In this page 10th grade geometry solution2 we are going to see solutions of some practice questions.
(2) In the figure AP = 3 cm, AR = 4.5 cm, AQ = 6 cm, AB = 5 cm and AC = 10 cm. Find the length of AD.
Solution:
From the given information we get, (AB/AP) = (AC/AQ)
In triangle ABC, (AB/AP) = (AC/AQ)
By using converse of “Thales theorem” PQ is parallel to BC.
RD = x
In triangle ABD, PR is parallel to BD
AD = AR + RD
AD = 4.5 + x
(AB/AP) = (AD/AR)
(5/3) = (4.5 + x)/4.5
(5 x 4.5)/3 = 4.5 + x
7.5 = 4.5 + x
x = 7.5 – 4.5
x = 3
Here we need to find the length of AD = 4.5 + x
= 4.5 + 3
= 7.5 cm
(3) E and F are points on the sides PQ and PR respectively, of a triangle PQR. For each of the following cases. Verify EF is parallel to QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
Solution:
First let us draw the picture for the above details
To verify whether EF is parallel to QR we have to check the condition
(PE/EQ) = (PF/FR) (3.9/3) = (3.6/2.4) 1.3 ≠ 1.5 So the sides EF and QR are not parallel. |
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Solution:
First let us draw the picture for the above details
To verify whether EF is parallel to QR we have to check the condition
(PE/EQ) = (PF/FR) (4/4.5) = (8/9) 0.88 = 0.88 So the sides EF and QR are parallel. |
(4) In the figure,AC is parallel to BD and CF is parallel to DF, if OA = 12 cm, AB = 9 cm, OC = 8 cm and EF= 4.5 cm, then find FQ.
Solution:
In triangle OBD, AC is parallel to BD
By using “Thales theorem” we get,
(OA/AB) = (OC/CD)
(12/9)= (8/CD)
CD = (8 x 9)/12
= 72/12
= 6 cm
In triangle ODF, the sides CE and DF are parallel, by using “Thales theorem” we get,
(OC/CD)= (OE/EF)
(8/6) = (OE/4.5)
OE = (8 x 4.5)/6
= 36/6
= 6 cm
So, OF = OE + EF
= 6 + 4.5
= 10.5 cm
10th grade geometry solution2 10th grade geometry solution2