In this page 10th grade geometry solution9 we are going to see solutions of some practice questions.

(5) P and Q
are points on sides AB and AC respectively, of triangle ABC. If AP = 3 cm, PB =
6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ.

Solution:

AP/PB = AQ/AC

3/9 = 5/15

1/3 = 1/3

In triangles
APQ, and ABC we get

(AP/AB) =
(AQ/AC)

∠ A = ∠A

By using SAS
criterion ∆ APQ ~ ∆ ABC

(AP/AB) =
(Q/AC) = (PQ/BC)

(AP/PB) =
(PQ/BC)

PQ/BC = 3/9

PQ/BC = 1/3

3 PQ = BC

BC = 3 PQ

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(6) In
triangle ABC, AB = AC and BC = 6 cm. D is a point on the side AC such that AD =
5 cm and CD = 4 cm. Show that triangle BCD and triangle ACB are congruent and
hence find BD.

Solution:

In triangle
ABC, AB = AC

(BC/AC) =
(CD/CB)

(6/9) = (4/6)

From the
triangles ∆ BCD, ∆ ACB we get

(BC/AC) =
(CD/CB)

∠ C = ∠ C
(Common angle)

By using SAS
criterion, we get

∆ BCD ~ ∆ ACB

(BD/AB) =
(BC/AC)

Here AB = AC

BD/AC =6/9

BD/9 = 6/9

BD = 6 cm

(7) The
points D and E are on the sides AB and AC of triangle ABC respectively, such
that DE and BC are parallel. If AB = 3 AD and the area triangle ABC is 72 cm², then find the area of the quadrilateral DBCE.

Solution:

From the
diagram he sides DE and BC are parallel and AB = 3 AD

(AD/AB) = 1/3

By taking the
two triangles ∆ ADE and ∆ ABC

∠ ADE = ∠ ABC

∠ A = ∠ A
(Common angle)

By using AA
similarity criterion ∆ ADE ~ ∆ ABC

Area of ∆
ADE/Area of ∆ ABC = AD²/AB²

Area of ∆
ADE/72 = 1/9

Area of ∆ ADE
= (1/9) x 72

= 8 cm

Area of
quadrilateral = area of ∆ ABC – Area of ∆ DE

= 72 – 8

= 64 cm²

Therefore the required area of quadrilateral is 64 cm²