Question

10. WHAT is any trapezoid. A straight line

parallel to the bases cuts [WT], [WA), and to
E, G, and F respectively.
H
a) Calculate WE and ET such that WT = 30 and
2FA = 3HF.
b) Calculate HF, FA, WE and ET such that
HA = 8, WT = 12 and GA = 2WG.
Please answer asap
Hint:using Thales theorem

Answer 1

(a) ET = 18 and WE = 12

(b) FA = 16/3

HF = 8/3

ET = 8

WE = 4

Step-by-step explanation:

WT = 30 ( WE + ET = 30)

WE = 30 - ET

2FA = 3HF

FA = 3/2HF

WE =?, ET = ?

Line WH, EF and TA are parallel to each other.

(a)

ΔWTA and ΔWHA are congruent to each other

Therefore, by congruency

`(WE)/(ET) = (HF)/(FA)`

So,

30 - ET/ ET = HF / 3/2HF

30 - ET / ET = 2HF / 3HF

30 - ET / ET = 2/3

On solving the above equation we get, ET = 18

WE = 30 - ET

WE = 30 - 18 = 12

Therefore, ET = 18 and WE = 12

(b)

HA = 8 ( HF + FA = 8)

HF = 8 - FA

WT = 12 ( WE + ET = 12 )

WE = 12 - ET

GA = 2WG

HF, WA, WE, ET = ?

ΔWHA and ΔGFA are congruent

So,

`(HF)/(FA) = (WG)/(GA)`

8 - FA / FA = WG / 2 WG

On solving the above equation we get,

FA = 16/3

HF + FA = 8

HF + 16/3 = 8

HF = 8/3

ΔWTA and ΔWEG are congruent

So,

`(WE)/(ET) = (WG)/(GA)`

12 - ET / ET = WG / 2WG

On solving the above equation, ET = 8

WE = 12 - ET

WE = 12 - 8

WE = 4

Therefore, FA = 16/3

HF = 8/3

ET = 8

WE = 4