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In this page 10th grade geometry solution10 we are going to see solutions of some practice questions.

(8) The
lengths of the three sides of triangle ABC are 6 cm, 4 cm and 9 cm. Triangle
PQR ad BC are congruent. One of the lengths of the sides of triangle PQR is 35
cm. What is the greatest perimeter possible for triangle PQR

Solution:

From the given information let us draw a rough diagram.

∆ PQR ~ ∆ ABC

PQ/AB = QR/BC
= PR/AC = perimeter of ∆ PQR/Perimeter of ∆ BC

Let QR = 35

The
corresponding sides must be QR and BC.

Perimeter of
∆ PQR/ Perimeter of ∆ ABC = QR/BC= 35/4

Perimeter of triangle PQR = (35/4) x 19

= 665/4

=
166.25

So,Perimeter of triangle PQR is 166.25 cm²

(9) In the
figure, the sides DE and BC are parallel and (AD/B) = 3/5, calculate the value
of

(i) area of
triangle ADE/are of triangle ABC

(ii) area of
trapezium BCED/area of triangle ABC

In triangle ABC, the sides DE and BC are parallel

∆ ADE ~ ∆ ABC

AD/BD = 3/5

AD = 3 k

BD = 5 k

Area of ∆ ADE/ Area of ∆ ABC = AD²/AB²

= (3k)²/(8k)²

= 9/64

(ii) Area of ∆ ADE = 9 k

Area of ∆ ADE = 64 k

Area of trapezium
BCDE = area of ∆ ABC – area of ∆ ADE

= 64 k – 9 k

= 55 k

Area of trapezium BCDE/Area of ∆ ABC = 55 k/64k

= 55/64

(10) The
government plans to develop a new industrial zone in an unused portion of land
in a city.