Mathematics
In the quadrilateral given below, AB || DC. E and F are mid-points of non-parallel sides AD and BC respectively. Calculate :
(i) EF if AB = 6 cm and DC = 4 cm
(ii) AB if DC = 8 cm and EF = 9 cm.
Mid-point Theorem
37 Likes
Answer
ABCD is a trapezium in which AB || DC and E, F are mid-points of AD and BC respectively.
Join CE and produce it to meet BA produced at G.
In △EDC and △EAG,
ED = EA (∵ E is mid-point of AD)
∠CED = ∠ GEA (Vertically opposite ∠s)
∠ECD = ∠EGA (Alternate ∠s)
∴ △EDC ≅ △EAG
⇒ CD = GA and EC = EG (c.p.c.t.)
In △CGB,
E is mid-point of CG
F is mid-point of BC
∴ By mid-point theorem, EF || AB and EF = GB.
But GB = GA + AB = CD + AB
∴ EF = (AB + CD) …..(1)
(i) Given,
AB = 6 cm and DC = 4 cm,
Putting these values in eq (1) we get,
EF = (6 + 4)
= x 10
= 5 cm
Hence, EF = 5 cm.
(ii) Given,
DC = 8 cm and EF = 9 cm
Putting these values in eq (1) we get,
9 = (AB + 8)
⇒ 18 = AB + 8
⇒ AB = 18 - 8 = 10 cm
Hence, AB = 10 cm.
Answered By
24 Likes
Related Questions
In the quadrilateral given below, AD = BC, P, Q, R and S are mid-points of AB, BD, CD and AC respectively. Prove that PQRS is a rhombus.
In the quadrilateral given below, AB || DC || EG. If E is mid-point of AD, prove that
(i) G is midpoint of BC
(ii) 2EG = AB + CD
In the figure given below, ABCD is a kite in which BC = CD, AB = AD. E, F, G are mid-points of CD, BC and AB respectively. Prove that :
(i) ∠EFG = 90°
(ii) The line drawn through G and parallel to FE bisects DA.
In the quadrilateral given below, AB || DC, E and F are mid-points of AD and BD respectively. Prove that
(i) G is the mid-point of BC
(ii) EG = (AB + DC).