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Subject > Math > Geometry > Triangles > Question and Answers 1

Triangles

Question (1): Say whether the following quadrilateral are similar or not.

(i)

(ii)

Answer: (i)ABCD similar to PQRS (ii)ABCD is not similar to PQRS.

Question (2): Give two examples of pairs of similar figures.

Answer: (i)

(ii)

Question (3): ABCD and EFGH are similar rectangles. What should be the measure of EF?

Answer: EF = 2 x 2.5 = 5cm.

Because GF = HE = 2 x BC = 2 x 1.8 = 3.6cm

Question (4): What is the difference between similarity and congruent?

Answer: Similarity: The shapes of two figures are same.

Congruent: The size and shape of two figures are same.

Question (5): If in a D ABC, DE||BC as shown in the figure. [3 Mark]

Show that

i.

ii.

Answer: If DE||BC (given)

... (1) (by basic proportionality theorem)

Adding 1 on both sides,

(On simplification)

.... (2)

Thus (i) is proved.

(Reciprocal ratio of (1))

Adding 1 on both sides,

Thus (ii) is proved.

Note:

The following results can be obtained from BPT

In a D ABC, DE||BC and intersects AB in D and AC in E, then

You should be familiar with these results to use them in the problems.

Question (6): Prove mid-point theorem from converse of Basic Proportionality Theorem. [2 Mark]

Answer:

Given:

D and E are the mid-points of AB and AC respectively of a triangle ABC.

To prove:

DE||BC

Proof:

.... (1) (Since D is the mid-point of AB, AD = DB)

From (1) and (2), we have

DE||BC (

If a line divides any two sides of a triangle in the same ratio, the line is parallel to the third side)

Question (7): Prove the converse of mid-point theorem using Basic Proportionality Theorem. That is, prove that the line drawn from the mid-point of one side of a triangle, parallel to the second side, bisects the third side. [2 Mark]

Answer:

Given:

ABC is a triangle where AD = DB and DE||BC.

To prove: AE = EC

Proof:

DE||BC (given)

AD = DB (given)

.... (1)

The Basic Proportionality Theorem states that 'If a line is drawn parallel to one side of a triangle, then other two sides are divided in the same ratio'.

.... (1) (since D is the mid-point of AB, AD = DB)

Since L.H.S of (1) and (2) are equal, their R.H.S are also equal

i.e., DE bisects AC .

Question (8): Given D ABC and DE||BC. [2 Mark]

AD = 4x - 3 DB = 3x - 1

AE = 8x - 7 EC = 5x - 3

Find the value of x.

Answer: Since DE||BC, using Basic Proportionality Theorem, we get

Discard the value

because the distance becomes negative on substituting this value, which is not possible.

\ x = 1

Question (9): P is the mid-point of side BC of a D ABC. If Q is the mid-point of AP and BQ when produced meets AC in L, then prove that

[3 Mark]

Answer:

Given:

In triangle D ABC, BP = PC and AQ = QP.

To prove:

Construction: Through P, draw PM parallel to BL. Proof: In D APM, QL||PM (by construction)

.... (1) (By BPT)

But

L.H.S of (1) and (2) are equal, therefore R.H.S of (1) and (2) must be equal

(3)

In D CBL, PM||BL

(4) (BPT)

But

(5) (given)

(From (4) and (5))

(6) From (3) and (6),

AL = LM = CM ...(7)

(from (7))

Question (10): Prove that the diagonals of a trapezium divide each other proportionally. [2 Mark]

Answer:

Given: ABCD is a trapezium. Diagonals AC and DB intersect at E.

To prove:

Construction:

Through E, draw EF||AB.

Proof:

In

(AB ll EF (by construction))

(1) (By BPT)

(2) (Taking the reciprocal ratio of (1))

In D DAB,

EF||AB (by construction)

(3) L.H.S of (2) and (3) are equal, therefore R.H.S must be equal

Question (11): In the given figure,

and YM = ZN, then prove that MN||YZ. [2 Mark]

Answer:

Given: XYZ is a

YM = ZN

To prove:

MN||YZ

Proof:

XY = YZ ... (1) (In a triangle, the sides opposite to equal angles are equal)

Now XM = XY - YM

= YZ - YM (From (1))

= YZ - ZN (given)

= XN

XM = XN

MN divides XY, XZ proportionally.

