10th grade geometry solution10

10th Grade Geometry Solution10

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

10th grade geometry solution10 10th grade geometry solution10

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