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

Prove that 9AD2 = 7AB2

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

- 3 Answers
- 13 Likes
- 93379 Views

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

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...

- 2 Answers
- 5 Likes
- 61312 Views

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

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...

Asked by Vaishali mathur

Aug 18

- 2 Answers
- 1 Like
- 56993 Views

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

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..

Answer.

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 ...

Asked by Tonykuttan

Aug 27

- 2 Answers
- 3 Likes
- 51074 Views

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

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

Answer.

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

- 2 Answers
- 1 Like
- 37669 Views

Ramesh
answered
3 year(s)
, 11 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. <...

- 1 Answer
- 3 Likes
- 32818 Views

Ramesh
answered
4 year(s)
, 7 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...

- 1 Answer
- 2 Likes
- 29742 Views

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

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

Answer.

Given that triangle ABC right angled at A.

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

Therefore, Area of the triangle = Base x Height.

A...

- 2 Answers
- 0 Like
- 27472 Views

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

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

Appolonius Therem:

Asked by Fakkufakir Mohammed

Sep 9

- 1 Answer
- 4 Likes
- 27279 Views

Agam
answered
1 year(s)
, 8 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

Hence AD = √3a/2

ΔABC �~ ΔADE [By...

- 2 Answers
- 1 Like
- 25041 Views

Agam
answered
1 year(s)
, 8 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,...

- 2 Answers
- 0 Like
- 24501 Views

Agam
answered
1 year(s)
, 8 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...

- 2 Answers
- 3 Likes
- 24291 Views

Agam
answered
1 year(s)
, 8 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.

- 2 Answers
- 1 Like
- 22039 Views

Ramesh
answered
4 year(s)
, 7 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...

- 1 Answer
- 0 Like
- 21831 Views

Agam
answered
1 year(s)
, 8 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 –...

- 2 Answers
- 0 Like
- 20830 Views

Syeda
answered
1 year(s)
, 9 month(s) ago

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

Answer.

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...

- 2 Answers
- 2 Likes
- 20672 Views

Agam
answered
1 year(s)
, 8 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)...

- 2 Answers
- 0 Like
- 19940 Views

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

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

Answer.

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 =

- 4 Answers
- 0 Like
- 18102 Views

Ramesh
answered
4 year(s)
, 7 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

...

- 1 Answer
- 1 Like
- 17520 Views

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

ABC is an isosceles triangle, right angled at B . similar triangles ACD & ABE

Given ΔABC is an isosceles triangle in which ∠B = 90°

⇒ AB = BC

By Pythagoras theorem, we have AC 2 = AB 2 + BC 2

⇒AC 2 = AB 2 + AB 2 [Since AB = BC]

∴ AC 2 = 2AB 2 → (1)...

- 2 Answers
- 2 Likes
- 17191 Views

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