In given figure, ∠ A C B = 90 ° and C D ⊥ A B , prove that C D 2 = B D × A D

Given, in ∆ A B C , ∠ A C B = 90 ° and C D ⊥ A B . To prove: C D 2 = B D × A D Proof: In ∆ C A D , C A 2 = C D 2 + A D 2 ------------(i) Similarly, In ∆ C D B , C B 2 = C D 2 + B D 2 -------(ii) Adding (i) and (ii), we get, C A 2 + C B 2 = C D 2 + A D 2 + C D 2 + B D 2 ⇒ A B 2 = 2 C D 2 + A D 2 + B D 2 ⇒ A B 2 - A D 2 = B D 2 + 2 C D 2 ⇒ A B + A D A B - A D - B D 2 = 2 C D 2 ⇒ A B + A D B D - B D 2 = 2 C D 2 ⇒ B D A B + A D - B D = 2 C D 2 ⇒ B D A B - B D + A D = 2 C D 2 ⇒ B D × 2 A D = 2 C D 2 ⇒ C D 2 = B D × A D Hence, proved.

In an equilateral △ A B C , D is a point on side B C , such that B D = 1 3 B C . Prove that 9 A D 2 = 7 A B 2 .

Given: An equilateral △ A B C , D is a point on B C such that B D = 1 3 B C To Prove: 9 A D 2 = 7 A B 2 Construction: Draw A P ⊥ B C . Proof: We have B D = 1 3 B C = 1 3 A B Since △ A B C is an equilateral and A P ⊥ B C B P = P C = 1 2 B C = 1 2 A B Using Pythagoras theorem in △ A B P , A P 2 + B P 2 = A B 2 A P 2 = A B 2 - B P 2 = A B 2 - A B 2 2 = 3 4 A B 2 --[1] Using Pythagoras theorem in △ A D P , A D 2 = A P 2 + D P 2 A D 2 = 3 4 A B 2 + B P - B D 2 A D 2 = 3 4 A B 2 + 1 2 A B - 1 3 A B 2 A D 2 = 3 4 A B 2 + 1 36 A B 2 A D 2 = 27 A B 2 + A B 2 36 A D 2 = 28 36 A B 2 A D 2 = 7 9 A B 2 9 A D 2 = 7 A B 2 Hence Proved.

HARD

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$P Q R$ is an equilateral triangle with each side of length $2 p .$ If $P S ⊥ Q R,$ and $P S = \sqrt{k} p$ , then find the value of $k$ .

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Important Questions on Pythagoras Theorem

HARD

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

A village has a circular wall around it, and the wall has four gates pointing north, south, east and west. A tree stands outside the village,

16 m

north of the north gate, and it can be just seen appearing on the horizon from a point

48 m

east of the south gate. What is the diameter, in meters, of the wall that surrounds the village?

MEDIUM

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

Triangle

P Q R

is a right-angled at

Q

. If

P Q = 6 cm

P R = 10 cm

, then

Q R

is equal to:

HARD

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

Triangles

A B C

and

D E F

are such that

∠ A = ∠ D, A B = D E = 17, B C = E F = 10

and

A C - D F = 12

. What is

A C + D F

HARD

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

A semicircular paper is folded along a chord such that the folded circular arc is tangent to the diameter of the semicircle. The radius of the semicircle is $4$ units and the point of tangency divides the diameter in the ratio $7 : 1 .$ If the length of the crease (the dotted line segment in the figure) is $l$ then determine $l^{2} .$

Question Image

MEDIUM

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

Six consecutive sides of an equiangular octagon are

6, 9, 8, 7, 10, 5

in that order. The integer nearest to the sum of the remaining two sides is

MEDIUM

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

A friction-less board has the shape of an equilateral triangle of side length

1

meter with bouncing walls along the sides. A tiny super bouncy ball is fired from vertex

A

towards the side

B C .

The ball bounces off the walls of the board nine times before it hits a vertex for the first time. The bounces are such that the angle of incidence equals the angle of reflection. The distance travelled by the ball in meters is of the form

\sqrt{N},

where

N

is an integer. What is the value of

N ?

EASY

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

A ladder leaning against a wall makes an angle

θ

with the horizontal ground such that

\cos (θ) = \frac{5}{13}

. If the height of the top of the ladder from the wall is

18 m

, then what is the distance (in

m

) of the foot of the ladder from the wall?

MEDIUM

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

In given figure, Question Image

$∠ A C B = 90 °$ and $C D ⊥ A B$ , prove that $C D^{2} = B D \times A D$

MEDIUM

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

From an external point

P

, a tangent

PQ

is drawn to a circle, with the centre

O

,touching the circle at

Q

. If the distance of

P

from the centre is

13 cm

and length of tangent

PQ

12 cm

, then the radius of the circle is

HARD

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

∆ A B C, ∠ B = 90 °

. If points

D

and

E

are on the side

B C

such that

B D = D E = E C

, then which of the following is true?

EASY

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

P

and

Q

are the points on side

C A

and

C B

respectively of triangle

△ A B C

, right angled at

C

, prove that

(A Q^{2} + B P^{2}) = (A B^{2} + P Q^{2})

HARD

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

A very tall tree breaks from the middle due to storm. The broken part of the tree did not separate from the tree and the top of the tree touched the ground at a distance of

12

metre from the base of the tree. If the tree broke at the height of

37

metre from the ground, then what is the length of the broken portion of the tree in metres?

EASY

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

Prove that in a right angle triangle, the square of the hypotenuse is equal the sum of squares of the other two sides.

EASY

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

Let

a b c

be a three-digit number with nonzero digits such that

a^{2} + b^{2} = c^{2}

. What is the largest possible prime factor of

\bar{a b c} ?

HARD

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

HARD

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

In an equilateral

△ A B C

D

is a point on side

B C

, such that

B D = \frac{1}{3} B C

. Prove that

9 A D^{2} = 7 A B^{2}

MEDIUM

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite to its first side is a right angle.

MEDIUM

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

△ ABC, ∠ C = 90 °, AC = 5 cm

and

BC = 12 cm

. The bisector of

∠ A

meets

BC

D

. What is the length of

AD

?

HARD

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

Let

P Q R

be a triangle in which

P Q = 3

. From the vertex

R,

draw the altitude

R S

to meet

P Q

S

. Assume that

R S = \sqrt{3}

and

P S = Q R

. Then,

P R

equals

EASY

Mathematics>Geometry>Pythagoras Theorem>Pythagoras Theorem

The circle of radius

4 c m

and centre

O

has a tangent

P R

. If

∠ P O R = 90^{o}

and

O R = 7 c m

and

O P = 24 c m

. Calculate the length (in

c m

.) of

P R

PQR is an equilateral triangle with each side of length 2p. If PS⊥QR, and PS=kp, then find the value of k.

Important Questions on Pythagoras Theorem

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$P Q R$ is an equilateral triangle with each side of length $2 p .$ If $P S ⊥ Q R,$ and $P S = \sqrt{k} p$ , then find the value of $k$ .