Subject > Math > Geometry > Triangles > Question and Answers 2

Triangles

Question (1): D, E and F are the mid-point of the sides BC, CA and AB respectively of a D ABC. Determine the ratio of the areas of D DEF and D ABC. [3 Mar />

Answer: By mid-point theorem DE||BC

....(1)

AFDE is a parallelogram. (Since opposite sides are parallel)

(Opposite angles of a parallelogram are equal0

Similarly

are equiangular and hence they are similar.

Question (2): ABC is a triangle and DE is a line segment meeting AB in D and AC in E such that DE||BC and divides D ABC into two parts of equal area. Find

[3 Mark]

Answer:

Given:

Area (D ADE) = Area (trapezium DECB) ... (1)

Area (D ABC) = Area (D ADE) + Area (trap DECB)

= Area (D ADE) + Area (D ADE)

Area (D ABC) = 2(Area (D ADE)) ... (2)

(By AA similarity criterion)

But

(From (2))

Question (3): In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, then prove that

[3 Mark]

Answer: Construction:

Draw AE

BC and DF

BC

Proof:

....(1)

Now in D AOE and D DOF, we have

(Vertically opposite angles)

(each angle = 90^o)

(AA similarity)

....(2)

From (1) and (2), we get

Note:

If two triangles have a common vertex and their bases are along the same straight line, the ratio between their areas is equal to the ratio between the lengths of their bases.

The triangles ABC, ABD, ADC, AEC have a common vertex A and their bases are along the same line BC.

Then, we have

i.

ii.

iii.

The proof of (i) is shown below

Question (4): In the figure, ABCD is a parallelogram. P is a point on BC such that BP:PC = 1:2. DP produced meets AB produced at Q. If area of D CPQ = 20 cm², find [3 Mark]

i. Area of D BPQ

ii. Area of D CDP

iii. Area of parallelogram ABCD.

Answer: D CPQ and D BPQ have the common vertex Q, bases on the same line, therefore

... (1)

ii.

(AAA Similarity)

(From 1)

= 4 x 10

= 40cm²

iii.

= 40 + 20 = 60cm²

Area of (parallelogram ABCD) = 2 x Area (D QDC) = 120 cm² (

If a parallelogram and a triangle are on the same base and between the same parallels; area of the parallelogram is twice the area of the triangle)

Question (5): In a triangle, AD^ BC, prove that AB²- BD²= AC²- CD². [2 Mark]

Answer:

Given:

In D ADB,

(Pythagoras Theorem)

... (1)

In D ADC,

(Pythagoras theorem)

...(2)

L.H.S of (1) and (2) are equal.

\ R.H.S of (1) and (2) must be equal.

Question (6): 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 (7): 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 (8): 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 (9): 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 (10): 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 (11): 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.

.... (1)

Now in

(From (1))

Question (12): 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): 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): 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): 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): 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): 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): 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): 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): 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.