Problems on Triangles - Test Questions

Question 21

Question:   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)

....(1)

Similarly, D ARQ ~ D ADC

....(2)

(From (1) and (2) both are equal to)

Question 22

Question:   In [3 Mark]

Answer:   

Given:



or



Proof:

In D ADC and D BDA,

(given)



(SAS similarity criterion)

The two triangles are equiangular.

....(1)

Now in D ABC,

(Angle sum property of a triangle)

(From (1))

Question 23

Question:   ABC is a triangle in which AB = AC and D is a point on the side AC such that BC² = AC x CD. Prove that BD = BC. [2 Mark]

Answer:   

Given:





AB = AC and BC² = AC x CD

i.e., To prove: BD = BC In D BCA and D DCB,

Note:

That we name triangle ABC as BCA to keep up the correspondence with D DCB. The corresponding angles at the vertices C in both triangles are equal.

Proof:

(given)

(common angle)

(SAS similarity criterion)

....(1)

But AB = AC (given)

BC = BD (FromNote:That we name triangle ABC as BCA to keep up the correspondence with D DCB. The corresponding angles at the vertices C in both triangles are equal.

Question 24

Question:   In the given figure, D ABC ~ D DEF. If AB = 2DE and area of D ABC = 56 sq.cm, then find the area of D DEF. [2 Mark]

Answer:    We know that, if D ABC ~ D DEF, then

[The ratio of areas of two similar is equal to the ratio of the squares of their corresponding sides]







\ Area of D DEF = 14 sq.cm

Question 25

Question:   In the given figure, DE ll BC and AD:DB = 5:4. find [2 Mark]

Answer:    (By AA similarity criterion)



....(1)

In D DFE and D CFB,

(Alternate angle)

(Vertically opposite angle)

(AA Similarity)