Triangles - free exam paper questions


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Question 1

Question:   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 />
image

Answer:    By mid-point theorem DE||BC

image

image....(1)

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

image (Opposite angles of a parallelogram are equal0

Similarly image

image

image are equiangular and hence they are similar.

image

Question 2

Question:   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
image [3 Mark]

Answer:    image

Given:

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

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

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

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

image (By AA similarity criterion)

image

But image (From (2))

image

image

image

image

Question 3

Question:   In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, then prove thatimage[3 Mark]

image

Answer:    Construction:

Draw AEimageBC and DFimageBC

Proof:

image

image....(1)

Now in D AOE and D DOF, we have

image (Vertically opposite angles)

image (each angle = 90o)

image (AA similarity)

image....(2)

From (1) and (2), we get

image

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.


image

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

Then, we have

i. image

ii. image

iii. image

The proof of (i) is shown below

image

imageNote: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.

Question 4

Question:   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.

image

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

image ... (1)

image

image

ii. image (AAA Similarity)

image

image (From 1)

image

image

= 4 x 10

= 40cm2

iii. image

= 40 + 20 = 60cm2

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

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

Answer:    image

Given:

In D ADB,

image (Pythagoras Theorem)

image ... (1)

In D ADC,

image (Pythagoras theorem)

image...(2)

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

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

image

Question 6

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

Answer:    image

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

image

image BD = DC

image

Given AB = BC = CA = a

D ADB is right angles triangle.

image

image

image

image

image

image

Area of image

image

image

Question 7

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 =imagecm, find the length of CE. [3 Mark]
image

Answer:    In D ABD, by Pythagoras Theorem

image....(1)

In D BEC,

image

image

image....(2)

Adding (1) and (2), we have

image

image

image

image

image

image

image

image

image

Question 8

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

Answer:    image

Draw AE ^ BC.

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

In D AEC, AC2 = AE2 + EC2 . . . (2)

Adding (1) and (2), we have

image

image

image

image

image

image

Question 9

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. image

ii. image

image

Answer:    Proof of (i)

image (AA Similarity)

image

image

image

Proof of (ii)

image (AA Similarity)

image

image

image....(1)

In D CDB,

image

image (From (1))

image

image

image

Question 10

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

image

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

image

image

image

image....(1)

Similarly CQimage....(2)

Proof of (i)

image (Applying Pythagoras theorem to right-angled imageACQ)

image (Multiplying by 9 throughout)

image (From 2)

image

image.... (3)

Proof of (ii)

Similarly in image

image

image (From (1))

image.... (4)

Proof of (iii)

Adding (3) and (4),

image

image

= 13 AB2 (from D ABC)



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