Triangles - free exam questions

Question 11

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 12

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 13

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 14

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 15

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))

Question 16

Question:   P is any point inside the rectangle ABCD, prove that [3 Mark]

PA² + PC² = PB² + PD².

Answer:   

ABCD is a rectangle. P is in the interior of ABCD

To prove:

PA² + PC² = PB² + PD²

Construction:

Through P, draw EF||BC.

Proof:

ADFE and EFCB are rectangles.

.... (1) (Pythagoras theorem)

.... (2) (Pythagoras theorem)

Adding (1) and (2), we get

Question 17

Question:   Each side of a rhombus is 10cm. If one of its diagonals is 16cm, find the length of the other diagonal. [2 Mark]

Answer:   

Since the diagonals in a rhombus bisect each other at right angles,

BO = 8cm.

AO² = AB² - BO²

= 100 - 64

= 36

AO = 6cm

AC = 2AO = 2 x 6 = 12cm

Question 18

Question:   In the given diagram, AB = 3CD = 18cm and 3BP = 4CP = 36cm. Show that the measure of angle APD = 90^o. [3 Mark]

Answer:    Given:

AB = 18cm









To prove:



Construction:

Draw DE || CB. Join DA.

Proof:

In triangle DCP,

DP² = DC² + CP² = 6² + 9²

= 36 + 81 = 117

PA² = PB² + BA²

= 12² + 18²

= 144 + 324 = 468

DP² + PA² = 117 + 468 = 585 ...(1)

DE = CB = 21cm

AE = 12cm







= 441 + 144

= 585 .... (2) From (1) and (2),

AD² = DP² + PA²

(Converse of Pythagoras theorem)

Question 19

Question:   In triangle ABC, AB = 8cm, BC = 6cm and AC = 3cm. Calculate the length of OC.[3 Mark]

Answer:    Let CO = 'x' cm.

In (Pythagoras theorem)

.... (1)

In



....(2)

Form (1) and (2), we get











= 1.58cm

Question 20

Question:   Prove that three times the sum of the squares on the sides of a triangle is equal to four times the sum of the squares on the medians of the triangle. [5 Mark]

Answer:   

Given:

ABC is a triangle in which AD, BE and CF are the medians. It has been proved earlier that in any triangle, the sum of the squares on two sides of a triangle is equal to twice the square of half the third side together with twice the square on the median bisecting the third side.





.... (1)

Similarly,

.... (2)

.... (3)

Adding (1) and (2) and (3), we get









Three times the sum of the squares on the sides of a triangle is equal to four times the sum of the square on the medians of the triangl