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