Problems on Pythagoras - Test Questions

Question 11

Question:   In an equilateral triangle with side 'a', prove that the length of the altitude is and its area is [3 Mark]

Answer:   

Let the altitude from A to BC be AD = h.



BD = DC



Given AB = BC = CA = a

D ADB is right angles triangle.













Area of

Question 12

Question:   In the given figure, ABC is a right triangle, right-angled at B. AD and CE are the two medians drawn from A and C respectively. If AC = 5cm and AD =cm, find the length of CE. [3 Mark]

Answer:    In D ABD, by Pythagoras Theorem

....(1)

In D BEC,





....(2)

Adding (1) and (2), we have

Question 13

Question:   In D ABC, if AD is the median, show that

Answer:   

Draw AE ^ BC.

In .... (1) (By Pythagoras Theorem)

In D AEC, AC² = AE² + EC² . . . (2)

Adding (1) and (2), we have

Question 14

Question:   ABC is a right triangle, right angled at C. If P is the length of the perpendicular drawn from C to AB and a, b, c have the usual meaning, then prove that [3 Mark]

i.

ii.

Answer:    Proof of (i)

(AA Similarity)







Proof of (ii)

(AA Similarity)





....(1)

In D CDB,



(From (1))

Question 15

Question:   In a right triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divides these sides in the ratio 1:2. Prove that [5 Mark]

i. 9AQ² = 9AC² + 4BC²

ii. 9BP² = 9BC² + 4AC²

iii. 9(AQ² + BP²) = 13AB²

Answer:    P divides CA in the ratio 1:2.







....(1)

Similarly CQ....(2)

Proof of (i)

(Applying Pythagoras theorem to right-angled ACQ)

(Multiplying by 9 throughout)

(From 2)



.... (3)

Proof of (ii)

Similarly in



(From (1))

.... (4)

Proof of (iii)

Adding (3) and (4),





= 13 AB² (from D ABC)

Question 16

Question:   ABC is an equilateral triangle P is a point on BC such that BP:PC = 2:1, prove that 9AP² = 7AB². [5 Mark]

Answer:   

Draw AD ^ BC

Let PC = x, then BP = 2x

BC = BP + PC = 3x



D is the mid-point of BC.





In







.... (1)

Now in



(From (1))

Question 17

Question:   In an isosceles triangles ABC, with AB = AC, BD is perpendicular from B to the side AC. Prove that BD² - CD² = 2CD.AD. [2 Mark]

Answer:   

From D ADB,

Question 18

Question:   In a quadrilateral ABCD, and AD² = AB² + BC² + CD², Prove that [3 Mark]

Answer:   

Given:



To prove:



Proof:

AD² = AB² + BC² + CD² (given)

But AC² = AB² + BC² (ABC is right angled; By Pythagoras Theorem)

AD² = AC² + CD²

(By converse of Pythagoras theorem)

Question 19

Question:   Prove that in any triangle the sum of the squares on any two sides is equal to twice the square on half the third side together with twice the square of the median. [5 Mark]

Answer:   

Given:

In D ABC, AD is the median.

To prove:

AB² + AC² = 2BD² + 2AD²

Construction:

Draw AM ^ BC.

Proof:

In D ABM,









....(1)

In D AMC,











Adding (1) and (2) , we get

(DC = BD)

Question 20

Question:   In the figure, D and E trisects BC, prove that 8AE² = 3AC² + 5AD². >[5 Mark]

Answer:    Given:

D and E trisects BC. Let BD = DE = EC = x

To prove:

8AE² = 3AC² + 5AD²

Proof:

BC = 3x, EB = 2x

In

....(1) (Pythagoras theorem)

Similarly in D ABE,



....(2)

In D ABC,



... (3)

Now









= 8AE² (From (2))