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Manvi Garg
Subject: Maths
, asked on 2/1/18
Plz answer 30th question
Answer
1
Harsh Gujrani
Subject: Maths
, asked on 2/1/18
Pls solve Q16
Answer
3
Jovi
Subject: Maths
, asked on 29/12/17
If in a triangle ABC,AD is the median then what can you say about the two triangles obtained?
Answer
3
Joselin John
Subject: Maths
, asked on 27/12/17
Triangle ABC is isosceles triangles in which AB=AC,Pand Q are points on AB and AC such that AP=AQ. Prove that
Answer
1
Joselin John
Subject: Maths
, asked on 26/12/17
Triangle ABC is isosceles triangles in which AB=AC,Pand Q are points on AB and AC such that AP=AQ. Prove that
Answer
1
Balaji
Subject: Maths
, asked on 19/12/17
Q). Arrange the sides of ∆ ABC in ascending order of lengths.
Answer
2
Admya Maini
Subject: Maths
, asked on 16/12/17
Solve this:
$Inthefigure,AB=BC=CD=DE=EF=FG=GA.Thenthemeasureof\angle DAEis:\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(A\right){\left(25\frac{5}{7}\right)}^{\xb0}\left(B\right){\left(51\frac{3}{7}\right)}^{\xb0}\phantom{\rule{0ex}{0ex}}\left(C\right){\left(25\frac{3}{7}\right)}^{\xb0}\left(D\right){\left(57\frac{5}{7}\right)}^{\xb0}$
Answer
1
Admya Maini
Subject: Maths
, asked on 16/12/17
Solve this:
Q.
ABC is a right triangle such that AB = AC and bisector of angle B intersects the side AC at D.
Then,
(A) AB + AD = CD (B) AB + AD = BD + CD
(C) AB + AD = BC (D) AB + AD = BD
Answer
2
Bitan
Subject: Maths
, asked on 9/12/17
Pls help this one
Q12. In the given figure, if PS is the median bisecting
$\angle $
P and PQ = PR then find the value of
$\angle $
QPS .
Answer
2
Serena Singh
Subject: Maths
, asked on 5/12/17
triangle ABC is a right angled triangle such that Ab=BC. Bisector of angle C intersects AB at D.Prove that AC+AD=BC.
Answer
1
Serena Singh
Subject: Maths
, asked on 5/12/17
Briefly solve and explain Q19
$\mathbf{19}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathit{I}\mathit{n}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}\mathit{g}\mathit{i}\mathit{v}\mathit{e}\mathit{n}\mathbf{}\mathit{f}\mathit{i}\mathit{g}\mathit{u}\mathit{r}\mathit{e}\mathbf{,}\mathbf{}\mathit{A}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathit{A}\mathit{C}\mathbf{.}\mathbf{}\mathit{D}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathit{a}\mathbf{}\mathit{p}\mathit{o}\mathit{i}\mathit{n}\mathit{t}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{A}\mathit{C}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{E}\mathbf{}\mathit{o}\mathit{n}\mathbf{}\mathit{A}\mathit{B}\mathbf{}\mathit{s}\mathit{u}\mathit{c}\mathit{h}\mathbf{}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathbf{}\mathit{A}\mathit{D}\mathbf{}\mathbf{=}\mathbf{}\mathit{E}\mathit{D}\mathbf{}\mathbf{=}\mathbf{}\mathit{E}\mathit{C}\mathbf{}\mathbf{=}\mathbf{}\mathit{B}\mathit{C}\mathbf{.}\mathbf{}\mathit{P}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathbf{}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathbf{}\mathbf{\angle}\mathit{A}\mathbf{}\mathbf{:}\mathbf{}\mathbf{\angle}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{:}\mathbf{}\mathbf{3}\mathbf{.}$
Answer
1
Rishi Verma
Subject: Maths
, asked on 4/12/17
Sir, I want to ask you that why we have to prove things that has already been proven in maths and why are there some questions that a student can't imagine how to prove that???
Please give me some advice that how can i overcome those hard or confusing proving questions!!!
Answer
1
Srishti Sinha
Subject: Maths
, asked on 29/11/17
solve this : In the following figure the side QR of
$\u2206$
PQR is produced to a point S. If the bisectors of
$\angle $
PQR and
$\angle $
PRS meet at a point T, Then prove that
$\angle $
QTR =
$\angle $
QPR.
Answer
2
Keerthana Dileep
Subject: Maths
, asked on 20/11/17
Solve fast
$InFig.9.42,sideBCof\u25b3ABCisproducedtopointDsuchthatbisectsat\angle ABCand\angle ACDmeetatpointE.If\angle BAC=68\xb0,find\angle BEC.$
Answer
1
Syed Sohail
Subject: Maths
, asked on 18/11/17
sir can u explain question no 2 in exercise 1
Answer
1
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$Inthefigure,AB=BC=CD=DE=EF=FG=GA.Thenthemeasureof\angle DAEis:\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(A\right){\left(25\frac{5}{7}\right)}^{\xb0}\left(B\right){\left(51\frac{3}{7}\right)}^{\xb0}\phantom{\rule{0ex}{0ex}}\left(C\right){\left(25\frac{3}{7}\right)}^{\xb0}\left(D\right){\left(57\frac{5}{7}\right)}^{\xb0}$

Q. ABC is a right triangle such that AB = AC and bisector of angle B intersects the side AC at D.

Then,

(A) AB + AD = CD (B) AB + AD = BD + CD

(C) AB + AD = BC (D) AB + AD = BD

Q12. In the given figure, if PS is the median bisecting $\angle $P and PQ = PR then find the value of $\angle $QPS .

$\mathbf{19}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathit{I}\mathit{n}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}\mathit{g}\mathit{i}\mathit{v}\mathit{e}\mathit{n}\mathbf{}\mathit{f}\mathit{i}\mathit{g}\mathit{u}\mathit{r}\mathit{e}\mathbf{,}\mathbf{}\mathit{A}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathit{A}\mathit{C}\mathbf{.}\mathbf{}\mathit{D}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathit{a}\mathbf{}\mathit{p}\mathit{o}\mathit{i}\mathit{n}\mathit{t}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{A}\mathit{C}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{E}\mathbf{}\mathit{o}\mathit{n}\mathbf{}\mathit{A}\mathit{B}\mathbf{}\mathit{s}\mathit{u}\mathit{c}\mathit{h}\mathbf{}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathbf{}\mathit{A}\mathit{D}\mathbf{}\mathbf{=}\mathbf{}\mathit{E}\mathit{D}\mathbf{}\mathbf{=}\mathbf{}\mathit{E}\mathit{C}\mathbf{}\mathbf{=}\mathbf{}\mathit{B}\mathit{C}\mathbf{.}\mathbf{}\mathit{P}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathbf{}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathbf{}\mathbf{\angle}\mathit{A}\mathbf{}\mathbf{:}\mathbf{}\mathbf{\angle}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{:}\mathbf{}\mathbf{3}\mathbf{.}$

Please give me some advice that how can i overcome those hard or confusing proving questions!!!

$InFig.9.42,sideBCof\u25b3ABCisproducedtopointDsuchthatbisectsat\angle ABCand\angle ACDmeetatpointE.If\angle BAC=68\xb0,find\angle BEC.$