Triangle and Trapezium Problems- Test Questions

Question 11

Question:   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

Question:   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

Question:   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

Question:   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

Question:   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

Question:   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

Question:   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

Question:   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

Question:   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

Question:   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)