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Yogita answered 1 year(s) , 3 month(s) ago

In this solution triangle ABC and ACE is taken in place of ABC, ABE should be taken.

Sep 1

• 13 Likes
• 94234 Views

Sep 1

D,E,F are the mid-points of the sides AB, BC and CA of triangle ABC

Given in ΔABC, D, E and F are midpoints of sides AB, BC and CA respectively.
BC = EC
Recall that the line joining the midpoints of two sides of a triangle is parallel to third side and ha...

Aug 9

• 5 Likes
• 62822 Views

Aug 9

Prove that the sum of the squares of the diagonals of parallelogram

In parallelogram ABCD, AB = CD, BC = AD
Draw perpendiculars from C and D on AB as shown.
In right angled ΔAEC, AC 2 = AE 2 + CE 2 [By Pythagoras theore...

Aug 18

• 1 Like
• 58145 Views

Aug 18

Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle describes on..

Sol:

Here ABCD is a square, AEB is an equilateral triangle described on the side of the square and DBF is an equilateral triangle described on diagonal BD of  ...

Aug 27

• 3 Likes
• 51756 Views

Aug 27

Prove that the sum of squares of of the sides of a rhombus

In rhombus ABCD, AB = BC = CD = DA
We know that diagonals of a rhombus bisect each other perpendicularly.
That is AC ⊥ BD, ∠AOB=∠BOC=∠COD=∠AOD=90° and

Consider right

...

Aug 7

• 1 Like
• 38591 Views

Aug 7

Ramesh answered 4 year(s) , 1 month(s) ago

Converse of Pythagoras Theorem

GIven a ΔABC, with BC = a, AC = b and AB = c.
Also given, c 2 = a 2 + b 2  --- (1)
Now construct a right angled ΔDEF, with sides EF = BC = a, AC = DF = b. <...

Apr 14

• 3 Likes
• 34601 Views

Apr 14

Ramesh answered 4 year(s) , 9 month(s) ago

Find the distance between the tops of two poles.

Sol:
Given:
Two poles AD and CE of height 6 m, 11 m respectively.

Distance between the feet of two poles (DC) = 12 m

AD = BC = 6 m

BE = CE...

Sep 6

• 2 Likes
• 30962 Views

Sep 6

In a triangle ABC right angled at A,If AD perpendicular BC then find AD.

Given that triangle ABC right angled at A.

Base = AB = 5cm, Height = AC = 12 cm

Therefore, Area of the triangle = Base x Height.

A...

Sep 18

• 0 Like
• 28193 Views

Sep 18

Ramesh answered 4 year(s) , 9 month(s) ago

in triangle abc if AD is the median then show that AB²+AC²= 2[AD²+BD²]

Appolonius Therem:

Sep 9

• 4 Likes
• 27540 Views

Sep 9

Agam answered 1 year(s) , 10 month(s) ago

AD is an altitude of an equilateral triangle ABC. On AD as a base

Given ABC and ADE are equilateral triangles.
Let AB=BC=CA = a
Recall that the altitude of an equilateral triangle is √3/2 times its side

Aug 10

• 1 Like
• 25615 Views

Aug 10

Agam answered 1 year(s) , 10 month(s) ago

Two trees of height a and b are p metres apart.

Let the height of the two trees AB and CD be b and a respectively.
Distance between the trees BC = p
Let EF = h be the height of point of intersection of the lines joining the top of each...

Sep 23

• 3 Likes
• 24978 Views

Sep 23

Agam answered 1 year(s) , 10 month(s) ago

Given a right triangle ABC, in which BC is trisected on points D and E

Given right DABC, right angled at B.
D and E are points of trisection of the side BC
Let BD = DE = EC = k
Hence we get BE = 2k and BC = 3k
In ΔABD, by Pythagoras theorem,...

Sep 9

• 0 Like
• 24959 Views

Sep 9

Agam answered 1 year(s) , 10 month(s) ago

The perimeter of two similar triangles ABC and PQR is 36 and 24

Given perimeter of�ΔABC = 36 units
Perimeter of�ΔPQR = 24 units and PQ = 10 units
Since�

Hence AB is 15 units.

Dec 14

• 1 Like
• 22833 Views

Dec 14

Ramesh answered 4 year(s) , 9 month(s) ago

Prove that three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of th..

Sol:

Apollonius theorem states that the sum of the squares of two sides of a triangle is equal to twice the square of the median on the third side plus half the square of the th...

Sep 1

• 0 Like
• 22285 Views

Sep 1

Agam answered 1 year(s) , 10 month(s) ago

In fig AD perpendicular to BC and BD=1/3 CD

In right DADB, AB 2 = AD 2 + BD 2 [By Pythagoras theorem] – (1)
In right DADC, AC 2 = AD 2 + CD 2 [By Pythagoras theorem] – (2)
Subtract (1) from (2), we get
AC 2 – AB 2 = AD 2 + CD 2 –...

Sep 4

• 0 Like
• 21268 Views

Sep 4

In a triangle PQR right angled at Q if QS = SR then prove that PR2 = 4PS2 - 3PQ2

Given in right triangle PQR, QS =  SR
By Pythagoras theorem, we have
PR 2 = PQ 2 + QR 2 → (1)
In right triangle PQS, we have...

Jul 22

• 3 Likes
• 21063 Views

Jul 22

Agam answered 1 year(s) , 10 month(s) ago

In a right triangle PQR right angled at Q the points S and T trisect the side QR

Let QR = 3a
Given QR is trisected at S and T, hence QS = ST = TR = a
In right DPQR, PR 2 �= PQ 2 �+ QR 2
⇒ PR 2 �= PQ 2 �+ (3a) 2
⇒�PR 2 �= PQ 2 �+ 9a 2 �� à (1)...

Jul 24

• 0 Like
• 20207 Views

Jul 24

ABCD is a square. F is the mid-point of AB. BE is one third of BC.

In square ABCD, let AB = BC = CD = DA = x
Since F is midpoint of AB, BF = x/2
Given that BE = (1/3) BC =(x/3)
Area of ΔFBE = 108 sq cm

Hence AB = BC =

...

Aug 12

• 0 Like
• 19684 Views

Aug 12

Ramesh answered 4 year(s) , 9 month(s) ago

P and Q are points on sides AB and AC respectively, of ΔABC. If AP = 3 cm,PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ.

Sol:

AP/PB = AQ/QC
3/6 = 5/10
1/2 = 1/2

Therefore, PQ || BC [Converse of basic proportionality theorem.]

ΔAPQ ~ ΔABC
...

Sep 1

• 1 Like
• 18082 Views

Sep 1

Raghunath answered 4 year(s) , 9 month(s) ago

Prove that AC2=AB2+ BC2+2 BC x BD.

Sol:

Given that in a ΔABC in which ∠B is obtuse and AD ⊥ BC.

To prove:  AC = AB 2 + BC + 2 BC X BD

<...

Sep 1

• 3 Likes
• 17775 Views

Sep 1

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