triangle. The height and hypotenuse of the triangle falls between these two parallel lines however their lengths differ due to the angle. Hope this clears the confusion.

@Lalit_Sharma_20 11 ай бұрын

The line parallel to the base and the base will have a certain distance between them right ? How can the distance between them vary ? I don't have anything to do with hypo and height of the triangle and also can you tell me in the given question why 10 and 20/3 is not coming same

@wwenabnas 11 ай бұрын

@@Lalit_Sharma_20The distance between them does not vary. The lengths of the lines between them can vary based on the angle the line makes. The steeper the line, the shorter the length. Which was what I tried explaining with the previous comment. Although the hypotunse and the height lies between the same two parallel lines, their lengths are different due to the angle it makes with the parallel lines. In the last question, it says that BC makes an angle of 90 with CD. But it doesn't talk about the angle of ED at all so we cannot assume the lengths of ED and BC to be the same.

@sauranilguha1968 4 жыл бұрын

Sir at last, why triangle AED area cannot be found by 1/2 x base x height formula ? Instead of s (s-a).....

@anjanasingh3404 4 жыл бұрын

height means perpendicular i.e 90 degree which is not given

@venugarg6991 4 жыл бұрын

@@anjanasingh3404 their is given 90 degree as line is parallel.

@hitensharma7155 3 жыл бұрын

@@venugarg6991 perpendicular is not on ED only BE||DC not ED||BC

@AdityaKumar-ez4pr Жыл бұрын

Same doubt cause CD is 90° acting as height for AED and ED 10 is base

@nihalgupta9224 3 жыл бұрын

I solved this question through different method, i.e Ar of shaded region= Total area of figure ABCDE - (ar of tri. ACD + ar. of tri.ABE - ar. of tri ABF) and I got a different answer (90 unit). Why is it so sir?

@rishavkhandelwal9323 3 жыл бұрын

Same bro, i tried similarity thing, and answer came out to be 90

@sojabhai5883 10 ай бұрын

Sir in last question ED should be equal to BC So BC =10 and AB = AC -BC SO AB= 16-10 =6 And now the ratio of AB & BC is not 7:5 is there any problem? Correct me if I m wrong

@prateekbirthal9649 5 ай бұрын

same doubt

@muneelay2814 Жыл бұрын

For last question everyone is imagining that ED//AC but they haven't mentioned that and angle FED also not mentioned, but everyone is assuming 90 degrees . I think this is the reason everyone is getting different answers. If there are any mistakes in my comment please let me know .

@ShivamVerma-cp4bk Жыл бұрын

Why can't we find the area of 🔼AED by 1/2 base height ??? we have DC as Height and 10 as base 18:00

@nisheshojha3539 Жыл бұрын

you are right. i have explained this in my other comments as well. "that is because DC is not the height of Triangle AED. You cant assume that. You must verify first. Check in Triangle AEB, By Pythagoras th., you get EB = 10.xx not equal to CD =12. => ED is not parallel to AC. => 12 cant be the height of triangle AED."

@ShivamVerma-cp4bk Жыл бұрын

@@nisheshojha3539 DC is perpendicular

@nisheshojha3539 Жыл бұрын

@@ShivamVerma-cp4bk perpendicular to AC not to ED.

@hussainpunawala6113 2 жыл бұрын

5:50 in this qsn if we draw straight line from point N to AB we will gt a line NP which is half of BC. So it would be easy

@kanakh7775 4 жыл бұрын

Why is the area of AED not half base(10) height(12) which is 60?

@TheScott689 4 жыл бұрын

Agreed. From vertex A to side ED, the way the figure is drawn ED is an Altitude Perpendicular to Base ED There seems to be other issues with the numbers in the question though. I just use it for the concept of the Proportionality Theorem and the multiple transversals from a single point. Main takeaway in my opinion.

@TheScott689 4 жыл бұрын

Meant CD is the Altitude. Oops

@kanakh7775 4 жыл бұрын

@@TheScott689 agree.. we should not look at the solution for this particular question. It can play with our heads 😅

@arunadityarajjain1129 3 жыл бұрын

The logic you are applying is that ED is parallel to BC, it is not given in the question, they are just appearing parallel as sir drew straight lines

@Diabolical3010 2 жыл бұрын

completely wrong, why assume stuff which is not given in the question? When all three sides are given , simply put the perimeter fomula, EB is not the height of triangle AED, the area is proof of that

@deepanjanghosh3693 4 жыл бұрын

Last ans should be 10/3✓66

@allandsouza9225 4 жыл бұрын

Right

@muhammedhamraznasarv4258 4 жыл бұрын

Why can't we solve like this Ar. of tri.FED = 5/12 of tri. AED = 5/12 *(1/2*10*12) = 25

@bijitht1115 4 жыл бұрын

i also got the same answer

@battusajeev 4 жыл бұрын

@@bijitht1115 bro a doubt, ratio of area of 2 triangles is in the ratio of squares of their sides right? so just because one side is in the ratio 5:12, how can area also be 5:12?

@AmanKumar-ww6ni 4 жыл бұрын

No since it's not given that E And D are at same distance because only DC nd EB are parallel not equal in length, that's not given so the the height is not known, ED is a slant line