MN||YZ (by converse of Basic Proportionality Theorem)

Question (12): In the given figure, PQRS is a trapezium in which PQ||SR||XY, then prove,

[2 Mark]

Answer:

Given:

PQ||XY||SR

To prove:

Construction:

Through P, draw a line parallel to QR which intersects XY and SR at M and N respectively.

Proof:

In the quadrilateral PQYM, PQ||MY (given)

PM||QY (construction)

PQYM is a parallelogram.

PM = QY ...(1) (opposite sides of a parallelogram are equal)

Similarly MYNR is a parallelogram.

MN = YR ... (2)

In D PSN, XM||SN

Question (13): D and E are points on the sides AB and AC of a D ABC. AB = 12cm, AD = 8cm, AE = 12cm and AC = 18cm, show that DE||BC. [2 Mark]

Answer:

Given:

AB = 12cm AD = 8cm, AE = 12cm, AC = 18cm

To prove:

DE || BC

Proof:

DB = AB - AD = 12 - 8 = 4cm

EC = AC - AE = 18 - 12 = 6cm

... (1)

...(2)

From (1) and (2), we get

DE||BC (Converse of BPT)

Question (14): In the given figure, XY||BC. Find the length of XY. [2 Mark]

Answer: In D AXY and D ABC,

(common angle)

(By definition of similarity)

= 1.5 cm

Question (15): In a right-angled triangle with sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x. Prove that ab = cx. [3 Mark]

Answer: In triangle PQR,

PR = c

PQ = a

QR = b

In DRSQ and D RQP,

(Common angle)

(AA Similarity)

Question (16): From the given figure, express x in terms of a, b and c [2 Mark]

Answer: In D KNP and D KML,

(each angle = 46^o)

(common angle)

(AA Similarity)

(By definition of similarity)

Question (17): In the given figure, PA, QB and RC are each perpendicular to AC, prove that

[3 Mark]

Answer: PA||QB||RC (each ^ AC) In D CBQ and D CAP,

(By definition of Similarity) or

....(1)

or

Similarly, D ACR ~ D ABQ

....(2)

Adding (1) and (2), we have

Question (18): In the given figure,

ABC = 90^o and BD ^ AC. If AB = 5.7cm, BD = 3.8cm and CD = 5.4cm, find BC. [2 Mark]

Answer:

(AA similarity)

= 8.1

\ BC = 8.1 cm

Question (19): If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median of another triangle, then prove that the two triangles are similar [5 Mark]

Answer:

Given: In D ABC and D DEF, AP and DQ are medians and

To prove:

Construction:

Produce AP to G such that AP = PG. Join CG.

Produce DQ to H such that DQ = QH. Join HF.

Proof:

In D APB and D GPC,

AP = PG (by construction)

(Vertically opposite angles) BP = PC (given)

....(1) (SAS)

....(2) (CPCT)

Similarly, D DQE ~ D HQF DE = FH ... (3)

.... (4) (CPCT)

Now

(given)

(From (1) and (2))

.... (5)

(by SSS similarity)

.... (6) (By definition of similarity)

.... (7) (same as above)

....(8) (From (2), (4), (7))

(From (6) and (8))

....(9)

In D ABC and D DEF,

(given)

(From (9))

(SAS similarity)

Question (20): ABCD is a trapezium in which AB is parallel to DC. If AC trisects BD, then prove that

[3 Mark]

Answer:

Given:

ABCD is a trapezium.AC and BD intersect at P. P trisects BD.

i.e.,

To prove:

Proof:

In D DPC and D BPA,

(Vertically opposite angles)

(Alternate angles)

(AA Similarity)

Question (21): Prove that if a line be drawn Parallel to the base of aD , the portion intercepted by the sides is bisected by the median to the base. [3 Mark]

Answer:

Given:

In D ABC, PQ||BC. AD is the median of BC and meets BC in D.

To prove:

PR = RQ

R is midpoint of PQ.

Proof:

Since PQ||BC,

D APR and D ABD are equiangular.

D APR ~ DABD (AAA)