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.

**Solution:**

By considering the lines AD and BC,the angles

∠ AEB = ∠ DEC (vertically opposite angles)

∠ EAB = ∠ EDC (alternate angles)

By using AA similarity criterion ∆ EAB ~ ∆ EDC

(AB/DC) = (EF/EG)

EF = (AB/DC) **x**
EG

= (3/1) **x** 1.4

= 4.2 km

Area of new industrial zone = Area of ∆ EAB

= (1/2) x AB x EF

= (1/2) x 3 x 4.2

= 6.3 km²

So the area of new industrial zone is 6.3 km²

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